# IB Math Analysis & Approaches Question bank-Topic: SL 5.12 Derivative of f(x) SL Paper 1

### Question

Let y = $$\frac{Inx}{x^{4}}$$ for x > 0.

(a)        Show that $$\frac{dy}{dx}= \frac{1-4Inx}{x^{5}}$$

Consider the function defined  by f (x) $$\frac{Inx}{x^{4}}$$ =  for x> 0 and its graph y = f (x) .

(b)        The graph of has a horizontal tangent at point P. Find the coordinates of P.                                                      [5]

(c)        Given that f ” (x) = $$\frac{20Lnx-9}{x^{6}}$$ show that P is a local maximum point.                                                [3]

(d)        Solve f (x) > 0 for x > 0.                                                                                                                                                          [2]

(e) Sketch the graph of f , showing clearly the value of the x-intercept and the approximate position of point P.    [3]

Ans

## Question

Consider $$f(x) = \frac{1}{3}{x^3} + 2{x^2} – 5x$$ . Part of the graph of f is shown below. There is a maximum point at M, and a point of inflexion at N.

Find $$f'(x)$$ .

[3]
a.

Find the x-coordinate of M.

[4]
b.

Find the x-coordinate of N.

[3]
c.

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

[4]
d.

## Markscheme

$$f'(x) = {x^2} + 4x – 5$$     A1A1A1     N3

[3 marks]

a.

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]

b.

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]

c.

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]

d.

## Question

A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.

The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is $$\theta$$ radians, where $$0 \le \theta \le \frac{\pi }{2}$$ .

Write down an expression in terms of $$\theta$$ for

(i)     $$x$$ ;

(ii)    $$y$$ .

[2]
a.

Let the area of the rectangle be A.

Show that $$A = 18\sin 2\theta$$ .

[3]
b.

(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.

[8]
c.

## Markscheme

(i) $$x = 3\cos \theta$$     A1     N1

(ii) $$y = 3\sin \theta$$     A1     N1

[2 marks]

a.

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]

b.

(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]

c.

## Question

Let $$f(x) = {{\rm{e}}^{ – 3x}}$$ and $$g(x) = \sin \left( {x – \frac{\pi }{3}} \right)$$ .

Write down

(i)     $$f'(x)$$ ;

(ii)    $$g'(x)$$ .

[2]
a.

Let $$h(x) = {{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right)$$ . Find the exact value of $$h’\left( {\frac{\pi }{3}} \right)$$ .

[4]
b.

## Markscheme

(i) $$– 3{{\rm{e}}^{ – 3x}}$$     A1     N1

(ii) $$\cos \left( {x – \frac{\pi }{3}} \right)$$     A1     N1

[4 marks]

a.

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]

b.

## Question

Consider $$f(x) = {x^2} + \frac{p}{x}$$ , $$x \ne 0$$ , where p is a constant.

Find $$f'(x)$$ .

[2]
a.

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

[4]
b.

## Markscheme

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

Note: Award A1 for $$2x$$ , A1 for $$– \frac{p}{{{x^2}}}$$ .

[2 marks]

a.

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]

b.

## Question

Let $$f(x) = 3 + \frac{{20}}{{{x^2} – 4}}$$ , for $$x \ne \pm 2$$ . The graph of f is given below.

The y-intercept is at the point A.

(i)     Find the coordinates of A.

(ii)    Show that $$f'(x) = 0$$ at A.

[7]
a.

The second derivative $$f”(x) = \frac{{40(3{x^2} + 4)}}{{{{({x^2} – 4)}^3}}}$$ . Use this to

(i)     justify that the graph of f has a local maximum at A;

(ii)    explain why the graph of f does not have a point of inflexion.

[6]
b.

Describe the behaviour of the graph of $$f$$ for large $$|x|$$ .

[1]
c.

Write down the range of $$f$$ .

[2]
d.

## Markscheme

(i) coordinates of A are $$(0{\text{, }} – 2)$$     A1A1     N2

(ii) derivative of $${x^2} – 4 = 2x$$ (seen anywhere)     (A1)

evidence of correct approach     (M1)

e.g. quotient rule, chain rule

finding $$f'(x)$$     A2

e.g. $$f'(x) = 20 \times ( – 1) \times {({x^2} – 4)^{ – 2}} \times (2x)$$ , $$\frac{{({x^2} – 4)(0) – (20)(2x)}}{{{{({x^2} – 4)}^2}}}$$

substituting $$x = 0$$ into $$f'(x)$$ (do not accept solving $$f'(x) = 0$$ )     M1

at A $$f'(x) = 0$$     AG     N0

[7 marks]

a.

(i) reference to $$f'(x) = 0$$ (seen anywhere)     (R1)

reference to $$f”(0)$$ is negative (seen anywhere)     R1

evidence of substituting $$x = 0$$ into $$f”(x)$$     M1

finding $$f”(0) = \frac{{40 \times 4}}{{{{( – 4)}^3}}}$$ $$\left( { = – \frac{5}{2}} \right)$$     A1

then the graph must have a local maximum     AG

(ii) reference to $$f”(x) = 0$$ at point of inflexion     (R1)

recognizing that the second derivative is never 0     A1     N2

e.g. $$40(3{x^2} + 4) \ne 0$$ , $$3{x^2} + 4 \ne 0$$ , $${x^2} \ne – \frac{4}{3}$$ , the numerator is always positive

Note: Do not accept the use of the first derivative in part (b).

[6 marks]

b.

correct (informal) statement, including reference to approaching $$y = 3$$     A1     N1

e.g. getting closer to the line $$y = 3$$ , horizontal asymptote at $$y = 3$$

[1 mark]

c.

correct inequalities, $$y \le – 2$$ , $$y > 3$$ , FT from (a)(i) and (c)     A1A1     N2

[2 marks]

d.

## Question

Let $$f(x) = \frac{1}{2}{x^3} – {x^2} – 3x$$ . Part of the graph of f is shown below.

There is a maximum point at A and a minimum point at B(3, − 9) .

Find the coordinates of A.

[8]
a.

Write down the coordinates of

(i)     the image of B after reflection in the y-axis;

(ii)    the image of B after translation by the vector $$\left( {\begin{array}{*{20}{c}} { – 2}\\ 5 \end{array}} \right)$$ ;

(iii)   the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor $$\frac{1}{2}$$ .

[6]
b(i), (ii) and (iii).

## Markscheme

$$f(x) = {x^2} – 2x – 3$$     A1A1A1

evidence of solving $$f'(x) = 0$$     (M1)

e.g. $${x^2} – 2x – 3 = 0$$

evidence of correct working     A1

e.g. $$(x + 1)(x – 3)$$ ,  $$\frac{{2 \pm \sqrt {16} }}{2}$$

$$x = – 1$$ (ignore $$x = 3$$ )     (A1)

evidence of substituting their negative x-value into $$f(x)$$     (M1)

e.g. $$\frac{1}{3}{( – 1)^3} – {( – 1)^2} – 3( – 1)$$ , $$– \frac{1}{3} – 1 + 3$$

$$y = \frac{5}{3}$$     A1

coordinates are $$\left( { – 1,\frac{5}{3}} \right)$$     N3

[8 marks]

a.

(i) $$( – 3{\text{, }} – 9)$$     A1     N1

(ii) $$(1{\text{, }} – 4)$$     A1A1    N2

(iii) reflection gives $$(3{\text{, }}9)$$     (A1)

stretch gives $$\left( {\frac{3}{2}{\text{, }}9} \right)$$     A1A1     N3

[6 marks]

b(i), (ii) and (iii).

## Question

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.

## Markscheme

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]

## Question

Let $$f(x) = {x^3}$$. The following diagram shows part of the graph of f .

The point $${\rm{P}}(a,f(a))$$ , where $$a > 0$$ , lies on the graph of f . The tangent at P crosses the x-axis at the point $${\rm{Q}}\left( {\frac{2}{3},0} \right)$$ . This tangent intersects the graph of f at the point R(−2, −8) .

The equation of the tangent at P is $$y = 3x – 2$$ . Let T be the region enclosed by the graph of f , the tangent [PR] and the line $$x = k$$ , between $$x = – 2$$ and $$x = k$$ where $$– 2 < k < 1$$ . This is shown in the diagram below.

(i)     Show that the gradient of [PQ] is $$\frac{{{a^3}}}{{a – \frac{2}{3}}}$$ .

(ii)    Find $$f'(a)$$ .

(iii)   Hence show that $$a = 1$$ .

[7]
a(i), (ii) and (iii).

Given that the area of T is $$2k + 4$$ , show that k satisfies the equation $${k^4} – 6{k^2} + 8 = 0$$ .

[9]
b.

## Markscheme

(i) substitute into gradient $$= \frac{{{y_1} – {y_2}}}{{{x_1} – {x_2}}}$$     (M1)

e.g. $$\frac{{f(a) – 0}}{{a – \frac{2}{3}}}$$

substituting $$f(a) = {a^3}$$

e.g. $$\frac{{{a^3} – 0}}{{a – \frac{2}{3}}}$$     A1

gradient $$\frac{{{a^3}}}{{a – \frac{2}{3}}}$$     AG     N0

e.g. $$3{a^2}$$ , $$f'(a) = 3$$ , $$f'(a) = \frac{{{a^3}}}{{a – \frac{2}{3}}}$$

(iii) METHOD 1

evidence of approach     (M1)

e.g. $$f'(a) = {\rm{gradient}}$$ , $$3{a^2} = \frac{{{a^3}}}{{a – \frac{2}{3}}}$$

simplify     A1

e.g. $$3{a^2}\left( {a – \frac{2}{3}} \right) = {a^3}$$

rearrange     A1

e.g. $$3{a^3} – 2{a^2} = {a^3}$$

evidence of solving     A1

e.g. $$2{a^3} – 2{a^2} = 2{a^2}(a – 1) = 0$$

$$a = 1$$     AG     N0

METHOD 2

gradient RQ $$= \frac{{ – 8}}{{ – 2 – \frac{2}{3}}}$$     A1

simplify     A1

e.g. $$\frac{{ – 8}}{{ – \frac{8}{3}}},3$$

evidence of approach     (M1)

e.g. $$f'(a) = {\rm{gradient}}$$ , $$3{a^2} = \frac{{ – 8}}{{ – 2 – \frac{2}{3}}}$$ , $$\frac{{{a^3}}}{{a – \frac{2}{3}}} = 3$$

simplify     A1

e.g. $$3{a^2} = 3$$ , $${a^2} = 1$$

$$a = 1$$     AG     N0

[7 marks]

a(i), (ii) and (iii).

approach to find area of T involving subtraction and integrals    (M1)

e.g. $$\int {f – (3x – 2){\rm{d}}x}$$ , $$\int_{ – 2}^k {(3x – 2) – \int_{ – 2}^k {{x^3}} }$$ , $$\int {({x^3} – 3x + 2)}$$

correct integration with correct signs     A1A1A1

e.g. $$\frac{1}{4}{x^4} – \frac{3}{2}{x^2} + 2x$$ , $$\frac{3}{2}{x^2} – 2x – \frac{1}{4}{x^4}$$

correct limits $$– 2$$ and k (seen anywhere)     A1

e.g. $$\int_{ – 2}^k {({x^3} – 3x + 2){\rm{d}}x}$$ , $$\left[ {\frac{1}{4}{x^4} – \frac{3}{2}{x^2} + 2x} \right]_{ – 2}^k$$

attempt to substitute k and $$– 2$$     (M1)

correct substitution into their integral if 2 or more terms     A1

e.g. $$\left( {\frac{1}{4}{k^4} – \frac{3}{2}{k^2} + 2k} \right) – (4 – 6 – 4)$$

setting their integral expression equal to $$2k + 4$$ (seen anywhere)     (M1)

simplifying     A1

e.g. $$\frac{1}{4}{k^4} – \frac{3}{2}{k^2} + 2 = 0$$

$${k^4} – 6{k^2} + 8 = 0$$     AG     N0

[9 marks]

b.

## Question

The following diagram shows part of the graph of a quadratic function f .

The x-intercepts are at $$( – 4{\text{, }}0)$$ and $$(6{\text{, }}0)$$ , and the y-intercept is at $$(0{\text{, }}240)$$ .

Write down $$f(x)$$ in the form $$f(x) = – 10(x – p)(x – q)$$ .

[2]
a.

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

[4]
b.

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

[2]
c.

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.

[7]
d(i) and (ii).

## Markscheme

$$f(x) = – 10(x + 4)(x – 6)$$     A1A1     N2

[2 marks]

a.

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]

b.

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]

c.

(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]

d(i) and (ii).

## Question

The following diagram shows the graph of $$f(x) = a\sin (b(x – c)) + d$$ , for $$2 \le x \le 10$$ .

There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .

Use the graph to write down the value of

(i)     a ;

(ii)    c ;

(iii)   d .

[3]
a(i), (ii) and (iii).

Show that $$b = \frac{\pi }{4}$$ .

[2]
b.

Find $$f'(x)$$ .

[3]
c.

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

[6]
d.

## Markscheme

(i) $$a = 8$$     A1     N1

(ii) $$c = 2$$     A1     N1

(iii) $$d = 4$$     A1     N1

[3 marks]

a(i), (ii) and (iii).

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]

b.

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]

c.

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]

d.

## Question

Let $$f(x) = {{\rm{e}}^{6x}}$$ .

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

[1]
a.

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 .

[4]
b(i) and (ii).

Hence, write down the equation of this tangent.

[1]
c.

## Markscheme

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

[1 mark]

a.

(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]

b(i) and (ii).

$$y = 6x + 1$$     A1     N1

[1 mark]

c.

## Question

Consider $$f(x) = {x^2}\sin x$$ .

Find $$f'(x)$$ .

[4]
a.

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

[3]
b.

## Markscheme

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]

a.

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]

b.

## Question

Let $$f(x) = \sin x + \frac{1}{2}{x^2} – 2x$$ , for $$0 \le x \le \pi$$ .

Let $$g$$ be a quadratic function such that $$g(0) = 5$$ . The line $$x = 2$$ is the axis of symmetry of the graph of $$g$$ .

The function $$g$$ can be expressed in the form $$g(x) = a{(x – h)^2} + 3$$ .

Find $$f'(x)$$ .

[3]
a.

Find $$g(4)$$ .

[3]
b.

(i)     Write down the value of $$h$$ .

(ii)     Find the value of $$a$$ .

[4]
c.

Find the value of $$x$$ for which the tangent to the graph of $$f$$ is parallel to the tangent to the graph of $$g$$ .

[6]
d.

## Markscheme

$$f'(x) = \cos x + x – 2$$     A1A1A1     N3

Note: Award A1 for each term.

[3 marks]

a.

recognizing $$g(0) = 5$$ gives the point ($$0$$, $$5$$)     (R1)

recognize symmetry     (M1)

eg vertex, sketch

$$g(4) = 5$$     A1     N3

[3 marks]

b.

(i)     $$h = 2$$     A1 N1

(ii)     substituting into $$g(x) = a{(x – 2)^2} + 3$$ (not the vertex)     (M1)

eg   $$5 = a{(0 – 2)^2} + 3$$ , $$5 = a{(4 – 2)^2} + 3$$

working towards solution     (A1)

eg   $$5 = 4a + 3$$ , $$4a = 2$$

$$a = \frac{1}{2}$$     A1     N2

[4 marks]

c.

$$g(x) = \frac{1}{2}{(x – 2)^2} + 3 = \frac{1}{2}{x^2} – 2x + 5$$

correct derivative of $$g$$     A1A1

eg   $$2 \times \frac{1}{2}(x – 2)$$ , $$x – 2$$

evidence of equating both derivatives     (M1)

eg   $$f’ = g’$$

correct equation     (A1)

eg   $$\cos x + x – 2 = x – 2$$

working towards a solution     (A1)

eg   $$\cos x = 0$$ , combining like terms

$$x = \frac{\pi }{2}$$    A1     N0

Note: Do not award final A1 if additional values are given.

[6 marks]

d.

## Question

Let $$f(x) = \frac{{{{(\ln x)}^2}}}{2}$$, for $$x > 0$$.

Let $$g(x) = \frac{1}{x}$$. The following diagram shows parts of the graphs of $$f’$$ and g.

The graph of $$f’$$ has an x-intercept at $$x = p$$.

Show that $$f'(x) = \frac{{\ln x}}{x}$$.

[2]
a.

There is a minimum on the graph of $$f$$. Find the $$x$$-coordinate of this minimum.

[3]
b.

Write down the value of $$p$$.

[2]
c.

The graph of $$g$$ intersects the graph of $$f’$$ when $$x = q$$.

Find the value of $$q$$.

[3]
d.

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 $$\frac{1}{2}$$.

[5]
e.

## Markscheme

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]

a.

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]

b.

intercept when $$f'(x) = 0$$     (M1)

$$p = 1$$     A1     N2

[2 marks]

c.

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]

d.

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]

e.

## Question

Let $$f(x) = p{x^3} + p{x^2} + qx$$.

Find $$f'(x)$$.

[2]
a.

Given that $$f'(x) \geqslant 0$$, show that $${p^2} \leqslant 3pq$$.

[5]
b.

## Markscheme

$$f'(x) = 3p{x^2} + 2px + q$$     A2     N2

Note:     Award A1 if only 1 error.

[2 marks]

a.

evidence of discriminant (must be seen explicitly, not in quadratic formula)     (M1)

eg     $${b^2} – 4ac$$

correct substitution into discriminant (may be seen in inequality)     A1

eg     $${(2p)^2} – 4 \times 3p \times q,{\text{ }}4{p^2} – 12pq$$

$$f'(x) \geqslant 0$$ then $$f’$$ has two equal roots or no roots     (R1)

recognizing discriminant less or equal than zero     R1

eg     $$\Delta \leqslant 0,{\text{ }}4{p^2} – 12pq \leqslant 0$$

eg     $${p^2} – 3pq \leqslant 0,{\text{ }}4{p^2} \leqslant 12pq$$

$${p^2} \leqslant 3pq$$     AG     N0

[5 marks]

b.

## Question

A function $$f$$ has its derivative given by $$f'(x) = 3{x^2} – 2kx – 9$$, where $$k$$ is a constant.

Find $$f”(x)$$.

[2]
a.

The graph of $$f$$ has a point of inflexion when $$x = 1$$.

Show that $$k = 3$$.

[3]
b.

Find $$f'( – 2)$$.

[2]
c.

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

[4]
d.

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

[3]
e.

## Markscheme

$$f”(x) = 6x – 2k$$     A1A1     N2

[2 marks]

a.

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]

b.

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

eg$$\;\;\;3{( – 2)^2} – 6( – 2) – 9$$

$$f'( – 2) = 15$$     A1     N2

[2 marks]

c.

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]

d.

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]

e.

## Question

Let $$f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}$$, for $$0 < x < 6$$.

The graph of $$f$$ has a maximum point at P.

The $$y$$-coordinate of P is $$\ln 27$$.

Find the $$x$$-coordinate of P.

[3]
a.

Find $$f(x)$$, expressing your answer as a single logarithm.

[8]
b.

The graph of $$f$$ is transformed by a vertical stretch with scale factor $$\frac{1}{{\ln 3}}$$. The image of P under this transformation has coordinates $$(a,{\text{ }}b)$$.

Find the value of $$a$$ and of $$b$$, where $$a,{\text{ }}b \in \mathbb{N}$$.

[[N/A]]
c.

## Markscheme

recognizing $$f'(x) = 0$$     (M1)

correct working     (A1)

eg$$\,\,\,\,\,$$$$6 – 2x = 0$$

$$x = 3$$    A1     N2

[3 marks]

a.

evidence of integration     (M1)

eg$$\,\,\,\,\,$$$$\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} }$$

using substitution     (A1)

eg$$\,\,\,\,\,$$$$\int {\frac{1}{u}{\text{d}}u}$$ where $$u = 6x – {x^2}$$

correct integral     A1

eg$$\,\,\,\,\,$$$$\ln (u) + c,{\text{ }}\ln (6x – {x^2})$$

substituting $$(3,{\text{ }}\ln 27)$$ into their integrated expression (must have $$c$$)     (M1)

eg$$\,\,\,\,\,$$$$\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$c = \ln 27 – \ln 9$$

EITHER

$$c = \ln 3$$    (A1)

attempt to substitute their value of $$c$$ into $$f(x)$$     (M1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln 3$$     A1     N4

OR

attempt to substitute their value of $$c$$ into $$f(x)$$     (M1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9$$

correct use of a log law     (A1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9$$

$$f(x) = \ln \left( {3(6x – {x^2})} \right)$$    A1     N4

[8 marks]

b.

$$a = 3$$    A1     N1

correct working     A1

eg$$\,\,\,\,\,$$$$\frac{{\ln 27}}{{\ln 3}}$$

correct use of log law     (A1)

eg$$\,\,\,\,\,$$$$\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27$$

$$b = 3$$    A1     N2

[4 marks]

c.

## Question

Fred makes an open metal container in the shape of a cuboid, as shown in the following diagram.

The container has height $$x{\text{ m}}$$, width $$x{\text{ m}}$$ and length $$y{\text{ m}}$$. The volume is $$36{\text{ }}{{\text{m}}^3}$$.

Let $$A(x)$$ be the outside surface area of the container.

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

[4]
a.

Find $$A'(x)$$.

[2]
b.

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

[5]
c.

[5 marks]

d.

## Question

Let $$f(x) = \cos x$$.

Let $$g(x) = {x^k}$$, where $$k \in {\mathbb{Z}^ + }$$.

Let $$k = 21$$ and $$h(x) = \left( {{f^{(19)}}(x) \times {g^{(19)}}(x)} \right)$$.

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

(ii)     Find $${f^{(19)}}(x)$$.

[4]
a.

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

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

[5]
b.

(i)     Find $$h'(x)$$.

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

[7]
c.

## Markscheme

(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]

a.

(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]

b.

(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]

c.

## Question

Let $$f(x) = {x^2}$$. The following diagram shows part of the graph of $$f$$.

The line $$L$$ is the tangent to the graph of $$f$$ at the point $${\text{A}}( – k,{\text{ }}{k^2})$$, and intersects the $$x$$-axis at point B. The point C is $$( – k,{\text{ }}0)$$.

The region $$R$$ is enclosed by $$L$$, the graph of $$f$$, and the $$x$$-axis. This is shown in the following diagram.

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

[1]
a.i.

Find the gradient of $$L$$.

[2]
a.ii.

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

[5]
b.

Find the area of triangle ABC, giving your answer in terms of $$k$$.

[2]
c.

Given that the area of triangle ABC is $$p$$ times the area of $$R$$, find the value of $$p$$.

[7]
d.

## Markscheme

$$f'(x) = 2x$$     A1     N1

[1 mark]

a.i.

attempt to substitute $$x = – k$$ into their derivative     (M1)

gradient of $$L$$ is $$– 2k$$     A1     N2

[2 marks]

a.ii.

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]

b.

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]

c.

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]

d.

## Question

Let $$f(x) = {x^2} – x$$, for $$x \in \mathbb{R}$$. The following diagram shows part of the graph of $$f$$.

The graph of $$f$$ crosses the $$x$$-axis at the origin and at the point $${\text{P}}(1,{\text{ }}0)$$.

The line L is the normal to the graph of f at P.

The line $$L$$ intersects the graph of $$f$$ at another point Q, as shown in the following diagram.

Show that $$f’(1) = 1$$.

[3]
a.

Find the equation of $$L$$ in the form $$y = ax + b$$.

[3]
b.

Find the $$x$$-coordinate of Q.

[4]
c.

Find the area of the region enclosed by the graph of $$f$$ and the line $$L$$.

[6]
d.

## Markscheme

$$f’(x) = 2x – 1$$     A1A1

correct substitution     A1

eg$$\,\,\,\,\,$$$$2(1) – 1,{\text{ }}2 – 1$$

$$f’(1) = 1$$     AG     N0

[3 marks]

a.

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]

b.

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]

c.

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]

d.

## Question

A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20$$\pi$$ cm3.

The material for the base and top of the can costs 10 cents per cm2 and the material for the curved side costs 8 cents per cm2. The total cost of the material, in cents, is C.

Express h in terms of r.

[2]
a.

Show that $$C = 20\pi {r^2} + \frac{{320\pi }}{r}$$.

[4]
b.

Given that there is a minimum value for C, find this minimum value in terms of $$\pi$$.

[9]
c.

## Markscheme

correct equation for volume      (A1)
eg  $$\pi {r^2}h = 20\pi$$

$$h = \frac{{20}}{{{r^2}}}$$     A1 N2

[2 marks]

a.

attempt to find formula for cost of parts      (M1)
eg  10 × two circles, 8 × curved side

correct expression for cost of two circles in terms of r (seen anywhere)      A1
eg  $$2\pi {r^2} \times 10$$

correct expression for cost of curved side (seen anywhere)      (A1)
eg  $$2\pi r \times h \times 8$$

correct expression for cost of curved side in terms of     A1
eg  $$8 \times 2\pi r \times \frac{{20}}{{{r^2}}},\,\,\frac{{320\pi }}{{{r^2}}}$$

$$C = 20\pi {r^2} + \frac{{320\pi }}{r}$$      AG N0

[4 marks]

b.

recognize $$C’ = 0$$ at minimum       (R1)
eg  $$C’ = 0,\,\,\frac{{{\text{d}}C}}{{{\text{d}}r}} = 0$$

correct differentiation (may be seen in equation)

$$C’ = 40\pi r – \frac{{320\pi }}{{{r^2}}}$$        A1A1

correct equation      A1
eg  $$40\pi r – \frac{{320\pi }}{{{r^2}}} = 0,\,\,40\pi r\frac{{320\pi }}{{{r^2}}}$$

correct working     (A1)
eg  $$40{r^3} = 320,\,\,{r^3} = 8$$

r = 2 (m)     A1

attempt to substitute their value of r into C
eg  $$20\pi \times 4 + 320 \times \frac{\pi }{2}$$     (M1)

correct working
eg  $$80\pi + 160\pi$$        (A1)

$$240\pi$$ (cents)      A1 N3

Note: Do not accept 753.6, 753.98 or 754, even if 240$$\pi$$ is seen.

[9 marks]

c.

## Question

Given that $$f(x) = \frac{1}{x}$$ , answer the following.

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

[4]
a.

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

[3]
b.

## Markscheme

$$f'(x) = – {x^{ – 2}}$$ (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]

a.

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

[3 marks]

b.

## Question

Let $$f(x) = \cos x + \sqrt 3 \sin x$$ , $$0 \le x \le 2\pi$$ . The following diagram shows the graph of $$f$$ .

The $$y$$-intercept is at ($$0$$, $$1$$) , there is a minimum point at A ($$p$$, $$q$$) and a maximum point at B.

Find $$f'(x)$$ .

[2]
a.

Hence

(i)     show that $$q = – 2$$ ;

(ii)    verify that A is a minimum point.

[10]
b(i) and (ii).

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

[3]
c.

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

Write down the value of r and of a .

[2]
d.

## Markscheme

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

[2 marks]

a.

(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]

b(i) and (ii).

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]

c.

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

[2 marks]

d.