# IB Math Analysis & Approaches Question bank-Topic: SL 5.8 Points of inflexion with zero and non-zero gradients SL Paper 1

## 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

The diagram shows part of the graph of $$y = f'(x)$$ . The x-intercepts are at points A and C. There is a minimum at B, and a maximum at D.

(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.

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

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

[1]
b.

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

[3]
c.

## Markscheme

(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(i) and (ii).

A     A1     N1

[1 mark]

b.

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]

c.

## Question

Let $$f(x) = \frac{{ax}}{{{x^2} + 1}}$$ , $$– 8 \le x \le 8$$ , $$a \in \mathbb{R}$$ .The graph of f is shown below.

The region between $$x = 3$$ and $$x = 7$$ is shaded.

Show that $$f( – x) = – f(x)$$ .

[2]
a.

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

[7]
b.

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 $$\int_4^8 {2f(x – 1){\rm{d}}x}$$ .

[7]
c.

## Markscheme

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]

a.

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]

b.

(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]

c.

## Question

A function f has its first derivative given by $$f'(x) = {(x – 3)^3}$$ .

Find the second derivative.

[2]
a.

Find $$f'(3)$$ and $$f”(3)$$ .

[1]
b.

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

[2]
c.

## Markscheme

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]

a.

$$f'(3) = 0$$ , $$f”(3) = 0$$     A1     N1

[1 mark]

b.

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]

c.

## 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 $$f(x) = \frac{{\cos x}}{{\sin x}}$$ , for $$\sin x \ne 0$$ .

In the following table, $$f’\left( {\frac{\pi }{2}} \right) = p$$ and $$f”\left( {\frac{\pi }{2}} \right) = q$$ . The table also gives approximate values of $$f'(x)$$ and $$f”(x)$$ near $$x = \frac{\pi }{2}$$ .

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

[5]
a.

Find $$f”(x)$$ .

[3]
b.

Find the value of p and of q.

[3]
c.

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

[2]
d.

## Markscheme

$$\frac{{\rm{d}}}{{{\rm{d}}x}}\sin x = \cos x$$ , $$\frac{{\rm{d}}}{{{\rm{d}}x}}\cos x = – \sin x$$ (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]

a.

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]

b.

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]

c.

second derivative is zero, second derivative changes sign     R1R1     N2

[2 marks]

d.

## Question

Consider $$f(x) = \ln ({x^4} + 1)$$ .

The second derivative is given by $$f”(x) = \frac{{4{x^2}(3 – {x^4})}}{{{{({x^4} + 1)}^2}}}$$ .

The equation $$f”(x) = 0$$ has only three solutions, when $$x = 0$$ , $$\pm \sqrt[4]{3}$$ $$( \pm 1.316 \ldots )$$ .

Find the value of $$f(0)$$ .

[2]
a.

Find the set of values of $$x$$ for which $$f$$ is increasing.

[5]
b.

(i)     Find $$f”(1)$$ .

(ii)     Hence, show that there is no point of inflexion on the graph of $$f$$ at $$x = 0$$ .

[5]
c.

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 $$x \ge 0$$ .

[3]
d.

## Markscheme

substitute $$0$$ into $$f$$     (M1)

eg   $$\ln (0 + 1)$$ , $$\ln 1$$

$$f(0) = 0$$     A1 N2

[2 marks]

a.

$$f'(x) = \frac{1}{{{x^4} + 1}} \times 4{x^3}$$ (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]

b.

(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]

c.

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]

d.

## Question

Consider the functions $$f(x)$$ , $$g(x)$$ and $$h(x)$$ . The following table gives some values associated with these functions.

The following diagram shows parts of the graphs of $$h$$ and $$h”$$ .

There is a point of inflexion on the graph of $$h$$ at P, when $$x = 3$$ .

Given that $$h(x) = f(x) \times g(x)$$ ,

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

[3]
a.

Explain why P is a point of inflexion.

[2]
b.

find the $$y$$-coordinate of P.

[2]
c.

find the equation of the normal to the graph of $$h$$ at P.

[7]
d.

## Markscheme

$$g(3) = – 18$$ , $$f'(3) = 1$$ , $$h”(2) = – 6$$     A1A1A1     N3

[3 marks]

a.

$$h”(3) = 0$$     (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]

b.

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]

c.

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]

d.

## 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

A function f (x) has derivative f ′(x) = 3x2 + 18x. The graph of f has an x-intercept at x = −1.

Find f (x).

[6]
a.

The graph of f has a point of inflexion at x = p. Find p.

[4]
b.

Find the values of x for which the graph of f is concave-down.

[3]
c.

## Markscheme

evidence of integration       (M1)

eg  $$\int {f’\left( x \right)}$$

correct integration (accept absence of C)       (A1)(A1)

eg  $${x^3} + \frac{{18}}{2}{x^2} + C,\,\,{x^3} + 9{x^2}$$

attempt to substitute x = −1 into their = 0 (must have C)      M1

eg  $${\left( { – 1} \right)^3} + 9{\left( { – 1} \right)^2} + C = 0,\,\, – 1 + 9 + C = 0$$

Note: Award M0 if they substitute into original or differentiated function.

correct working       (A1)

eg  $$8 + C = 0,\,\,\,C = – 8$$

$$f\left( x \right) = {x^3} + 9{x^2} – 8$$      A1 N5

[6 marks]

a.

METHOD 1 (using 2nd derivative)

recognizing that f” = 0 (seen anywhere)      M1

correct expression for f”      (A1)

eg   6x + 18, 6p + 18

correct working      (A1)

6+ 18 = 0

p = −3       A1 N3

METHOD 1 (using 1st derivative)

recognizing the vertex of f′ is needed       (M2)

eg   $$– \frac{b}{{2a}}$$ (must be clear this is for f′)

correct substitution      (A1)

eg   $$\frac{{ – 18}}{{2 \times 3}}$$

p = −3       A1 N3

[4 marks]

b.

valid attempt to use f” (x) to determine concavity      (M1)

eg   f” (x) < 0, f” (−2), f” (−4),  6x + 18 ≤ 0

correct working       (A1)

eg   6x + 18 < 0, f” (−2) = 6, f” (−4) = −6

f concave down for x < −3 (do not accept ≤ −3)       A1 N2

[3 marks]

c.