# IB Math Analysis & Approaches Question bank-Topic: SL 5.12 The product and quotient rules SL Paper 1

## 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 $$f(x) = {{\rm{e}}^{ – 3x}}$$ 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

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

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

Use the quotient rule to show that $$g'(x) = \frac{{1 – 2\ln x}}{{{x^3}}}$$ .

[4]
a.

The graph of g has a maximum point at A. Find the x-coordinate of A.

[3]
b.

## Markscheme

$$\frac{{\rm{d}}}{{{\rm{d}}x}}\ln x = \frac{1}{x}$$ , $$\frac{{\rm{d}}}{{{\rm{d}}x}}{x^2} = 2x$$ (seen anywhere)     A1A1

attempt to substitute into the quotient rule (do not accept product rule)     M1

e.g. $$\frac{{{x^2}\left( {\frac{1}{x}} \right) – 2x\ln x}}{{{x^4}}}$$

correct manipulation that clearly leads to result     A1

e.g. $$\frac{{x – 2x\ln x}}{{{x^4}}}$$ , $$\frac{{x(1 – 2\ln x)}}{{{x^4}}}$$ , $$\frac{x}{{{x^4}}}$$ , $$\frac{{2x\ln x}}{{{x^4}}}$$

$$g'(x) = \frac{{1 – 2\ln x}}{{{x^3}}}$$     AG     N0

[4 marks]

a.

evidence of setting the derivative equal to zero     (M1)

e.g. $$g'(x) = 0$$ , $$1 – 2\ln x = 0$$

$$\ln x = \frac{1}{2}$$     A1

$$x = {{\rm{e}}^{\frac{1}{2}}}$$     A1     N2

[3 marks]

b.

## Question

Let $$h(x) = \frac{{6x}}{{\cos x}}$$ . Find $$h'(0)$$ .

## Markscheme

METHOD 1 (quotient)

derivative of numerator is 6     (A1)

derivative of denominator is $$– \sin x$$     (A1)

attempt to substitute into quotient rule     (M1)

correct substitution     A1

e.g. $$\frac{{(\cos x)(6) – (6x)( – \sin x)}}{{{{(\cos x)}^2}}}$$

substituting $$x = 0$$     (A1)

e.g. $$\frac{{(\cos 0)(6) – (6 \times 0)( – \sin 0)}}{{{{(\cos 0)}^2}}}$$

$$h'(0) = 6$$     A1     N2

METHOD 2 (product)

$$h(x) = 6x \times {(\cos x)^{ – 1}}$$

derivative of 6x is 6     (A1)

derivative of $${(\cos x)^{ – 1}}$$ is $$( – {(\cos x)^{ – 2}}( – \sin x))$$     (A1)

attempt to substitute into product rule     (M1)

correct substitution     A1

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

substituting $$x = 0$$    (A1)

e.g. $$(6 \times 0)( – {(\cos 0)^{ – 2}}( – \sin 0)) + (6){(\cos 0)^{ – 1}}$$

$$h'(0) = 6$$     A1     N2

[6 marks]

## Question

Let  $$f(x) = \frac{x}{{ – 2{x^2} + 5x – 2}}$$ for $$– 2 \le x \le 4$$ , $$x \ne \frac{1}{2}$$ , $$x \ne 2$$ . The graph of $$f$$ is given below.

The graph of $$f$$ has a local minimum at A($$1$$, $$1$$) and a local maximum at B.

Use the quotient rule to show that $$f'(x) = \frac{{2{x^2} – 2}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$ .

[6]
a.

Hence find the coordinates of B.

[7]
b.

Given that the line $$y = k$$ does not meet the graph of f , find the possible values of k .

[3]
c.

## Markscheme

correct derivatives applied in quotient rule     (A1)A1A1

$$1$$, $$– 4x + 5$$

Note: Award (A1) for 1, A1 for $$– 4x$$ and A1 for $$5$$, only if it is clear candidates are using the quotient rule.

correct substitution into quotient rule     A1

e.g. $$\frac{{1 \times ( – 2{x^2} + 5x – 2) – x( – 4x + 5)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$ , $$\frac{{ – 2{x^2} + 5x – 2 – x( – 4x + 5)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$

correct working     (A1)

e.g. $$\frac{{ – 2{x^2} + 5x – 2 – ( – 4{x^2} + 5x)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$

expression clearly leading to the answer     A1

e.g. $$\frac{{ – 2{x^2} + 5x – 2 + 4{x^2} – 5x}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$

$$f'(x) = \frac{{2{x^2} – 2}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$    AG     N0

[6 marks]

a.

evidence of attempting to solve $$f'(x) = 0$$     (M1)

e.g. $$2{x^2} – 2 = 0$$

evidence of correct working     A1

e.g. $${x^2} = 1,\frac{{ \pm \sqrt {16} }}{4}{\text{, }}2(x – 1)(x + 1)$$

correct solution to quadratic     (A1)

e.g. $$x = \pm 1$$

correct x-coordinate $$x = – 1$$ (may be seen in coordinate form $$\left( { – 1,\frac{1}{9}} \right)$$ )    A1     N2

attempt to substitute $$– 1$$ into f (do not accept any other value)     (M1)

e.g. $$f( – 1) = \frac{{ – 1}}{{ – 2 \times {{( – 1)}^2} + 5 \times ( – 1) – 2}}$$

correct working

e.g. $$\frac{{ – 1}}{{ – 2 – 5 – 2}}$$     A1

correct y-coordinate $$y = \frac{1}{9}$$ (may be seen in coordinate form $$\left( { – 1,\frac{1}{9}} \right)$$ )    A1     N2

[7 marks]

b.

recognizing values between max and min     (R1)

$$\frac{1}{9} < k < 1$$     A2     N3

[3 marks]

c.

## Examiners report

While most candidates answered part (a) correctly, there were some who did not show quite enough work for a “show that” question. A very small number of candidates did not follow the instruction to use the quotient rule.

a.

In part (b), most candidates knew that they needed to solve the equation $$f'(x) = 0$$ , and many were successful in answering this question correctly. However, some candidates failed to find both values of x, or made other algebraic errors in their solutions. One common error was to find only one solution for $${x^2} = 1$$ ; another was to work with the denominator equal to zero, rather than the numerator.

b.

In part (c), a significant number of candidates seemed to think that the line $$y = k$$ was a vertical line, and attempted to find the vertical asymptotes. Others tried looking for a horizontal asymptote. Fortunately, there were still a good number of intuitive candidates who recognized the link with the graph and with part (b), and realized that the horizontal line must pass through the space between the given local minimum and the local maximum they had found in part (b).

c.

## Question

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

Find $$f'(x)$$ .

[5]
a.

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

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

[4]
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 .

[7]
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. $$\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \frac{6}{{{{(x + 1)}^2}}}$$ , $$\left( {\frac{6}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{{6x}}} \right)$$

working that clearly leads to the answer     A1

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

working that clearly leads to the answer     A1

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

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

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 $$f(x) = \frac{{2x}}{{{x^2} + 5}}$$.

Use the quotient rule to show that $$f'(x) = \frac{{10 – 2{x^2}}}{{{{({x^2} + 5)}^2}}}$$.

[4]
a.

Find $$\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x}$$.

[4]
b.

The following diagram shows part of the graph of $$f$$.

The shaded region is enclosed by the graph of $$f$$, the $$x$$-axis, and the lines $$x = \sqrt 5$$ and $$x = q$$. This region has an area of $$\ln 7$$. Find the value of $$q$$.

[7]
c.

## Markscheme

derivative of $$2x$$ is $$2$$ (must be seen in quotient rule)     (A1)

derivative of $${x^2} + 5$$ is $$2x$$ (must be seen in quotient rule)     (A1)

correct substitution into quotient rule     A1

eg     $$\frac{{({x^2} + 5)(2) – (2x)(2x)}}{{{{({x^2} + 5)}^2}}},{\text{ }}\frac{{2({x^2} + 5) – 4{x^2}}}{{{{({x^2} + 5)}^2}}}$$

correct working which clearly leads to given answer   A1

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

$$f'(x) = \frac{{10 – 2{x^2}}}{{{{({x^2} + 5)}^2}}}$$     AG     N0

[4 marks]

a.

valid approach using substitution or inspection     (M1)

eg     $$u = {x^2} + 5,{\text{ d}}u = 2x{\text{d}}x,{\text{ }}\frac{1}{2}\ln ({x^2} + 5)$$

$$\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x = \int {\frac{1}{u}{\text{d}}u} }$$     (A1)

$$\int {\frac{1}{u}{\text{d}}u = \ln u + c}$$     (A1)

$$\ln ({x^2} + 5) + c$$     A1     N4

[4 marks]

b.

correct expression for area     (A1)

eg     $$\left[ {\ln \left( {{x^2} + 5} \right)} \right]_{\sqrt 5 }^q,{\text{ }}\int\limits_{\sqrt 5 }^q {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x}$$

substituting limits into their integrated function and subtracting (in either order)     (M1)

eg     $$\ln ({q^2} + 5) – \ln \left( {{{\sqrt 5 }^2} + 5} \right)$$

correct working     (A1)

eg     $$\ln \left( {{q^2} + 5} \right) – \ln 10,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}}$$

equating their expression to $$\ln 7$$ (seen anywhere)     (M1)

eg     $$\ln \left( {{q^2} + 5} \right) – \ln 10 = \ln 7,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}} = \ln 7,{\text{ }}\ln ({q^2} + 5) = \ln 7 + \ln 10$$

correct equation without logs     (A1)

eg     $$\frac{{{q^2} + 5}}{{10}} = 7,{\text{ }}{q^2} + 5 = 70$$

$${q^2} = 65$$     (A1)

$$q = \sqrt {65}$$     A1     N3

Note: Award A0 for $$q = \pm \sqrt {65}$$.

[7 marks]

c.

## 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

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

The values of the functions $$f$$ and $$g$$ and their derivatives for $$x = 1$$ and $$x = 8$$ are shown in the following table.

Let $$h(x) = f(x)g(x)$$.

Find $$h(1)$$.

[2]
a.

Find $$h'(8)$$.

[3]
b.

## Markscheme

expressing $$h(1)$$ as a product of $$f(1)$$ and  $$g(1)$$     (A1)

eg$$\,\,\,\,\,$$$$f(1) \times g(1),{\text{ }}2(9)$$

$$h(1) = 18$$     A1     N2

[2 marks]

a.

attempt to use product rule (do not accept $$h’ = f’ \times g’$$)     (M1)

eg$$\,\,\,\,\,$$$$h’ = fg’ + gf’,{\text{ }}h'(8) = f'(8)g(8) + g’(8)f(8)$$

correct substitution of values into product rule     (A1)

eg$$\,\,\,\,\,$$$$h’(8) = 4(5) + 2( – 3),{\text{ }} – 6 + 20$$

$$h’(8) = 14$$     A1 N2

[3 marks]

b.