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HL Paper 3

This question investigates some applications of differential equations to modeling population growth.

One model for population growth is to assume that the rate of change of the population is proportional to the population, i.e. $\frac{{{\text{d}}P}}{{{\text{d}}t}} = kP$ , where $k \in \mathbb{R}$ , $t$ is the time (in years) and $P$ is the population

The initial population is 1000.

Given that $k = 0.003$ , use your answer from part (a) to find

Consider now the situation when $k$ is not a constant, but a function of time.

Given that $k = 0.003 + 0.002t$ , find

Another model for population growth assumes

there is a maximum value for the population, $L$ .
that $k$ is not a constant, but is proportional to $\left( {1 - \frac{P}{L}} \right)$ .

Show that the general solution of this differential equation is $P = A{{\text{e}}^{kt}}$ , where $A \in \mathbb{R}$ .

[5]

the population after 10 years

[2]

b.i.

the number of years it will take for the population to triple.

[2]

b.ii.

$\mathop {{\text{lim}}}\limits_{t \to \infty } P$

[1]

b.iii.

the solution of the differential equation, giving your answer in the form $P = f\left( t \right)$ .

[5]

c.i.

the number of years it will take for the population to triple.

[4]

c.ii.

Show that $\frac{{{\text{d}}P}}{{{\text{d}}t}} = \frac{m}{L}P\left( {L - P} \right)$ , where $m \in \mathbb{R}$ .

[2]

Solve the differential equation $\frac{{{\text{d}}P}}{{{\text{d}}t}} = \frac{m}{L}P\left( {L - P} \right)$ , giving your answer in the form $P = g\left( t \right)$ .

[10]

Given that the initial population is 1000, $L = 10000$ and $m = 0.003$ , find the number of years it will take for the population to triple.

[4]

In this question you will explore some of the properties of special functions $f$ and $g$ and their relationship with the trigonometric functions, sine and cosine.

Functions $f$ and $g$ are defined as $f (z) = \frac{e^{z} + e^{- z}}{2}$ and $g (z) = \frac{e^{z} - e^{- z}}{2}$ , where $z \in ℂ$ .

Consider $t$ and $u$ , such that $t, u \in ℝ$ .

Using $e^{i u} = \cos u + i \sin u$ , find expressions, in terms of $\sin u$ and $\cos u$ , for

The functions $\cos x$ and $\sin x$ are known as circular functions as the general point ( $\cos θ, \sin θ$ ) defines points on the unit circle with equation $x^{2} + y^{2} = 1$ .

The functions $f (x)$ and $g (x)$ are known as hyperbolic functions, as the general point ( $f (θ), g (θ)$ ) defines points on a curve known as a hyperbola with equation $x^{2} - y^{2} = 1$ . This hyperbola has two asymptotes.

Verify that $u = f (t)$ satisfies the differential equation $\frac{d^{2} u}{d t^{2}} = u$ .

[2]

Show that ${(f (t))}^{2} + {(g (t))}^{2} = f (2 t)$ .

[3]

$f (i u)$ .

[3]

c.i.

$g (i u)$ .

[2]

c.ii.

Hence find, and simplify, an expression for ${(f (i u))}^{2} + {(g (i u))}^{2}$ .

[2]

Show that ${(f (t))}^{2} - {(g (t))}^{2} = {(f (i u))}^{2} - {(g (i u))}^{2}$ .

[4]

Sketch the graph of $x^{2} - y^{2} = 1$ , stating the coordinates of any axis intercepts and the equation of each asymptote.

[4]

The hyperbola with equation $x^{2} - y^{2} = 1$ can be rotated to coincide with the curve defined by $x y = k, k \in ℝ$ .

Find the possible values of $k$ .

[5]

This question asks you to investigate some properties of the sequence of functions of the form ${f_n}(x) = {\text{cos}}\left( {n\,{\text{arccos}}\,x} \right)$ , −1 ≤ $x$ ≤ 1 and $n \in {\mathbb{Z}^ + }$ .

Important: When sketching graphs in this question, you are not required to find the coordinates of any axes intercepts or the coordinates of any stationary points unless requested.

For odd values of $n$ > 2, use your graphic display calculator to systematically vary the value of $n$ . Hence suggest an expression for odd values of $n$ describing, in terms of $n$ , the number of

For even values of $n$ > 2, use your graphic display calculator to systematically vary the value of $n$ . Hence suggest an expression for even values of $n$ describing, in terms of $n$ , the number of

The sequence of functions, ${f_n}(x)$ , defined above can be expressed as a sequence of polynomials of degree $n$ .

Consider ${f_{n + 1}}(x) = {\text{cos}}\left( {\left( {n + 1} \right)\,{\text{arccos}}\,x} \right)$ .

On the same set of axes, sketch the graphs of $y = {f_1}(x)$ and $y = {f_3}(x)$ for −1 ≤ $x$ ≤ 1.

[2]

local maximum points;

[3]

b.i.

local minimum points;

[1]

b.ii.

On a new set of axes, sketch the graphs of $y = {f_2}(x)$ and $y = {f_4}(x)$ for −1 ≤ $x$ ≤ 1.

[2]

local maximum points;

[3]

d.i.

local minimum points.

[1]

d.ii.

Solve the equation ${f_n}^\prime (x) = 0$ and hence show that the stationary points on the graph of $y = {f_n}(x)$ occur at $x = {\text{cos}}\frac{{k\pi }}{n}$ where $k \in {\mathbb{Z}^ + }$ and 0 < $k$ < $n$ .

[4]

Use an appropriate trigonometric identity to show that ${f_2}(x) = 2{x^2} - 1$ .

[2]

Use an appropriate trigonometric identity to show that ${f_{n + 1}}(x) = {\text{cos}}\left( {n\,{\text{arccos}}\,x} \right){\text{cos}}\left( {{\text{arccos}}\,x} \right) - {\text{sin}}\left( {n\,{\text{arccos}}\,x} \right){\text{sin}}\left( {{\text{arccos}}\,x} \right)$ .

[2]

Hence show that ${f_{n + 1}}(x) + {f_{n - 1}}(x) = 2x{f_n}\left( x \right)$ , $n \in {\mathbb{Z}^ + }$ .

[3]

h.i.

Hence express ${f_3}(x)$ as a cubic polynomial.

[2]

h.ii.

Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 + \frac{y}{x}$ , where $x \ne 0$ .

Consider the family of curves which satisfy the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 + \frac{y}{x}$ , where $x \ne 0$ .

Given that $y\left( 1 \right) = 1$ , use Euler’s method with step length $h$ = 0.25 to find an approximation for $y\left( 2 \right)$ . Give your answer to two significant figures.

[4]

Solve the equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 + \frac{y}{x}$ for $y\left( 1 \right) = 1$ .

[6]

Find the percentage error when $y\left( 2 \right)$ is approximated by the final rounded value found in part (a). Give your answer to two significant figures.

[3]

Find the equation of the isocline corresponding to $\frac{{{\text{d}}y}}{{{\text{d}}x}} = k$ , where $k \ne 0$ , $k \in \mathbb{R}$ .

[1]

d.i.

Show that such an isocline can never be a normal to any of the family of curves that satisfy the differential equation.

[4]

d.ii.

This question will investigate power series, as an extension to the Binomial Theorem for negative and fractional indices.

A power series in $x$ is defined as a function of the form $f\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...$ where the ${a_i} \in \mathbb{R}$ .

It can be considered as an infinite polynomial.

This is an example of a power series, but is only a finite power series, since only a finite number of the ${a_i}$ are non-zero.

We will now attempt to generalise further.

Suppose ${\left( {1 + x} \right)^q}{\text{,}}\,\,q \in \mathbb{Q}$ can be written as the power series ${a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...$ .

Expand ${\left( {1 + x} \right)^5}$ using the Binomial Theorem.

[2]

Consider the power series $1 - x + {x^2} - {x^3} + {x^4} - ...$

By considering the ratio of consecutive terms, explain why this series is equal to ${\left( {1 + x} \right)^{ - 1}}$ and state the values of $x$ for which this equality is true.

[4]

Differentiate the equation obtained part (b) and hence, find the first four terms in a power series for ${\left( {1 + x} \right)^{ - 2}}$ .

[2]

Repeat this process to find the first four terms in a power series for ${\left( {1 + x} \right)^{ - 3}}$ .

[2]

Hence, by recognising the pattern, deduce the first four terms in a power series for ${\left( {1 + x} \right)^{ - n}}$ , $n \in {\mathbb{Z}^ + }$ .

[3]

By substituting $x = 0$ , find the value of ${a_0}$ .

[1]

By differentiating both sides of the expression and then substituting $x = 0$ , find the value of ${a_1}$ .

[2]

Repeat this procedure to find ${a_2}$ and ${a_3}$ .

[4]

Hence, write down the first four terms in what is called the Extended Binomial Theorem for ${\left( {1 + x} \right)^q}{\text{,}}\,\,q \in \mathbb{Q}$ .

[1]

Write down the power series for $\frac{1}{{1 + {x^2}}}$ .

[2]

Hence, using integration, find the power series for ${\text{arctan}}\,x$ , giving the first four non-zero terms.

[4]

The function $f$ is defined by $f(x){\text{ }}={\text{ }}{(\arcsin{\text{ }}x)^2},{\text{ }} - 1 \leqslant x \leqslant 1$ .

The function $f$ satisfies the equation $\left( {1 - {x^2}} \right)f''\left( x \right) - xf'\left( x \right) - 2 = 0$ .

Show that $f'\left( 0 \right) = 0$ .

[2]

By differentiating the above equation twice, show that

$\left( {1 - {x^2}} \right){f^{\left( 4 \right)}}\left( x \right) - 5x{f^{\left( 3 \right)}}\left( x \right) - 4f''\left( x \right) = 0$

where ${f^{\left( 3 \right)}}\left( x \right)$ and ${f^{\left( 4 \right)}}\left( x \right)$ denote the 3rd and 4th derivative of $f\left( x \right)$ respectively.

[4]

Hence show that the Maclaurin series for $f\left( x \right)$ up to and including the term in ${x^4}$ is ${x^2} + \frac{1}{3}{x^4}$ .

[3]

Use this series approximation for $f\left( x \right)$ with $x = \frac{1}{2}$ to find an approximate value for ${\pi ^2}$ .

[2]

Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{4{x^2} + {y^2} - xy}}{{{x^2}}}$ , with $y = 2$ when $x = 1$ .

Use Euler’s method, with step length $h = 0.1$ , to find an approximate value of $y$ when $x = 1.4$ .

[5]

Sketch the isoclines for $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4$ .

[3]

Express ${m^2} - 2m + 4$ in the form ${\left( {m - a} \right)^2} + b$ , where $a{\text{, }}b \in \mathbb{Z}$ .

[1]

c.i.

Solve the differential equation, for $x > 0$ , giving your answer in the form $y = f\left( x \right)$ .

[10]

c.ii.

Sketch the graph of $y = f\left( x \right)$ for $1 \leqslant x \leqslant 1.4$ .

[1]

c.iii.

With reference to the curvature of your sketch in part (c)(iii), and without further calculation, explain whether you conjecture $f\left( {1.4} \right)$ will be less than, equal to, or greater than your answer in part (a).

[2]

c.iv.

Consider the differential equation $2xy\frac{{{\text{d}}y}}{{{\text{d}}x}} = {y^2} - {x^2}$ , where $x > 0$ .

Solve the differential equation and show that a general solution is ${x^2} + {y^2} = cx$ where $c$ is a positive constant.

[11]

Prove that there are two horizontal tangents to the general solution curve and state their equations, in terms of $c$ .

[5]

Find the value of $\int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x}$ .

[3]

Illustrate graphically the inequality $\sum\limits_{n = 5}^\infty {\frac{1}{{{n^3}}}} < \int\limits_4^\infty {\frac{1}{{{x^3}}}{\text{d}}x} < \sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}}$ .

[4]

Hence write down a lower bound for $\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}}$ .

[1]

Find an upper bound for $\sum\limits_{n = 4}^\infty {\frac{1}{{{n^3}}}}$ .

[3]

This question asks you to explore cubic polynomials of the form $(x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $x \in ℝ$ and corresponding cubic equations with one real root and two complex roots of the form $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ for $z \in ℂ$ .

In parts (a), (b) and (c), let $r = 1, a = 4$ and $b = 1$ .

Consider the equation $(z - 1) (z^{2} - 8 z + 17) = 0$ for $z \in ℂ$ .

Consider the function $f (x) = (x - 1) (x^{2} - 8 x + 17)$ for $x \in ℝ$ .

Consider the function $g (x) = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $x \in ℝ$ where $r, a \in ℝ$ and $b \in ℝ, b > 0$ .

The equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ for $z \in ℂ$ has roots $r$ and $a \pm b i$ where $r, a \in ℝ$ and $b \in ℝ, b > 0$ .

On the Cartesian plane, the points $C_{1} (a, \sqrt{g' (a)})$ and $C_{2} (a, - \sqrt{g' (a)})$ represent the real and imaginary parts of the complex roots of the equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ .

The following diagram shows a particular curve of the form $y = (x - r) (x^{2} - 2 a x + a^{2} + 16)$ and the tangent to the curve at the point $A (a, 80)$ . The curve and the tangent both intersect the $x$ -axis at the point $R (- 2, 0)$ . The points $C_{1}$ and $C_{2}$ are also shown.

Consider the curve $y = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $a \neq r, b > 0$ . The points $A (a, g (a))$ and $R (r, 0)$ are as defined in part (d)(ii). The curve has a point of inflexion at point $P$ .

Consider the special case where $a = r$ and $b > 0$ .

Given that $1$ and $4 + i$ are roots of the equation, write down the third root.

[1]

a.i.

Verify that the mean of the two complex roots is $4$ .

[1]

a.ii.

Show that the line $y = x - 1$ is tangent to the curve $y = f (x)$ at the point $A (4, 3)$ .

[4]

Sketch the curve $y = f (x)$ and the tangent to the curve at point $A$ , clearly showing where the tangent crosses the $x$ -axis.

[2]

Show that $g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$ .

[2]

d.i.

Hence, or otherwise, prove that the tangent to the curve $y = g (x)$ at the point $A (a, g (a))$ intersects the $x$ -axis at the point $R (r, 0)$ .

[6]

d.ii.

Deduce from part (d)(i) that the complex roots of the equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ can be expressed as $a \pm i \sqrt{g' (a)}$ .

[1]

Use this diagram to determine the roots of the corresponding equation of the form $(z - r) (z^{2} - 2 a z + a^{2} + 16) = 0$ for $z \in ℂ$ .

[4]

f.i.

State the coordinates of $C_{2}$ .

[1]

f.ii.

Show that the $x$ -coordinate of $P$ is $\frac{1}{3} (2 a + r)$ .

You are not required to demonstrate a change in concavity.

[2]

g.i.

Hence describe numerically the horizontal position of point $P$ relative to the horizontal positions of the points $R$ and $A$ .

[1]

g.ii.

Sketch the curve $y = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $a = r = 1$ and $b = 2$ .

[2]

h.i.

For $a = r$ and $b > 0$ , state in terms of $r$ , the coordinates of points $P$ and $A$ .

[1]

h.ii.

This question will investigate methods for finding definite integrals of powers of trigonometrical functions.

Let ${I_n} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{si}}{{\text{n}}^n}} x\,dx{\text{,}}\,\,n \in \mathbb{N}$ .

Let ${J_n} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{co}}{{\text{s}}^n}} x\,dx{\text{,}}\,\,n \in \mathbb{N}.$

Let ${T_n} = \int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^n}} x\,dx{\text{,}}\,\,n \in \mathbb{N}$ .

Find the exact values of ${I_0}$ , ${I_1}$ and ${I_2}$ .

[6]

Use integration by parts to show that ${I_n} = \frac{{n - 1}}{n}{I_{n - 2}}{\text{,}}\,\,n \geqslant 2$ .

[5]

b.i.

Explain where the condition $n \geqslant 2$ was used in your proof.

[1]

b.ii.

Hence, find the exact values of ${I_3}$ and ${I_4}$ .

[2]

Use the substitution $x = \frac{\pi }{2} - u$ to show that ${J_n} = {I_n}$ .

[4]

Hence, find the exact values of ${J_{5}}$ and ${J_{6}}$

[2]

Find the exact values of ${T_{0}}$ and ${T_{1}}$ .

[3]

Use the fact that ${\text{ta}}{{\text{n}}^2}x = {\text{se}}{{\text{c}}^2}x - 1$ to show that ${T_n} = \frac{1}{{n - 1}} - {T_{n - 2}}{\text{,}}\,\,n \geqslant 2$ .

[3]

g.i.

Explain where the condition $n \geqslant 2$ was used in your proof.

[1]

g.ii.

Hence, find the exact values of ${T_{2}}$ and ${T_{3}}$ .

[2]

This question asks you to investigate regular $n$ -sided polygons inscribed and circumscribed in a circle, and the perimeter of these as $n$ tends to infinity, to make an approximation for $\pi$ .

Let ${P_i}\left( n \right)$ represent the perimeter of any $n$ -sided regular polygon inscribed in a circle of radius 1 unit.

Consider an equilateral triangle ABC of side length, $x$ units, circumscribed about a circle of radius 1 unit and centre O as shown in the following diagram.

Let ${P_c}\left( n \right)$ represent the perimeter of any $n$ -sided regular polygon circumscribed about a circle of radius 1 unit.

Consider an equilateral triangle ABC of side length, $x$ units, inscribed in a circle of radius 1 unit and centre O as shown in the following diagram.

The equilateral triangle ABC can be divided into three smaller isosceles triangles, each subtending an angle of $\frac{{2\pi }}{3}$ at O, as shown in the following diagram.

Using right-angled trigonometry or otherwise, show that the perimeter of the equilateral triangle ABC is equal to $3\sqrt 3$ units.

[3]

Consider a square of side length, $x$ units, inscribed in a circle of radius 1 unit. By dividing the inscribed square into four isosceles triangles, find the exact perimeter of the inscribed square.

[3]

Find the perimeter of a regular hexagon, of side length, $x$ units, inscribed in a circle of radius 1 unit.

[2]

Show that ${P_i}\left( n \right) = 2n\,{\text{sin}}\left( {\frac{\pi }{n}} \right)$ .

[3]

Use an appropriate Maclaurin series expansion to find $\mathop {{\text{lim}}}\limits_{n \to \infty } {P_i}\left( n \right)$ and interpret this result geometrically.

[5]

Show that ${P_c}\left( n \right) = 2n\,{\text{tan}}\left( {\frac{\pi }{n}} \right)$ .

[4]

By writing ${P_c}\left( n \right)$ in the form $\frac{{2\,{\text{tan}}\left( {\frac{\pi }{n}} \right)}}{{\frac{1}{n}}}$ , find $\mathop {{\text{lim}}}\limits_{n \to \infty } {P_c}\left( n \right)$ .

[5]

Use the results from part (d) and part (f) to determine an inequality for the value of $\pi$ in terms of $n$ .

[2]

The inequality found in part (h) can be used to determine lower and upper bound approximations for the value of $\pi$ .

Determine the least value for $n$ such that the lower bound and upper bound approximations are both within 0.005 of $\pi$ .

[3]

Consider the differential equation $x\frac{{{\text{d}}y}}{{{\text{d}}x}} - y = {x^p} + 1$ where $x \in \mathbb{R},\,x \ne 0$ and $p$ is a positive integer, $p > 1$ .

Solve the differential equation given that $y = - 1$ when $x = 1$ . Give your answer in the form $y = f\left( x \right)$ .

[8]

Show that the $x$ -coordinate(s) of the points on the curve $y = f\left( x \right)$ where $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ satisfy the equation ${x^{p - 1}} = \frac{1}{p}$ .

[2]

b.i.

Deduce the set of values for $p$ such that there are two points on the curve $y = f\left( x \right)$ where $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ . Give a reason for your answer.

[2]

b.ii.

This question asks you to explore properties of a family of curves of the type $y^{2} = x^{3} + a x + b$ for various values of $a$ and $b$ , where $a, b \in ℕ$ .

On the same set of axes, sketch the following curves for $- 2 \leq x \leq 2$ and $- 2 \leq y \leq 2$ , clearly indicating any points of intersection with the coordinate axes.

Now, consider curves of the form $y^{2} = x^{3} + b$ , for $x \geq - \sqrt[3]{b}$ , where $b \in ℤ^{+}$ .

Next, consider the curve $y^{2} = x^{3} + x, x \geq 0$ .

The curve $y^{2} = x^{3} + x$ has two points of inflexion. Due to the symmetry of the curve these points have the same $x$ -coordinate.

$P (x, y)$ is defined to be a rational point on a curve if $x$ and $y$ are rational numbers.

The tangent to the curve $y^{2} = x^{3} + a x + b$ at a rational point $P$ intersects the curve at another rational point $Q$ .

Let $C$ be the curve $y^{2} = x^{3} + 2$ , for $x \geq - \sqrt[3]{2}$ . The rational point $P (- 1, - 1)$ lies on $C$ .

$y^{2} = x^{3}, x \geq 0$

[2]

a.i.

$y^{2} = x^{3} + 1, x \geq - 1$

[2]

a.ii.

Write down the coordinates of the two points of inflexion on the curve $y^{2} = x^{3} + 1$ .

[1]

b.i.

By considering each curve from part (a), identify two key features that would distinguish one curve from the other.

[1]

b.ii.

By varying the value of $b$ , suggest two key features common to these curves.

[2]

Show that $\frac{d y}{d x} = \pm \frac{3 x^{2} + 1}{2 \sqrt{x^{3} + x}}$ , for $x > 0$ .

[3]

d.i.

Hence deduce that the curve $y^{2} = x^{3} + x$ has no local minimum or maximum points.

[1]

d.ii.

Find the value of this $x$ -coordinate, giving your answer in the form $x = \sqrt{\frac{p \sqrt{3} + q}{r}}$ , where $p, q, r \in ℤ$ .

[7]

Find the equation of the tangent to $C$ at $P$ .

[2]

f.i.

Hence, find the coordinates of the rational point $Q$ where this tangent intersects $C$ , expressing each coordinate as a fraction.

[2]

f.ii.

The point $S (- 1, 1)$ also lies on $C$ . The line $[QS]$ intersects $C$ at a further point. Determine the coordinates of this point.

[5]

The function $f$ is defined by $f (x) = \ln (1 + x^{2})$ where $- 1 < x < 1$ .

The seventh derivative of $f$ is given by $f^{(7)} (x) = \frac{1440 x (x^{6} - 21 x^{4} + 35 x^{2} - 7)}{{(1 + x^{2})}^{7}}$ .

Use the Maclaurin series for $\ln (1 + x)$ to write down the first three non-zero terms of the Maclaurin series for $f (x)$ .

[2]

a.i.

Hence find the first three non-zero terms of the Maclaurin series for $\frac{x}{1 + x^{2}}$ .

[4]

a.ii.

Use your answer to part (a)(i) to write down an estimate for $f (0.4)$ .

[1]

Use the Lagrange form of the error term to find an upper bound for the absolute value of the error in calculating $f (0.4)$ , using the first three non-zero terms of the Maclaurin series for $f (x)$ .

[6]

c.i.

With reference to the Lagrange form of the error term, explain whether your answer to part (b) is an overestimate or an underestimate for $f (0.4)$ .

[2]

c.ii.

The curve $y = f (x)$ has a gradient function given by

$\frac{d y}{d x} = x - y$ .

The curve passes through the point $(1, 1)$ .

On the same set of axes, sketch and label isoclines for $\frac{d y}{d x} = - 1, 0$ and $1$ , and clearly indicate the value of each $y$ -intercept.

[3]

a.i.

Hence or otherwise, explain why the point $(1, 1)$ is a local minimum.

[3]

a.ii.

Find the solution of the differential equation $\frac{d y}{d x} = x - y$ , which passes through the point $(1, 1)$ . Give your answer in the form $y = f (x)$ .

[8]

Explain why the graph of $y = f (x)$ does not intersect the isocline $\frac{d y}{d x} = 1$ .

[2]

c.i.

Sketch the graph of $y = f (x)$ on the same set of axes as part (a)(i).

[2]

c.ii.

Consider the differential equation

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = f\left( {\frac{y}{x}} \right),{\text{ }}x > 0.$

Use the substitution $y = vx$ to show that the general solution of this differential equation is

$\int {\frac{{{\text{d}}v}}{{f(v) - v}} = \ln x + } {\text{ Constant.}}$

[3]

Hence, or otherwise, solve the differential equation

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{x^2} + 3xy + {y^2}}}{{{x^2}}},{\text{ }}x > 0,$

given that $y = 1$ when $x = 1$ . Give your answer in the form $y = g(x)$ .

[10]

Use L’Hôpital’s rule to determine the value of

$\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{{\text{e}}^{ - 3x}}^{^2} + 3\,{\text{cos}}\left( {2x} \right) - 4}}{{3{x^2}}}} \right)$

[5]

Hence find $\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{\int_0^x {\left( {{{\text{e}}^{ - 3t}}^{^2} + 3\,{\text{cos}}\left( {2t} \right) - 4} \right)} \,{\text{d}}t}}{{\int_0^x {3{t^2}} \,{\text{d}}t}}} \right)$ .

[3]

Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{x}{{{x^2} + 1}}y = x$ where $y = 1$ when $x = 0$ .

Show that $\sqrt {{x^2} + 1}$ is an integrating factor for this differential equation.

[4]

Solve the differential equation giving your answer in the form $y = f(x)$ .

[6]

This question asks you to explore the behaviour and key features of cubic polynomials of the form $x^{3} - 3 c x + d$ .

Consider the function $f (x) = x^{3} - 3 c x + 2$ for $x \in ℝ$ and where $c$ is a parameter, $c \in ℝ$ .

The graphs of $y = f (x)$ for $c = - 1$ and $c = 0$ are shown in the following diagrams.

$c = - 1$ $c = 0$

On separate axes, sketch the graph of $y = f (x)$ showing the value of the $y$ -intercept and the coordinates of any points with zero gradient, for

Hence, or otherwise, find the set of values of $c$ such that the graph of $y = f (x)$ has

Given that the graph of $y = f (x)$ has one local maximum point and one local minimum point, show that

Hence, for $c > 0$ , find the set of values of $c$ such that the graph of $y = f (x)$ has

$c = 1$ .

[3]

a.i.

$c = 2$ .

[3]

a.ii.

Write down an expression for $f' (x)$ .

[1]

a point of inflexion with zero gradient.

[1]

c.i.

one local maximum point and one local minimum point.

[2]

c.ii.

no points where the gradient is equal to zero.

[1]

c.iii.

the $y$ -coordinate of the local maximum point is $2 c^{\frac{3}{2}} + 2$ .

[3]

d.i.

the $y$ -coordinate of the local minimum point is $- 2 c^{\frac{3}{2}} + 2$ .

[1]

d.ii.

exactly one $x$ -axis intercept.

[2]

e.i.

exactly two $x$ -axis intercepts.

[2]

e.ii.

exactly three $x$ -axis intercepts.

[2]

e.iii.

Consider the function $g (x) = x^{3} - 3 c x + d$ for $x \in ℝ$ and where $c, d \in ℝ$ .

Find all conditions on $c$ and $d$ such that the graph of $y = g (x)$ has exactly one $x$ -axis intercept, explaining your reasoning.

[6]

This question asks you to examine various polygons for which the numerical value of the area is the same as the numerical value of the perimeter. For example, a $3$ by $6$ rectangle has an area of $18$ and a perimeter of $18$ .

For each polygon in this question, let the numerical value of its area be $A$ and let the numerical value of its perimeter be $P$ .

An $n$ -sided regular polygon can be divided into $n$ congruent isosceles triangles. Let $x$ be the length of each of the two equal sides of one such isosceles triangle and let $y$ be the length of the third side. The included angle between the two equal sides has magnitude $\frac{2 π}{n}$ .

Part of such an $n$ -sided regular polygon is shown in the following diagram.

Consider a $n$ -sided regular polygon such that $A = P$ .

The Maclaurin series for $\tan x$ is $x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + \dots$

Consider a right-angled triangle with side lengths $a, b$ and $\sqrt{a^{2} + b^{2}}$ , where $a \geq b$ , such that $A = P$ .

Find the side length, $s$ , where $s > 0$ , of a square such that $A = P$ .

[3]

Write down, in terms of $x$ and $n$ , an expression for the area, $A_{T}$ , of one of these isosceles triangles.

[1]

Show that $y = 2 x \sin \frac{π}{n}$ .

[2]

Use the results from parts (b) and (c) to show that $A = P = 4 n \tan \frac{π}{n}$ .

[7]

Use the Maclaurin series for $\tan x$ to find $\lim_{n \to \infty} (4 n \tan \frac{π}{n})$ .

[3]

e.i.

Interpret your answer to part (e)(i) geometrically.

[1]

e.ii.

Show that $a = \frac{8}{b - 4} + 4$ .

[7]

By using the result of part (f) or otherwise, determine the three side lengths of the only two right-angled triangles for which $a, b, A, P \in ℤ$ .

[3]

g.i.

Determine the area and perimeter of these two right-angled triangles.

[1]

g.ii.

This question asks you to explore the behaviour and some key features of the function $f_{n} (x) = x^{n} (a - x)^{n}$ , where $a \in ℝ^{+}$ and $n \in ℤ^{+}$ .

In parts (a) and (b), only consider the case where $a = 2$ .

Consider $f_{1} (x) = x (2 - x)$ .

Consider $f_{n} (x) = x^{n} {(2 - x)}^{n}$ , where $n \in ℤ^{+}, n > 1$ .

Now consider $f_{n} (x) = x^{n} {(a - x)}^{n}$ where $a \in ℝ^{+}$ and $n \in ℤ^{+}, n > 1$ .

By using the result from part (f) and considering the sign of ${f_{n}}^{'} (- 1)$ , show that the point $(0, 0)$ on the graph of $y = f_{n} (x)$ is

Sketch the graph of $y = f_{1} (x)$ , stating the values of any axes intercepts and the coordinates of any local maximum or minimum points.

[3]

Use your graphic display calculator to explore the graph of $y = f_{n} (x)$ for

• the odd values $n = 3$ and $n = 5$ ;

• the even values $n = 2$ and $n = 4$ .

Hence, copy and complete the following table.

[6]

Show that ${f_{n}}^{'} (x) = n x^{n - 1} (a - 2 x) {(a - x)}^{n - 1}$ .

[5]

State the three solutions to the equation ${f_{n}}^{'} (x) = 0$ .

[2]

Show that the point $(\frac{a}{2}, f_{n} (\frac{a}{2}))$ on the graph of $y = f_{n} (x)$ is always above the horizontal axis.

[3]

Hence, or otherwise, show that ${f_{n}}^{'} (\frac{a}{4}) > 0$ , for $n \in ℤ^{+}$ .

[2]

a local minimum point for even values of $n$ , where $n > 1$ and $a \in ℝ^{+}$ .

[3]

g.i.

a point of inflexion with zero gradient for odd values of $n$ , where $n > 1$ and $a \in ℝ^{+}$ .

[2]

g.ii.

Consider the graph of $y = x^{n} {(a - x)}^{n} - k$ , where $n \in ℤ^{+}$ , $a \in ℝ^{+}$ and $k \in ℝ$ .

State the conditions on $n$ and $k$ such that the equation $x^{n} {(a - x)}^{n} = k$ has four solutions for $x$ .

[5]

In this question you will be exploring the strategies required to solve a system of linear differential equations.

Consider the system of linear differential equations of the form:

$\frac{d x}{d t} = x - y$ and $\frac{d y}{d t} = a x + y$ ,

where $x, y, t \in ℝ^{+}$ and $a$ is a parameter.

First consider the case where $a = 0$ .

Now consider the case where $a = - 1$ .

Now consider the case where $a = - 4$ .

From previous cases, we might conjecture that a solution to this differential equation is $y = F e^{λ t}$ , $λ \in ℝ$ and $F$ is a constant.

By solving the differential equation $\frac{d y}{d t} = y$ , show that $y = A e^{t}$ where $A$ is a constant.

[3]

a.i.

Show that $\frac{d x}{d t} - x = - A e^{t}$ .

[1]

a.ii.

Solve the differential equation in part (a)(ii) to find $x$ as a function of $t$ .

[4]

a.iii.

By differentiating $\frac{d y}{d t} = - x + y$ with respect to $t$ , show that $\frac{d^{2} y}{d t^{2}} = 2 \frac{d y}{d t}$ .

[3]

b.i.

By substituting $Y = \frac{d y}{d t}$ , show that $Y = B e^{2 t}$ where $B$ is a constant.

[3]

b.ii.

Hence find $y$ as a function of $t$ .

[2]

b.iii.

Hence show that $x = - \frac{B}{2} e^{2 t} + C$ , where $C$ is a constant.

[3]

b.iv.

Show that $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .

[3]

c.i.

Find the two values for $λ$ that satisfy $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .

[4]

c.ii.

Let the two values found in part (c)(ii) be $λ_{1}$ and $λ_{2}$ .

Verify that $y = F e^{λ_{1} t} + G e^{λ_{2} t}$ is a solution to the differential equation in (c)(i),where $G$ is a constant.

[4]

c.iii.

The function $f$ is defined by $f\left( x \right) = {\text{arcsin}}\left( {2x} \right)$ , where $- \frac{1}{2} \leqslant x \leqslant \frac{1}{2}$ .

By finding a suitable number of derivatives of $f$ , find the first two non-zero terms in the Maclaurin series for $f$ .

[8]

Hence or otherwise, find $\mathop {{\text{lim}}}\limits_{x \to 0} \frac{{{\text{arcsin}}\left( {2x} \right) - 2x}}{{{{\left( {2x} \right)}^3}}}$ .

[3]

Using L’Hôpital’s rule, find $\mathop {{\text{lim}}}\limits_{x \to 0} \left( {\frac{{{\text{tan}}\,3x - 3\,{\text{tan}}\,x}}{{{\text{sin}}\,3x - 3\,{\text{sin}}\,x}}} \right)$ .

A simple model to predict the population of the world is set up as follows. At time $t$ years the population of the world is $x$ , which can be assumed to be a continuous variable. The rate of increase of $x$ due to births is 0.056 $x$ and the rate of decrease of $x$ due to deaths is 0.035 $x$ .

Show that $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.021x$ .

[1]

Find a prediction for the number of years it will take for the population of the world to double.

[6]

Use l’Hôpital’s rule to determine the value of

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{x\ln (1 + x)}}.$

Use l’Hôpital’s rule to find

$\lim_{x \to 1} \frac{\cos (x^{2} - 1) - 1}{e^{x - 1} - x}$ .

Consider the differential equation $\frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {\frac{{2x}}{{1 + {x^2}}}} \right)y = {x^2}$ , given that $y = 2$ when $x = 0$ .

Show that $1 + {x^2}$ is an integrating factor for this differential equation.

[5]

Hence solve this differential equation. Give the answer in the form $y = f(x)$ .

[6]