Write down the gradient of the curve of \(f\) at P.

Find the equation of the normal to the curve of \(f\) at P.

Determine the concavity of the graph of \(f\) when \(4 < x < 5\) and justify your answer.

\( - 2\)     A1     N1

[1 mark]

gradient of normal \( = \frac{1}{2}\)     (A1)

attempt to substitute their normal gradient and coordinates of P (in any order)     (M1)

eg\(\,\,\,\,\,\)\(y - 4 = \frac{1}{2}(x - 3),{\text{ }}3 = \frac{1}{2}(4) + b,{\text{ }}b = 1\)

\(y - 3 = \frac{1}{2}(x - 4),{\text{ }}y = \frac{1}{2}x + 1,{\text{ }}x - 2y + 2 = 0\)     A1     N3

[3 marks]

correct answer and valid reasoning     A2     N2

answer:     eg     graph of \(f\) is concave up, concavity is positive (between \(4 < x < 5\))

reason:     eg     slope of \(f’\) is positive, \(f’\) is increasing, \(f’’ > 0\),

sign chart (must clearly be for \(f’’\) and show A and B)

Note:     The reason given must refer to a specific function/graph. Referring to “the graph” or “it” is not sufficient.

[2 marks]

Find \(h(1)\).

Find \(h'(8)\).

expressing \(h(1)\) as a product of \(f(1)\) and  \(g(1)\)     (A1)

eg\(\,\,\,\,\,\)\(f(1) \times g(1),{\text{ }}2(9)\)

\(h(1) = 18\)     A1     N2

[2 marks]

attempt to use product rule (do not accept \(h’ = f' \times g’\))     (M1)

eg\(\,\,\,\,\,\)\(h’ = fg' + gf',{\text{ }}h'(8) = f'(8)g(8) + g’(8)f(8)\)

correct substitution of values into product rule     (A1) 

eg\(\,\,\,\,\,\)\(h’(8) = 4(5) + 2( - 3),{\text{ }} - 6 + 20\)

\(h’(8) = 14\)     A1 N2

[3 marks]

Write down \(f(x)\) in the form \(f(x) = - 10(x - p)(x - q)\) .

Find another expression for \(f(x)\) in the form \(f(x) = - 10{(x - h)^2} + k\) .

Show that \(f(x)\) can also be written in the form \(f(x) = 240 + 20x - 10{x^2}\) .

A particle moves along a straight line so that its velocity, \(v{\text{ m}}{{\text{s}}^{ - 1}}\) , at time t seconds is given by \(v = 240 + 20t - 10{t^2}\) , for \(0 \le t \le 6\) .

(i)     Find the value of t when the speed of the particle is greatest.

(ii)    Find the acceleration of the particle when its speed is zero.

\(f(x) = - 10(x + 4)(x - 6)\)     A1A1     N2

[2 marks]

METHOD 1

attempting to find the x-coordinate of maximum point     (M1)

e.g. averaging the x-intercepts, sketch, \(y' = 0\) , axis of symmetry

attempting to find the y-coordinate of maximum point     (M1)

e.g. \(k = - 10(1 + 4)(1 - 6)\)

\(f(x) = - 10{(x - 1)^2} + 250\)     A1A1     N4

METHOD 2

attempt to expand \(f(x)\)     (M1)

e.g. \( - 10({x^2} - 2x - 24)\)

attempt to complete the square     (M1)

e.g. \( - 10({(x - 1)^2} - 1 - 24)\)

\(f(x) = - 10{(x - 1)^2} + 250\)     A1A1     N4

[4 marks]

attempt to simplify     (M1)

e.g. distributive property, \( - 10(x - 1)(x - 1) + 250\)

correct simplification     A1

e.g. \( - 10({x^2} - 6x + 4x - 24)\) , \( - 10({x^2} - 2x + 1) + 250\)

\(f(x) = 240 + 20x - 10{x^2}\)     AG     N0

[2 marks]

(i) valid approach     (M1)

e.g. vertex of parabola, \(v'(t) = 0\)

\(t = 1\)     A1     N2

(ii) recognizing \(a(t) = v'(t)\)     (M1)

\(a(t) = 20 - 20t\)     A1A1

speed is zero \( \Rightarrow t = 6\)     (A1)

\(a(6) = - 100\) (\({\text{m}}{{\text{s}}^{ - 2}}\))     A1     N3

[7 marks]

Parts (a) and (c) of this question were very well done by most candidates.

In part (b), many candidates attempted to use the method of completing the square, but were unsuccessful dealing with the coefficient of \( - 10\). Candidates who recognized that the x-coordinate of the vertex was 1, then substituted this value into the function from part (a), were generally able to earn full marks here.

Parts (a) and (c) of this question were very well done by most candidates.

In part (d), it was clear that many candidates were not familiar with the relationship between velocity and acceleration, and did not understand how those concepts were related to the graph which was given. A large number of candidates used time \(t = 1\) in part b(ii), rather than \(t = 6\) . To find the acceleration, some candidates tried to integrate the velocity function, rather than taking the derivative of velocity. Still others found the derivative in part b(i), but did not realize they needed to use it in part b(ii), as well.

Find \(g'(x)\) .

Find the gradient of the graph of g at \(x = \pi \) .

evidence of choosing the product rule     (M1)

e.g. \(uv' + vu'\)

correct derivatives \(\cos x\) , 2     (A1)(A1)

\(g'(x) = 2x\cos x + 2\sin x\)     A1     N4

[4 marks]

attempt to substitute into gradient function     (M1)

e.g. \(g'(\pi )\)

correct substitution     (A1)

e.g. \(2\pi \cos \pi  + 2\sin \pi \)

\({\text{gradient}} = - 2\pi \)     A1     N2

[3 marks]

Most candidates answered part (a) correctly, using the product rule to find the derivative, and earned full marks here. There were some who did not know to use the product rule, and of course did not find the correct derivative.

In part (b), many candidates substituted correctly into their derivatives, but then used incorrect values for \(\sin x\) and \(\cos x\) , leading to the wrong gradient in their final answers.

Show that \(f( - x) = - f(x)\) .

Given that \(f''(x) = \frac{{2ax({x^2} - 3)}}{{{{({x^2} + 1)}^3}}}\) , find the coordinates of all points of inflexion.

It is given that \(\int {f(x){\rm{d}}x = \frac{a}{2}} \ln ({x^2} + 1) + C\) .

(i)     Find the area of the shaded region, giving your answer in the form \(p\ln q\) .

(ii)    Find the value of \(\int_4^8 {2f(x - 1){\rm{d}}x} \) .

METHOD 1

evidence of substituting \( - x\) for \(x\)     (M1)

\(f( - x) = \frac{{a( - x)}}{{{{( - x)}^2} + 1}}\)     A1

\(f( - x) = \frac{{ - ax}}{{{x^2} + 1}}\) \(( = - f(x))\)     AG     N0

METHOD 2

\(y = - f(x)\) is reflection of \(y = f(x)\) in x axis

and \(y = f( - x)\) is reflection of \(y = f(x)\) in y axis     (M1)

sketch showing these are the same     A1

\(f( - x) = \frac{{ - ax}}{{{x^2} + 1}}\) \(( = - f(x))\)     AG     N0

[2 marks]

evidence of appropriate approach     (M1)

e.g. \(f''(x) = 0\)

to set the numerator equal to 0     (A1)

e.g. \(2ax({x^2} - 3) = 0\) ; \(({x^2} - 3) = 0\)

(0, 0) , \(\left( {\sqrt 3 ,\frac{{a\sqrt 3 }}{4}} \right)\) , \(\left( { - \sqrt 3 , - \frac{{a\sqrt 3 }}{4}} \right)\) (accept \(x = 0\) , \(y = 0\) etc)      A1A1A1A1A1     N5

[7 marks]

(i) correct expression     A2

e.g. \(\left[ {\frac{a}{2}\ln ({x^2} + 1)} \right]_3^7\) , \(\frac{a}{2}\ln 50 - \frac{a}{2}\ln 10\) , \(\frac{a}{2}(\ln 50 - \ln 10)\)

area = \(\frac{a}{2}\ln 5\)     A1A1     N2

(ii) METHOD 1

recognizing the shift that does not change the area     (M1)

e.g. \(\int_4^8 {f(x - 1){\rm{d}}x}  = \int_3^7 {f(x){\rm{d}}x} \) , \(\frac{a}{2}\ln 5\)

recognizing that the factor of 2 doubles the area     (M1)

e.g. \(\int_4^8 {2f(x - 1){\rm{d}}x = } 2\int_4^8 {f(x - 1){\rm{d}}x} \) \(\left( { = 2\int_3^7 {f(x){\rm{d}}x} } \right)\)

\(\int_4^8 {2f(x - 1){\rm{d}}x = a\ln 5} \) (i.e. \(2 \times \) their answer to (c)(i))     A1     N3

METHOD 2

changing variable

let \(w = x - 1\) , so \(\frac{{{\rm{d}}w}}{{{\rm{d}}x}} = 1\)

\(2\int {f(w){\rm{d}}w = } \frac{{2a}}{2}\ln ({w^2} + 1) + c\)     (M1)

substituting correct limits

e.g. \(\left[ {a\ln \left[ {{{(x - 1)}^2} + 1} \right]} \right]_4^8\) , \(\left[ {a\ln ({w^2} + 1)} \right]_3^7\) , \(a\ln 50 - a\ln 10\)     (M1)

\(\int_4^8 {2f(x - 1){\rm{d}}x = a\ln 5} \)     A1     N3

[7 marks]

Part (a) was achieved by some candidates, although brackets around the \( - x\) were commonly neglected. Some attempted to show the relationship by substituting a specific value for x . This earned no marks as a general argument is required.

Although many recognized the requirement to set the second derivative to zero in (b), a majority neglected to give their answers as ordered pairs, only writing the x-coordinates. Some did not consider the negative root.

For those who found a correct expression in (c)(i), many finished by calculating \(\ln 50 - \ln 10 = \ln 40\) . Few recognized that the translation did not change the area, although some factored the 2 from the integrand, appreciating that the area is double that in (c)(i).

Show that \(f’(1) = 1\).

Find the equation of \(L\) in the form \(y = ax + b\).

Find the \(x\)-coordinate of Q.

Find the area of the region enclosed by the graph of \(f\) and the line \(L\).

\(f’(x) = 2x - 1\)     A1A1

correct substitution     A1

eg\(\,\,\,\,\,\)\(2(1) - 1,{\text{ }}2 - 1\)

\(f’(1) = 1\)     AG     N0

[3 marks]

correct approach to find the gradient of the normal     (A1)

eg\(\,\,\,\,\,\)\(\frac{{ - 1}}{{f'(1)}},{\text{ }}{m_1}{m_2} =  - 1,{\text{ slope}} =  - 1\)

attempt to substitute correct normal gradient and coordinates into equation of a line     (M1)

eg\(\,\,\,\,\,\)\(y - 0 =  - 1(x - 1),{\text{ }}0 =  - 1 + b,{\text{ }}b = 1,{\text{ }}L =  - x + 1\)

\(y =  - x + 1\)     A1     N2

[3 marks]

equating expressions     (M1)

eg\(\,\,\,\,\,\)\(f(x) = L,{\text{ }} - x + 1 = {x^2} - x\)

correct working (must involve combining terms)     (A1)

eg\(\,\,\,\,\,\)\({x^2} - 1 = 0,{\text{ }}{x^2} = 1,{\text{ }}x = 1\)

\(x =  - 1\,\,\,\,\,\left( {{\text{accept }}Q( - 1,{\text{ }}2)} \right)\)     A2     N3

[4 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\int {L - f,{\text{ }}\int_{ - 1}^1 {(1 - {x^2}){\text{d}}x} } \), splitting area into triangles and integrals

correct integration     (A1)(A1)

eg\(\,\,\,\,\,\)\(\left[ {x - \frac{{{x^3}}}{3}} \right]_{ - 1}^1,{\text{ }} - \frac{{{x^3}}}{3} - \frac{{{x^2}}}{2} + \frac{{{x^2}}}{2} + x\)

substituting their limits into their integrated function and subtracting (in any order)     (M1)

eg\(\,\,\,\,\,\)\(1 - \frac{1}{3} - \left( { - 1 - \frac{{ - 1}}{3}} \right)\)

Note:     Award M0 for substituting into original or differentiated function.

area \( = \frac{4}{3}\)     A2     N3

[6 marks]

Use the quotient rule to show that \(f'(x) = \frac{{ - 1}}{{{{\sin }^2}x}}\) .

Find \(f''(x)\) .

Find the value of p and of q.

Use information from the table to explain why there is a point of inflexion on the graph of f where \(x = \frac{\pi }{2}\) .

\(\frac{{\rm{d}}}{{{\rm{d}}x}}\sin x = \cos x\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}\cos x = - \sin x\) (seen anywhere)     (A1)(A1)

evidence of using the quotient rule     M1

correct substitution     A1

e.g. \(\frac{{\sin x( - \sin x) - \cos x(\cos x)}}{{{{\sin }^2}x}}\) , \(\frac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}}\)

\(f'(x) = \frac{{ - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x}}\)     A1

\(f'(x) = \frac{{ - 1}}{{{{\sin }^2}x}}\)     AG      N0

[5 marks]

METHOD 1

appropriate approach     (M1)

e.g. \(f'(x) = - {(\sin x)^{ - 2}}\)

\(f''(x) = 2({\sin ^{ - 3}}x)(\cos x)\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\)     A1A1     N3

Note: Award A1 for \(2{\sin ^{ - 3}}x\) , A1 for \(\cos x\) .

METHOD 2

derivative of \({\sin ^2}x = 2\sin x\cos x\) (seen anywhere)     A1

evidence of choosing quotient rule     (M1)

e.g. \(u = - 1\) ,  \(v = {\sin ^2}x\) , \(f'' = \frac{{{{\sin }^2}x \times 0 - ( - 1)2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\)

\(f''(x) = \frac{{2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\)     A1     N3

[3 marks]

evidence of substituting \(\frac{\pi }{2}\)     M1

e.g. \(\frac{{ - 1}}{{{{\sin }^2}\frac{\pi }{2}}}\) , \(\frac{{2\cos \frac{\pi }{2}}}{{{{\sin }^3}\frac{\pi }{2}}}\)

\(p = - 1\) ,  \(q = 0\)    A1A1     N1N1

[3 marks]

second derivative is zero, second derivative changes sign     R1R1     N2

[2 marks]

Many candidates comfortably applied the quotient rule, although some did not completely show that the Pythagorean identity achieves the numerator of the answer given. Whether changing to \( - {(\sin x)^{ - 2}}\) , or applying the quotient rule a second time, most candidates neglected the chain rule in finding the second derivative.

Whether changing to \( - {(\sin x)^{ - 2}}\) , or applying the quotient rule a second time, most candidates neglected the chain rule in finding the second derivative.

Those who knew the trigonometric ratios at \(\frac{\pi }{2}\)typically found the values of p and of q, sometimes in follow-through from an incorrect \(f''(x)\) .

Few candidates gave two reasons from the table that supported the existence of a point of inflexion. Most stated that the second derivative is zero and neglected to consider the sign change to the left and right of q. Some discussed a change of concavity, but without supporting this statement by referencing the change of sign in \(f''(x)\) , so no marks were earned.

Find \(f'(x)\) .

There is a minimum value of \(f(x)\) when \(x = - 2\) . Find the value of \(p\) .

\(f'(x) = 2x - \frac{p}{{{x^2}}}\)     A1A1     N2

Note: Award A1 for \(2x\) , A1 for \( - \frac{p}{{{x^2}}}\) .

[2 marks]

evidence of equating derivative to 0 (seen anywhere)     (M1)

evidence of finding \(f'( - 2)\) (seen anywhere)     (M1)

correct equation     A1

e.g. \( - 4 - \frac{p}{4} = 0\) , \( - 16 - p = 0\)

\(p = - 16\)     A1     N3

[4 marks]

Candidates did well on (a).

For (b), a great number of candidates substituted into the function instead of into the derivative.

The derivate of \({x^2}\) was calculated without difficulties, but there were numerous problems regarding the derivative of \(\frac{p}{x}\) . There were several candidates who considered both p and x as variables; some tried to use the quotient rule and had difficulties, others used negative exponents and were not successful.

Complete the following table by noting which graph A, B or C corresponds to each function.

Write down the value of t when the velocity is greatest.

     A2A2     N4

[4 marks]

\(t = 3\)     A2     N2

[2 marks]

Many candidates answered this question completely and correctly, showing a good understanding of the graphical relationship between displacement, velocity and acceleration. When done incorrectly, many answered with the displacement as graph B and acceleration as graph C.

Many candidates found the value of t which gave a maximum in the remaining graph, and were awarded follow through marks.

Find \(f'(x)\) .

Hence

(i)     show that \(q = - 2\) ;

(ii)    verify that A is a minimum point.

Find the maximum value of \(f(x)\) .

The function \(f(x)\) can be written in the form \(r\cos (x - a)\) .

Write down the value of r and of a .

\(f'(x) = - \sin x + \sqrt 3 \cos x\)     A1A1     N2

[2 marks]

(i) at A, \(f'(x) = 0\)     R1

correct working     A1

e.g. \(\sin x = \sqrt 3 \cos x\)

\(\tan x = \sqrt 3 \)     A1

\(x = \frac{\pi }{3}\) , \(\frac{{4\pi }}{3}\)     A1

attempt to substitute their x into \(f(x)\)     M1

e.g. \(\cos \left( {\frac{{4\pi }}{3}} \right) + \sqrt 3 \sin \left( {\frac{{4\pi }}{3}} \right)\)

correct substitution     A1

e.g. \( - \frac{1}{2} + \sqrt 3 \left( { - \frac{{\sqrt 3 }}{2}} \right)\)

correct working that clearly leads to \( - 2\)     A1

e.g. \( - \frac{1}{2} - \frac{3}{2}\)

 \(q = - 2\)     AG     N0

(ii) correct calculations to find \(f'(x)\) either side of \(x = \frac{{4\pi }}{3}\)     A1A1

e.g. \(f'(\pi ) = 0 - \sqrt 3 \) ,  \(f'(2\pi ) = 0 + \sqrt 3 \)
  

\(f'(x)\) changes sign from negative to positive     R1

so A is a minimum     AG     N0

[10 marks]

max when \(x = \frac{\pi }{3}\)     R1

correctly substituting \(x = \frac{\pi }{3}\) into \(f(x)\)     A1

e.g. \(\frac{1}{2} + \sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right)\)

max value is 2     A1     N1

[3 marks]

\(r = 2\) , \(a = \frac{\pi }{3}\)     A1A1     N2

[2 marks]

Write down \(f'\left( 2 \right)\).

Find \(f\left( 2 \right)\).

Show that the graph of g has a gradient of 6 at P.

Let L₂ be the tangent to the graph of g at P. L₁ intersects L₂ at the point Q.

Find the y-coordinate of Q.

recognize that \(f'\left( x \right)\) is the gradient of the tangent at \(x\)     (M1)

eg   \(f'\left( x \right) = m\)

\(f'\left( 2 \right) = 3\)  (accept m = 3)     A1 N2

[2 marks]

recognize that \(f\left( 2 \right) = y\left( 2 \right)\)     (M1)

eg  \(f\left( 2 \right) = 3 \times 2 + 1\)

\(f\left( 2 \right) = 7\)     A1 N2

[2 marks]

recognize that the gradient of the graph of g is \(g'\left( x \right)\)      (M1)

choosing chain rule to find \(g'\left( x \right)\)      (M1)

eg  \(\frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},\,\,u = {x^2} + 1,\,\,u' = 2x\)

\(g'\left( x \right) = f'\left( {{x^2} + 1} \right) \times 2x\)     A2

\(g'\left( 1 \right) = 3 \times 2\)     A1

\(g'\left( 1 \right) = 6\)     AG N0

[5 marks]

 at Q, L₁ = L₂ (seen anywhere)      (M1)

recognize that the gradient of L₂ is g'(1)  (seen anywhere)     (M1)
eg  m = 6

finding g (1)  (seen anywhere)      (A1)
eg  \(g\left( 1 \right) = f\left( 2 \right),\,\,g\left( 1 \right) = 7\)

attempt to substitute gradient and/or coordinates into equation of a straight line      M1
eg  \(y - g\left( 1 \right) = 6\left( {x - 1} \right),\,\,y - 1 = g'\left( 1 \right)\left( {x - 7} \right),\,\,7 = 6\left( 1 \right) + {\text{b}}\)

correct equation for L₂ 

eg  \(y - 7 = 6\left( {x - 1} \right),\,\,y = 6x + 1\)     A1

correct working to find Q       (A1)
eg   same y-intercept, \(3x = 0\)

\(y = 1\)     A1 N2

[7 marks]

Use the graph to write down the value of

(i)     a ;

(ii)    c ;

(iii)   d .

Show that \(b = \frac{\pi }{4}\) .

Find \(f'(x)\) .

At a point R, the gradient is \( - 2\pi \) . Find the x-coordinate of R.

(i) \(a = 8\)     A1     N1

(ii) \(c = 2\)     A1     N1

(iii) \(d = 4\)     A1     N1

[3 marks]

METHOD 1

recognizing that period \( = 8\)     (A1)

correct working     A1

e.g. \(8 = \frac{{2\pi }}{b}\) , \(b = \frac{{2\pi }}{8}\)

\(b = \frac{\pi }{4}\)     AG     N0

METHOD 2

attempt to substitute     M1

e.g. \(12 = 8\sin (b(4 - 2)) + 4\)

correct working     A1

e.g. \(\sin 2b = 1\)

\(b = \frac{\pi }{4}\)     AG     N0

[2 marks]

evidence of attempt to differentiate or choosing chain rule     (M1)

e.g. \(\cos \frac{\pi }{4}(x - 2)\) , \(\frac{\pi }{4} \times 8\)

\(f'(x) = 2\pi \cos \left( {\frac{\pi }{4}(x - 2)} \right)\) (accept \(2\pi \cos \frac{\pi }{4}(x - 2)\) )     A2     N3

[3 marks]

recognizing that gradient is \(f'(x)\)     (M1)

e.g. \(f'(x) = m\)

correct equation     A1

e.g. \( - 2\pi  = 2\pi \cos \left( {\frac{\pi }{4}(x - 2)} \right)\) , \( - 1 = \cos \left( {\frac{\pi }{4}(x - 2)} \right)\)

correct working     (A1)

e.g. \({\cos ^{ - 1}}( - 1) = \frac{\pi }{4}(x - 2)\)

using \({\cos ^{ - 1}}( - 1) = \pi \) (seen anywhere)     (A1)

e.g. \(\pi  = \frac{\pi }{4}(x - 2)\)

simplifying     (A1)

e.g. \(4 = (x - 2)\)

\(x = 6\)     A1     N4

[6 marks]

Part (a) of this question proved challenging for most candidates.

Although a good number of candidates recognized that the period was 8 in part (b), there were some who did not seem to realize that this period could be found using the given coordinates of the maximum and minimum points.

In part (c), not many candidates found the correct derivative using the chain rule.

For part (d), a good number of candidates correctly set their expression equal to \( - 2\pi \) , but errors in their previous values kept most from correctly solving the equation. Most candidates who had the correct equation were able to gain full marks here.

Find the first four derivatives of \(f(x)\) .

Write an expression for \({f^{(n)}}(x)\) in terms of x and n .

\(f'(x) = - {x^{ - 2}}\) (or \( - \frac{1}{{{x^2}}}\) )     A1     N1

\(f''(x) = 2{x^{ - 3}}\) (or \(\frac{2}{{{x^3}}}\) )     A1     N1

\(f'''(x) = - 6{x^{ - 4}}\) (or \( - \frac{6}{{{x^4}}}\) )     A1     N1

\({f^{(4)}}(x) = 24{x^{ - 5}}\) (or \(\frac{{24}}{{{x^5}}}\) )     A1     N1

[4 marks]

\({f^{(n)}}(x) = \frac{{{{( - 1)}^n}n!}}{{{x^{n + 1}}}}\) or \({( - 1)^n}n!({x^{ - (n + 1)}})\)     A1A1A1     N3

[3 marks]

Solve for \(0 \le x < 2\pi \)

(i)     \(6 + 6\sin x = 6\) ;

(ii)    \(6 + 6\sin x = 0\) .

Write down the exact value of the x-intercept of f , for \(0 \le x < 2\pi \) .

The area of the shaded region is k . Find the value of k , giving your answer in terms of \(\pi \) .

Let \(g(x) = 6 + 6\sin \left( {x - \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.

Give a full geometric description of this transformation.

Let \(g(x) = 6 + 6\sin \left( {x - \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.

Given that \(\int_p^{p + \frac{{3\pi }}{2}} {g(x){\rm{d}}x}  = k\) and \(0 \le p < 2\pi \) , write down the two values of p.

(i) \(\sin x = 0\)     A1

\(x = 0\) , \(x = \pi \)     A1A1     N2

(ii) \(\sin x = - 1\)     A1

\(x = \frac{{3\pi }}{2}\)    A1     N1

[5 marks]

\(\frac{{3\pi }}{2}\)     A1     N1

[1 mark]

evidence of using anti-differentiation     (M1)

e.g. \(\int_0^{\frac{{3\pi }}{2}} {(6 + 6\sin x){\rm{d}}x} \)

correct integral \(6x - 6\cos x\) (seen anywhere)     A1A1

correct substitution     (A1)

e.g. \(6\left( {\frac{{3\pi }}{2}} \right) - 6\cos \left( {\frac{{3\pi }}{2}} \right) - ( - 6\cos 0)\) , \(9\pi  - 0 + 6\)

\(k = 9\pi + 6\)     A1A1     N3

[6 marks]

translation of \(\left( {\begin{array}{*{20}{c}}
{\frac{\pi }{2}}\\
0
\end{array}} \right)\)     A1A1     N2

[2 marks]

recognizing that the area under g is the same as the shaded region in f     (M1)

\(p = \frac{\pi }{2}\) , \(p = 0\)     A1A1     N3

[3 marks]

Many candidates again had difficulty finding the common angles in the trigonometric equations. In part (a), some did not show sufficient working in solving the equations. Others obtained a single solution in (a)(i) and did not find another. Some candidates worked in degrees; the majority worked in radians.

While some candidates appeared to use their understanding of the graph of the original function to find the x-intercept in part (b), most used their working from part (a)(ii) sometimes with follow-through on an incorrect answer.

Most candidates recognized the need for integration in part (c) but far fewer were able to see the solution through correctly to the end. Some did not show the full substitution of the limits, having incorrectly assumed that evaluating the integral at 0 would be 0; without this working, the mark for evaluating at the limits could not be earned. Again, many candidates had trouble working with the common trigonometric values.

While there was an issue in the wording of the question with the given domains, this did not appear to bother candidates in part (d). This part was often well completed with candidates using a variety of language to describe the horizontal translation to the right by \(\frac{\pi }{2}\) .

Most candidates who attempted part (e) realized that the integral was equal to the value that they had found in part (c), but a majority tried to integrate the function g without success. Some candidates used sketches to find one or both values for p. The problem in the wording of the question did not appear to have been noticed by candidates in this part either.

Write down \(f'(x)\).

Find the gradient of \(L\).

Show that the \(x\)-coordinate of B is \( - \frac{k}{2}\).

Find the area of triangle ABC, giving your answer in terms of \(k\).

Given that the area of triangle ABC is \(p\) times the area of \(R\), find the value of \(p\).

\(f'(x) = 2x\)     A1     N1

[1 mark]

attempt to substitute \(x =  - k\) into their derivative     (M1)

gradient of \(L\) is \( - 2k\)     A1     N2

[2 marks]

METHOD 1 

attempt to substitute coordinates of A and their gradient into equation of a line     (M1)

eg\(\,\,\,\,\,\)\({k^2} =  - 2k( - k) + b\)

correct equation of \(L\) in any form     (A1)

eg\(\,\,\,\,\,\)\(y - {k^2} =  - 2k(x + k),{\text{ }}y =  - 2kx - {k^2}\)

valid approach     (M1)

eg\(\,\,\,\,\,\)\(y = 0\)

correct substitution into \(L\) equation     A1

eg\(\,\,\,\,\,\)\( - {k^2} =  - 2kx - 2{k^2},{\text{ }}0 =  - 2kx - {k^2}\)

correct working     A1

eg\(\,\,\,\,\,\)\(2kx =  - {k^2}\)

\(x =  - \frac{k}{2}\)     AG     N0

METHOD 2

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{gradient}} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}},{\text{ }} - 2k = \frac{{{\text{rise}}}}{{{\text{run}}}}\)

recognizing \(y = 0\) at B     (A1)

attempt to substitute coordinates of A and B into slope formula     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{k^2} - 0}}{{ - k - x}},{\text{ }}\frac{{ - {k^2}}}{{x + k}}\)

correct equation     A1

eg\(\,\,\,\,\,\)\(\frac{{{k^2} - 0}}{{ - k - x}} =  - 2k,{\text{ }}\frac{{ - {k^2}}}{{x + k}} =  - 2k,{\text{ }} - {k^2} =  - 2k(x + k)\)

correct working     A1

eg\(\,\,\,\,\,\)\(2kx =  - {k^2}\)

\(x =  - \frac{k}{2}\)     AG     N0

[5 marks]

valid approach to find area of triangle     (M1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}({k^2})\left( {\frac{k}{2}} \right)\)

area of \({\text{ABC}} = \frac{{{k^3}}}{4}\)     A1     N2

[2 marks]

METHOD 1 (\(\int {f - {\text{triangle}}} \))

valid approach to find area from \( - k\) to 0     (M1)

eg\(\,\,\,\,\,\)\(\int_{ - k}^0 {{x^2}{\text{d}}x,{\text{ }}\int_0^{ - k} f } \)

correct integration (seen anywhere, even if M0 awarded)     A1

eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3},{\text{ }}\left[ {\frac{1}{3}{x^3}} \right]_{ - k}^0\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(0 - \frac{{{{( - k)}^3}}}{3}\), area from \( - k\) to 0 is \(\frac{{{k^3}}}{3}\)

Note:     Award M0 for substituting into original or differentiated function.

attempt to find area of \(R\)     (M1)

eg\(\,\,\,\,\,\)\(\int_{ - k}^0 {f(x){\text{d}}x - {\text{ triangle}}} \)

correct working for \(R\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{3} - \frac{{{k^3}}}{4},{\text{ }}R = \frac{{{k^3}}}{{12}}\)

correct substitution into \({\text{triangle}} = pR\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{3} - \frac{{{k^3}}}{4}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)

\(p = 3\)     A1     N2

METHOD 2 (\(\int {(f - L)} \))

valid approach to find area of \(R\)     (M1)

eg\(\,\,\,\,\,\)\(\int_{ - k}^{ - \frac{k}{2}} {{x^2} - ( - 2kx - {k^2}){\text{d}}x + \int_{ - \frac{k}{2}}^0 {{x^2}{\text{d}}x,{\text{ }}\int_{ - k}^{ - \frac{k}{2}} {(f - L) + \int_{ - \frac{k}{2}}^0 f } } } \)

correct integration (seen anywhere, even if M0 awarded)     A2

eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3} + k{x^2} + {k^2}x,{\text{ }}\left[ {\frac{{{x^3}}}{3} + k{x^2} + {k^2}x} \right]_{ - k}^{ - \frac{k}{2}} + \left[ {\frac{{{x^3}}}{3}} \right]_{ - \frac{k}{2}}^0\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(\left( {\frac{{{{\left( { - \frac{k}{2}} \right)}^3}}}{3} + k{{\left( { - \frac{k}{2}} \right)}^2} + {k^2}\left( { - \frac{k}{2}} \right)} \right) - \left( {\frac{{{{( - k)}^3}}}{3} + k{{( - k)}^2} + {k^2}( - k)} \right) + (0) - \left( {\frac{{{{\left( { - \frac{k}{2}} \right)}^3}}}{3}} \right)\)

Note:     Award M0 for substituting into original or differentiated function.

correct working for \(R\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}},{\text{ }} - \frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{4} - \frac{{{k^3}}}{2} + \frac{{{k^3}}}{3} - {k^3} + {k^3} + \frac{{{k^3}}}{{24}},{\text{ }}R = \frac{{{k^3}}}{{12}}\)

correct substitution into \({\text{triangle}} = pR\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)

\(p = 3\)     A1     N2

[7 marks]

Find \(f'(1)\).

Write down \(g'(1)\).

Find \(g(1)\).

Let \(h(x) = f(x) \times g(x)\). Find the equation of the tangent to the graph of \(h\) at the point where \(x = 1\).

choosing chain rule     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},{\text{ }}u = 4x + 5,{\text{ }}u' = 4\)

correct derivative of \(f\)     A2

eg\(\,\,\,\,\,\)\(\frac{1}{2}{(4x + 5)^{ - \frac{1}{2}}} \times 4,{\text{ }}f'(x) = \frac{2}{{\sqrt {4x + 5} }}\)

\(f'(1) = \frac{2}{3}\)    A1     N2

[4 marks]

recognize that \(g'(x)\) is the gradient of the tangent     (M1)

eg\(\,\,\,\,\,\)\(g'(x) = m\)

\(g'(1) = 3\)    A1     N2

[2 marks]

recognize that R is on the tangent     (M1)

eg\(\,\,\,\,\,\)\(g(1) = 3 \times 1 + 6\), sketch

\(g(1) = 9\)    A1     N2

[2 marks]

\(f(1) = \sqrt {4 + 5} {\text{ }}( = 3)\) (seen anywhere)     A1

\(h(1) = 3 \times 9{\text{ }}( = 27)\) (seen anywhere)     A1

choosing product rule to find \(h'(x)\)     (M1)

eg\(\,\,\,\,\,\)\(uv' + u'v\)

correct substitution to find \(h'(1)\)     (A1)

eg\(\,\,\,\,\,\)\(f(1) \times g'(1) + f'(1) \times g(1)\)

\(h'(1) = 3 \times 3 + \frac{2}{3} \times 9{\text{ }}( = 15)\)     A1

EITHER

attempt to substitute coordinates (in any order) into the equation of a straight line     (M1)

eg\(\,\,\,\,\,\)\(y - 27 = h'(1)(x - 1),{\text{ }}y - 1 = 15(x - 27)\)

\(y - 27 = 15(x - 1)\)     A1     N2

OR

attempt to substitute coordinates (in any order) to find the \(y\)-intercept     (M1)

eg\(\,\,\,\,\,\)\(27 = 15 \times 1 + b,{\text{ }}1 = 15 \times 27 + b\)

\(y = 15x + 12\)     A1     N2

[7 marks]

Part a) was relatively well answered – the obvious errors seen were not using the chain rule correctly and simple fraction calculations being wrong.

In parts b) and c) it seemed that the students did not have a good conceptual understanding of what was actually happening in this question. There was lack of understanding of tangents, gradients and their relationship to the original function, \(g\). A working sketch may have been beneficial but few were seen and many did a lot more work than required.

In parts b) and c) it seemed that the students did not have a good conceptual understanding of what was actually happening in this question. There was lack of understanding of tangents, gradients and their relationship to the original function, \(g\). A working sketch may have been beneficial but few were seen and many did a lot more work than required.

In part d) although candidates recognized \(h(x)\) as a product and may have correctly found \(h(1)\), they did not necessarily use the product rule to find \(h'(x)\), instead incorrectly using \(h'(x) = f'(x) \times g'(x)\). It was rare for a candidate to get as far as finding the equation of a straight line but those who did usually gained full marks.

(i)     Find the first four derivatives of \(f(x)\).

(ii)     Find \({f^{(19)}}(x)\).

(i)     Find the first three derivatives of \(g(x)\).

(ii)     Given that \({g^{(19)}}(x) = \frac{{k!}}{{(k - p)!}}({x^{k - 19}})\), find \(p\).

(i)     Find \(h'(x)\).

(ii)     Hence, show that \(h'(\pi ) = \frac{{ - 21!}}{2}{\pi ^2}\).

(i)     \(f'(x) =  - \sin x,{\text{ }}f''(x) =  - \cos x,{\text{ }}{f^{(3)}}(x) = \sin x,{\text{ }}{f^{(4)}}(x) = \cos x\)     A2     N2

(ii)     valid approach     (M1)

eg\(\,\,\,\,\,\)recognizing that 19 is one less than a multiple of 4, \({f^{(19)}}(x) = {f^{(3)}}(x)\)

\({f^{(19)}}(x) = \sin x\)     A1     N2

[4 marks]

(i)     \(g'(x) = k{x^{k - 1}}\)

\(g''(x) = k(k - 1){x^{k - 2}},{\text{ }}{g^{(3)}}(x) = k(k - 1)(k - 2){x^{k - 3}}\)     A1A1     N2

(ii)     METHOD 1

correct working that leads to the correct answer, involving the correct expression for the 19th derivative     A2

eg\(\,\,\,\,\,\)\(k(k - 1)(k - 2) \ldots (k - 18) \times \frac{{(k - 19)!}}{{(k - 19)!}},{{\text{ }}_k}{P_{19}}\)

\(p = 19\) (accept \(\frac{{k!}}{{(k - 19)!}}{x^{k - 19}}\))     A1     N1

METHOD 2

correct working involving recognizing patterns in coefficients of first three derivatives (may be seen in part (b)(i)) leading to a general rule for 19th coefficient     A2

eg\(\,\,\,\,\,\)\(g'' = 2!\left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right),{\text{ }}k(k - 1)(k - 2) = \frac{{k!}}{{(k - 3)!}},{\text{ }}{g^{(3)}}(x){ = _k}{P_3}({x^{k - 3}})\)

\({g^{(19)}}(x) = 19!\left( {\begin{array}{*{20}{c}} k \\ {19} \end{array}} \right),{\text{ }}19! \times \frac{{k!}}{{(k - 19)! \times 19!}},{{\text{ }}_k}{P_{19}}\)

\(p = 19\) (accept \(\frac{{k!}}{{(k - 19)!}}{x^{k - 19}}\))     A1     N1

[5 marks]

(i)     valid approach using product rule     (M1)

eg\(\,\,\,\,\,\)\(uv' + vu',{\text{ }}{f^{(19)}}{g^{(20)}} + {f^{(20)}}{g^{(19)}}\)

correct 20th derivatives (must be seen in product rule)     (A1)(A1)

eg\(\,\,\,\,\,\)\({g^{(20)}}(x) = \frac{{21!}}{{(21 - 20)!}}x,{\text{ }}{f^{(20)}}(x) = \cos x\)

\(h'(x) = \sin x(21!x) + \cos x\left( {\frac{{21!}}{2}{x^2}} \right){\text{ }}\left( {{\text{accept }}\sin x\left( {\frac{{21!}}{{1!}}x} \right) + \cos x\left( {\frac{{21!}}{{2!}}{x^2}} \right)} \right)\)    A1     N3

(ii)     substituting \(x = \pi \) (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\({f^{(19)}}(\pi ){g^{(20)}}(\pi ) + {f^{(20)}}(\pi ){g^{(19)}}(\pi ),{\text{ }}\sin \pi \frac{{21!}}{{1!}}\pi  + \cos \pi \frac{{21!}}{{2!}}{\pi ^2}\)

evidence of one correct value for \(\sin \pi \) or \(\cos \pi \) (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(\sin \pi  = 0,{\text{ }}\cos \pi  =  - 1\)

evidence of correct values substituted into \(h'(\pi )\)     A1

eg\(\,\,\,\,\,\)\(21!(\pi )\left( {0 - \frac{\pi }{{2!}}} \right),{\text{ }}21!(\pi )\left( { - \frac{\pi }{2}} \right),{\text{ }}0 + ( - 1)\frac{{21!}}{2}{\pi ^2}\)

Note: If candidates write only the first line followed by the answer, award A1A0A0.

\(\frac{{ - 21!}}{2}{\pi ^2}\)     AG     N0

[7 marks]

(i)     Write down the value of \(f'(x)\) at C.

(ii)    Hence, show that C corresponds to a minimum on the graph of f , i.e. it has the same x-coordinate.

Which of the points A, B, D corresponds to a maximum on the graph of f ?

Show that B corresponds to a point of inflexion on the graph of f .

(i) \(f'(x) = 0\)     A1     N1

(ii) METHOD 1

 \(f'(x) < 0\) to the left of C, \(f'(x) > 0\) to the right of C     R1R1     N2

METHOD 2

\(f''(x) > 0\)     R2     N2

[3 marks]

A     A1     N1

[1 mark]

METHOD 1

\(f''(x) = 0\)     R2

discussion of sign change of \(f''(x)\)     R1

e.g. \(f''(x) < 0\) to the left of B and \(f''(x) > 0\) to the right of B; \(f''(x)\) changes sign either side of B

B is a point of inflexion     AG     N0

METHOD 2

B is a minimum on the graph of the derivative \({f'}\)     R2

discussion of sign change of \(f''(x)\)     R1

e.g. \(f''(x) < 0\) to the left of B and \(f''(x) > 0\) to the right of B; \(f''(x)\) changes sign either side of B

B is a point of inflexion     AG     N0

[3 marks]

The variation in successful and unsuccessful responses to this question was remarkable. Many candidates did not even attempt it. Candidates could often determine from the graph, the minimum and maximum values of the original function, but few could correctly use the graph to analyse and justify these results. Responses indicated that some candidates did not realize that they were looking at the graph of \({f'}\) and not the graph of \(f\) .

The variation in successful and unsuccessful responses to this question was remarkable. Many candidates did not even attempt it. Candidates could often determine from the graph, the minimum and maximum values of the original function, but few could correctly use the graph to analyse and justify these results. Responses indicated that some candidates did not realize that they were looking at the graph of \({f'}\) and not the graph of \(f\) .

In part (c), many candidates once more failed to respect the command term "show" and often provided an incomplete answer. Candidates should be encouraged to refer to the number of marks available for a particular part when deciding how much information should be given.

Write down \(f'(x)\) .

The tangent to the graph of f at the point \({\text{P}}(0{\text{, }}b)\) has gradient m .

(i)     Show that \(m = 6\) .

(ii)    Find b .

Hence, write down the equation of this tangent.

\(f'(x) = 6{{\rm{e}}^{6x}}\)     A1     N1

[1 mark]

(i) evidence of valid approach     (M1)

e.g. \(f'(0)\) ,  \(6{{\rm{e}}^{6 \times 0}}\)

correct manipulation     A1

e.g. \(6{{\rm{e}}^0}\) , \(6 \times 1\)

\(m = 6\)    AG     N0

(ii) evidence of finding \(f(0)\)     (M1)

e.g. \(y = {{\rm{e}}^{6(0)}}\)

\(b = 1\)     A1     N2

[4 marks]

\(y = 6x + 1\)     A1     N1

[1 mark]

On the whole, candidates handled this question quite well with most candidates correctly applying the chain rule to an exponential function and successfully finding the equation of the tangent line.

On the whole, candidates handled this question quite well with most candidates correctly applying the chain rule to an exponential function and successfully finding the equation of the tangent line. Some candidates lost a mark in (b)(i) for not showing sufficient working leading to the given answer.

On the whole, candidates handled this question quite well.

Write down the value of \(g(3)\) , of \(f'(3)\) , and of \(h''(2)\) .

Explain why P is a point of inflexion.

find the \(y\)-coordinate of P.

find the equation of the normal to the graph of \(h\) at P.

\(g(3) = - 18\) , \(f'(3) = 1\) , \(h''(2) = - 6\)     A1A1A1     N3

[3 marks]

\(h''(3) = 0\)     (A1)

valid reasoning     R1

eg   \({h''}\) changes sign at \(x = 3\) , change in concavity of \(h\) at \(x = 3\)

so P is a point of inflexion     AG     N0

[2 marks]

writing \(h(3)\) as a product of \(f(3)\) and \(g(3)\)     A1

eg   \(f(3) \times g(3)\) , \(3 \times ( - 18)\)

\(h(3) = - 54\)     A1 N1

[2 marks]

recognizing need to find derivative of \(h\)     (R1)

eg   \({h'}\) , \(h'(3)\)

attempt to use the product rule (do not accept \(h' = f' \times g'\) )     (M1)

eg   \(h' = fg' + gf'\) ,  \(h'(3) = f(3) \times g'(3) + g(3) \times f'(3)\)

correct substitution     (A1)

eg   \(h'(3) = 3( - 3) + ( - 18) \times 1\)

\(h'(3) = - 27\)    A1

attempt to find the gradient of the normal     (M1)

eg   \( - \frac{1}{m}\) , \( - \frac{1}{{27}}x\) 

attempt to substitute their coordinates and their normal gradient into the equation of a line     (M1)

eg   \( - 54 = \frac{1}{{27}}(3) + b\) , \(0 = \frac{1}{{27}}(3) + b\) , \(y + 54 = 27(x - 3)\) , \(y - 54 = \frac{1}{{27}}(x + 3)\)

correct equation in any form     A1     N4

eg   \(y + 54 = \frac{1}{{27}}(x - 3)\) , \(y = \frac{1}{{27}}x - 54\frac{1}{9}\)

[7 marks]

Nearly all candidates who attempted to answer parts (a) and (c) did so correctly, as these questions simply required them to understand the notation being used and to read the values from the given table.

In part (b), the majority of candidates earned one mark for stating that \(h''(x) = 0\) at point P. As this is not enough to determine a point of inflexion, very few candidates earned full marks on this question.

Nearly all candidates who attempted to answer parts (a) and (c) did so correctly, as these questions simply required them to understand the notation being used and to read the values from the given table.

Part (d) proved to be quite challenging for even the strongest candidates, as almost none of them used the product rule to find \(h'(3)\). The most common error was to say \(h'(3) = f'(3) \times g'(3)\). Despite this error, many candidates were able to earn further method marks for their work in finding the equation of the normal. There were also a small number of candidates who were able to find the equation for \(h'(x)\) , and from that \(h''(x)\). These candidates were often successful in earning full marks, although this method was quite time-consuming.

Find  \(\int {\frac{1}{{2x + 3}}} {\rm{d}}x\) .

Given that \(\int_0^3 {\frac{1}{{2x + 3}}} {\rm{d}}x = \ln \sqrt P \) , find the value of P.

\(\int {\frac{1}{{2x + 3}}} {\rm{d}}x = \frac{1}{2}\ln (2x + 3) + C\)  (accept \(\frac{1}{2}\ln |(2x + 3)| + C\) )    A1A1     N2

[2 marks]

\(\int_0^3 {\frac{1}{{2x + 3}}} {\rm{d}}x = \left[ {\frac{1}{2}\ln (2x + 3)} \right]_0^3\)

evidence of substitution of limits     (M1)

e.g.\(\frac{1}{2}\ln 9 - \frac{1}{2}\ln 3\)

evidence of correctly using \(\ln a - \ln b = \ln \frac{a}{b}\) (seen anywhere)     (A1)

e.g. \(\frac{1}{2}\ln 3\)

evidence of correctly using \(a\ln b = \ln {b^a}\) (seen anywhere)     (A1)

e.g. \(\ln \sqrt {\frac{9}{3}} \)

\(P = 3\) (accept \(\ln \sqrt 3 \) )     A1     N2

[4 marks]

Many candidates were unable to correctly integrate but did recognize that the integral involved the natural log function; they most often missed the factor \(\frac{1}{2}\) or replaced it with 2.

Part (b) proved difficult as many were unable to use the basic rules of logarithms.

The function can be written in the form \(f(x) = a{(x - h)^2} + k\) .

(i)     Write down the value of h and of k .

(ii)    Show that \(a = 3\) .

Find \(f(x)\)  , giving your answer in the form \(A{x^2} + Bx + C\) .

Calculate the area enclosed by the graph of f , the x-axis, and the lines \(x = 2\) and \(x = 4\) .

(i) \(h = 2\) , \(k = 1\)     A1A1     N2

(ii) attempt to substitute coordinates of any point (except the vertex) on the graph into f     M1

e.g. \(13 = a{(0 - 2)^2} + 1\)

working towards solution     A1

e.g. \(13 = 4a + 1\)

\(a = 3\)     AG     N0

[4 marks]

attempting to expand their binomial     (M1)

e.g. \(f(x) = 3({x^2} - 2 \times 2x + 4) + 1\) , \({(x - 2)^2} = {x^2} - 4x + 4\)

correct working     (A1)

e.g. \(f(x) = 3{x^2} - 12x + 12 + 1\)

\(f(x) = 3{x^2} - 12x + 13\) (accept \(A = 3\) , \(B = - 12\) , \(C = 13\) )     A1     N2

[3 marks]

METHOD 1

integral expression     (A1)

e.g. \(\int_2^4 {(3{x^2}}  - 12x + 13)\) , \(\int {f{\rm{d}}x} \)

\({\rm{Area}} = [{x^3} - 6{x^2} + 13x]_2^4\)     A1A1A1

Note: Award A1 for \({x^3}\) , A1 for \( - 6{x^2}\) , A1 for \(13x\) .

correct substitution of correct limits into their expression     A1A1

e.g. \(({4^3} - 6 \times {4^2} + 13 \times 4) - ({2^3} - 6 \times {2^2} + 13 \times 2)\) , \(64 - 96 + 52 - (8 - 24 + 26)\)

Note: Award A1 for substituting 4, A1 for substituting 2.

correct working     (A1)

e.g. \(64 - 96 + 52 - 8 + 24 - 26,20 - 10\)

\({\rm{Area}} = 10\)     A1     N3

[8 marks]

METHOD 2

integral expression     (A1)

e.g. \(\int_2^4 {(3{{(x - 2)}^2}}  + 1)\) , \(\int {f{\rm{d}}x} \)

\({\rm{Area}} = [{(x - 2)^3} + x]_2^4\)     A2A1

Note: Award A2 for \({(x - 2)^3}\) , A1 for \(x\) .

correct substitution of correct limits into their expression     A1A1

e.g. \({(4 - 2)^3} + 4 - [{(2 - 2)^3} + 2]\) , \({2^3} + 4 - ({0^3} + 2)\) , \({2^3} + 4 - 2\)

Note: Award A1 for substituting 4, A1 for substituting 2.

correct working     (A1)

e.g. \(8 + 4 - 2\)

\({\rm{Area}} = 10\)    A1     N3

[8 marks]

METHOD 3

recognizing area from 0 to 2 is same as area from 2 to 4     (R1)

e.g. sketch, \(\int_2^4 {f = \int_0^2 f } \)

integral expression     (A1)

e.g. \(\int_0^2 {(3{x^2}}  - 12x + 13)\) , \(\int {f{\rm{d}}x} \)

\({\rm{Area}} = [{x^3} - 6{x^2} + 13x]_0^2\)     A1A1A1

Note: Award A1 for \({x^3}\) , A1 for \( - 6{x^2}\) , A1 for \(13x\) .

correct substitution of correct limits into their expression     A1(A1)

e.g. \(({2^3} - 6 \times {2^2} + 13 \times 2) - ({0^3} - 6 \times {0^2} + 13 \times 0)\) , \(8 - 24 + 26\)

Note: Award A1 for substituting 2, (A1) for substituting 0.

\({\rm{Area}} = 10\)     A1     N3

[8 marks]

In part (a), nearly all the candidates recognized that h and k were the coordinates of the vertex of the parabola, and most were able to successfully show that \(a = 3\) . Unfortunately, a few candidates did not understand the "show that" command, and simply verified that \(a = 3\) would work, rather than showing how to find \(a = 3\) .

In part (b), most candidates were able to find \(f(x)\) in the required form. For a few candidates, algebraic errors kept them from finding the correct function, even though they started with correct values for a, h and k.

In part (c), nearly all candidates knew that they needed to integrate to find the area, but errors in integration, and algebraic and arithmetic errors prevented many from finding the correct area.

(i)     If \(s = 100\) when \(t = 0\) , find an expression for s in terms of a and t.

(ii)    If \(s = 0\) when \(t = 0\) , write down an expression for s in terms of a and t.

A train M slows down so that it comes to a stop at the station.

(i)     Find the time it takes train M to come to a stop, giving your answer in terms of a.

(ii)    Hence show that \(a = \frac{8}{5}\) .

For a different train N, the value of a is 4.

Show that this train will stop before it reaches the station.

Note: In this question, do not penalize absence of units.

(i) \(s = \int {(40 - at){\rm{d}}t} \)     (M1)

\(s = 40t - \frac{1}{2}a{t^2} + c\)     (A1)(A1)

substituting \(s = 100\) when \(t = 0\) (\(c = 100\) )     (M1)

\(s = 40t - \frac{1}{2}a{t^2} + 100\)     A1     N5

(ii) \(s = 40t - \frac{1}{2}a{t^2}\)     A1     N1

[6 marks]

(i) stops at station, so \(v = 0\)     (M1)

\(t = \frac{{40}}{a}\) (seconds)     A1     N2

(ii) evidence of choosing formula for s from (a) (ii)     (M1)

substituting \(t = \frac{{40}}{a}\)     (M1)

e.g. \(40 \times \frac{{40}}{a} - \frac{1}{2}a \times \frac{{{{40}^2}}}{{{a^2}}}\)

setting up equation     M1

e.g. \(500 = s\) , \(500 = 40 \times \frac{{40}}{a} - \frac{1}{2}a \times \frac{{{{40}^2}}}{{{a^2}}}\) , \(500 = \frac{{1600}}{a} - \frac{{800}}{a}\)

evidence of simplification to an expression which obviously leads to \(a = \frac{8}{5}\)     A1

e.g. \(500a = 800\) , \(5 = \frac{8}{a}\) , \(1000a = 3200 - 1600\)

\(a = \frac{8}{5}\)     AG     N0

[6 marks]

METHOD 1

\(v = 40 - 4t\) , stops when \(v = 0\)

\(40 - 4t = 0\)     (A1)

\(t = 10\)     A1

substituting into expression for s     M1

\(s = 40 \times 10 - \frac{1}{2} \times 4 \times {10^2}\)

\(s = 200\)     A1

since \(200 < 500\) (allow FT on their s, if \(s < 500\) )     R1

train stops before the station     AG     N0

METHOD 2

from (b) \(t = \frac{{40}}{4} = 10\)     A2

substituting into expression for s

e.g. \(s = 40 \times 10 - \frac{1}{2} \times 4 \times {10^2}\)     M1

\(s = 200\)     A1

since \(200 < 500\)      R1

train stops before the station     AG     N0

METHOD 3

a is deceleration     A2

\(4 > \frac{8}{5}\)    A1

so stops in shorter time     (A1)

so less distance travelled     R1

so stops before station     AG     N0

[5 marks]

Part (a) proved accessible for most.

Part (b), simple as it is, proved elusive as many candidates did not make the connection that \(v = 0\) when the train stops. Instead, many attempted to find the value of t using \(a = \frac{8}{5}\) .

Few were successful in part (c).

Find \(\int {x{{\text{e}}^{{x^2} - 1}}{\text{d}}x} \).

Find \(f(x)\), given that \(f’(x) = x{{\text{e}}^{{x^2} - 1}}\) and \(f( - 1) = 3\).

valid approach to set up integration by substitution/inspection     (M1)

eg\(\,\,\,\,\,\)\(u = {x^2} - 1,{\text{ d}}u = 2x,{\text{ }}\int {2x{{\text{e}}^{{x^2} - 1}}{\text{d}}x} \)

correct expression     (A1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}\int {2x{{\text{e}}^{{x^2} - 1}}{\text{d}}x,{\text{ }}\frac{1}{2}\int {{{\text{e}}^u}{\text{d}}u} } \)

\(\frac{1}{2}{{\text{e}}^{{x^2} - 1}} + c\)     A2     N4

Notes: Award A1 if missing “\( + c\)”.

[4 marks]

substituting \(x =  - 1\) into their answer from (a)     (M1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}{{\text{e}}^0},{\text{ }}\frac{1}{2}{{\text{e}}^{1 - 1}} = 3\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\frac{1}{2} + c = 3,{\text{ }}c = 2.5\)

\(f(x) = \frac{1}{2}{{\text{e}}^{{x^2} - 1}} + 2.5\)     A1     N2

[3 marks]

Let \(f(x) = k{x^4}\) . The point \({\text{P}}(1{\text{, }}k)\) lies on the curve of f . At P, the normal to the curve is parallel to \(y = - \frac{1}{8}x\) . Find the value of k.

gradient of tangent \(= 8\) (seen anywhere)     (A1)

\(f'(x) = 4k{x^3}\) (seen anywhere)     A1

recognizing the gradient of the tangent is the derivative     (M1)

setting the derivative equal to 8     (A1)

e.g. \(4k{x^3} = 8\) , \(k{x^3} = 2\)

substituting \(x = 1\) (seen anywhere)     (M1)

\(k = 2\)     A1    N4

[6 marks]

Candidates‟ success with this question was mixed. Those who understood the relationship between the derivative and the gradient of the normal line were not bothered by the lack of structure in the question, solving clearly with only a few steps, earning full marks. Those who were unclear often either gained a few marks for finding the derivative and substituting \(x = 1\) , or no marks for working that did not employ the derivative. Misunderstandings included simply finding the equation of the tangent or normal line, setting the derivative equal to the gradient of the normal, and equating the function with the normal or tangent line equation. Among the candidates who demonstrated greater understanding, more used the gradient of the normal (the equation \( - \frac{1}{4}k = - \frac{1}{8}\) ) than the gradient of the tangent (\(4k = 8\) ) ; this led to more algebraic errors in obtaining the final answer of \(k = 2\) . A number of unsuccessful candidates wrote down a lot of irrelevant mathematics with no plan in mind and earned no marks.

The graph of the function \(y = f(x)\) passes through the point \(\left( {\frac{3}{2},4} \right)\) . The gradient function of f is given as \(f'(x) = \sin (2x - 3)\) . Find \(f(x)\) .

evidence of integration

e.g. \(f(x) = \int {\sin (2x - 3){\rm{d}}x} \)     (M1)

\( = - \frac{1}{2}\cos (2x - 3) + C\)     A1A1

substituting initial condition into their expression (even if C is missing)     M1

e.g. \(4 = - \frac{1}{2}\cos 0 + C\)

\(C = 4.5\)     (A1)

\(f(x) = - \frac{1}{2}\cos (2x - 3) + 4.5\)     A1     N5

[6 marks]

While most candidates realized they needed to integrate in this question, many did so unsuccessfully. Many did not account for the coefficient of x, and failed to multiply by \(\frac{1}{2}\). Some of the candidates who substituted the initial condition into their integral were not able to solve for "c", either because of arithmetic errors or because they did not know the correct value for \(\cos 0\) .

On the following axes, sketch the graph of \(y = f'(x)\).

Write down the following in order from least to greatest: \(f(0),{\text{ }}f'(6),{\text{ }}f''( - 2)\).

     A1A1A1A1     N4

Note: Award A1 for x-intercept in circle at \(-2\), A1 for x-intercept in circle at \(6\).

     Award A1 for approximately correct shape.

     Only if this A1 is awarded, award A1 for a negative y-intercept.

[4 marks]

\(f''( - 2),{\text{ }}f'(6),{\text{ }}f(0)\)     A2     N2

[2 marks]

Explain why the graph of \(f\) has a local minimum when \(x = 5\).

Find the set of values of \(x\) for which the graph of \(f\) is concave down.

The following diagram shows the shaded regions \(A\), \(B\) and \(C\).

The regions are enclosed by the graph of \(f'\), the \(x\)-axis, the \(y\)-axis, and the line \(x = 6\).

The area of region \(A\) is 12, the area of region \(B\) is 6.75 and the area of region \(C\) is 6.75.

Given that \(f(0) = 14\), find \(f(6)\).

The following diagram shows the shaded regions \(A\), \(B\) and \(C\).

The regions are enclosed by the graph of \(f'\), the x-axis, the y-axis, and the line \(x = 6\).

The area of region \(A\) is 12, the area of region \(B\) is 6.75 and the area of region \(C\) is 6.75.

Let \(g(x) = {\left( {f(x)} \right)^2}\). Given that \(f'(6) = 16\), find the equation of the tangent to the graph of \(g\) at the point where \(x = 6\).

METHOD 1

\(f'(5) = 0\)     (A1)

valid reasoning including reference to the graph of \(f'\)     R1

eg\(\;\;\;f'\) changes sign from negative to positive at \(x = 5\), labelled sign chart for \(f'\)

so \(f\) has a local minimum at \(x = 5\)     AG     N0

Note:     It must be clear that any description is referring to the graph of \(f'\), simply giving the conditions for a minimum without relating them to \(f'\) does not gain the R1.

METHOD 2

\(f'(5) = 0\)     A1

valid reasoning referring to second derivative     R1

eg\(\;\;\;f''(5) > 0\)

so \(f\) has a local minimum at \(x = 5\)     AG     N0

[2 marks]

attempt to find relevant interval     (M1)

eg\(\;\;\;f'\) is decreasing, gradient of \(f'\) is negative, \(f'' < 0\)

\(2 < x < 4\;\;\;\)(accept “between 2 and 4”)     A1     N2

Notes:     If no other working shown, award M1A0 for incorrect inequalities such as \(2 \le \) \(x\) \( \le \) 4, or “from 2 to 4”

[2 marks]

METHOD 1 (one integral)

correct application of Fundamental Theorem of Calculus     (A1)

eg\(\;\;\;\int_0^6 {f'(x){\text{d}}x = } f(6) - f(0),{\text{ }}f(6) = 14 + \int_0^6 {f'(x){\text{d}}x} \)

attempt to link definite integral with areas     (M1)

eg\(\;\;\;\int_0^6 {f'(x){\text{d}}x =  - 12 - 6.75 + 6.75,{\text{ }}\int_0^6 {f'(x){\text{d}}x = {\text{Area }}A + {\text{Area }}B + {\text{ Area }}C} } \)

correct value for \(\int_0^6 {f'(x){\text{d}}x} \)     (A1)

eg\(\;\;\;\int_0^6 {f'(x){\text{d}}x}  =  - 12\)

correct working     A1

eg\(\;\;\;f(6) - 14 =  - 12,{\text{ }}f(6) =  - 12 + f(0)\)

\(f(6) = 2\)     A1     N3

METHOD 2 (more than one integral)

correct application of Fundamental Theorem of Calculus     (A1)

eg\(\;\;\;\int_0^2 {f'(x){\text{d}}x}  = f(2) - f(0),{\text{ }}f(2) = 14 + \int_0^2 {f'(x)} \)

attempt to link definite integrals with areas     (M1)

eg\(\;\;\;\int_0^2 {f'(x){\text{d}}x}  = 12,{\text{ }}\int_2^5 {f'(x){\text{d}}x =  - 6.75} ,{\text{ }}\int_0^6 {f'(x)}  = 0\)

correct values for integrals     (A1)

eg\(\;\;\;\int_0^2 {f'(x){\text{d}}x}  =  - 12,{\text{ }}\int_5^2 {f'(x)} {\text{d}}x = 6.75,{\text{ }}f(6) - f(2) = 0\)

one correct intermediate value     A1

eg\(\;\;\;f(2) = 2,{\text{ }}f(5) =  - 4.75\)

\(f(6) = 2\)     A1     N3

[5 marks]

correct calculation of \(g(6)\) (seen anywhere)     A1

eg\(\;\;\;{2^2},{\text{ }}g(6) = 4\)

choosing chain rule or product rule     (M1)

eg\(\;\;\;g'\left( {f(x)} \right)f'(x),{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},{\text{ }}f(x)f'(x) + f'(x)f(x)\)

correct derivative     (A1)

eg\(\;\;\;g'(x) = 2f(x)f'(x),{\text{ }}f(x)f'(x) + f'(x)f(x)\)

correct calculation of \(g'(6)\) (seen anywhere)     A1

eg\(\;\;\;2(2)(16),{\text{ }}g'(6) = 64\)

attempt to substitute their values of \(g'(6)\) and \(g(6)\) (in any order) into equation of a line     (M1)

eg\(\;\;\;{2^2} = (2 \times 2 \times 16)6 + b,{\text{ }}y - 6 = 64(x - 4)\)

correct equation in any form     A1     N2

eg\(\;\;\;y - 4 = 64(x - 6),{\text{ }}y = 64x - 380\)

[6 marks]

[Total 15 marks]

Show that \(\int_5^1 {f(x){\rm{d}}x = - 4} \) .

Find the value of  \(\int_1^2 {(x + f(x)){\rm{d}}x + } \int_2^5 {(x + f(x)){\rm{d}}x} \) .

evidence of factorising 3/division by 3     A1

e.g. \(\int_1^5 {3f(x){\rm{d}}x = 3\int_1^5 {f(x){\rm{d}}x} } \) , \(\frac{{12}}{3}\) , \(\int_1^5 {\frac{{3f(x){\rm{d}}x}}{3}} \) (do not accept 4 as this is show that)

evidence of stating that reversing the limits changes the sign     A1

e.g. \(\int_5^1 {f(x){\rm{d}}x = } - \int_1^5 {f(x){\rm{d}}x} \)

\(\int_5^1 {f(x){\rm{d}}x = } - 4\)     AG     N0

[2 marks]

evidence of correctly combining the integrals (seen anywhere)     (A1)

e.g. \(I = \int_1^2 {(x + f(x)){\rm{d}}x + } \int_2^5 {(x + f(x)){\rm{d}}x = } \int_1^5 {(x + f(x)){\rm{d}}x} \)

evidence of correctly splitting the integrals (seen anywhere)     (A1)

e.g. \(I = \int_1^5 {x{\rm{d}}x + } \int_1^5 {f(x){\rm{d}}x} \) 

\(\int {x{\rm{d}}x = } \frac{{{x^2}}}{2}\) (seen anywhere)     A1

\(\int_1^5 {x{\rm{d}}x = } \left[ {\frac{{{x^2}}}{2}} \right]_1^5 = \frac{{25}}{2} - \frac{1}{2}\) \(\left( { = \frac{{24}}{2},12} \right)\)     A1

\(I =16\)     A1     N3

[5 marks]

This question was very poorly done. Very few candidates provided proper justification for part (a), a common error being to write \(\int_1^5 {f(x){\rm{d}}x = } f(5) - f(1)\) . What was being looked for was that \(\int_1^5 {3f(x){\rm{d}}x = } 3\int_1^5 {f(x){\rm{d}}x} \) and \(\int_5^1 {f(x){\rm{d}}x = } - \int_1^5 {f(x){\rm{d}}x} \) .

Part (b) had similar problems with neither the combining of limits nor the splitting of integrals being done very often. A common error was to treat \(f(x)\) as 1 in order to make \(\int_1^5 {f(x){\rm{d}}x = 4} \) and then write  \(\int_1^5 {(x + f(x)){\rm{d}}x = \left[ {x + 1} \right]} _1^5\) .

Find \(f''(x)\).

The graph of \(f\) has a point of inflexion when \(x = 1\).

Show that \(k = 3\).

Find \(f'( - 2)\).

Find the equation of the tangent to the curve of \(f\) at \(( - 2,{\text{ }}1)\), giving your answer in the form \(y = ax + b\).

Given that \(f'( - 1) = 0\), explain why the graph of \(f\) has a local maximum when \(x =  - 1\).

\(f''(x) = 6x - 2k\)     A1A1     N2

[2 marks]

substituting \(x = 1\) into \(f''\)     (M1)

eg\(\;\;\;f''(1),{\text{ }}6(1) - 2k\)

recognizing \(f''(x) = 0\;\;\;\)(seen anywhere)     M1

correct equation     A1

eg\(\;\;\;6 - 2k = 0\)

\(k = 3\)     AG     N0

[3 marks]

correct substitution into \(f'(x)\)     (A1)

eg\(\;\;\;3{( - 2)^2} - 6( - 2) - 9\)

\(f'( - 2) = 15\)     A1     N2

[2 marks]

recognizing gradient value (may be seen in equation)     M1

eg\(\;\;\;a = 15,{\text{ }}y = 15x + b\)

attempt to substitute \(( - 2,{\text{ }}1)\) into equation of a straight line     M1

eg\(\;\;\;1 = 15( - 2) + b,{\text{ }}(y - 1) = m(x + 2),{\text{ }}(y + 2) = 15(x - 1)\)

correct working     (A1)

eg\(\;\;\;31 = b,{\text{ }}y = 15x + 30 + 1\)

\(y = 15x + 31\)     A1     N2

[4 marks]

METHOD 1 (\({{\text{2}}^{{\text{nd}}}}\) derivative)

recognizing \(f'' < 0\;\;\;\)(seen anywhere)     R1

substituting \(x =  - 1\) into \(f''\)     (M1)

eg\(\;\;\;f''( - 1),{\text{ }}6( - 1) - 6\)

\(f''( - 1) =  - 12\)     A1

therefore the graph of \(f\) has a local maximum when \(x =  - 1\)     AG     N0

METHOD 2 (\({{\text{1}}^{{\text{st}}}}\) derivative)

recognizing change of sign of \(f'(x)\;\;\;\)(seen anywhere)     R1

eg\(\;\;\;\)sign chart\(\;\;\;\)

correct value of \(f'\) for \( - 1 < x < 3\)     A1

eg\(\;\;\;f'(0) =  - 9\)

correct value of \(f'\) for \(x\) value to the left of \( - 1\)     A1

eg\(\;\;\;f'( - 2) = 15\)

therefore the graph of \(f\) has a local maximum when \(x =  - 1\)     AG     N0

[3 marks]

Total [14 marks]

Well answered and candidates coped well with \(k\) in the expression.

Mostly answered well with the common error being to substitute into \(f'\) instead of \(f''\).

A straightforward question that was typically answered correctly.

Some candidates recalculated the gradient, not realising this had already been found in part c). Many understood they were finding a linear equation but were hampered by arithmetic errors.

Using change of sign of the first derivative was the most common approach used with a sign chart or written explanation. However, few candidates then supported their approach by calculating suitable values for \(f'(x)\). This was necessary because the question already identified a local maximum, hence candidates needed to explain why this was so. Some candidates did not mention the ‘first derivative’ just that ‘it’ was increasing/decreasing. Few candidates used the more efficient second derivative test.

Find \(f'(x)\) .

Find the gradient of the curve of \(f\) at \(x = \frac{\pi }{2}\) .

evidence of choosing product rule     (M1)

eg   \(uv' + vu'\)

correct derivatives (must be seen in the product rule) \(\cos x\) , \(2x\)     (A1)(A1)

\(f'(x) = {x^2}\cos x + 2x\sin x\)     A1 N4

[4 marks]

substituting \(\frac{\pi }{2}\) into their \(f'(x)\)     (M1)

eg   \(f'\left( {\frac{\pi }{2}} \right)\) , \({\left( {\frac{\pi }{2}} \right)^2}\cos \left( {\frac{\pi }{2}} \right) + 2\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{2}} \right)\)

correct values for both \(\sin \frac{\pi }{2}\) and \(\cos \frac{\pi }{2}\) seen in \(f'(x)\)     (A1)

eg   \(0 + 2\left( {\frac{\pi }{2}} \right) \times 1\) 

\(f'\left( {\frac{\pi }{2}} \right) = \pi \)     A1 N2

[3 marks]

Many candidates correctly applied the product rule for the derivative, although a common error was to answer \(f'(x) = 2x\cos x\) .

Candidates generally understood that the gradient of the curve uses the derivative, although in some cases the substitution was made in the original function. Some candidates did not know the values of sine and cosine at \({\frac{\pi }{2}}\) .

Find \(f'(x)\) .

Find the x-coordinate of M.

Find the x-coordinate of N.

The line L is the tangent to the curve of f at \((3{\text{, }}12)\). Find the equation of L in the form \(y = ax + b\) .

\(f'(x) = {x^2} + 4x - 5\)     A1A1A1     N3

[3 marks]

evidence of attempting to solve \(f'(x) = 0\)     (M1)

evidence of correct working     A1

e.g. \((x + 5)(x - 1)\) , \(\frac{{ - 4 \pm \sqrt {16 + 20} }}{2}\) , sketch

\(x = - 5\), \(x = 1\)     (A1)

so \(x = - 5\)     A1     N2

[4 marks]

METHOD 1

\(f''(x) = 2x + 4\) (may be seen later)     A1

evidence of setting second derivative = 0     (M1)

e.g. \(2x + 4 = 0\)

\(x = - 2\)     A1     N2

METHOD 2

evidence of use of symmetry     (M1)

e.g. midpoint of max/min, reference to shape of cubic

correct calculation     A1

e.g. \(\frac{{ - 5 + 1}}{2}\)

\(x = - 2\)     A1     N2

[3 marks]

attempting to find the value of the derivative when \(x = 3\)     (M1)

\(f'(3) = 16\)     A1

valid approach to finding the equation of a line     M1

e.g. \(y - 12 = 16(x - 3)\) , \(12 = 16 \times 3 + b\)

\(y = 16x - 36\)     A1     N2

[4 marks]

This question was very well done with most candidates showing their work in an orderly manner. There were a number of candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression rather than an equation.

This question was very well done with most candidates showing their work in an orderly manner. There were a number of candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression rather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was a maximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which was unnecessary and used up valuable time on the exam.

This question was very well done with most candidates showing their work in an orderly manner. There were a number of candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression rather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was a maximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which was unnecessary and used up valuable time on the exam.

This question was very well done with most candidates showing their work in an orderly manner. There were a number of candidates, however, who were a bit sloppy in indicating when a function was being equated to zero and they “solved” an expression rather than an equation. Many candidates went through first and second derivative tests to verify that the point they found was a maximum or an inflexion point; this was unnecessary since the graph was given. Many also found the y-coordinate which was unnecessary and used up valuable time on the exam.

Find the equation of L .

Find the area of the region enclosed by the curve of g , the x-axis, and the lines \(x = 2\) and \(x = 12\) . Give your answer in the form \(a\ln b\) , where \(a,b \in \mathbb{Z}\) .

The graph of g is reflected in the x-axis to give the graph of h . The area of the region enclosed by the lines L , \(x = 2\) , \(x = 12\) and the x-axis is 120 \(120{\text{ c}}{{\text{m}}^2}\) .

Find the area enclosed by the lines L , \(x = 2\) , \(x = 12\) and the graph of h .

finding \(f'(x) = \frac{1}{2}x\)     A1

attempt to find \(f'(4)\)     (M1)

correct value \(f'(4) = 2\)     A1

correct equation in any form     A1     N2

e.g. \(y - 6 = 2(x - 4)\) , \(y = 2x - 2\)

[4 marks]

\({\rm{area}} = \int_2^{12} {\frac{{90}}{{3x + 4}}} {\rm{d}}x\)

correct integral     A1A1

e.g. \(30\ln (3x + 4)\)

substituting limits and subtracting     (M1)

e.g. \(30\ln (3 \times 12 + 4) - 30\ln (3 \times 2 + 4)\) , \(30\ln 40 - 30\ln 10\)

correct working     (A1)

e.g. \(30(\ln 40 - \ln 10)\)

correct application of \(\ln b - \ln a\)     (A1)

e.g. \(30\ln \frac{{40}}{{10}}\)

\({\rm{area}} = 30\ln 4\)     A1     N4

[6 marks]

valid approach     (M1)

e.g. sketch, area h = area g , 120 + their answer from (b)

\({\rm{area}} = 120 + 30\ln 4\)     A2     N3

[3 marks]

While most candidates answered part (a) correctly, finding the equation of the tangent, there were some who did not consider the value of their derivative when \(x = 4\) .

In part (b), most candidates knew that they needed to integrate to find the area, but errors in integration, and misapplication of the rules of logarithms kept many from finding the correct area.

In part (c), it was clear that a significant number of candidates understood the idea of the reflected function, and some recognized that the integral was the negative of the integral from part (b), but only a few recognized the relationship between the areas. Many thought the area between h and the x-axis was 120.

Show that \(A(x) = \frac{{108}}{x} + 2{x^2}\).

Find \(A'(x)\).

Given that the outside surface area is a minimum, find the height of the container.

Fred paints the outside of the container. A tin of paint covers a surface area of \({\text{10 }}{{\text{m}}^{\text{2}}}\) and costs $20. Find the total cost of the tins needed to paint the container.

correct substitution into the formula for volume     A1

eg\(\,\,\,\,\,\)\(36 = y \times x \times x\)

valid approach to eliminate \(y\) (may be seen in formula/substitution)     M1

eg\(\,\,\,\,\,\)\(y = \frac{{36}}{{{x^2}}},{\text{ }}xy = \frac{{36}}{x}\)

correct expression for surface area     A1

eg\(\,\,\,\,\,\)\(xy + xy + xy + {x^2} + {x^2},{\text{ area}} = 3xy + 2{x^2}\)

correct expression in terms of \(x\) only     A1

eg\(\,\,\,\,\,\)\(3x\left( {\frac{{36}}{{{x^2}}}} \right) + 2{x^2},{\text{ }}{x^2} + {x^2} + \frac{{36}}{x} + \frac{{36}}{x} + \frac{{36}}{x},{\text{ }}2{x^2} + 3\left( {\frac{{36}}{x}} \right)\)

\(A(x) = \frac{{108}}{x} + 2{x^2}\)    AG     N0

[4 marks]

\(A'(x) =  - \frac{{108}}{{{x^2}}} + 4x,{\text{ }}4x - 108{x^{ - 2}}\)    A1A1     N2

Note:     Award A1 for each term.

[2 marks]

recognizing that minimum is when \(A'(x) = 0\)     (M1)

correct equation     (A1)

eg\(\,\,\,\,\,\)\( - \frac{{108}}{{{x^2}}} + 4x = 0,{\text{ }}4x = \frac{{108}}{{{x^2}}}\)

correct simplification     (A1)

eg\(\,\,\,\,\,\)\( - 108 + 4{x^3} = 0,{\text{ }}4{x^3} = 108\)

correct working     (A1)

eg\(\,\,\,\,\,\)\({x^3} = 27\)

\({\text{height}} = 3{\text{ (m) }}({\text{accept }}x = 3)\)    A1     N2

[5 marks]

attempt to find area using their height     (M1)

eg\(\,\,\,\,\,\)\(\frac{{108}}{3} + 2{(3)^2},{\text{ }}9 + 9 + 12 + 12 + 12\)

minimum surface area \( = 54{\text{ }}{{\text{m}}^{\text{2}}}\) (may be seen in part (c))     A1

attempt to find the number of tins     (M1)

eg\(\,\,\,\,\,\)\(\frac{{54}}{{10}},{\text{ }}5.4\)

6 (tins)     (A1)

$120     A1     N3

[5 marks]

Many candidates answered part (a) of this question correctly, though some seemed to be working backwards from the given expression for area, which is not the intention of a "show that" question.

In part (b), while many candidates found the correct derivative, some did so using cumbersome methods such as the quotient rule, rather than using the simpler power rule.

It was disappointing to see the number of candidates who did not recognize that the derivative they had just found in part (b) would have to be equal to zero in order for the surface area to be a minimum. For the candidates who did set their derivative equal to zero, most were able to find the correct height.

In part (d) of this question, there were some arithmetic errors which kept candidates from finding the correct area. The most common error here, by far, was not considering that the number of tins purchased must be an integer.

(i)     Given that \(f'(x) = \frac{{2{a^2} - 4{x^2}}}{{\sqrt {{a^2} - {x^2}} }}\), for \( - 1 \leqslant x < a\), find the equation of \(L\).

(ii)     Hence or otherwise, find an expression for \(b\) in terms of \(a\).

Show that \({A_R} = \frac{2}{3}{a^3}\).

Let \({A_T}\) be the area of the triangle OPQ. Given that \({A_T} = k{A_R}\), find the value of \(k\).

(i)     recognizing the need to find the gradient when \(x = 0\) (seen anywhere)     R1

eg\(\,\,\,\,\,\)\(f'(0)\)

correct substitution     (A1)

\(f'(0) = \frac{{2{a^2} - 4(0)}}{{\sqrt {{a^2} - 0} }}\)

\(f'(0) = 2a\)    (A1)

correct equation with gradient 2\(a\) (do not accept equations of the form \(L = 2ax\))     A1     N3

eg\(\,\,\,\,\,\)\(y = 2ax,{\text{ }}y - b = 2a(x - a),{\text{ }}y = 2ax - 2{a^2} + b\)

(ii)     METHOD 1

attempt to substitute \(x = a\) into their equation of \(L\)     (M1)

eg\(\,\,\,\,\,\)\(y = 2a \times a\)

\(b = 2{a^2}\)     A1     N2

METHOD 2

equating gradients     (M1)

eg\(\,\,\,\,\,\)\(\frac{b}{a} = 2a\)

\(b = 2{a^2}\)    A1     N2

[6 marks]

METHOD 1

recognizing that area \( = \int_0^a {f(x){\text{d}}x} \) (seen anywhere)     R1

valid approach using substitution or inspection     (M1)

eg\(\,\,\,\,\,\)\(\int {2x\sqrt u {\text{d}}x,{\text{ }}u = {a^2} - {x^2},{\text{ d}}u =  - 2x{\text{d}}x,{\text{ }}\frac{2}{3}{{({a^2} - {x^2})}^{\frac{3}{2}}}} \)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\int {2x\sqrt {{a^2} - {x^2}} {\text{d}}x = \int { - \sqrt u {\text{d}}u} } \)

\(\int { - \sqrt u {\text{d}}u =  - \frac{{{u^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \)    (A1)

\(\int {f(x){\text{d}}x =  - \frac{2}{3}{{({a^2} - {x^2})}^{\frac{3}{2}}} + c} \)    (A1)

substituting limits and subtracting     A1

eg\(\,\,\,\,\,\)\({A_R} =  - \frac{2}{3}{({a^2} - {a^2})^{\frac{3}{2}}} + \frac{2}{3}{({a^2} - 0)^{\frac{3}{2}}},{\text{ }}\frac{2}{3}{({a^2})^{\frac{3}{2}}}\)

\({A_R} = \frac{2}{3}{a^3}\)    AG     N0

METHOD 2

recognizing that area \( = \int_0^a {f(x){\text{d}}x} \) (seen anywhere)     R1

valid approach using substitution or inspection     (M1)

eg\(\,\,\,\,\,\)\(\int {2x\sqrt u {\text{d}}x,{\text{ }}u = {a^2} - {x^2},{\text{ d}}u =  - 2x{\text{d}}x,{\text{ }}\frac{2}{3}{{({a^2} - {x^2})}^{\frac{3}{2}}}} \)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\int {2x\sqrt {{a^2} - {x^2}} {\text{d}}x = \int { - \sqrt u {\text{d}}u} } \)

\(\int { - \sqrt u {\text{d}}u =  - \frac{{{u^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \)    (A1)

new limits for u (even if integration is incorrect)     (A1)

eg\(\,\,\,\,\,\)\(u = 0{\text{ and }}u = {a^2},{\text{ }}\int_0^{{a^2}} {{u^{\frac{1}{2}}}{\text{d}}u,{\text{ }}\left[ { - \frac{2}{3}{u^{\frac{3}{2}}}} \right]} _{{a^2}}^0\)

substituting limits and subtracting     A1

eg\(\,\,\,\,\,\)\({A_R} =  - \left( {0 - \frac{2}{3}{a^3}} \right),{\text{ }}\frac{2}{3}{({a^2})^{\frac{3}{2}}}\)

\({A_R} = \frac{2}{3}{a^3}\)    AG     N0

[6 marks]

METHOD 1

valid approach to find area of triangle     (M1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}({\text{OQ)(PQ), }}\frac{1}{2}ab\)

correct substitution into formula for \({A_T}\) (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\({A_T} = \frac{1}{2} \times a \times 2{a^2},{\text{ }}{a^3}\)

valid attempt to find \(k\) (must be in terms of \(a\))     (M1)

eg\(\,\,\,\,\,\)\({a^3} = k\frac{2}{3}{a^3},{\text{ }}k = \frac{{{a^3}}}{{\frac{2}{3}{a^3}}}\)

\(k = \frac{3}{2}\)     A1     N2

METHOD 2

valid approach to find area of triangle     (M1)

eg\(\,\,\,\,\,\)\(\int_0^a {(2ax){\text{d}}x} \)

correct working     (A1)

eg\(\,\,\,\,\,\)\([a{x^2}]_0^a,{\text{ }}{a^3}\)

valid attempt to find \(k\) (must be in terms of \(a\))     (M1)

eg\(\,\,\,\,\,\)\({a^3} = k\frac{2}{3}{a^3},{\text{ }}k = \frac{{{a^3}}}{{\frac{2}{3}{a^3}}}\)

\(k = \frac{3}{2}\)     A1     N2

[4 marks]

As is typically the case with question 10, this proved to be quite a challenging question for many candidates. In part (a), while many candidates seemed to recognize that there was some relationship between the given derivative and the gradient of the tangent line, most did not substitute zero for the \(x\)-value, and were unable to find the correct gradient of the line.

In part (b), nearly every candidate understood that the area was equal to the integral of \(f\) from 0 to \(a\), very few were able to integrate correctly using either substitution or inspection. Many candidates did not even attempt to integrate, stopping after writing the integral expression.

In part (c), most candidates started with a correct expression for the area of the triangle such as \(\frac{{ab}}{2}\). However, very few were able to substitute their expression for \(b\) from part (a)(ii), and therefore did not find a value for \(k\).

Use the quotient rule to show that \(f'(x) = \frac{{10 - 2{x^2}}}{{{{({x^2} + 5)}^2}}}\).

Find \(\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x} \).

The following diagram shows part of the graph of \(f\).

The shaded region is enclosed by the graph of \(f\), the \(x\)-axis, and the lines \(x = \sqrt 5 \) and \(x = q\). This region has an area of \(\ln 7\). Find the value of \(q\).

derivative of \(2x\) is \(2\) (must be seen in quotient rule)     (A1)

derivative of \({x^2} + 5\) is \(2x\) (must be seen in quotient rule)     (A1)

correct substitution into quotient rule     A1

eg     \(\frac{{({x^2} + 5)(2) - (2x)(2x)}}{{{{({x^2} + 5)}^2}}},{\text{ }}\frac{{2({x^2} + 5) - 4{x^2}}}{{{{({x^2} + 5)}^2}}}\)

correct working which clearly leads to given answer   A1

eg    \(\frac{{2{x^2} + 10 - 4{x^2}}}{{{{({x^2} + 5)}^2}}},{\text{ }}\frac{{2{x^2} + 10 - 4{x^2}}}{{{x^4} + 10{x^2} + 25}}\)

\(f'(x) = \frac{{10 - 2{x^2}}}{{{{({x^2} + 5)}^2}}}\)     AG     N0

[4 marks]

valid approach using substitution or inspection     (M1)

eg     \(u = {x^2} + 5,{\text{ d}}u = 2x{\text{d}}x,{\text{ }}\frac{1}{2}\ln ({x^2} + 5)\)

\(\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x = \int {\frac{1}{u}{\text{d}}u} } \)     (A1)

\(\int {\frac{1}{u}{\text{d}}u = \ln u + c} \)     (A1)

\(\ln ({x^2} + 5) + c\)     A1     N4

[4 marks]

correct expression for area     (A1)

eg     \(\left[ {\ln \left( {{x^2} + 5} \right)} \right]_{\sqrt 5 }^q,{\text{ }}\int\limits_{\sqrt 5 }^q {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x} \)

substituting limits into their integrated function and subtracting (in either order)     (M1)

eg     \(\ln ({q^2} + 5) - \ln \left( {{{\sqrt 5 }^2} + 5} \right)\)

correct working     (A1)

eg     \(\ln \left( {{q^2} + 5} \right) - \ln 10,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}}\)

equating their expression to \(\ln 7\) (seen anywhere)     (M1)

eg     \(\ln \left( {{q^2} + 5} \right) - \ln 10 = \ln 7,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}} = \ln 7,{\text{ }}\ln ({q^2} + 5) = \ln 7 + \ln 10\)

correct equation without logs     (A1)

eg     \(\frac{{{q^2} + 5}}{{10}} = 7,{\text{ }}{q^2} + 5 = 70\)

\({q^2} = 65\)     (A1)

\(q = \sqrt {65} \)     A1     N3

Note: Award A0 for \(q =  \pm \sqrt {65} \).

[7 marks]

Let \(\int_\pi ^a {\cos 2x{\text{d}}x}  = \frac{1}{2}{\text{, where }}\pi  < a < 2\pi \). Find the value of \(a\).

correct integration (ignore absence of limits and “\(+C\)”)     (A1)

eg     \(\frac{{\sin (2x)}}{2},{\text{ }}\int_\pi ^a {\cos 2x = \left[ {\frac{1}{2}\sin (2x)} \right]_\pi ^a} \)

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg     \(\frac{1}{2}\sin (2a) - \frac{1}{2}\sin (2\pi ),{\text{ }}\sin (2\pi ) - \sin (2a)\)

\(\sin (2\pi ) = 0\)     (A1)

setting their result from an integrated function equal to \(\frac{1}{2}\)     M1

eg     \(\frac{1}{2}\sin 2a = \frac{1}{2},{\text{ }}\sin (2a) = 1\)

recognizing \({\sin ^{ - 1}}1 = \frac{\pi }{2}\)     (A1)

eg     \(2a = \frac{\pi }{2},{\text{ }}a = \frac{\pi }{4}\)

correct value     (A1)

eg     \(\frac{\pi }{2} + 2\pi ,{\text{ }}2a = \frac{{5\pi }}{2},{\text{ }}a = \frac{\pi }{4} + \pi \)

\(a = \frac{{5\pi }}{4}\)     A1     N3

[7 marks]

A rocket moving in a straight line has velocity \(v\) km s^–1 and displacement \(s\) km at time \(t\) seconds. The velocity \(v\) is given by \(v(t) = 6{{\rm{e}}^{2t}} + t\) . When \(t = 0\) , \(s = 10\) .

Find an expression for the displacement of the rocket in terms of \(t\) .

evidence of anti-differentiation     (M1)

eg   \(\int {(6{{\rm{e}}^{2t}} + t)} \)

\(s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + C\)     A2A1

Note: Award A2 for \(3{{\rm{e}}^{2t}}\) , A1 for \(\frac{{{t^2}}}{2}\) .

attempt to substitute (\(0\), \(10\)) into their integrated expression (even if \(C\) is missing)     (M1) 

correct working     (A1)

eg   \(10 = 3 + C\) , \(C = 7\)

\(s = 3{{\rm{e}}^{2t}} + \frac{{{t^2}}}{2} + 7\)     A1     N6

Note: Exception to the FT rule. If working shown, allow full FT on incorrect integration which must involve a power of \({\rm{e}}\).

[7 marks]

Find the \(x\)-coordinate of \(A\).

The \(y\)-intercept of the graph is at (\(0,6\)). Find an expression for \(f(x)\).

The graph of a function \(g\) is obtained by reflecting the graph of \(f\) in the \(y\)-axis, followed by a translation of \(\left({\begin{array}{*{20}{c}}m\\n\end{array}}\right)\).

Find the \(x\)-coordinate of the local minimum point on the graph of \(g\).

recognizing that the local minimum occurs when \(f'(x) = 0\)     (M1)

valid attempt to solve \(3{x^2} - 8x - 3 = 0\)     (M1)

eg\(\;\;\;\)factorization, formula

correct working     A1

\((3x + 1)(x - 3),{\text{ }}x = \frac{{8 \pm \sqrt {64 + 36} }}{6}\)

\(x = 3\)     A2     N3

Note:     Award A1 if both values \(x = \frac{{ - 1}}{3},{\text{ }}x = 3\) are given.

[5 marks]

valid approach     (M1)

\(f(x) = \int {f'(x){\text{d}}x} \)

\(f(x) = {x^3} - 4{x^2} - 3x + c\;\;\;\)(do not penalize for missing “\( + c\)”)     A1A1A1

\(c = 6\)     (A1)

\(f(x) = {x^3} - 4{x^2} - 3x + 6\)     A1     N6

[6 marks]

applying reflection     (A1)

eg\(\;\;\;f( - x)\)

recognizing that the minimum is the image of \(A\)     (M1)

eg\(\;\;\;x =  - 3\)

correct expression for \(x\)     A1     N3

eg\(\;\;\; - 3 + m,{\text{ }}\left( {\begin{array}{*{20}{c}} { - 3 + m} \\ { - 12 + n} \end{array}} \right),{\text{ }}(m - 3,{\text{ }}n - 12)\)

[3 marks]

Total [14 marks]

The majority of candidates approached part (a) correctly, and most recognized that only one solution was possible within the given domain.

Nearly all candidates answered part (b) correctly, earning all the available marks for integrating the polynomial and solving for \(C\).

Part (c) proved to be much more difficult for candidates, who either did not know how to apply the transformations correctly, or who engaged in lengthy and unnecessary manipulations of the function, rather than simply finding the image of the local minimum point \(A\).

Find \((g \circ f)(x)\).

Given that \(\mathop {\lim }\limits_{x \to  + \infty } (g \circ f)(x) =  - 3\), find the value of \(b\).

attempt to form composite     (M1)

eg\(\,\,\,\,\,\)\(g(1 + {{\text{e}}^{ - x}})\)

correct function     A1     N2

eg\(\,\,\,\,\,\)\((g \circ f)(x) = 2 + b + 2{{\text{e}}^{ - x}},{\text{ }}2(1 + {{\text{e}}^{ - x}}) + b\)

[2 marks]

evidence of \(\mathop {\lim }\limits_{x \to \infty } (2 + b + 2{{\text{e}}^{ - x}}) = 2 + b + \mathop {\lim }\limits_{x \to \infty } (2{{\text{e}}^{ - x}})\)     (M1)

eg\(\,\,\,\,\,\)\(2 + b + 2{{\text{e}}^{ - \infty }}\), graph with horizontal asymptote when \(x \to \infty \)

Note:     Award M0 if candidate clearly has incorrect limit, such as \(x \to 0,{\text{ }}{{\text{e}}^\infty },{\text{ }}2{{\text{e}}^0}\).

evidence that \({{\text{e}}^{ - x}} \to 0\) (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(\mathop {\lim }\limits_{x \to \infty } ({{\text{e}}^{ - x}}) = 0,{\text{ }}1 + {{\text{e}}^{ - x}} \to 1,{\text{ }}2(1) + b =  - 3,{\text{ }}{{\text{e}}^{{\text{large negative number}}}} \to 0\), graph of \(y = {{\text{e}}^{ - x}}\) or

\(y = 2{{\text{e}}^{ - x}}\) with asymptote \(y = 0\), graph of composite function with asymptote \(y =  - 3\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(2 + b =  - 3\)

\(b =  - 5\)     A1     N2

[4 marks]

Let \(f’(x) = \frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}\). Given that \(f(0) = 1\), find \(f(x)\).

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\int {f'{\text{d}}x,{\text{ }}\int {\frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}{\text{d}}x} } \)

correct integration by substitution/inspection     A2

eg\(\,\,\,\,\,\)\(f(x) =  - \frac{1}{4}{({x^3} + 1)^{ - 4}} + c,{\text{ }}\frac{{ - 1}}{{4{{({x^3} + 1)}^4}}}\)

correct substitution into their integrated function (must include \(c\))     M1

eg\(\,\,\,\,\,\)\(1 = \frac{{ - 1}}{{4{{({0^3} + 1)}^4}}} + c,{\text{ }} - \frac{1}{4} + c = 1\)

Note:     Award M0 if candidates substitute into \(f’\) or \(f’’\).

\(c = \frac{5}{4}\)     (A1)

\(f(x) =  - \frac{1}{4}{({x^3} + 1)^{ - 4}} + \frac{5}{4}{\text{ }}\left( { = \frac{{ - 1}}{{4{{({x^3} + 1)}^4}}} + \frac{5}{4},{\text{ }}\frac{{5{{({x^3} + 1)}^4} - 1}}{{4{{({x^3} + 1)}^4}}}} \right)\)     A1     N4

[6 marks]

Find the second derivative.

Find \(f'(3)\) and \(f''(3)\) .

The point P on the graph of f has x-coordinate \(3\). Explain why P is not a point of inflexion.

METHOD 1

\(f''(x) = 3{(x - 3)^2}\)     A2     N2

METHOD 2

attempt to expand \({(x - 3)^3}\)     (M1)

e.g. \(f'(x) = {x^3} - 9{x^2} + 27x - 27\)

\(f''(x) = 3{x^2} - 18x + 27\)     A1     N2

[2 marks]

\(f'(3) = 0\) , \(f''(3) = 0\)     A1     N1

[1 mark]

METHOD 1

\({f''}\) does not change sign at P     R1

evidence for this     R1     N0

METHOD 2

\({f'}\) changes sign at P so P is a maximum/minimum (i.e. not inflexion)     R1

evidence for this     R1     N0

METHOD 3

finding \(f(x) = \frac{1}{4}{(x - 3)^4} + c\) and sketching this function     R1

indicating minimum at \(x = 3\)     R1     N0 

[2 marks]

Many candidates completed parts (a) and (b) successfully.

Many candidates completed parts (a) and (b) successfully.

A rare few earned any marks in part (c) - most justifying the point of inflexion with the zero answers in part (b), not thinking that there is more to consider.

Find \(f'(x)\) .

Let \(g(x) = \ln \left( {\frac{{6x}}{{x + 1}}} \right)\) , for \(x > 0\) .

Show that \(g'(x) = \frac{1}{{x(x + 1)}}\) .

Let \(h(x) = \frac{1}{{x(x + 1)}}\) . The area enclosed by the graph of h , the x-axis and the lines \(x = \frac{1}{5}\)  and \(x = k\) is \(\ln 4\) . Given that \(k > \frac{1}{5}\) , find the value of k .

METHOD 1

evidence of choosing quotient rule     (M1)

e.g. \(\frac{{u'v - uv'}}{{{v^2}}}\)

evidence of correct differentiation (must be seen in quotient rule)     (A1)(A1)

e.g. \(\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}(x + 1) = 1\)

correct substitution into quotient rule     A1

e.g. \(\frac{{(x + 1)6 - 6x}}{{{{(x + 1)}^2}}}\) , \(\frac{{6x + 6 - 6x}}{{{{(x + 1)}^2}}}\)

\(f'(x) = \frac{6}{{{{(x + 1)}^2}}}\)    A1     N4

[5 marks]

METHOD 2

evidence of choosing product rule     (M1)

e.g. \(6x{(x + 1)^{ - 1}}\) , \(uv' + vu'\)

evidence of correct differentiation (must be seen in product rule)     (A1)(A1)

e.g. \(\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}{(x + 1)^{ - 1}} = - 1{(x + 1)^{ - 2}} \times 1\)

correct working     A1

e.g. \(6x \times - {(x + 1)^{ - 2}} + {(x + 1)^{ - 1}} \times 6\) , \(\frac{{ - 6x + 6(x + 1)}}{{{{(x + 1)}^2}}}\)

\(f'(x) = \frac{6}{{{{(x + 1)}^2}}}\)   A1     N4

[5 marks]

METHOD 1

evidence of choosing chain rule     (M1)

e.g. formula, \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \left( {\frac{{6x}}{{x + 1}}} \right)\)

correct reciprocal of \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}}\) is \(\frac{{x + 1}}{{6x}}\) (seen anywhere)     A1

correct substitution into chain rule     A1

e.g. \(\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \frac{6}{{{{(x + 1)}^2}}}\) , \(\left( {\frac{6}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{{6x}}} \right)\)

working that clearly leads to the answer     A1

e.g. \(\left( {\frac{6}{{(x + 1)}}} \right)\left( {\frac{1}{{6x}}} \right)\) , \(\left( {\frac{1}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{x}} \right)\) , \(\frac{{6(x + 1)}}{{6x{{(x + 1)}^2}}}\)

\(g'(x) = \frac{1}{{x(x + 1)}}\)     AG     N0

[4 marks]

METHOD 2

attempt to subtract logs     (M1)

e.g. \(\ln a - \ln b\) , \(\ln 6x - \ln (x + 1)\)

correct derivatives (must be seen in correct expression)     A1A1

e.g. \(\frac{6}{{6x}} - \frac{1}{{x + 1}}\) , \(\frac{1}{x} - \frac{1}{{x + 1}}\)

working that clearly leads to the answer     A1

e.g. \(\frac{{x + 1 - x}}{{x(x + 1)}}\) , \(\frac{{6x + 6 - 6x}}{{6x(x + 1)}}\) , \(\frac{{6(x + 1 - x)}}{{6x(x + 1)}}\)

\(g'(x) = \frac{1}{{x(x + 1)}}\)     AG     N0

[4 marks]

valid method using integral of  h(x) (accept missing/incorrect limits or missing \({\text{d}}x\) )     (M1)

e.g. \({\rm{area}} = \int_{\frac{1}{5}}^k {h(x){\rm{d}}x} \) , \(\int{\left( {\frac{1}{{x(x + 1)}}} \right)} \) 

recognizing that integral of derivative will give original function     (R1)

e.g. \(\int{\left( {\frac{1}{{x(x + 1)}}} \right)} {\rm{d}}x = \ln \left( {\frac{{6x}}{{x + 1}}} \right)\)

correct substitution and subtraction     A1

e.g. \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) - \ln \left( {\frac{{6 \times \frac{1}{5}}}{{\frac{1}{5} + 1}}} \right)\) , \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) - \ln (1)\)

setting their expression equal to \(\ln 4\)     (M1) 

e.g. \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) - \ln (1) = \ln 4\) , \(\ln \left( {\frac{{6k}}{{k + 1}}} \right) = \ln 4\) , \(\int_{\frac{1}{5}}^k {h(x){\rm{d}}x = \ln 4} \)

correct equation without logs     A1

e.g.\(\frac{{6k}}{{k + 1}} = 4\) , \(6k = 4(k + 1)\) 

correct working     (A1)

e.g. \(6k = 4k + 4\) , \(2k = 4\)

\(k = 2\)    A1     N4

[7 marks]

In part (a), most candidates recognized the need to apply the quotient rule to find the derivative, and many were successful in earning full marks here.

In part (b), many candidates struggled with the chain rule, or did not realize the chain rule was necessary to find the derivative. Again, some candidates attempted to work backward from the given answer, which is not allowed in a "show that" question. A few clever candidates simplified the situation by applying properties of logarithms before finding their derivative.

For part (c), many candidates recognized the need to integrate the function, and that their integral would equal \(\ln 4\) . However, many did not recognize that the integral of h is g . Those candidates who made this link between the parts (b) and (c) often carried on correctly to find the value of k , with a few candidates having errors in working with logarithms.

Write down the car’s velocity at \(t = 3\) .

Write down the value of \(f( - 3)\);

Find the car’s acceleration at \(t = 1.5\) .

Find the total distance travelled.

\(4{\text{ (m}}{{\text{s}}^{ - 1}}{\text{)}}\)     A1     N1

[1 mark]

\(f( - 3) =  - 1\)     A1     N1

[1 mark]

recognizing that acceleration is the gradient     M1

e.g. \(a(1.5) = \frac{{4 - 0}}{{2 - 0}}\)

\(a = 2\) \({\text{(m}}{{\text{s}}^{ - 2}}{\text{)}}\)     A1     N1

[2 marks]

recognizing area under curve     M1

e.g. trapezium, triangles, integration

correct substitution     A1

e.g. \(\frac{1}{2}(3 + 6)4\) , \(\int_0^6 {\left| {v(t)} \right|} {\rm{d}}t\)

distance 18 (m)     A1     N2

[3 marks]

Write down the equation of \({L_1}\).

A line \({L_a}\) crosses the \(y\)-axis at a point \(P\).

Show that \(P\) has coordinates \(\left( {0,{\text{ }}\frac{4}{a}} \right)\).

The line \({L_a}\) crosses the \(x\)-axis at \({\text{Q}}(2a,{\text{ }}0)\). Let \(d = {\text{P}}{{\text{Q}}^2}\).

Show that \(d = 4{a^2} + \frac{{16}}{{{a^2}}}\).

There is a minimum value for \(d\). Find the value of \(a\) that gives this minimum value.

attempt to substitute \(x = 1\)     (M1)

eg\(\;\;\;\)r \( = \left( {\begin{array}{*{20}{c}} 1 \\ {\frac{2}{1}} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} {{1^2}} \\ { - 2} \end{array}} \right),{\text{ }}{L_1} = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \end{array}} \right)\)

correct equation (vector or Cartesian, but do not accept “\({L_1}\)”)

eg\(\;\;\;\)r \( = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { - 2} \end{array}} \right),{\text{ }}y =  - 2x + 4\;\;\;\)(must be an equation)     A1     N2

[2 marks]

appropriate approach     (M1)

eg\(\;\;\;\left( {\begin{array}{*{20}{c}} 0 \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} {{a^2}} \\ { - 2} \end{array}} \right)\)

correct equation for \(x\)-coordinate     A1

eg\(\;\;\;0 = a + t{a^2}\)

\(t = \frac{{ - 1}}{a}\)     A1

substituting their parameter to find \(y\)     (M1)

eg\(\;\;\;y = \frac{2}{a} - 2\left( {\frac{{ - 1}}{a}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) - \frac{1}{a}\left( {\begin{array}{*{20}{c}} {{a^2}} \\ { - 2} \end{array}} \right)\)

correct working     A1

eg\(\;\;\;y = \frac{2}{a} + \frac{2}{a},{\text{ }}\left( {\begin{array}{*{20}{c}} a \\ {\frac{2}{a}} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} a \\ { - \frac{2}{a}} \end{array}} \right)\)

finding correct expression for \(y\)     A1

eg\(\;\;\;y = \frac{4}{a},{\text{ }}\left( {\begin{array}{*{20}{c}} 0 \\ {\frac{4}{a}} \end{array}} \right)\) \({\text{P}}\left( {0,{\text{ }}\frac{4}{a}} \right)\)     AG     N0

[6 marks]

valid approach M1

eg\(\;\;\;\)distance formula, Pythagorean Theorem, \(\overrightarrow {{\text{PQ}}}  = \left( {\begin{array}{*{20}{c}} {2a} \\ { - \frac{4}{a}} \end{array}} \right)\)

correct simplification     A1

eg\(\;\;\;{(2a)^2} + {\left( {\frac{4}{a}} \right)^2}\)

\(d = 4{a^2} + \frac{{16}}{{{a^2}}}\)     AG     N0

[2 marks]

recognizing need to find derivative     (M1)

eg\(\;\;\;d',{\text{ }}d'(a)\)

correct derivative     A2

eg\(\;\;\;8a - \frac{{32}}{{{a^3}}},{\text{ }}8x - \frac{{32}}{{{x^3}}}\)

setting their derivative equal to \(0\)     (M1)

eg\(\;\;\;8a - \frac{{32}}{{{a^3}}} = 0\)

correct working     (A1)

eg\(\;\;\;8a = \frac{{32}}{{{a^3}}},{\text{ }}8{a^4} - 32 = 0\)

working towards solution     (A1)

eg\(\;\;\;{a^4} = 4,{\text{ }}{a^2} = 2,{\text{ }}a =  \pm \sqrt 2 \)

\(a = \sqrt[4]{4}\;\;\;(a = \sqrt 2 )\;\;\;({\text{do not accept }} \pm \sqrt 2 )\)     A1     N3

[7 marks]

Total [17 marks]

In part (a), most candidates correctly substituted 1 for \(x\), although many of them did not earn full marks for their work here, as they wrote their vector equation using \({L_1} = \), not understanding that \({L_1}\) is the name of the line, and not a vector.

Very few candidates answered parts (b) and (c) correctly, often working backwards from the given answer, which is not appropriate in "show that" questions. In these types of questions, candidates are required to clearly show their working and reasoning, which will hopefully lead them to the given answer.

Very few candidates answered parts (b) and (c) correctly, often working backwards from the given answer, which is not appropriate in "show that" questions. In these types of questions, candidates are required to clearly show their working and reasoning, which will hopefully lead them to the given answer.

Fortunately, a good number of candidates recognized the need to find the derivative of the given expression for \(d\) in part (d) of the question, and so were able to earn at least some of the available marks in the final part.

Find the area of the triangle OPB, in terms of \(\theta \) .

Explain why the area of triangle OPA is the same as the area triangle OPB.

Show that \(S = 2(\pi  - 2\sin \theta )\) .

Find the value of \(\theta \) when S is a local minimum, justifying that it is a minimum.

Find a value of \(\theta \) for which S has its greatest value.

evidence of using area of a triangle     (M1)

e.g. \(A = \frac{1}{2} \times 2 \times 2 \times \sin \theta \)

\(A = 2\sin \theta \)     A1     N2

[2 marks]

METHOD 1

\({\rm{P}}\widehat {\rm{O}}{\rm{A = }}\pi  - \theta \)     (A1)

\({\text{area }}\Delta {\rm{OPA}} = \frac{1}{2}2 \times 2 \times \sin (\pi  - \theta )\) \(( = 2\sin (\pi  - \theta ))\)     A1

since \(\sin (\pi  - \theta ) = \sin \theta \)     R1

then both triangles have the same area     AG     N0

METHOD 2

triangle OPA has the same height and the same base as triangle OPB     R3

then both triangles have the same area     AG     N0

[3 marks]

area semicircle \( = \frac{1}{2} \times \pi {(2)^2}\) \(( = 2\pi )\)     A1

\({\text{area }}\Delta {\rm{APB}} = 2\sin \theta + 2\sin \theta \) \(( = 4\sin \theta )\)     A1

\(S{\text{ = area of semicircle}} - {\text{area }}\Delta {\rm{APB}}\) \(( = 2\pi - 4\sin \theta )\)     M1

\(S = 2(\pi - 2\sin \theta )\)     AG     N0

[3 marks]

METHOD 1

attempt to differentiate     (M1)

e.g. \(\frac{{{\rm{d}}S}}{{{\rm{d}}\theta }} = - 4\cos \theta \)

setting derivative equal to 0     (M1)

correct equation     A1

e.g. \( - 4\cos \theta = 0\) , \(\cos \theta = 0\) , \(4\cos \theta = 0\)

\(\theta = \frac{\pi }{2}\)     A1     N3

EITHER

evidence of using second derivative     (M1)

\(S''(\theta ) = 4\sin \theta \)   A1

\(S''\left( {\frac{\pi }{2}} \right) = 4\)     A1

it is a minimum because \(S''\left( {\frac{\pi }{2}} \right) > 0\)     R1     N0

OR

evidence of using first derivative      (M1)

for \(\theta < \frac{\pi }{2},S'(\theta ) < 0\) (may use diagram)     A1

for \(\theta > \frac{\pi }{2},S'(\theta ) > 0\) (may use diagram)    A1

it is a minimum since the derivative goes from negative to positive     R1     N0

METHOD 2

\(2\pi - 4\sin \theta \) is minimum when \(4\sin \theta \) is a maximum     R3

\(4\sin \theta \) is a maximum when \(\sin \theta = 1\)     (A2)

\(\theta = \frac{\pi }{2}\)     A3     N3

[8 marks]

S is greatest when \(4\sin \theta \) is smallest (or equivalent)     (R1)

\(\theta = 0\) (or \(\pi \) )     A1     N2

[2 marks]

Most candidates could obtain the area of triangle OPB as equal to \(2\sin \theta \) , though \(2\theta \) was given quite often as the area.

A minority recognized the equality of the sines of supplementary angles and the term complementary was frequently used instead of supplementary. Only a handful of candidates used the simple equal base and altitude argument.

Many candidates seemed to see why \(S = 2(\pi - 2\sin \theta )\) but the arguments presented for showing why this result was true were not very convincing in many cases. Explicit evidence of why the area of the semicircle was \(2\pi \) was often missing as was an explanation for \(2(2\sin \theta )\) and for subtraction.

Only a small number of candidates recognized the fact S would be minimum when sin was maximum, leading to a simple non-calculus solution. Those who chose the calculus route often had difficulty finding the derivative of S, failing in a significant number of cases to recognize that the derivative of a constant is 0, and also going through painstaking application of the product rule to find the simple derivative. When it came to justify a minimum, there was evidence in some cases of using some form of valid test, but explanation of the test being used was generally poor.

Candidates who answered part (d) correctly generally did well in part (e) as well, though answers outside the domain of \(\theta \) were frequently seen.

Find \(\int_1^6 {2f(x){\text{d}}x} \).

Find \(\int_1^6 {\left( {f(x) + 2} \right){\text{d}}x} \).

appropriate approach     (M1)

eg     \(2\int {f(x),{\text{ }}2(8)} \)

\(\int_1^6 {2f(x){\text{d}}x = 16} \)     A1     N2

[2 marks]

appropriate approach     (M1)

eg     \(\int {f(x) + \int {2,{\text{ }}8 + \int 2 } } \)

\(\int {2{\text{d}}x = 2x} \)   (seen anywhere)     (A1)

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg     \(2(6) - 2(1),{\text{ }}8 + 12 - 2\)

\(\int_1^6 {\left( {f(x) + 2} \right){\text{d}}x = 18} \)     A1     N3

[4 marks] 

Write down an expression in terms of \(\theta \) for

(i)     \(x\) ;

(ii)    \(y\) .

Let the area of the rectangle be A.

Show that \(A = 18\sin 2\theta \) .

(i)     Find \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }}\) .

(ii)    Hence, find the exact value of \(\theta \) which maximizes the area of the rectangle.

(iii)   Use the second derivative to justify that this value of \(\theta \) does give a maximum.

(i) \(x = 3\cos \theta \)     A1     N1 

(ii) \(y = 3\sin \theta \)     A1     N1

[2 marks]

finding area     (M1)

e.g. \(A = 2x \times 2y\) , \(A = 8 \times \frac{1}{2}bh\) 

substituting     A1

e.g. \(A = 4 \times 3\sin \theta  \times 3\cos \theta \) , \(8 \times \frac{1}{2} \times 3\cos \theta  \times 3\sin \theta \)

\(A = 18(2\sin \theta \cos \theta )\)    A1

\(A = 18\sin 2\theta \)     AG     N0

[3 marks]

(i) \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 36\cos 2\theta \)     A2     N2 

(ii) for setting derivative equal to 0     (M1)

e.g. \(36\cos 2\theta  = 0\) , \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 0\)

\(2\theta  = \frac{\pi }{2}\)     (A1)

\(\theta  = \frac{\pi }{4}\)     A1     N2

(iii) valid reason (seen anywhere)     R1

e.g. at \(\frac{\pi }{4}\), \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} < 0\) ; maximum when \(f''(x) < 0\)

finding second derivative \(\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} =  - 72\sin 2\theta \)     A1

evidence of substituting \(\frac{\pi }{4}\)     M1

e.g. \( - 72\sin \left( {2 \times \frac{\pi }{4}} \right)\) , \( - 72\sin \left( {\frac{\pi }{2}} \right)\) , \( - 72\)

\(\theta  = \frac{\pi }{4}\) produces the maximum area     AG     N0

[8 marks]

Candidates familiar with the circular nature of sine and cosine found part (a) accessible. However, a good number of candidates left this part blank, which suggests that there was difficulty interpreting the meaning of the x and y in the diagram. 

Those with answers from (a) could begin part (b), but many worked backwards and thus earned no marks. In a "show that" question, a solution cannot begin with the answer given. The area of the rectangle could be found by using \(2x \times 2y\) , or by using the eight small triangles, but it was essential that the substitution of the double-angle formula was shown before writing the given answer.

As the area function was given in part (b), many candidates correctly found the derivative in (c) and knew to set this derivative to zero for a maximum value. Many gave answers in degrees, however, despite the given domain in radians.

Although some candidates found the second derivative function correctly, few stated that the second derivative must be negative at a maximum value. Simply calculating a negative value is not sufficient for a justification.

Show that \(f'(x) = \frac{{\ln x}}{x}\).

There is a minimum on the graph of \(f\). Find the \(x\)-coordinate of this minimum.

Write down the value of \(p\).

The graph of \(g\) intersects the graph of \(f'\) when \(x = q\).

Find the value of \(q\).

The graph of \(g\) intersects the graph of \(f'\) when \(x = q\).

Let \(R\) be the region enclosed by the graph of \(f'\), the graph of \(g\) and the line \(x = p\).

Show that the area of \(R\) is \(\frac{1}{2}\).

METHOD 1

correct use of chain rule     A1A1

eg     \(\frac{{2\ln x}}{2} \times \frac{1}{x},{\text{ }}\frac{{2\ln x}}{{2x}}\)

Note: Award A1 for \(\frac{{2\ln x}}{{2x}}\), A1 for \( \times \frac{1}{x}\).

\(f'(x) = \frac{{\ln x}}{x}\)     AG     N0

[2 marks]

METHOD 2

correct substitution into quotient rule, with derivatives seen     A1

eg     \(\frac{{2 \times 2\ln x \times \frac{1}{x} - 0 \times {{(\ln x)}^2}}}{4}\)

correct working     A1

eg     \(\frac{{4\ln x \times \frac{1}{x}}}{4}\)

\(f'(x) = \frac{{\ln x}}{x}\)     AG     N0

[2 marks]

setting derivative \( = 0\)     (M1)

eg     \(f'(x) = 0,{\text{ }}\frac{{\ln x}}{x} = 0\)

correct working     (A1)

eg     \(\ln x = 0,{\text{ }}x = {{\text{e}}^0}\)

\(x = 1\)     A1     N2

[3 marks] 

intercept when \(f'(x) = 0\)     (M1)

\(p = 1\)     A1     N2

[2 marks]

equating functions     (M1)

eg     \(f' = g,{\text{ }}\frac{{\ln x}}{x} = \frac{1}{x}\)

correct working     (A1)

eg     \(\ln x = 1\)

\(q = {\text{e   (accept }}x = {\text{e)}}\)     A1     N2

[3 marks]

evidence of integrating and subtracting functions (in any order, seen anywhere)     (M1)

eg     \(\int_q^e {\left( {\frac{1}{x} - \frac{{\ln x}}{x}} \right){\text{d}}x{\text{, }}\int {f' - g} } \)

correct integration \(\ln x - \frac{{{{(\ln x)}^2}}}{2}\)     A2

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg     \((\ln {\text{e}} - \ln 1) - \left( {\frac{{{{(\ln {\text{e}})}^2}}}{2} - \frac{{{{(\ln 1)}^2}}}{2}} \right)\)

Note: Do not award M1 if the integrated function has only one term.

correct working     A1

eg     \((1 - 0) - \left( {\frac{1}{2} - 0} \right),{\text{ }}1 - \frac{1}{2}\)

\({\text{area}} = \frac{1}{2}\)     AG     N0

Notes: Candidates may work with two separate integrals, and only combine them at the end. Award marks in line with the markscheme.

[5 marks]

(a)     Find \(\int_0^2 {f(x){\rm{d}}x} \) .

(b)     The shaded region is enclosed by the graph of \(f\) , the \(x\)-axis, the \(y\)-axis and the line \(x = 6\) . The area of this region is \(3\pi \) .

Find \(\int_2^6 {f(x){\rm{d}}x} \) .

Find \(\int_0^2 {f(x){\rm{d}}x} \) .

The shaded region is enclosed by the graph of \(f\) , the \(x\)-axis, the \(y\)-axis and the line \(x = 6\) . The area of this region is \(3\pi \) .

Find \(\int_2^6 {f(x){\rm{d}}x} \) .

(a)     attempt to find quarter circle area     (M1)

eg   \(\frac{1}{4}(4\pi )\) , \(\frac{{\pi {r^2}}}{4}\) , \(\int_0^2 {\sqrt {4 - {x^2}{\rm{d}}x} } \)

area of region \( = \pi \)     (A1)

\(\int_0^2 {f(x){\rm{d}}x = - \pi } \)     A2     N3

[4 marks]

(b)     attempted set up with both regions     (M1)

eg   \({\text{shaded area}} - {\text{quarter circle}}\) , \(3\pi  - \pi \) , \(3\pi  - \int_0^2 {f = \int_2^6 f } \) 

\(\int_2^6 {f(x){\rm{d}}x = 2\pi } \)     A2     N2

[3 marks]

Total [7 marks]

attempt to find quarter circle area     (M1)

eg   \(\frac{1}{4}(4\pi )\) , \(\frac{{\pi {r^2}}}{4}\) , \(\int_0^2 {\sqrt {4 - {x^2}{\rm{d}}x} } \)

area of region \( = \pi \)     (A1)

\(\int_0^2 {f(x){\rm{d}}x = - \pi } \)     A2     N3

[4 marks]

attempted set up with both regions     (M1)

eg   \({\text{shaded area}} - {\text{quarter circle}}\) , \(3\pi  - \pi \) , \(3\pi  - \int_0^2 {f = \int_2^6 f } \) 

\(\int_2^6 {f(x){\rm{d}}x = 2\pi } \)     A2     N2

[3 marks]

Total [7 marks]

There was a minor error on the diagram, where the point on the \(y\)-axis was labelled \(2\) (to indicate the length of the radius), rather than \( - 2\). Examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect.

While most candidates were able to correctly find the area of the quarter circle in part (a), very few considered that the value of the definite integral is negative for the part of the function below the \(x\)-axis. In part (b), most went on to earn full marks by subtracting the area of the quarter circle from \(3 \pi\).

Candidates who did not understand the connection between area and the value of the integral often tried to find a function to integrate. These candidates were not successful using this method.

There was a minor error on the diagram, where the point on the \(y\)-axis was labelled \(2\) (to indicate the length of the radius), rather than \( - 2\). Examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect.

While most candidates were able to correctly find the area of the quarter circle in part (a), very few considered that the value of the definite integral is negative for the part of the function below the \(x\)-axis. In part (b), most went on to earn full marks by subtracting the area of the quarter circle from \(3 \pi\).

Candidates who did not understand the connection between area and the value of the integral often tried to find a function to integrate. These candidates were not successful using this method.

There was a minor error on the diagram, where the point on the \(y\)-axis was labelled \(2\) (to indicate the length of the radius), rather than \( - 2\). Examiners were instructed to notify the IB assessment centre of any candidates adversely affected. Candidate scripts did not indicate any adverse effect.

While most candidates were able to correctly find the area of the quarter circle in part (a), very few considered that the value of the definite integral is negative for the part of the function below the \(x\)-axis. In part (b), most went on to earn full marks by subtracting the area of the quarter circle from \(3 \pi\).

Candidates who did not understand the connection between area and the value of the integral often tried to find a function to integrate. These candidates were not successful using this method.

Find the value of \(f(0)\) .

Find the set of values of \(x\) for which \(f\) is increasing.

(i)     Find \(f''(1)\) .

(ii)     Hence, show that there is no point of inflexion on the graph of \(f\) at \(x = 0\) .

There is a point of inflexion on the graph of \(f\) at \(x = \sqrt[4]{3}\) \((x = 1.316 \ldots )\) .

Sketch the graph of \(f\) , for \(x \ge 0\) .

substitute \(0\) into \(f\)     (M1)

eg   \(\ln (0 + 1)\) , \(\ln 1\)

\(f(0) = 0\)     A1 N2

[2 marks]

\(f'(x) = \frac{1}{{{x^4} + 1}} \times 4{x^3}\) (seen anywhere)     A1A1

Note: Award A1 for \(\frac{1}{{{x^4} + 1}}\) and A1 for \(4{x^3}\) .

recognizing \(f\) increasing where \(f'(x) > 0\) (seen anywhere)     R1

eg   \(f'(x) > 0\) , diagram of signs

attempt to solve \(f'(x) > 0\)     (M1)

eg   \(4{x^3} = 0\) , \({x^3} > 0\)

\(f\) increasing for \(x > 0\) (accept \(x \ge 0\) )     A1     N1

[5 marks]

(i)     substituting \(x = 1\) into \(f''\)     (A1)

eg   \(\frac{{4(3 - 1)}}{{{{(1 + 1)}^2}}}\) , \(\frac{{4 \times 2}}{4}\)

\(f''(1) = 2\)     A1     N2

(ii)     valid interpretation of point of inflexion (seen anywhere)     R1

eg   no change of sign in \(f''(x)\) , no change in concavity,

\(f'\) increasing both sides of zero

attempt to find \(f''(x)\) for \(x < 0\)     (M1)

eg   \(f''( - 1)\) , \(\frac{{4{{( - 1)}^2}(3 - {{( - 1)}^4})}}{{{{({{( - 1)}^4} + 1)}^2}}}\) , diagram of signs

correct working leading to positive value     A1

eg   \(f''( - 1) = 2\) , discussing signs of numerator and denominator

there is no point of inflexion at \(x = 0\)     AG     N0

[5 marks]

     A1A1A1     N3

Notes: Award A1 for shape concave up left of POI and concave down right of POI.

    Only if this A1 is awarded, then award the following:

    A1 for curve through (\(0\), \(0\)) , A1 for increasing throughout.

    Sketch need not be drawn to scale. Only essential features need to be clear.

[3 marks]

Many candidates left their answer to part (a) as \(\ln 1\). While this shows an understanding for substituting a value into a function, it leaves an unfinished answer that should be expressed as an integer.

Candidates who attempted to consider where \(f\) is increasing generally understood the derivative is needed. However, a number of candidates did not apply the chain rule, which commonly led to answers such as “increasing for all \(x\)”. Many set their derivative equal to zero, while neglecting to indicate in their working that \(f'(x) > 0\) for an increasing function. Some created a diagram of signs, which provides appropriate evidence as long as it is clear that the signs represent \(f’\).

Finding \(f''(1)\) proved no challenge, however, using this value to show that no point of inflexion exists proved elusive for many. Some candidates recognized the signs must not change in the second derivative. Few candidates presented evidence in the form of a calculation, which follows from the “hence” command of the question. In this case, a sign diagram without numerical evidence was not sufficient.

Few candidates created a correct graph from the information given or found in the question. This included the point (\(0\), \(0\)), the fact that the function is always increasing for \(x > 0\) , the concavity at \(x = 1\) and the change in concavity at the given point of inflexion. Many incorrect attempts showed a graph concave down to the right of \(x = 0\) , changing to concave up.

Find the acceleration of the particle at \(t = 0\) .

Find the velocity, v, at time t, given that the initial velocity of the particle is \({\text{m}}{{\text{s}}^{ - 1}}\) .

Find \(\int_0^3 {v{\rm{d}}t} \) , giving your answer in the form \(p - q\cos 3\) .

What information does the answer to part (c) give about the motion of the particle?

substituting \(t = 0\)     (M1)

e.g. \(a(0) = 0 + \cos 0\)

\(a(0) = 1\)     A1     N2

[2 marks]

evidence of integrating the acceleration function     (M1)

e.g. \(\int {(2t + \cos t){\text{d}}t} \)

correct expression \({t^2} + \sin t + c\)     A1A1

Note: If "\( + c\)" is omitted, award no further marks.

evidence of substituting (2,0) into indefinite integral     (M1)

e.g. \(2 = 0 + \sin 0 + c\) , \(c = 2\)

\(v(t) = {t^2} + \sin t + 2\)     A1     N3

[5 marks]

\(\int {({t^2} + \sin t + 2)} {\rm{d}}t = \frac{{{t^3}}}{3} - \cos t + 2t\)     A1A1A1

Note: Award A1 for each correct term.

evidence of using \(v(3) - v(0)\)     (M1)

correct substitution     A1

e.g. \((9 - \cos 3 + 6) - (0 - \cos 0 + 0)\) , \((15 - \cos 3) - ( - 1)\)

\(16 - \cos 3\) (accept \(p = 16\) , \(q = - 1\) )     A1A1     N3

[7 marks]

reference to motion, reference to first 3 seconds     R1R1     N2

e.g. displacement in 3 seconds, distance travelled in 3 seconds

[2 marks]

Parts (a) and (b) of this question were generally well done.

Parts (a) and (b) of this question were generally well done.

Problems arose in part (c) with many candidates not substituting \(s(3) - s(0)\) correctly, leading to only a partially correct final answer. There were also a notable few who were not aware that \(\cos 0 = 1\) in both parts (a) and (c).

There were a variety of interesting answers about the motion of the particle, few being able to give both parts of the answer correctly.

(i) Write down the range of the function f .

(ii) Consider \(f(x) = 1\) , \(0 \le x \le 2\pi \) . Write down the number of solutions to this equation. Justify your answer.

Find \(f'(x)\) , giving your answer in the form \(a{\sin ^p}x{\cos ^q}x\) where \(a{\text{, }}p{\text{, }}q \in \mathbb{Z}\) .

Let \(g(x) = \sqrt 3 \sin x{(\cos x)^{\frac{1}{2}}}\) for \(0 \le x \le \frac{\pi }{2}\) . Find the volume generated when the curve of g is revolved through \(2\pi \) about the x-axis.

(i) range of f is \([ - 1{\text{, }}1]\) , \(( - 1 \le f(x) \le 1)\)     A2     N2

(ii) \({\sin ^3}x \Rightarrow 1 \Rightarrow \sin x = 1\)     A1

justification for one solution on \([0{\text{, }}2\pi ]\)    R1

e.g. \(x = \frac{\pi }{2}\) , unit circle, sketch of \(\sin x\)

1 solution (seen anywhere)     A1     N1

[5 marks]

\(f'(x) = 3{\sin ^2}x\cos x\)     A2     N2

[2 marks]

using \(V = \int_a^b {\pi {y^2}{\rm{d}}x} \)     (M1)

\(V = \int_0^{\frac{\pi }{2}} {\pi (\sqrt 3 } \sin x{\cos ^{\frac{1}{2}}}x{)^2}{\rm{d}}x\)     (A1)

\( = \pi \int_0^{\frac{\pi }{2}} {3{{\sin }^2}x\cos x{\rm{d}}x} \)     A1

\(V = \pi \left[ {{{\sin }^3}x} \right]_0^{\frac{\pi }{2}}\) \(\left( { = \pi \left( {{{\sin }^3}\left( {\frac{\pi }{2}} \right) - {{\sin }^3}0} \right)} \right)\)     A2

evidence of using \(\sin \frac{\pi }{2} = 1\) and \(\sin 0 = 0\)     (A1)

e.g. \(\pi \left( {1 - 0} \right)\)

\(V = \pi \)     A1     N1

[7 marks]

This question was not done well by most candidates. No more than one-third of them could correctly give the range of \(f(x) = {\sin ^3}x\) and few could provide adequate justification for there being exactly one solution to \(f(x) = 1\) in the interval \([0{\text{, }}2\pi ]\) .

This question was not done well by most candidates.

This question was not done well by most candidates. No more than one-third of them could correctly give the range of \(f(x) = {\sin ^3}x\) and few could provide adequate justification for there being exactly one solution to \(f(x) = 1\) in the interval \([0{\text{, }}2\pi ]\) . Finding the derivative of this function also presented major problems, thus making part (c) of the question much more difficult. In spite of the formula for volume of revolution being given in the Information Booklet, fewer than half of the candidates could correctly put the necessary function and limits into \(\pi \int_a^b {{y^2}{\rm{d}}x} \) and fewer still could square \(\sqrt 3 \sin x{\cos ^{\frac{1}{2}}}x\) correctly. From those who did square correctly, the correct antiderivative was not often recognized. All manner of antiderivatives were suggested instead.

Write down the x-intercepts of the graph of the derivative function, \(f'\) .

Write down all values of x for which \(f'(x)\) is positive.

At point D on the graph of f , the x-coordinate is \( - 0.5\). Explain why \(f''(x) < 0\) at D.

x-intercepts at \( - 3\), 0, 2     A2     N2

[2 marks]

\( - 3 < x < 0\) , \(2 < x < 3\)     A1A1    N2

[2 marks]

correct reasoning     R2

e.g. the graph of f is concave-down (accept convex), the first derivative is decreasing

therefore the second derivative is negative     AG

[2 marks]

Candidates had mixed success with parts (a) and (b). Weaker candidates either incorrectly used the x-intercepts of f or left this question blank. Some wrote down only two of the three values in part (a). Candidates who answered part (a) correctly often had trouble writing the set of values in part (b); difficulties included poor notation and incorrectly including the endpoints. Other candidates listed individual x-values here rather than a range of values.

Candidates had mixed success with parts (a) and (b). Weaker candidates either incorrectly used the x-intercepts of f or left this question blank. Some wrote down only two of the three values in part (a). Candidates who answered part (a) correctly often had trouble writing the set of values in part (b); difficulties included poor notation and incorrectly including the endpoints. Other candidates listed individual x-values here rather than a range of values.

Many candidates had difficulty explaining why the second derivative is negative in part (c). A number claimed that since the point D was “close” to a maximum value, the second derivative must be negative; this incorrect appeal to the second derivative test indicates a lack of understanding of how the test works and the relative concept of closeness. Some candidates claimed D was a point of inflexion, again demonstrating poor understanding of the second derivative. Among candidates who answered part (c) correctly, some stated that f was concave down while others gave well-formed arguments for why the first derivative was decreasing. A few candidates provided nicely sketched graphs of \(f'\) and \(f''\) and used them in their explanations.

The graph of a function h passes through the point \(\left( {\frac{\pi }{{12}}, 5} \right)\).

Given that \(h'(x) = 4\cos 2x\), find \(h(x)\).

evidence of anti-differentiation     (M1)

eg     \(\int {h'(x), \int {4\cos 2x{\text{d}}x} } \)

correct integration     (A2)

eg     \(h(x) = 2\sin 2x + c, \frac{{4\sin 2x}}{2}\)

attempt to substitute \(\left( {\frac{\pi }{{12}},5} \right)\) into their equation     (M1)

eg     \(2\sin \left( {2 \times \frac{\pi }{{12}}} \right) + c = 5,{\text{ }}2\sin \left( {\frac{\pi }{6}} \right) = 5\)

correct working     (A1)

eg     \(2\left( {\frac{1}{2}} \right) + c = 5,{\text{ }}c = 4\)

\(h(x) = 2\sin 2x + 4\)     A1     N5

[6 marks]

Write down

(i)     \(f'(x)\) ;

(ii)    \(g'(x)\) .

Let \(h(x) = {{\rm{e}}^{ - 3x}}\sin \left( {x - \frac{\pi }{3}} \right)\) . Find the exact value of \(h'\left( {\frac{\pi }{3}} \right)\) .

(i) \( - 3{{\rm{e}}^{ - 3x}}\)     A1     N1

(ii) \(\cos \left( {x - \frac{\pi }{3}} \right)\)     A1     N1

[4 marks]

evidence of choosing product rule     (M1)

e.g. \(uv' + vu'\)

correct expression     A1

e.g. \( - 3{{\rm{e}}^{ - 3x}}\sin \left( {x - \frac{\pi }{3}} \right) + {{\rm{e}}^{ - 3x}}\cos \left( {x - \frac{\pi }{3}} \right)\)

complete correct substitution of \(x = \frac{\pi }{3}\)     (A1)

e.g. \( - 3{{\rm{e}}^{ - 3\frac{\pi }{3}}}\sin \left( {\frac{\pi }{3} - \frac{\pi }{3}} \right) + {{\rm{e}}^{ - 3\frac{\pi }{3}}}\cos \left( {\frac{\pi }{3} - \frac{\pi }{3}} \right)\)        

\(h'\left( {\frac{\pi }{3}} \right) = {{\rm{e}}^{ - \pi }}\)     A1     N3

[4 marks]

A good number of candidates found the correct derivative expressions in (a). Many applied the product rule, although with mixed success.

Often the substitution of \({\frac{\pi }{3}}\) was incomplete or not done at all.

Write down the velocity of the particle when \(t = 0\) .

When \(t = k\) , the acceleration is zero.

(i)     Show that \(k = \frac{\pi }{4}\) .

(ii)    Find the exact velocity when \(t = \frac{\pi }{4}\) .

When \(t < \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\) and when \(t > \frac{\pi }{4}\) , \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} > 0\)  .

Sketch a graph of v against t .

Let d be the distance travelled by the particle for \(0 \le t \le 1\) .

(i)     Write down an expression for d .

(ii)    Represent d on your sketch.

\(v = 1\)     A1     N1

[1 mark]

(i) \(\frac{{\rm{d}}}{{{\rm{d}}t}}(2t) = 2\)     A1

\(\frac{{\rm{d}}}{{{\rm{d}}t}}(\cos 2t) = - 2\sin 2t\)     A1A1

Note: Award A1 for coefficient 2 and A1 for \( - \sin 2t\) .

evidence of considering acceleration = 0     (M1)

e.g. \(\frac{{{\rm{d}}v}}{{{\rm{d}}t}} = 0\) , \(2 - 2\sin 2t = 0\)

correct manipulation     A1

e.g. \(\sin 2k = 1\) , \(\sin 2t = 1\)

\(2k = \frac{\pi }{2}\) (accept \(2t = \frac{\pi }{2}\) )     A1

\(k = \frac{\pi }{4}\)     AG     N0

(ii) attempt to substitute \(t = \frac{\pi }{4}\) into v     (M1)

e.g. \(2\left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{{2\pi }}{4}} \right)\)

\(v = \frac{\pi }{2}\)     A1     N2

[8 marks]

     A1A1A2      N4

Notes: Award A1 for y-intercept at \((0{\text{, }}1)\) , A1 for curve having zero gradient at \(t = \frac{\pi }{4}\) , A2 for shape that is concave down to the left of \(\frac{\pi }{4}\) and concave up to the right of \(\frac{\pi }{4}\) . If a correct curve is drawn without indicating \(t = \frac{\pi }{4}\) , do not award the second A1 for the zero gradient, but award the final A2 if appropriate. Sketch need not be drawn to scale. Only essential features need to be clear.

[4 marks]

(i) correct expression     A2

e.g. \(\int_0^1 {(2t + \cos 2t){\rm{d}}t} \) , \(\left[ {{t^2} + \frac{{\sin 2t}}{2}} \right]_0^1\) , \(1 + \frac{{\sin 2}}{2}\) , \(\int_0^1 {v{\rm{d}}t} \)

(ii)

     A1

Note: The line at \(t = 1\) needs to be clearly after \(t = \frac{\pi }{4}\) .

[3 marks]

Many candidates gave a correct initial velocity, although a substantial number of candidates answered that \(0 + \cos 0 = 0\) .

For (b), students commonly applied the chain rule correctly to achieve the derivative, and many recognized that the acceleration must be zero. Occasionally a student would use a double-angle identity on the velocity function before differentiating. This is not incorrect, but it usually caused problems when trying to show \(k = \frac{\pi }{4}\) . At times students would reach the equation \(\sin 2k = 1\) and then substitute the \(\frac{\pi }{4}\) , which does not satisfy the “show that” instruction.

The challenge in this question is sketching the graph using the information achieved and provided. This requires students to make graphical interpretations, and as typical in section B, to link the early parts of the question with later parts. Part (a) provides the y-intercept, and part (b) gives a point with a horizontal tangent. Plotting these points first was a helpful strategy. Few understood either the notation or the concept that the function had to be increasing on either side of the \(\frac{\pi }{4}\) , with most thinking that the point was either a max or min. It was the astute student who recognized that the derivatives being positive on either side of \(\frac{\pi }{4}\) creates a point of inflexion.

Additionally, important points should be labelled in a sketch. Indicating the \(\frac{\pi }{4}\) on the x-axis is a requirement of a clear graph. Although students were not penalized for not labelling the \(\frac{\pi }{2}\) on the y-axis, there should be a recognition that the point is higher than the y-intercept.

While some candidates recognized that the distance is the area under the velocity graph, surprisingly few included neither the limits of integration in their expression, nor the “dt”. Most unnecessarily attempted to integrate the function, often giving an answer with “+C”, and only earned marks if the limits were included with their result. Few recognized that a shaded area is an adequate representation of distance on the sketch, with most fruitlessly attempting to graph a new curve.

Find \(s'(t)\) .

In this interval, there are only two values of t for which the object is not moving. One value is \(t = \frac{\pi }{6}\) .

Find the other value.

Show that \(s'(t) > 0\) between these two values of t .

Find the distance travelled between these two values of t .

\(s'(t) = 1 - 2\cos 2t\)    A1A2     N3

Note: Award A1 for 1, A2 for \(- 2\cos 2t\) .

[3 marks]

evidence of valid approach     (M1)

e.g. setting \(s'(t) = 0\)

correct working     A1

e.g. \(2\cos 2t = 1\) , \(\cos 2t = \frac{1}{2}\)

\(2t = \frac{\pi }{3}\) , \(\frac{{5\pi }}{3}\) , \(\ldots \)     (A1)

\(t = \frac{{5\pi }}{6}\)     A1     N3 

[4 marks]

evidence of valid approach     (M1)

e.g. choosing a value in the interval \(\frac{\pi }{6} < t < \frac{{5\pi }}{6}\)

correct substitution     A1

e.g. \(s'\left( {\frac{\pi }{2}} \right) = 1 - 2\cos \pi \)

\(s'\left( {\frac{\pi }{2}} \right) = 3\)     A1

\(s'(t) > 0\)     AG     N0

[3 marks]

evidence of approach using s or integral of \(s'\)     (M1)

e.g. \(\int {s'(t){\rm{d}}t} \) ; \(s\left( {\frac{{5\pi }}{6}} \right)\) , \(s\left( {\frac{\pi }{6}} \right)\) ; \(\left[ {t - \sin 2t} \right]_{\frac{\pi }{6}}^{\frac{{5\pi }}{6}}\)

substituting values and subtracting     (M1)

e.g. \(s\left( {\frac{{5\pi }}{6}} \right) - s\left( {\frac{\pi }{6}} \right)\) , \(\left( {\frac{\pi }{6} - \frac{{\sqrt 3 }}{2}} \right) - \left( {\frac{{5\pi }}{6} - \left( { - \frac{{\sqrt 3 }}{2}} \right)} \right)\)

correct substitution     A1

e.g. \(\frac{{5\pi }}{6} - \sin \frac{{5\pi }}{3} - \left[ {\frac{\pi }{6} - \sin \frac{\pi }{3}} \right]\) , \(\left( {\frac{{5\pi }}{6} - \left( { - \frac{{\sqrt 3 }}{2}} \right)} \right) - \left( {\frac{\pi }{6} - \frac{{\sqrt 3 }}{2}} \right)\)

distance is \(\frac{{2\pi }}{3} + \sqrt 3 \)     A1A1     N3

Note: Award A1 for \(\frac{{2\pi }}{3}\) , A1 for \(\sqrt 3 \) .

[5 marks]

The derivative in part (a) was reasonably well done, but errors here often caused trouble in later parts. Candidates occasionally attempted to use the double angle identity for \(\sin 2t\) before differentiating, but they rarely were successful in then applying the product rule.

In part (b), most candidates understood that they needed to set their derivative equal to zero, but fewer were able to take the next step to solve the resulting double angle equation. Again, some candidates over-complicated the equation by using the double angle identity. Few ended up with the correct answer \(\frac{{5\pi }}{6}\) .

In part (c), many candidates knew they needed to test a value between \(\pi /6\) and their value from part (b), but fewer were able to successfully complete that calculation. Some candidates simply tested their boundary values while others unsuccessfully attempted to make use of the second derivative.

Although many candidates did not attempt part (d), those who did often demonstrated a good understanding of how to use the displacement function s or the integral of their derivative from part (a). Candidates who had made an error in part (b) often could not finish, as \(\sin (2t)\) could not be evaluated at their value without a calculator. Of those who had successfully found the other boundary of \(5\pi /6\) , a common error was giving the incorrect sign of the value of \(\sin (5\pi /3)\) . Again, this part was a good discriminator between the grade 6 and 7 candidates.

The following diagram shows the graph of \(f(x) = \frac{x}{{{x^2} + 1}}\), for \(0 \le x \le 4\), and the line \(x = 4\).

Let \(R\) be the region enclosed by the graph of \(f\) , the \(x\)-axis and the line \(x = 4\).

Find the area of \(R\).

substitution of limits or function     (A1)

eg\(\;\;\;A = \int_0^4 {f(x),{\text{ }}\int {\frac{x}{{{x^2} + 1}}{\text{d}}x} } \)

correct integration by substitution/inspection     A2

\(\frac{1}{2}\ln ({x^2} + 1)\)

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg\(\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) - \ln ({0^2} + 1)} \right)\)

correct working     A1

eg\(\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) - \ln ({0^2} + 1)} \right),{\text{ }}\frac{1}{2}\left( {\ln (17) - \ln (1)} \right),{\text{ }}\frac{1}{2}\ln 17 - 0\)

\(A = \frac{1}{2}\ln (17)\)     A1     N3

Note:     Exception to FT rule. Allow full FT on incorrect integration involving a \(\ln \) function.

[6 marks]

Very few candidates earned full marks in this question. While most candidates knew to integrate, many seemed unfamiliar with integrating using substitution or inspection. This topic is part of the syllabus, but it did not occur to many candidates to use a substitution method. A large number of them tried to integrate the individual terms in the numerator and denominator as though this were a polynomial function. While there were some candidates who knew the integral would involve a natural log function and substituted 4 and 0 into their function, many ended up with undefined values such as or did not know what to do with expressions containing \(\ln 1\).

Show that the equation of L is \(y = - 4x + 18\) .

Point A is the x-intercept of L . Find the x-coordinate of A.

Find an expression for the area of R .

The region R is rotated \(360^\circ \) about the x-axis. Find the volume of the solid formed, giving your answer in terms of \(\pi \) .

finding derivative     (A1)

e.g. \(f'(x) = \frac{1}{2}{x^{\frac{1}{2}}},\frac{{1}}{{2\sqrt x }}\)

correct value of derivative or its negative reciprocal (seen anywhere)     A1

e.g. \(\frac{1}{{2\sqrt 4 }}\) , \(\frac{1}{4}\)

gradient of normal =  \(\frac{1}{{{\text{gradient of tangent}}}}\) (seen anywhere)     A1

e.g. \( - \frac{1}{{f'(4)}} = - 4\) , \( - 2\sqrt x \)

substituting into equation of line (for normal)     M1

e.g. \(y - 2 = - 4(x - 4)\)

\(y = - 4x + 18\)     AG     N0

[4 marks]

recognition that \(y = 0\) at A     (M1)

e.g. \( - 4x + 18 = 0\)

\(x = \frac{{18}}{4}\) \(\left( { = \frac{9}{2}} \right)\)     A1     N2

[2 marks]

splitting into two appropriate parts (areas and/or integrals)     (M1)

correct expression for area of R     A2     N3

e.g. area of R = \(\int_0^4 {\sqrt x } {\rm{d}}x + \int_4^{4.5} {( - 4x + 18){\rm{d}}x} \) , \(\int_0^4 {\sqrt x } {\rm{d}}x + \frac{1}{2} \times 0.5 \times 2\) (triangle)

Note: Award A1 if dx is missing.

[3 marks]

correct expression for the volume from \(x = 0\) to \(x = 4\)     (A1)

e.g. \(V = \int_0^4 {\pi \left[ {f{{(x)}^2}} \right]} {\rm{d}}x\) , \({\int_0^4 {\pi \sqrt x } ^2}{\rm{d}}x\) , \(\int_0^4 {\pi x{\rm{d}}x} \)

\(V = \left[ {\frac{1}{2}\pi {x^2}} \right]_0^4\)     A1

\(V = \pi \left( {\frac{1}{2} \times 16 - \frac{1}{2} \times 0} \right)\)     (A1)

\(V = 8\pi \)     A1

finding the volume from \(x = 4\) to \(x = 4.5\)

EITHER

recognizing a cone     (M1)

e.g. \(V = \frac{1}{3}\pi {r^2}h\)