# IB Math Analysis & Approaches Question bank-Topic: SL 5.6 The chain rule for composite functions SL Paper 1

## Question

Let  $$f:x \mapsto {\sin ^3}x$$ .

(i) Write down the range of the function f .

(ii) Consider $$f(x) = 1$$ , $$0 \le x \le 2\pi$$ . Write down the number of solutions to this equation. Justify your answer.


a.

Find $$f'(x)$$ , giving your answer in the form $$a{\sin ^p}x{\cos ^q}x$$ where $$a{\text{, }}p{\text{, }}q \in \mathbb{Z}$$ .


b.

Let $$g(x) = \sqrt 3 \sin x{(\cos x)^{\frac{1}{2}}}$$ for $$0 \le x \le \frac{\pi }{2}$$ . Find the volume generated when the curve of g is revolved through $$2\pi$$ about the x-axis.


c.

## Markscheme

(i) range of f is $$[ – 1{\text{, }}1]$$ , $$( – 1 \le f(x) \le 1)$$     A2     N2

(ii) $${\sin ^3}x \Rightarrow 1 \Rightarrow \sin x = 1$$     A1

justification for one solution on $$[0{\text{, }}2\pi ]$$    R1

e.g. $$x = \frac{\pi }{2}$$ , unit circle, sketch of $$\sin x$$

1 solution (seen anywhere)     A1     N1

[5 marks]

a.

$$f'(x) = 3{\sin ^2}x\cos x$$     A2     N2

[2 marks]

b.

using $$V = \int_a^b {\pi {y^2}{\rm{d}}x}$$     (M1)

$$V = \int_0^{\frac{\pi }{2}} {\pi (\sqrt 3 } \sin x{\cos ^{\frac{1}{2}}}x{)^2}{\rm{d}}x$$     (A1)

$$= \pi \int_0^{\frac{\pi }{2}} {3{{\sin }^2}x\cos x{\rm{d}}x}$$     A1

$$V = \pi \left[ {{{\sin }^3}x} \right]_0^{\frac{\pi }{2}}$$ $$\left( { = \pi \left( {{{\sin }^3}\left( {\frac{\pi }{2}} \right) – {{\sin }^3}0} \right)} \right)$$     A2

evidence of using $$\sin \frac{\pi }{2} = 1$$ and $$\sin 0 = 0$$     (A1)

e.g. $$\pi \left( {1 – 0} \right)$$

$$V = \pi$$     A1     N1

[7 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)    $$g'(x)$$ .


a.

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


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

Let $$g(x) = 2x\sin x$$ .

Find $$g'(x)$$ .


a.

Find the gradient of the graph of g at $$x = \pi$$ .


b.

## Markscheme

evidence of choosing the product rule     (M1)

e.g. $$uv’ + vu’$$

correct derivatives $$\cos x$$ , 2     (A1)(A1)

$$g'(x) = 2x\cos x + 2\sin x$$     A1     N4

[4 marks]

a.

attempt to substitute into gradient function     (M1)

e.g. $$g'(\pi )$$

correct substitution     (A1)

e.g. $$2\pi \cos \pi + 2\sin \pi$$

$${\text{gradient}} = – 2\pi$$     A1     N2

[3 marks]

b.

## 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 .


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

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


b.

Find $$f'(x)$$ .


c.

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


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 $$f(x) = \frac{{6x}}{{x + 1}}$$ , for $$x > 0$$ .

Find $$f'(x)$$ .


a.

Let $$g(x) = \ln \left( {\frac{{6x}}{{x + 1}}} \right)$$ , for $$x > 0$$ .

Show that $$g'(x) = \frac{1}{{x(x + 1)}}$$ .


b.

Let $$h(x) = \frac{1}{{x(x + 1)}}$$ . The area enclosed by the graph of h , the x-axis and the lines $$x = \frac{1}{5}$$  and $$x = k$$ is $$\ln 4$$ . Given that $$k > \frac{1}{5}$$ , find the value of k .


c.

## Markscheme

METHOD 1

evidence of choosing quotient rule     (M1)

e.g. $$\frac{{u’v – uv’}}{{{v^2}}}$$

evidence of correct differentiation (must be seen in quotient rule)     (A1)(A1)

e.g. $$\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6$$ , $$\frac{{\rm{d}}}{{{\rm{d}}x}}(x + 1) = 1$$

correct substitution into quotient rule     A1

e.g. $$\frac{{(x + 1)6 – 6x}}{{{{(x + 1)}^2}}}$$ , $$\frac{{6x + 6 – 6x}}{{{{(x + 1)}^2}}}$$

$$f'(x) = \frac{6}{{{{(x + 1)}^2}}}$$    A1     N4

[5 marks]

METHOD 2

evidence of choosing product rule     (M1)

e.g. $$6x{(x + 1)^{ – 1}}$$ , $$uv’ + vu’$$

evidence of correct differentiation (must be seen in product rule)     (A1)(A1)

e.g. $$\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6$$ , $$\frac{{\rm{d}}}{{{\rm{d}}x}}{(x + 1)^{ – 1}} = – 1{(x + 1)^{ – 2}} \times 1$$

correct working     A1

e.g. $$6x \times – {(x + 1)^{ – 2}} + {(x + 1)^{ – 1}} \times 6$$ , $$\frac{{ – 6x + 6(x + 1)}}{{{{(x + 1)}^2}}}$$

$$f'(x) = \frac{6}{{{{(x + 1)}^2}}}$$   A1     N4

[5 marks]

a.

METHOD 1

evidence of choosing chain rule     (M1)

e.g. formula, $$\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \left( {\frac{{6x}}{{x + 1}}} \right)$$

correct reciprocal of $$\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}}$$ is $$\frac{{x + 1}}{{6x}}$$ (seen anywhere)     A1

correct substitution into chain rule     A1

e.g. $$\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \frac{6}{{{{(x + 1)}^2}}}$$ , $$\left( {\frac{6}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{{6x}}} \right)$$

e.g. $$\left( {\frac{6}{{(x + 1)}}} \right)\left( {\frac{1}{{6x}}} \right)$$ , $$\left( {\frac{1}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{x}} \right)$$ , $$\frac{{6(x + 1)}}{{6x{{(x + 1)}^2}}}$$

$$g'(x) = \frac{1}{{x(x + 1)}}$$     AG     N0

[4 marks]

METHOD 2

attempt to subtract logs     (M1)

e.g. $$\ln a – \ln b$$ , $$\ln 6x – \ln (x + 1)$$

correct derivatives (must be seen in correct expression)     A1A1

e.g. $$\frac{6}{{6x}} – \frac{1}{{x + 1}}$$ , $$\frac{1}{x} – \frac{1}{{x + 1}}$$

e.g. $$\frac{{x + 1 – x}}{{x(x + 1)}}$$ , $$\frac{{6x + 6 – 6x}}{{6x(x + 1)}}$$ , $$\frac{{6(x + 1 – x)}}{{6x(x + 1)}}$$

$$g'(x) = \frac{1}{{x(x + 1)}}$$     AG     N0

[4 marks]

b.

valid method using integral of  h(x) (accept missing/incorrect limits or missing $${\text{d}}x$$ )     (M1)

e.g. $${\rm{area}} = \int_{\frac{1}{5}}^k {h(x){\rm{d}}x}$$ , $$\int{\left( {\frac{1}{{x(x + 1)}}} \right)}$$

recognizing that integral of derivative will give original function     (R1)

e.g. $$\int{\left( {\frac{1}{{x(x + 1)}}} \right)} {\rm{d}}x = \ln \left( {\frac{{6x}}{{x + 1}}} \right)$$

correct substitution and subtraction     A1

e.g. $$\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln \left( {\frac{{6 \times \frac{1}{5}}}{{\frac{1}{5} + 1}}} \right)$$ , $$\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln (1)$$

setting their expression equal to $$\ln 4$$     (M1)

e.g. $$\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln (1) = \ln 4$$ , $$\ln \left( {\frac{{6k}}{{k + 1}}} \right) = \ln 4$$ , $$\int_{\frac{1}{5}}^k {h(x){\rm{d}}x = \ln 4}$$

correct equation without logs     A1

e.g.$$\frac{{6k}}{{k + 1}} = 4$$ , $$6k = 4(k + 1)$$

correct working     (A1)

e.g. $$6k = 4k + 4$$ , $$2k = 4$$

$$k = 2$$    A1     N4

[7 marks]

c.

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


a.

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


b.

Write down the value of $$p$$.


c.

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

Find the value of $$q$$.


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


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 $$y = f(x)$$, 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 $$x = 5$$.


a.

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


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 $$f(6)$$.


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 $$x = 6$$.


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 $$f'(1)$$.


a.

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


b.

Find $$g(1)$$.


c.

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


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

Consider f(x), g(x) and h(x), for x∈$$\mathbb{R}$$ where h(x) = $$\left( {f \circ g} \right)$$(x).

Given that g(3) = 7 , g′ (3) = 4 and f ′ (7) = −5 , find the gradient of the normal to the curve of h at x = 3.

## Markscheme

recognizing the need to find h′      (M1)

recognizing the need to find h′ (3) (seen anywhere)      (M1)

evidence of choosing chain rule        (M1)

eg   $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}u}} \times \frac{{{\text{d}}u}}{{{\text{d}}x}},\,\,f’\left( {g\left( 3 \right)} \right) \times g’\left( 3 \right),\,\,f’\left( g \right) \times g’$$

correct working       (A1)

eg  $$f’\left( 7 \right) \times 4,\,\, – 5 \times 4$$

$$h’\left( 3 \right) = – 20$$      (A1)

evidence of taking their negative reciprocal for normal       (M1)

eg  $$– \frac{1}{{h’\left( 3 \right)}},\,\,{m_1}{m_2} = – 1$$

gradient of normal is $$\frac{1}{{20}}$$      A1 N4

[7 marks]

## 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 $$f’\left( 2 \right)$$.


a.i.

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


a.ii.

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


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.


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.