If z is a non-zero complex number, we define \(L(z)\) by the equation

\[L(z) = \ln \left| z \right| + {\text{i}}\arg (z),{\text{ }}0 \leqslant \arg (z) < 2\pi .\]

(a)     Show that when z is a positive real number, \(L(z) = \ln z\) .

(b)     Use the equation to calculate

(i)     \(L( - 1)\) ;

(ii)     \(L(1 - {\text{i}})\) ;

(iii)     \(L( - 1 + {\text{i}})\) .

(c)     Hence show that the property \(L({z_1}{z_2}) = L({z_1}) + L({z_2})\) does not hold for all values of \({z_1}\) and \({z_2}\) .

Let f be a function with domain \(\mathbb{R}\) that satisfies the conditions,

\(f(x + y) = f(x)f(y)\) , for all x and y and \(f(0) \ne 0\) .

(a)     Show that \(f(0) = 1\).

(b)     Prove that \(f(x) \ne 0\) , for all \(x \in \mathbb{R}\) .

(c)     Assuming that \(f'(x)\) exists for all \(x \in \mathbb{R}\) , use the definition of derivative to show that \(f(x)\) satisfies the differential equation \(f'(x) = k{\text{ }}f(x)\) , where \(k = f'(0)\) .

(d)     Solve the differential equation to find an expression for \(f(x)\) .

(a)     \(\left| z \right| = z\) , \(\arg (z) = 0\)     A1A1

so \(L(z) = \ln z\)     AG     N0

[2 marks]

(b)     (i)     \(L( - 1) = \ln 1 + {\text{i}}\pi = {\text{i}}\pi \)     A1A1     N2

(ii)     \(L(1 - {\text{i}}) = \ln \sqrt 2 + {\text{i}}\frac{{7\pi }}{4}\)     A1A1     N2

(iii)     \(L( - 1 + {\text{i}}) = \ln \sqrt 2 + {\text{i}}\frac{{3\pi }}{4}\)     A1     N1

[5 marks]

(c)     for comparing the product of two of the above results with the third     M1

for stating the result \( - 1 + {\text{i}} = - 1 \times (1 - {\text{i}})\) and \(L( - 1 + {\text{i}}) \ne L( - 1) + L(1 - {\text{i}})\)     R1

hence, the property \(L({z_1}{z_2}) = L({z_1}) + L({z_2})\)

does not hold for all values of \({z_1}\) and \({z_2}\)     AG     N0

[2 marks]

Total [9 marks]

(a)     from \(f(x + y) = f(x)f(y)\)

for x = y = 0     M1

we have \(f(0 + 0) = f(0)f(0) \Leftrightarrow f(0) = {\left( {f(0)} \right)^2}\)     A1

as \(f(0) \ne 0\), this implies that \(f(0) = 1\)     R1AG     N0

[3 marks]

(b)     METHOD 1

from \(f(x + y) = f(x)f(y)\)

for y = –x , we have \(f(x - x) = f(x)f( - x) \Leftrightarrow f(0) = f(x)f( - x)\)     M1A1

as \(f(0) \ne 0\) this implies that \(f(x) \ne 0\)     R1AG     N0

METHOD 2

suppose that, for a value of x, \(f(x) = 0\)     M1

from \(f(x + y) = f(x)f(y)\)

for \(y = - x\), we have \(f(x - x) = f(x)f( - x) \Leftrightarrow f(0) = f(x)f( - x)\)     A1

substituting \(f(x)\) by 0 gives \(f(0) = 0\) which contradicts part (a)     R1

therefore \(f(x) \ne 0\) for all x.     AG     N0

[3 marks]

(c)     by the definition of derivative

\(f'(x) = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x + h) - f(x)}}{h}} \right)\)     (M1)

\( = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x)f(h) - f(x)f(0)}}{h}} \right)\)     A1(A1)

\( = \mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(h) - f(0)}}{h}} \right)f(x)\)     A1

\( = f'(0)f(x)\,\,\,\,\,\left( { = k{\text{ }}f(x)} \right)\)     AG     N0

[4 marks]

(d)     \(\int {\frac{{f'(x)}}{{f(x)}}{\text{d}}x = \int {k{\text{d}}x \Rightarrow \ln f(x) = kx + C} } \)     M1A1

\(\ln f(0) = C \Rightarrow C = 0\)     A1

\(f(x) = {{\text{e}}^{kx}}\)     A1     N1

Note: Award M1A0A0A0 if no arbitrary constant C .

[4 marks]

Total [14 marks]

Part A was answered well by a fair amount of candidates, with some making mistakes in calculating the arguments of complex numbers, as well as careless mistakes in finding the products of complex numbers.

Part B proved demanding for most candidates, particularly parts (c) and (d). A surprising number of candidates did not seem to know what was meant by the ‘definition of derivative’ in part (c) as they attempted to use quotient rule rather than first principles.

A gourmet chef is renowned for her spherical shaped soufflé. Once it is put in the oven, its volume increases at a rate proportional to its radius.

(a)     Show that the radius r cm of the soufflé, at time t minutes after it has been put in the oven, satisfies the differential equation \(\frac{{{\text{d}}r}}{{{\text{d}}t}} = \frac{k}{r}\), where k is a constant.

(b)     Given that the radius of the soufflé is 8 cm when it goes in the oven, and 12 cm when it’s cooked 30 minutes later, find, to the nearest cm, its radius after 15 minutes in the oven.

(a)     \(\frac{{dV}}{{{\text{d}}t}} = cr\)     A1

\(V = \frac{4}{3}\pi {r^3}\)

\(\frac{{dV}}{{{\text{d}}t}} = 4\pi {r^2}\frac{{{\text{d}}r}}{{{\text{d}}t}}\)     M1A1

\( \Rightarrow 4\pi {r^2}\frac{{{\text{d}}r}}{{{\text{d}}t}} = cr\)     M1

\( \Rightarrow \frac{{{\text{d}}r}}{{{\text{d}}t}} = \frac{c}{{4\pi r}}\)     A1

\( = \frac{k}{r}\)     AG

[5 marks]

(b)     \(\frac{{{\text{d}}r}}{{{\text{d}}t}} = \frac{k}{r}\)

\( \Rightarrow \int {r{\text{d}}r = \int {k{\text{d}}t} } \)     M1

\(\frac{{{r^2}}}{2} = kt + d\)     A1

An attempt to substitute either t = 0, r = 8 or t = 30, r = 12     M1

When t = 0, r = 8

\( \Rightarrow d = 32\)     A1

\( \Rightarrow \frac{{{r^2}}}{2} = kt + 32\)

When t = 30, r = 12

\( \Rightarrow \frac{{{{12}^2}}}{2} = 30k + 32\)

\( \Rightarrow k = \frac{4}{3}\)     A1

\(\therefore \frac{{{r^2}}}{2} = \frac{4}{3}t + 32\)

When t = 15 , \(\frac{{{r^2}}}{2} = \frac{4}{3}15 + 32\)     M1

\( \Rightarrow {r^2} = 104\)     A1

\(r \approx 10{\text{ cm}}\)     A1

Note: Award M0 to incorrect methods using proportionality which give solution r = 10 cm .

[8 marks]

Total [13 marks]

Candidates found this question quite difficult, with only the better students making appreciable progress on part (a). Relatively few candidates recognised that part (b) was asking them to solve a differential equation. Many students tried methods involving direct proportion, which did not lead anywhere.

Find y in terms of x, given that \((1 + {x^3})\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2{x^2}\tan y\) and \(y = \frac{\pi }{2}\) when x = 0.

\((1 + {x^3})\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2{x^2}\tan y \Rightarrow \int {\frac{{{\text{d}}y}}{{\tan y}} = \int {\frac{{2{x^2}}}{{1 + {x^3}}}{\text{d}}x} } \)     M1

\(\int {\frac{{\cos y}}{{\sin y}}{\text{d}}y = \frac{2}{3}\int {\frac{{3{x^2}}}{{1 + {x^3}}}{\text{d}}x} } \)     (A1)(A1)

\(\ln \left| {\sin y} \right| = \frac{2}{3}\ln \left| {1 + {x^3}} \right| + C\)     A1A1

Notes: Do not penalize omission of modulus signs.

Do not penalize omission of constant at this stage.

EITHER

\(\ln \left| {\sin \frac{\pi }{2}} \right| = \frac{2}{3}\ln \left| 1 \right| + C \Rightarrow C = 0\)     M1

OR

\(\left| {\sin y} \right| = A{\left| {1 + {x^3}} \right|^{\frac{2}{3}}},{\text{ }}A = {{\text{e}}^C}\)

\(\left| {\sin \frac{\pi }{2}} \right| = A{\left| {1 + {0^3}} \right|^{\frac{2}{3}}} \Rightarrow A = 1\)     M1

THEN

\(y = \arcsin \left( {{{(1 + {x^3})}^{\frac{2}{3}}}} \right)\)     A1

Note: Award M0A0 if constant omitted earlier.

[7 marks]

Many candidates separated the variables correctly but were then unable to perform the integrations.

A certain population can be modelled by the differential equation \(\frac{{{\text{d}}y}}{{{\text{d}}t}} = k\,y\cos kt\) , where y is the population at time t hours and k is a positive constant.

(a)     Given that \(y = {y_0}\) when t = 0 , express y in terms of k , t and \({y_0}\) .

(b)     Find the ratio of the minimum size of the population to the maximum size of the population.

(a)     \(\frac{{{\text{d}}y}}{{{\text{d}}t}} = ky\cos (kt)\)

\(\frac{{{\text{d}}y}}{y} = k\cos (kt){\text{d}}t\)     (M1)

\(\int {\frac{{{\text{d}}y}}{y} = \int {k\cos (kt){\text{d}}t} } \)     M1

\(\ln y = \sin (kt) + c\)     A1

\(y = A{{\text{e}}^{\sin (kt)}}\)

\(t = 0 \Rightarrow {y_0} = A\)     (M1)

\( \Rightarrow y = {y_0}{{\text{e}}^{\sin kt}}\)     A1

(b)     \( - 1 \leqslant \sin kt \leqslant 1\)     (M1)

\({y_0}{{\text{e}}^{ - 1}} \leqslant y \leqslant {y_0}{{\text{e}}^1}\)

so the ratio is \(\frac{1}{{\text{e}}}:{\text{e}}\,\,\,\,\,{\text{or }}1:{{\text{e}}^2}\)     A1

[7 marks]

Part (a) was done successfully by many candidates. However, very few attempted part (b).

Find the equation of the tangent to C at the point (2, e).

Find \(f(x)\).

Determine the largest possible domain of f.

Show that the equation \(f(x) = f'(x)\) has no solution.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{\text{e}}}{{\ln {\text{e}}}}(2 + 2) = 4{\text{e}}\)     A1

at (2, e) the tangent line is \(y - {\text{e}} = 4{\text{e}}(x - 2)\)     M1

hence \(y = 4{\text{e}}x - 7{\text{e}}\)     A1

[3 marks]

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{\ln y}}(x + 2) \Rightarrow \frac{{\ln y}}{y}{\text{d}}y = (x + 2){\text{d}}x\)     M1

\(\int {\frac{{\ln y}}{y}{\text{d}}y = \int {(x + 2){\text{d}}x} } \)

using substitution \(u = \ln y;{\text{ d}}u = \frac{1}{y}{\text{d}}y\)     (M1)(A1)

\( \Rightarrow \int {\frac{{\ln y}}{y}{\text{d}}y = \int {u{\text{d}}u = \frac{1}{2}{u^2}} } \)     (A1)

\( \Rightarrow \frac{{{{(\ln y)}^2}}}{2} = \frac{{{x^2}}}{2} + 2x + c\)     A1A1

at (2, e), \(\frac{{{{(\ln {\text{e}})}^2}}}{2} = 6 + c\)     M1

\( \Rightarrow c = - \frac{{11}}{2}\)     A1

\( \Rightarrow \frac{{{{(\ln y)}^2}}}{2} = \frac{{{x^2}}}{2} + 2x - \frac{{11}}{2} \Rightarrow {(\ln y)^2} = {x^2} + 4x - 11\)

\(\ln y = \pm \sqrt {{x^2} + 4x - 11}  \Rightarrow y = {{\text{e}}^{ \pm \sqrt {{x^2} + 4x - 11} }}\)     M1A1

since y > 1, \(f(x) = {{\text{e}}^{\sqrt {{x^2} + 4x - 11} }}\)     R1

Note:M1 for attempt to make y the subject.

[11 marks]

EITHER

\({x^2} + 4x - 11 > 0\)     A1

using the quadratic formula     M1

critical values are \(\frac{{ - 4 \pm \sqrt {60} }}{2}{\text{ }}\left( { = - 2 \pm \sqrt {15} } \right)\)     A1

using a sign diagram or algebraic solution     M1

\(x < - 2 - \sqrt {15} ;{\text{ }}x > - 2 + \sqrt {15} \)     A1A1

OR

\({x^2} + 4x - 11 > 0\)     A1

by methods of completing the square     M1

\({(x + 2)^2} > 15\)     A1

\( \Rightarrow x + 2 < - \sqrt {15} {\text{ or }}x + 2 > \sqrt {15} \)     (M1)

\(x < - 2 - \sqrt {15} ;{\text{ }}x > - 2 + \sqrt {15} \)     A1A1

[6 marks]

\(f(x) = f'(x) \Rightarrow f(x) = \frac{{f(x)}}{{\ln f(x)}}(x + 2)\)     M1

\( \Rightarrow \ln \left( {f(x)} \right) = x + 2\,\,\,\,\,\left( { \Rightarrow x + 2 = \sqrt {{x^2} + 4x - 11} } \right)\)     A1

\( \Rightarrow {(x + 2)^2} = {x^2} + 4x - 11 \Rightarrow {x^2} + 4x + 4 = {x^2} + 4x - 11\)     A1

\( \Rightarrow 4 = - 11,{\text{ hence }}f(x) \ne f'(x)\)     R1AG

[4 marks]

Nearly always correctly answered.

Most candidates separated the variables and attempted the integrals. Very few candidates made use of the condition y > 1, so losing 2 marks.

Part (c) was often well answered, sometimes with follow through.

Only the best candidates were successful on part (d).