# IB Math Analysis & Approaches Question bank-Topic SL 5.4- Tangents and normals, and their equations 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 the functions f (x) = -(x – h)2 + 2k and g(x) = e x-2 + k where h , k ∈ R.
(a) Find f ′(x).   [1]
The graphs of f and g have a common tangent at x = 3

(b) Show that h + = $$\frac{e+6}{2}$$ [3]

(c) Hence, show that  k = $$e+ \frac{e^{2}}{4}$$ [3]

Ans

(a) f ‘( x) = −2( x − h)

## 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

Let $$f(x) = {{\rm{e}}^x}\cos x$$ . Find the gradient of the normal to the curve of f at $$x = \pi$$ .

## Markscheme

evidence of choosing the product rule     (M1)

$$f'(x) = {{\rm{e}}^x} \times ( – \sin x) + \cos x \times {{\rm{e}}^x}$$ $$( = {{\rm{e}}^x}\cos x – {{\rm{e}}^x}\sin x)$$     A1A1

substituting $$\pi$$     (M1)

e.g.  $$f'(\pi ) = {{\rm{e}}^\pi }\cos \pi – {{\rm{e}}^\pi }\sin \pi$$ , $${{\rm{e}}^\pi }( – 1 – 0)$$ , $$– {{\rm{e}}^\pi }$$

taking negative reciprocal      (M1)

e.g. $$– \frac{1}{{f'(\pi )}}$$

gradient is $$\frac{1}{{{{\rm{e}}^\pi }}}$$     A1     N3

[6 marks]

## Question

Let $$f(x) = \sqrt x$$ . Line L is the normal to the graph of f at the point (4, 2) .

In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the line L .

Show that the equation of L is $$y = – 4x + 18$$ .

[4]
a.

Point A is the x-intercept of L . Find the x-coordinate of A.

[2]
b.

Find an expression for the area of R .

[3]
c.

The region R is rotated $$360^\circ$$ about the x-axis. Find the volume of the solid formed, giving your answer in terms of $$\pi$$ .

[8]
d.

## Markscheme

finding derivative     (A1)

e.g. $$f'(x) = \frac{1}{2}{x^{\frac{1}{2}}},\frac{{1}}{{2\sqrt x }}$$

correct value of derivative or its negative reciprocal (seen anywhere)     A1

e.g. $$\frac{1}{{2\sqrt 4 }}$$ , $$\frac{1}{4}$$

gradient of normal =  $$\frac{1}{{{\text{gradient of tangent}}}}$$ (seen anywhere)     A1

e.g. $$– \frac{1}{{f'(4)}} = – 4$$ , $$– 2\sqrt x$$

substituting into equation of line (for normal)     M1

e.g. $$y – 2 = – 4(x – 4)$$

$$y = – 4x + 18$$     AG     N0

[4 marks]

a.

recognition that $$y = 0$$ at A     (M1)

e.g. $$– 4x + 18 = 0$$

$$x = \frac{{18}}{4}$$ $$\left( { = \frac{9}{2}} \right)$$     A1     N2

[2 marks]

b.

splitting into two appropriate parts (areas and/or integrals)     (M1)

correct expression for area of R     A2     N3

e.g. area of R = $$\int_0^4 {\sqrt x } {\rm{d}}x + \int_4^{4.5} {( – 4x + 18){\rm{d}}x}$$ , $$\int_0^4 {\sqrt x } {\rm{d}}x + \frac{1}{2} \times 0.5 \times 2$$ (triangle)

Note: Award A1 if dx is missing.

[3 marks]

c.

correct expression for the volume from $$x = 0$$ to $$x = 4$$     (A1)

e.g. $$V = \int_0^4 {\pi \left[ {f{{(x)}^2}} \right]} {\rm{d}}x$$ , $${\int_0^4 {\pi \sqrt x } ^2}{\rm{d}}x$$ , $$\int_0^4 {\pi x{\rm{d}}x}$$

$$V = \left[ {\frac{1}{2}\pi {x^2}} \right]_0^4$$     A1

$$V = \pi \left( {\frac{1}{2} \times 16 – \frac{1}{2} \times 0} \right)$$     (A1)

$$V = 8\pi$$     A1

finding the volume from $$x = 4$$ to $$x = 4.5$$

EITHER

recognizing a cone     (M1)

e.g. $$V = \frac{1}{3}\pi {r^2}h$$

$$V = \frac{1}{3}\pi {(2)^2} \times \frac{1}{2}$$     (A1)

$$= \frac{{2\pi }}{3}$$     A1

total volume is $$8\pi + \frac{2}{3}\pi$$ $$\left( { = \frac{{26}}{3}\pi } \right)$$     A1     N4

OR

$$V = \pi \int_4^{4.5} {{{( – 4x + 18)}^2}{\rm{d}}x}$$     (M1)

$$= \int_4^{4.5} {\pi (16{x^2} – 144x + 324){\rm{d}}x}$$

$$= \pi \left[ {\frac{{16}}{3}{x^3} – 72{x^2} + 324x} \right]_4^{4.5}$$     A1

$$= \frac{{2\pi }}{3}$$     A1

total volume is $$8\pi + \frac{2}{3}\pi$$ $$\left( { = \frac{{26}}{3}\pi } \right)$$     A1     N4

[8 marks]

d.

## 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

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

The line T is the tangent to the graph of f at $$x = 1$$ .

Show that the equation of T is $$y = 4x – 2$$ .

[5]
a.

Find the x-intercept of T .

[2]
b.

The shaded region R is enclosed by the graph of f , the line T , and the x-axis.

(i)     Write down an expression for the area of R .

(ii)    Find the area of R .

[9]
c(i) and (ii).

## Markscheme

$$f(1) = 2$$     (A1)

$$f'(x) = 4x$$     A1

evidence of finding the gradient of f at $$x = 1$$     M1

e.g. substituting $$x = 1$$ into $$f'(x)$$

finding gradient of f at $$x = 1$$     A1

e.g. $$f'(1) = 4$$

evidence of finding equation of the line     M1

e.g. $$y – 2 = 4(x – 1)$$ , $$2 = 4(1) + b$$

$$y = 4x – 2$$     AG     N0

[5 marks]

a.

appropriate approach     (M1)

e.g. $$4x – 2 = 0$$

$$x = \frac{1}{2}$$     A1     N2

[2 marks]

b.

(i) bottom limit $$x = 0$$ (seen anywhere)     (A1)

approach involving subtraction of integrals/areas     (M1)

e.g. $$\int {f(x) – {\text{area of triangle}}}$$ , $$\int {f – \int l }$$

correct expression     A2     N4

e.g. $$\int_0^1 {2{x^2}{\rm{d}}x – } \int_{0.5}^1 {(4x – 2){\rm{d}}x}$$ , $$\int_0^1 {f(x){\rm{d}}x – \frac{1}{2}}$$ , $$\int_0^{0.5} {2{x^2}{\rm{d}}x} + \int_{0.5}^1 {(f(x) – (4x – 2)){\rm{d}}x}$$

(ii) METHOD 1 (using only integrals)

correct integration     (A1)(A1)(A1)

$$\int {2{x^2}{\rm{d}}x} = \frac{{2{x^3}}}{3}$$ , $$\int {(4x – 2){\rm{d}}x = } 2{x^2} – 2x$$

substitution of limits     (M1)

e.g. $$\frac{1}{{12}} + \frac{2}{3} – 2 + 2 – \left( {\frac{1}{{12}} – \frac{1}{2} + 1} \right)$$

area = $$\frac{1}{6}$$     A1     N4

METHOD 2 (using integral and triangle)

area of triangle = $$\frac{1}{2}$$     (A1)

correct integration     (A1)

$$\int {2{x^2}{\rm{d}}x = } \frac{{2{x^3}}}{3}$$

substitution of limits     (M1)

e.g. $$\frac{2}{3}{(1)^3} – \frac{2}{3}{(0)^3}$$ , $$\frac{2}{3} – 0$$

correct simplification     (A1)

e.g. $$\frac{2}{3} – \frac{1}{2}$$

area = $$\frac{1}{6}$$     A1     N4

[9 marks]

c(i) and (ii).

## Question

Let $$f(x) = \frac{1}{4}{x^2} + 2$$  . The line L is the tangent to the curve of f at (4, 6) .

Let $$g(x) = \frac{{90}}{{3x + 4}}$$ , for $$2 \le x \le 12$$ . The following diagram shows the graph of g .

Find the equation of L .

[4]
a.

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

[6]
b.

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

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

[3]
c.

## Markscheme

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

attempt to find $$f'(4)$$     (M1)

correct value $$f'(4) = 2$$     A1

correct equation in any form     A1     N2

e.g. $$y – 6 = 2(x – 4)$$ , $$y = 2x – 2$$

[4 marks]

a.

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

correct integral     A1A1

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

substituting limits and subtracting     (M1)

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

correct working     (A1)

e.g. $$30(\ln 40 – \ln 10)$$

correct application of $$\ln b – \ln a$$     (A1)

e.g. $$30\ln \frac{{40}}{{10}}$$

$${\rm{area}} = 30\ln 4$$     A1     N4

[6 marks]

b.

valid approach     (M1)

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

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

[3 marks]

c.

## 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

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

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 $$f(x) = {{\text{e}}^{2x}}$$. The line $$L$$ is the tangent to the curve of $$f$$ at $$(1,{\text{ }}{{\text{e}}^2})$$.

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

## Markscheme

recognising need to differentiate (seen anywhere)     R1

eg     $$f’,{\text{ }}2{{\text{e}}^{2x}}$$

attempt to find the gradient when $$x = 1$$     (M1)

eg     $$f'(1)$$

$$f'(1) = 2{{\text{e}}^2}$$     (A1)

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

eg     $$y – {{\text{e}}^2} = 2{{\text{e}}^2}(x – 1),{\text{ }}{{\text{e}}^2} = 2{{\text{e}}^2}(1) + b$$

correct working     (A1)

eg     $$y – {{\text{e}}^2} = 2{{\text{e}}^2}x – 2{{\text{e}}^2},{\text{ }}b = – {{\text{e}}^2}$$

$$y = 2{{\text{e}}^2}x – {{\text{e}}^2}$$     A1     N3

[6 marks]

## 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 $$y = f(x)$$, for $$– 0.5 \le$$ x $$\le$$ $$6.5$$. The following diagram shows the graph of $$f’$$, the derivative of $$f$$.

The graph of $$f’$$ has a local maximum when $$x = 2$$, a local minimum when $$x = 4$$, and it crosses the $$x$$-axis at the point $$(5,{\text{ }}0)$$.

Explain why the graph of $$f$$ has a local minimum when $$x = 5$$.

[2]
a.

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

[2]
b.

The following diagram shows the shaded regions $$A$$, $$B$$ and $$C$$.

The regions are enclosed by the graph of $$f’$$, the $$x$$-axis, the $$y$$-axis, and the line $$x = 6$$.

The area of region $$A$$ is 12, the area of region $$B$$ is 6.75 and the area of region $$C$$ is 6.75.

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

[5]
c.

The following diagram shows the shaded regions $$A$$, $$B$$ and $$C$$.

The regions are enclosed by the graph of $$f’$$, the x-axis, the y-axis, and the line $$x = 6$$.

The area of region $$A$$ is 12, the area of region $$B$$ is 6.75 and the area of region $$C$$ is 6.75.

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

[6]
d.

## Markscheme

METHOD 1

$$f'(5) = 0$$     (A1)

valid reasoning including reference to the graph of $$f’$$     R1

eg$$\;\;\;f’$$ changes sign from negative to positive at $$x = 5$$, labelled sign chart for $$f’$$

so $$f$$ has a local minimum at $$x = 5$$     AG     N0

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

METHOD 2

$$f'(5) = 0$$     A1

valid reasoning referring to second derivative     R1

eg$$\;\;\;f”(5) > 0$$

so $$f$$ has a local minimum at $$x = 5$$     AG     N0

[2 marks]

a.

attempt to find relevant interval     (M1)

eg$$\;\;\;f’$$ is decreasing, gradient of $$f’$$ is negative, $$f” < 0$$

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

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

[2 marks]

b.

METHOD 1 (one integral)

correct application of Fundamental Theorem of Calculus     (A1)

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

attempt to link definite integral with areas     (M1)

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

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

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

correct working     A1

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

$$f(6) = 2$$     A1     N3

METHOD 2 (more than one integral)

correct application of Fundamental Theorem of Calculus     (A1)

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

attempt to link definite integrals with areas     (M1)

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

correct values for integrals     (A1)

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

one correct intermediate value     A1

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

$$f(6) = 2$$     A1     N3

[5 marks]

c.

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

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

choosing chain rule or product rule     (M1)

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

correct derivative     (A1)

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

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

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

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

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

correct equation in any form     A1     N2

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

[6 marks]

[Total 15 marks]

d.

## Question

Let $$f(x) = \sqrt {4x + 5}$$, for $$x \geqslant – 1.25$$.

Consider another function $$g$$. Let R be a point on the graph of $$g$$. The $$x$$-coordinate of R is 1. The equation of the tangent to the graph at R is $$y = 3x + 6$$.

Find $$f'(1)$$.

[4]
a.

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

[2]
b.

Find $$g(1)$$.

[2]
c.

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

[7]
d.

## Markscheme

choosing chain rule     (M1)

eg$$\,\,\,\,\,$$$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},{\text{ }}u = 4x + 5,{\text{ }}u’ = 4$$

correct derivative of $$f$$     A2

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

$$f'(1) = \frac{2}{3}$$    A1     N2

[4 marks]

a.

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

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

$$g'(1) = 3$$    A1     N2

[2 marks]

b.

recognize that R is on the tangent     (M1)

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

$$g(1) = 9$$    A1     N2

[2 marks]

c.

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

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

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

eg$$\,\,\,\,\,$$$$uv’ + u’v$$

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

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

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

EITHER

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

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

$$y – 27 = 15(x – 1)$$     A1     N2

OR

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

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

$$y = 15x + 12$$     A1     N2

[7 marks]

d.

## Question

The following diagram shows the graph of $$f(x) = 2x\sqrt {{a^2} – {x^2}}$$, for $$– 1 \leqslant x \leqslant a$$, where $$a > 1$$.

The line $$L$$ is the tangent to the graph of $$f$$ at the origin, O. The point $${\text{P}}(a,{\text{ }}b)$$ lies on $$L$$.

The point $${\text{Q}}(a,{\text{ }}0)$$ lies on the graph of $$f$$. Let $$R$$ be the region enclosed by the graph of $$f$$ and the $$x$$-axis. This information is shown in the following diagram.

Let $${A_R}$$ be the area of the region $$R$$.

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

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

[6]
a.

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

[6]
b.

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

[4]
c.

## Markscheme

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

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

correct substitution     (A1)

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

$$f'(0) = 2a$$    (A1)

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

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

(ii)     METHOD 1

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

eg$$\,\,\,\,\,$$$$y = 2a \times a$$

$$b = 2{a^2}$$     A1     N2

METHOD 2

eg$$\,\,\,\,\,$$$$\frac{b}{a} = 2a$$

$$b = 2{a^2}$$    A1     N2

[6 marks]

a.

METHOD 1

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

valid approach using substitution or inspection     (M1)

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

correct working     (A1)

eg$$\,\,\,\,\,$$$$\int {2x\sqrt {{a^2} – {x^2}} {\text{d}}x = \int { – \sqrt u {\text{d}}u} }$$

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

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

substituting limits and subtracting     A1

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

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

METHOD 2

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

valid approach using substitution or inspection     (M1)

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

correct working     (A1)

eg$$\,\,\,\,\,$$$$\int {2x\sqrt {{a^2} – {x^2}} {\text{d}}x = \int { – \sqrt u {\text{d}}u} }$$

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

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

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

substituting limits and subtracting     A1

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

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

[6 marks]

b.

METHOD 1

valid approach to find area of triangle     (M1)

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

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

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

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

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

$$k = \frac{3}{2}$$     A1     N2

METHOD 2

valid approach to find area of triangle     (M1)

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

correct working     (A1)

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

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

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

$$k = \frac{3}{2}$$     A1     N2

[4 marks]

c.

## Question

The following diagram shows the graph of $$f’$$, the derivative of $$f$$.

The graph of $$f’$$ has a local minimum at A, a local maximum at B and passes through $$(4,{\text{ }} – 2)$$.

The point $${\text{P}}(4,{\text{ }}3)$$ lies on the graph of the function, $$f$$.

Write down the gradient of the curve of $$f$$ at P.

[1]
a.i.

Find the equation of the normal to the curve of $$f$$ at P.

[3]
a.ii.

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

[2]
b.

## Markscheme

$$– 2$$     A1     N1

[1 mark]

a.i.

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

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

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

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

[3 marks]

a.ii.

correct answer and valid reasoning     A2     N2

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

reason:     eg     slope of $$f’$$ is positive, $$f’$$ is increasing, $$f’’ > 0$$,

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

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

[2 marks]

b.

## Question

A quadratic function $$f$$ can be written in the form $$f(x) = a(x – p)(x – 3)$$. The graph of $$f$$ has axis of symmetry $$x = 2.5$$ and $$y$$-intercept at $$(0,{\text{ }} – 6)$$

Find the value of $$p$$.

[3]
a.

Find the value of $$a$$.

[3]
b.

The line $$y = kx – 5$$ is a tangent to the curve of $$f$$. Find the values of $$k$$.

[8]
c.

## Markscheme

METHOD 1 (using x-intercept)

determining that 3 is an $$x$$-intercept     (M1)

eg$$\,\,\,\,\,$$$$x – 3 = 0$$,

valid approach     (M1)

eg$$\,\,\,\,\,$$$$3 – 2.5,{\text{ }}\frac{{p + 3}}{2} = 2.5$$

$$p = 2$$     A1     N2

METHOD 2 (expanding f (x))

correct expansion (accept absence of $$a$$)     (A1)

eg$$\,\,\,\,\,$$$$a{x^2} – a(3 + p)x + 3ap,{\text{ }}{x^2} – (3 + p)x + 3p$$

valid approach involving equation of axis of symmetry     (M1)

eg$$\,\,\,\,\,$$$$\frac{{ – b}}{{2a}} = 2.5,{\text{ }}\frac{{a(3 + p)}}{{2a}} = \frac{5}{2},{\text{ }}\frac{{3 + p}}{2} = \frac{5}{2}$$

$$p = 2$$     A1     N2

METHOD 3 (using derivative)

correct derivative (accept absence of $$a$$)     (A1)

eg$$\,\,\,\,\,$$$$a(2x – 3 – p),{\text{ }}2x – 3 – p$$

valid approach     (M1)

eg$$\,\,\,\,\,$$$$f’(2.5) = 0$$

$$p = 2$$     A1     N2

[3 marks]

a.

attempt to substitute $$(0,{\text{ }} – 6)$$     (M1)

eg$$\,\,\,\,\,$$$$– 6 = a(0 – 2)(0 – 3),{\text{ }}0 = a( – 8)( – 9),{\text{ }}a{(0)^2} – 5a(0) + 6a = – 6$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$– 6 = 6a$$

$$a = – 1$$     A1     N2

[3 marks]

b.

METHOD 1 (using discriminant)

recognizing tangent intersects curve once     (M1)

recognizing one solution when discriminant = 0     M1

attempt to set up equation     (M1)

eg$$\,\,\,\,\,$$$$g = f,{\text{ }}kx – 5 = – {x^2} + 5x – 6$$

rearranging their equation to equal zero     (M1)

eg$$\,\,\,\,\,$$$${x^2} – 5x + kx + 1 = 0$$

correct discriminant (if seen explicitly, not just in quadratic formula)     A1

eg$$\,\,\,\,\,$$$${(k – 5)^2} – 4,{\text{ }}25 – 10k + {k^2} – 4$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$k – 5 = \pm 2,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}$$

$$k = 3,{\text{ }}7$$     A1A1     N0

METHOD 2 (using derivatives)

attempt to set up equation     (M1)

eg$$\,\,\,\,\,$$$$g = f,{\text{ }}kx – 5 = – {x^2} + 5x – 6$$

recognizing derivative/slope are equal     (M1)

eg$$\,\,\,\,\,$$$$f’ = {m_T},{\text{ }}f’ = k$$

correct derivative of $$f$$     (A1)

eg$$\,\,\,\,\,$$$$– 2x + 5$$

attempt to set up equation in terms of either $$x$$ or $$k$$     M1

eg$$\,\,\,\,\,$$$$( – 2x + 5)x – 5 = – {x^2} + 5x – 6,{\text{ }}k\left( {\frac{{5 – k}}{2}} \right) – 5 = – {\left( {\frac{{5 – k}}{2}} \right)^2} + 5\left( {\frac{{5 – k}}{2}} \right) – 6$$

rearranging their equation to equal zero     (M1)

eg$$\,\,\,\,\,$$$${x^2} – 1 = 0,{\text{ }}{k^2} – 10k + 21 = 0$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$x = \pm 1,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}$$

$$k = 3,{\text{ }}7$$     A1A1     N0

[8 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

Consider a function $$f$$. The line L1 with equation $$y = 3x + 1$$ is a tangent to the graph of $$f$$ when $$x = 2$$

Let $$g\left( x \right) = f\left( {{x^2} + 1} \right)$$ and P be the point on the graph of $$g$$ where $$x = 1$$.

Write down $$f’\left( 2 \right)$$.

[2]
a.i.

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

[2]
a.ii.

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

[5]
b.

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

Find the y-coordinate of Q.

[7]
c.

## Markscheme

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

eg   $$f’\left( x \right) = m$$

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

[2 marks]

a.i.

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

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

$$f\left( 2 \right) = 7$$     A1 N2

[2 marks]

a.ii.

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

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

eg  $$\frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},\,\,u = {x^2} + 1,\,\,u’ = 2x$$

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

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

$$g’\left( 1 \right) = 6$$     AG N0

[5 marks]

b.

at Q, L1L2 (seen anywhere)      (M1)

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

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

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

correct equation for L2

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

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

$$y = 1$$     A1 N2

[7 marks]

c.