# IB Math AA Question bank-Topic: SL 5.11 Definite integrals, including analytical approach SL Paper 1

### Question

Let f (x) =  $$\sqrt{12-2x}$$ , x ≤ a . 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 y-axis. x

The graph of f intersects the x-axis at the point (a , 0) .

1. Find the value of a . 

2. Find the volume of the solid formed when the shaded region is revolved 360° about the x-axis. 

Ans:

(a)

recognize f (x) = 0

eg $$\sqrt{12-2x}= 0, 2x=12$$

a= 6 (accept x= 6, (6,0))

(b)

attempt to substitute either their limits or the function into volume formula (must involve $$f^{2})$$

eg $$\int_{0}^{6}f^{2}dx , \pi \int (\sqrt{12-2x})^{2}, \pi \int_{0}^{6}$$(12-2x) dx

correct integaration of each term

eg $$12x-x^{2}, 12x-x^{2}$$+c, $$\left [ 12x-x^{2} \right ]_0^6$$

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

eg $$\pi (12(6)-(6)^{2})-\pi (0),72\pi -36\pi ,(12(6)-(6)^{2})-(0)$$

volume = $$36\pi$$

### Question

Find  $$\int {\frac{1}{{2x + 3}}} {\rm{d}}x$$ .

a.

Given that $$\int_0^3 {\frac{1}{{2x + 3}}} {\rm{d}}x = \ln \sqrt P$$ , find the value of P.

b.

## Markscheme

$$\int {\frac{1}{{2x + 3}}} {\rm{d}}x = \frac{1}{2}\ln (2x + 3) + C$$  (accept $$\frac{1}{2}\ln |(2x + 3)| + C$$ )    A1A1     N2

[2 marks]

a.

$$\int_0^3 {\frac{1}{{2x + 3}}} {\rm{d}}x = \left[ {\frac{1}{2}\ln (2x + 3)} \right]_0^3$$

evidence of substitution of limits     (M1)

e.g.$$\frac{1}{2}\ln 9 – \frac{1}{2}\ln 3$$

evidence of correctly using $$\ln a – \ln b = \ln \frac{a}{b}$$ (seen anywhere)     (A1)

e.g. $$\frac{1}{2}\ln 3$$

evidence of correctly using $$a\ln b = \ln {b^a}$$ (seen anywhere)     (A1)

e.g. $$\ln \sqrt {\frac{9}{3}}$$

$$P = 3$$ (accept $$\ln \sqrt 3$$ )     A1     N2

[4 marks]

b.

## Question

Let  $$\int_1^5 {3f(x){\rm{d}}x = 12}$$ .

Show that $$\int_5^1 {f(x){\rm{d}}x = – 4}$$ .


a.

Find the value of  $$\int_1^2 {(x + f(x)){\rm{d}}x + } \int_2^5 {(x + f(x)){\rm{d}}x}$$ .


b.

## Markscheme

evidence of factorising 3/division by 3     A1

e.g. $$\int_1^5 {3f(x){\rm{d}}x = 3\int_1^5 {f(x){\rm{d}}x} }$$ , $$\frac{{12}}{3}$$ , $$\int_1^5 {\frac{{3f(x){\rm{d}}x}}{3}}$$ (do not accept 4 as this is show that)

evidence of stating that reversing the limits changes the sign     A1

e.g. $$\int_5^1 {f(x){\rm{d}}x = } – \int_1^5 {f(x){\rm{d}}x}$$

$$\int_5^1 {f(x){\rm{d}}x = } – 4$$     AG     N0

[2 marks]

a.

evidence of correctly combining the integrals (seen anywhere)     (A1)

e.g. $$I = \int_1^2 {(x + f(x)){\rm{d}}x + } \int_2^5 {(x + f(x)){\rm{d}}x = } \int_1^5 {(x + f(x)){\rm{d}}x}$$

evidence of correctly splitting the integrals (seen anywhere)     (A1)

e.g. $$I = \int_1^5 {x{\rm{d}}x + } \int_1^5 {f(x){\rm{d}}x}$$

$$\int {x{\rm{d}}x = } \frac{{{x^2}}}{2}$$ (seen anywhere)     A1

$$\int_1^5 {x{\rm{d}}x = } \left[ {\frac{{{x^2}}}{2}} \right]_1^5 = \frac{{25}}{2} – \frac{1}{2}$$ $$\left( { = \frac{{24}}{2},12} \right)$$     A1

$$I =16$$     A1     N3

[5 marks]

b.

## Question

The graph of $$f(x) = \sqrt {16 – 4{x^2}}$$ , for $$– 2 \le x \le 2$$ , is shown below. The region enclosed by the curve of f and the x-axis is rotated $$360^\circ$$ about the x-axis.

Find the volume of the solid formed.

## Markscheme

attempt to set up integral expression     M1

e.g. $$\pi {\int {\sqrt {16 – 4{x^2}} } ^2}{\rm{d}}x$$ , $$2\pi \int_0^2 {(16 – 4{x^2})}$$ ,  $${\int {\sqrt {16 – 4{x^2}} } ^2}{\rm{d}}x$$

$$\int {16} {\rm{d}}x = 16x$$ , $$\int {4{x^2}{\rm{d}}x = } \frac{{4{x^3}}}{3}$$ (seen anywhere)     A1A1

evidence of substituting limits into the integrand     (M1)

e.g. $$\left( {32 – \frac{{32}}{3}} \right) – \left( { – 32 + \frac{{32}}{3}} \right)$$ , $$64 – \frac{{64}}{3}$$

volume $$= \frac{{128\pi }}{3}$$     A2     N3

[6 marks]

## 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 .


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 $$a,b \in \mathbb{Z}$$ .


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 .


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

Given that $$\int_0^5 {\frac{2}{{2x + 5}}} {\rm{d}}x = \ln k$$ , find the value of k .

## Markscheme

correct integration, $$2 \times \frac{1}{2}\ln (2x + 5)$$     A1A1

Note: Award A1 for $$2 \times \frac{1}{2}( = 1)$$ and A1 for $$\ln (2x + 5)$$ .

evidence of substituting limits into integrated function and subtracting     (M1)

e.g. $$\ln (2 \times 5 + 5) – \ln (2 \times 0 + 5)$$

correct substitution     A1

e.g. $$\ln 15 – \ln 5$$

correct working     (A1)

e.g. $$\ln \frac{{15}}{5},\ln 3$$

$$k = 3$$     A1     N3

[6 marks]

## Question

Find $$\int_4^{10} {(x – 4){\rm{d}}x}$$ .


a.

Part of the graph of $$f(x) = \sqrt {{x^{}} – 4}$$ , for $$x \ge 4$$ , is shown below. The shaded region R is enclosed by the graph of $$f$$ , the line $$x = 10$$ , and the x-axis. The region R is rotated $${360^ \circ }$$ about the x-axis. Find the volume of the solid formed.


b.

## Markscheme

correct integration     A1A1

e.g. $$\frac{{{x^2}}}{2} – 4x$$, $$\left[ {\frac{{{x^2}}}{2} – 4x} \right]_4^{10}$$, $$\frac{{{{(x – 4)}^2}}}{2}$$

Notes: In the first 2 examples, award A1 for each correct term.

In the third example, award A1 for $$\frac{1}{2}$$ and A1 for $${(x – 4)^2}$$.

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

e.g. $$\left( {\frac{{{{10}^2}}}{2} – 4(10)} \right) – \left( {\frac{{{4^2}}}{2} – 4(4)} \right),10 – ( – 8),\frac{1}{2}({6^2} – 0)$$

$$\int_4^{10} {(x – 4){\rm{d}}x = 18}$$     A1     N2

a.

attempt to substitute either limits or the function into volume formula     (M1)

e.g. $$\pi \int_4^{10} {{f^2}} {\rm{d}}x{\text{, }}{\int_a^b {(\sqrt {x – 4} )} ^2}{\text{, }}\pi \int_4^{10} {\sqrt {x – 4} }$$

Note: Do not penalise for missing $$\pi$$ or dx.

correct substitution (accept absence of dx and $$\pi$$)     (A1)

e.g. $$\pi {\int_4^{10} {(\sqrt {x – 4} )} ^2}{\text{, }}\pi \int_4^{10} {(x – 4){\rm{d}}x} {\text{, }}\int_4^{10} {(x – 4){\rm{d}}x}$$

volume = $$18\pi$$     A1     N2

[3 marks]

b.

## Question

Consider a function $$f(x)$$ such that $$\int_1^6 {f(x){\text{d}}x = 8}$$.

Find $$\int_1^6 {2f(x){\text{d}}x}$$.


a.

Find $$\int_1^6 {\left( {f(x) + 2} \right){\text{d}}x}$$.


b.

## Markscheme

appropriate approach     (M1)

eg     $$2\int {f(x),{\text{ }}2(8)}$$

$$\int_1^6 {2f(x){\text{d}}x = 16}$$     A1     N2

[2 marks]

a.

appropriate approach     (M1)

eg     $$\int {f(x) + \int {2,{\text{ }}8 + \int 2 } }$$

$$\int {2{\text{d}}x = 2x}$$   (seen anywhere)     (A1)

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

eg     $$2(6) – 2(1),{\text{ }}8 + 12 – 2$$

$$\int_1^6 {\left( {f(x) + 2} \right){\text{d}}x = 18}$$     A1     N3

[4 marks]

b.

## Question

Let $$f(x) = {x^2}$$.

Find $$\int_1^2 {{{\left( {f(x)} \right)}^2}{\text{d}}x}$$.


a.

The following diagram shows part of the graph of $$f$$. The shaded region $$R$$ is enclosed by the graph of $$f$$, the $$x$$-axis and the lines $$x = 1$$ and $$x = 2$$.

Find the volume of the solid formed when $$R$$ is revolved $${360^ \circ }$$ about the $$x$$-axis.


b.

## Markscheme

substituting for $${\left( {f(x)} \right)^2}$$ (may be seen in integral)     A1

eg     $${\left( {{x^2}} \right)^2}{\text{, }}{x^4}$$

correct integration, $$\int {{x^4}{\text{d}}x = \frac{1}{5}{x^5}}$$     (A1)

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

eg     $$\frac{{{2^5}}}{5} – \frac{1}{5}{\text{, }}\frac{1}{5}(1 – 4)$$

$$\int_1^2 {{{\left( {f(x)} \right)}^2}{\text{d}}x} = \frac{{31}}{5}( = 6.2)$$     A1     N2

[4 marks]

a.

attempt to substitute limits or function into formula involving $${f^2}$$     (M1)

eg     $$\int_1^2 {{{\left( {f(x)} \right)}^2}{\text{d}}x{\text{, }}\pi \int {{x^4}{\text{d}}x} }$$

$$\frac{{31}}{5}\pi {\text{ }}( = 6.2\pi )$$     A1     N2

[2 marks]

b.

## Question

Let $$\int_\pi ^a {\cos 2x{\text{d}}x} = \frac{1}{2}{\text{, where }}\pi < a < 2\pi$$. Find the value of $$a$$.

## Markscheme

correct integration (ignore absence of limits and “$$+C$$”)     (A1)

eg     $$\frac{{\sin (2x)}}{2},{\text{ }}\int_\pi ^a {\cos 2x = \left[ {\frac{1}{2}\sin (2x)} \right]_\pi ^a}$$

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

eg     $$\frac{1}{2}\sin (2a) – \frac{1}{2}\sin (2\pi ),{\text{ }}\sin (2\pi ) – \sin (2a)$$

$$\sin (2\pi ) = 0$$     (A1)

setting their result from an integrated function equal to $$\frac{1}{2}$$     M1

eg     $$\frac{1}{2}\sin 2a = \frac{1}{2},{\text{ }}\sin (2a) = 1$$

recognizing $${\sin ^{ – 1}}1 = \frac{\pi }{2}$$     (A1)

eg     $$2a = \frac{\pi }{2},{\text{ }}a = \frac{\pi }{4}$$

correct value     (A1)

eg     $$\frac{\pi }{2} + 2\pi ,{\text{ }}2a = \frac{{5\pi }}{2},{\text{ }}a = \frac{\pi }{4} + \pi$$

$$a = \frac{{5\pi }}{4}$$     A1     N3

[7 marks]

## Question

The following diagram shows the graph of $$f(x) = \frac{x}{{{x^2} + 1}}$$, for $$0 \le x \le 4$$, and the line $$x = 4$$. Let $$R$$ be the region enclosed by the graph of $$f$$ , the $$x$$-axis and the line $$x = 4$$.

Find the area of $$R$$.

## Markscheme

substitution of limits or function     (A1)

eg$$\;\;\;A = \int_0^4 {f(x),{\text{ }}\int {\frac{x}{{{x^2} + 1}}{\text{d}}x} }$$

correct integration by substitution/inspection     A2

$$\frac{1}{2}\ln ({x^2} + 1)$$

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

eg$$\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) – \ln ({0^2} + 1)} \right)$$

correct working     A1

eg$$\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) – \ln ({0^2} + 1)} \right),{\text{ }}\frac{1}{2}\left( {\ln (17) – \ln (1)} \right),{\text{ }}\frac{1}{2}\ln 17 – 0$$

$$A = \frac{1}{2}\ln (17)$$     A1     N3

Note:     Exception to FT rule. Allow full FT on incorrect integration involving a $$\ln$$ function.

[6 marks]

## 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 $$f(x) = {x^2}$$. 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 $$f'(x)$$.


a.i.

Find the gradient of $$L$$.


a.ii.

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


b.

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


c.

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


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 $$f(x) = \frac{{ax}}{{{x^2} + 1}}$$ , $$– 8 \le x \le 8$$ , $$a \in \mathbb{R}$$ .The graph of f is shown below. The region between $$x = 3$$ and $$x = 7$$ is shaded.

Show that $$f( – x) = – f(x)$$ .


a.

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


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 $$\int_4^8 {2f(x – 1){\rm{d}}x}$$ .


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

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 $$y = – 4x + 18$$ .


a.

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


b.

Find an expression for the area of R .


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 $$\pi$$ .


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 $$f(x) = 6 + 6\sin x$$ . Part of the graph of f is shown below. The shaded region is enclosed by the curve of f , the x-axis, and the y-axis.

Solve for $$0 \le x < 2\pi$$

(i)     $$6 + 6\sin x = 6$$ ;

(ii)    $$6 + 6\sin x = 0$$ .


a(i) and (ii).

Write down the exact value of the x-intercept of f , for $$0 \le x < 2\pi$$ .


b.

The area of the shaded region is k . Find the value of k , giving your answer in terms of $$\pi$$ .


c.

Let $$g(x) = 6 + 6\sin \left( {x – \frac{\pi }{2}} \right)$$ . The graph of f is transformed to the graph of g.

Give a full geometric description of this transformation.


d.

Let $$g(x) = 6 + 6\sin \left( {x – \frac{\pi }{2}} \right)$$ . The graph of f is transformed to the graph of g.

Given that $$\int_p^{p + \frac{{3\pi }}{2}} {g(x){\rm{d}}x} = k$$ and $$0 \le p < 2\pi$$ , write down the two values of p.


e.

## Markscheme

(i) $$\sin x = 0$$     A1

$$x = 0$$ , $$x = \pi$$     A1A1     N2

(ii) $$\sin x = – 1$$     A1

$$x = \frac{{3\pi }}{2}$$    A1     N1

[5 marks]

a(i) and (ii).

$$\frac{{3\pi }}{2}$$     A1     N1

[1 mark]

b.

evidence of using anti-differentiation     (M1)

e.g. $$\int_0^{\frac{{3\pi }}{2}} {(6 + 6\sin x){\rm{d}}x}$$

correct integral $$6x – 6\cos x$$ (seen anywhere)     A1A1

correct substitution     (A1)

e.g. $$6\left( {\frac{{3\pi }}{2}} \right) – 6\cos \left( {\frac{{3\pi }}{2}} \right) – ( – 6\cos 0)$$ , $$9\pi – 0 + 6$$

$$k = 9\pi + 6$$     A1A1     N3

[6 marks]

c.

translation of $$\left( {\begin{array}{*{20}{c}} {\frac{\pi }{2}}\\ 0 \end{array}} \right)$$     A1A1     N2

[2 marks]

d.

recognizing that the area under g is the same as the shaded region in f     (M1)

$$p = \frac{\pi }{2}$$ , $$p = 0$$     A1A1     N3

[3 marks]

e.

## 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 $$y = 4x – 2$$ .


a.

Find the x-intercept of T .


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 .


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 .


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 $$a,b \in \mathbb{Z}$$ .


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 .


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

The following diagram shows the graph of a quadratic function f , for $$0 \le x \le 4$$ . The graph passes through the point P(0, 13) , and its vertex is the point V(2, 1) .

The function can be written in the form $$f(x) = a{(x – h)^2} + k$$ .

(i)     Write down the value of h and of k .

(ii)    Show that $$a = 3$$ .


a(i) and (ii).

Find $$f(x)$$  , giving your answer in the form $$A{x^2} + Bx + C$$ .


b.

Calculate the area enclosed by the graph of f , the x-axis, and the lines $$x = 2$$ and $$x = 4$$ .


c.

## Markscheme

(i) $$h = 2$$ , $$k = 1$$     A1A1     N2

(ii) attempt to substitute coordinates of any point (except the vertex) on the graph into f     M1

e.g. $$13 = a{(0 – 2)^2} + 1$$

working towards solution     A1

e.g. $$13 = 4a + 1$$

$$a = 3$$     AG     N0

[4 marks]

a(i) and (ii).

attempting to expand their binomial     (M1)

e.g. $$f(x) = 3({x^2} – 2 \times 2x + 4) + 1$$ , $${(x – 2)^2} = {x^2} – 4x + 4$$

correct working     (A1)

e.g. $$f(x) = 3{x^2} – 12x + 12 + 1$$

$$f(x) = 3{x^2} – 12x + 13$$ (accept $$A = 3$$ , $$B = – 12$$ , $$C = 13$$ )     A1     N2

[3 marks]

b.

METHOD 1

integral expression     (A1)

e.g. $$\int_2^4 {(3{x^2}} – 12x + 13)$$ , $$\int {f{\rm{d}}x}$$

$${\rm{Area}} = [{x^3} – 6{x^2} + 13x]_2^4$$     A1A1A1

Note: Award A1 for $${x^3}$$ , A1 for $$– 6{x^2}$$ , A1 for $$13x$$ .

correct substitution of correct limits into their expression     A1A1

e.g. $$({4^3} – 6 \times {4^2} + 13 \times 4) – ({2^3} – 6 \times {2^2} + 13 \times 2)$$ , $$64 – 96 + 52 – (8 – 24 + 26)$$

Note: Award A1 for substituting 4, A1 for substituting 2.

correct working     (A1)

e.g. $$64 – 96 + 52 – 8 + 24 – 26,20 – 10$$

$${\rm{Area}} = 10$$     A1     N3

[8 marks]

METHOD 2

integral expression     (A1)

e.g. $$\int_2^4 {(3{{(x – 2)}^2}} + 1)$$ , $$\int {f{\rm{d}}x}$$

$${\rm{Area}} = [{(x – 2)^3} + x]_2^4$$     A2A1

Note: Award A2 for $${(x – 2)^3}$$ , A1 for $$x$$ .

correct substitution of correct limits into their expression     A1A1

e.g. $${(4 – 2)^3} + 4 – [{(2 – 2)^3} + 2]$$ , $${2^3} + 4 – ({0^3} + 2)$$ , $${2^3} + 4 – 2$$

Note: Award A1 for substituting 4, A1 for substituting 2.

correct working     (A1)

e.g. $$8 + 4 – 2$$

$${\rm{Area}} = 10$$    A1     N3

[8 marks]

METHOD 3

recognizing area from 0 to 2 is same as area from 2 to 4     (R1)

e.g. sketch, $$\int_2^4 {f = \int_0^2 f }$$

integral expression     (A1)

e.g. $$\int_0^2 {(3{x^2}} – 12x + 13)$$ , $$\int {f{\rm{d}}x}$$

$${\rm{Area}} = [{x^3} – 6{x^2} + 13x]_0^2$$     A1A1A1

Note: Award A1 for $${x^3}$$ , A1 for $$– 6{x^2}$$ , A1 for $$13x$$ .

correct substitution of correct limits into their expression     A1(A1)

e.g. $$({2^3} – 6 \times {2^2} + 13 \times 2) – ({0^3} – 6 \times {0^2} + 13 \times 0)$$ , $$8 – 24 + 26$$

Note: Award A1 for substituting 2, (A1) for substituting 0.

$${\rm{Area}} = 10$$     A1     N3

[8 marks]

c.

## 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.

## Examiners report

In part (a), most candidates recognized the need to apply the quotient rule to find the derivative, and many were successful in earning full marks here.

a.

In part (b), many candidates struggled with the chain rule, or did not realize the chain rule was necessary to find the derivative. Again, some candidates attempted to work backward from the given answer, which is not allowed in a “show that” question. A few clever candidates simplified the situation by applying properties of logarithms before finding their derivative.

b.

For part (c), many candidates recognized the need to integrate the function, and that their integral would equal $$\ln 4$$ . However, many did not recognize that the integral of h is g . Those candidates who made this link between the parts (b) and (c) often carried on correctly to find the value of k , with a few candidates having errors in working with logarithms.

c.

## Question

The following is the graph of a function $$f$$ , for $$0 \le x \le 6$$ . The first part of the graph is a quarter circle of radius $$2$$ with centre at the origin.

(a)     Find $$\int_0^2 {f(x){\rm{d}}x}$$ .

(b)     The shaded region is enclosed by the graph of $$f$$ , the $$x$$-axis, the $$y$$-axis and the line $$x = 6$$ . The area of this region is $$3\pi$$ .

Find $$\int_2^6 {f(x){\rm{d}}x}$$ .


.

Find $$\int_0^2 {f(x){\rm{d}}x}$$ .


a.

The shaded region is enclosed by the graph of $$f$$ , the $$x$$-axis, the $$y$$-axis and the line $$x = 6$$ . The area of this region is $$3\pi$$ .

Find $$\int_2^6 {f(x){\rm{d}}x}$$ .


b.

## Markscheme

(a)     attempt to find quarter circle area     (M1)

eg   $$\frac{1}{4}(4\pi )$$ , $$\frac{{\pi {r^2}}}{4}$$ , $$\int_0^2 {\sqrt {4 – {x^2}{\rm{d}}x} }$$

area of region $$= \pi$$     (A1)

$$\int_0^2 {f(x){\rm{d}}x = – \pi }$$     A2     N3

[4 marks]

(b)     attempted set up with both regions     (M1)

eg   $${\text{shaded area}} – {\text{quarter circle}}$$ , $$3\pi – \pi$$ , $$3\pi – \int_0^2 {f = \int_2^6 f }$$

$$\int_2^6 {f(x){\rm{d}}x = 2\pi }$$     A2     N2

[3 marks]

Total [7 marks]

.

attempt to find quarter circle area     (M1)

eg   $$\frac{1}{4}(4\pi )$$ , $$\frac{{\pi {r^2}}}{4}$$ , $$\int_0^2 {\sqrt {4 – {x^2}{\rm{d}}x} }$$

area of region $$= \pi$$     (A1)

$$\int_0^2 {f(x){\rm{d}}x = – \pi }$$     A2     N3

[4 marks]

a.

attempted set up with both regions     (M1)

eg   $${\text{shaded area}} – {\text{quarter circle}}$$ , $$3\pi – \pi$$ , $$3\pi – \int_0^2 {f = \int_2^6 f }$$

$$\int_2^6 {f(x){\rm{d}}x = 2\pi }$$     A2     N2

[3 marks]

Total [7 marks]

b.

## Question

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

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


a.

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


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 $$q$$.


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

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

The following diagram shows the graph of $$f(x) = \frac{x}{{{x^2} + 1}}$$, for $$0 \le x \le 4$$, and the line $$x = 4$$. Let $$R$$ be the region enclosed by the graph of $$f$$ , the $$x$$-axis and the line $$x = 4$$.

Find the area of $$R$$.

## Markscheme

substitution of limits or function     (A1)

eg$$\;\;\;A = \int_0^4 {f(x),{\text{ }}\int {\frac{x}{{{x^2} + 1}}{\text{d}}x} }$$

correct integration by substitution/inspection     A2

$$\frac{1}{2}\ln ({x^2} + 1)$$

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

eg$$\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) – \ln ({0^2} + 1)} \right)$$

correct working     A1

eg$$\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) – \ln ({0^2} + 1)} \right),{\text{ }}\frac{1}{2}\left( {\ln (17) – \ln (1)} \right),{\text{ }}\frac{1}{2}\ln 17 – 0$$

$$A = \frac{1}{2}\ln (17)$$     A1     N3

Note:     Exception to FT rule. Allow full FT on incorrect integration involving a $$\ln$$ function.

[6 marks]

## Question

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

There are $$x$$-intercepts at $$x = \frac{\pi }{2},{\text{ }}\frac{{3\pi }}{2}$$. The shaded region $$R$$ is enclosed by the graph of $$f$$, the line $$x = b$$, where $$b > \frac{{3\pi }}{2}$$, and the $$x$$-axis. The area of $$R$$ is $$\left( {1 – \frac{{\sqrt 3 }}{2}} \right)$$. Find the value of $$b$$.

## Markscheme

attempt to set up integral (accept missing or incorrect limits and missing $${\text{d}}x$$)     M1

eg$$\;\;\;\int_{\frac{{3\pi }}{2}}^b {\cos x{\text{d}}x,{\text{ }}\int_a^b {\cos x{\text{d}}x,{\text{ }}\int_{\frac{{3\pi }}{2}}^b {f{\text{d}}x,{\text{ }}\int {\cos x} } } }$$

correct integration (accept missing or incorrect limits)     (A1)

eg$$\;\;\;[\sin x]_{\frac{{3\pi }}{2}}^b,{\text{ }}\sin x$$

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

eg$$\;\;\;\sin b – \sin \left( {\frac{{3\pi }}{2}} \right),{\text{ }}\sin \left( {\frac{{3\pi }}{2}} \right) – \sin b$$

$$\sin \left( {\frac{{3\pi }}{2}} \right) = – 1\;\;\;$$(seen anywhere)     (A1)

setting their result from an integrated function equal to $$\left( {1 – \frac{{\sqrt 3 }}{2}} \right)$$     M1

eg$$\;\;\;\sin b = – \frac{{\sqrt 3 }}{2}$$

evaluating $${\sin ^{ – 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{3}$$ or $${\sin ^{ – 1}}\left( { – \frac{{\sqrt 3 }}{2}} \right) = – \frac{\pi }{3}$$     (A1)

eg$$\;\;\;b = \frac{\pi }{3},{\text{ }} – 60^\circ$$

identifying correct value     (A1)

eg$$\;\;\;2\pi – \frac{\pi }{3},{\text{ }}360 – 60$$

$$b = \frac{{5\pi }}{3}$$     A1     N3

[8 marks]

## 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

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 $$a$$.


a.

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


b.

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


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

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 $$f'(x)$$.


a.i.

Find the gradient of $$L$$.


a.ii.

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


b.

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


c.

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


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 $$f\left( x \right) = 6{x^2} – 3x$$. The graph of $$f$$ is shown in the following diagram. Find $$\int {\left( {6{x^2} – 3x} \right){\text{d}}x}$$.


a.

Find the area of the region enclosed by the graph of $$f$$, the x-axis and the lines x = 1 and x = 2 .


b.

## Markscheme

$$2{x^3} – \frac{{3{x^2}}}{2} + c\,\,\,\left( {{\text{accept}}\,\,\frac{{6{x^3}}}{3} – \frac{{3{x^2}}}{2} + c} \right)$$     A1A1 N2

Notes: Award A1A0 for both correct terms if +c is omitted.
Award A1A0 for one correct term eg $$2{x^3} + c$$.
Award A1A0 if both terms are correct, but candidate attempts further working to solve for c.

[2 marks]

a.

substitution of limits or function (A1)

eg  $$\int_1^2 {f\left( x \right)} \,{\text{d}}x,\,\,\left[ {2{x^3} – \frac{{3{x^2}}}{2}} \right]_1^2$$

substituting limits into their integrated function and subtracting     (M1)

eg  $$\frac{{6 \times {2^3}}}{3} – \frac{{3 \times {2^2}}}{2} – \left( {\frac{{6 \times {1^3}}}{3} + \frac{{3 \times {1^2}}}{2}} \right)$$

Note: Award M0 if substituted into original function.

correct working      (A1)

eg  $$\frac{{6 \times 8}}{3} – \frac{{3 \times 4}}{2} – \frac{{6 \times 1}}{3} + \frac{{3 \times 1}}{2},\,\,\left( {16 – 6} \right) – \left( {2 – \frac{3}{2}} \right)$$

$$\frac{{19}}{2}$$     A1 N3

[4 marks]

b.

## Question

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

The equation of the tangent at P is $$y = 3x – 2$$ . Let T be the region enclosed by the graph of f , the tangent [PR] and the line $$x = k$$ , between $$x = – 2$$ and $$x = k$$ where $$– 2 < k < 1$$ . This is shown in the diagram below. (i)     Show that the gradient of [PQ] is $$\frac{{{a^3}}}{{a – \frac{2}{3}}}$$ .

(ii)    Find $$f'(a)$$ .

(iii)   Hence show that $$a = 1$$ .


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

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


b.

## Markscheme

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

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

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

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

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

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

(iii) METHOD 1

evidence of approach     (M1)

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

simplify     A1

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

rearrange     A1

e.g. $$3{a^3} – 2{a^2} = {a^3}$$

evidence of solving     A1

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

$$a = 1$$     AG     N0

METHOD 2

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

simplify     A1

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

evidence of approach     (M1)

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

simplify     A1

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

$$a = 1$$     AG     N0

[7 marks]

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

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

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

correct integration with correct signs     A1A1A1

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

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

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

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

correct substitution into their integral if 2 or more terms     A1

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

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

simplifying     A1

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

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

[9 marks]

b.

## Question

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 $$f(x) = {x^2} – x$$, 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 $$f’(1) = 1$$.


a.

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


b.

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


c.

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


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.