# IB Math Analysis & Approaches Question bank-Topic: SL 5.8 Local maximum and minimum points SL Paper 1

## Question

Consider $$f(x) = \frac{1}{3}{x^3} + 2{x^2} – 5x$$ . Part of the graph of f is shown below. There is a maximum point at M, and a point of inflexion at N.

Find $$f'(x)$$ .

[3]
a.

Find the x-coordinate of M.

[4]
b.

Find the x-coordinate of N.

[3]
c.

The line L is the tangent to the curve of f at $$(3{\text{, }}12)$$. Find the equation of L in the form $$y = ax + b$$ .

[4]
d.

## Markscheme

$$f'(x) = {x^2} + 4x – 5$$     A1A1A1     N3

[3 marks]

a.

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

evidence of correct working     A1

e.g. $$(x + 5)(x – 1)$$ , $$\frac{{ – 4 \pm \sqrt {16 + 20} }}{2}$$ , sketch

$$x = – 5$$, $$x = 1$$     (A1)

so $$x = – 5$$     A1     N2

[4 marks]

b.

METHOD 1

$$f”(x) = 2x + 4$$ (may be seen later)     A1

evidence of setting second derivative = 0     (M1)

e.g. $$2x + 4 = 0$$

$$x = – 2$$     A1     N2

METHOD 2

evidence of use of symmetry     (M1)

e.g. midpoint of max/min, reference to shape of cubic

correct calculation     A1

e.g. $$\frac{{ – 5 + 1}}{2}$$

$$x = – 2$$     A1     N2

[3 marks]

c.

attempting to find the value of the derivative when $$x = 3$$     (M1)

$$f'(3) = 16$$     A1

valid approach to finding the equation of a line     M1

e.g. $$y – 12 = 16(x – 3)$$ , $$12 = 16 \times 3 + b$$

$$y = 16x – 36$$     A1     N2

[4 marks]

d.

## Question

The following diagram shows a semicircle centre O, diameter [AB], with radius 2.

Let P be a point on the circumference, with $${\rm{P}}\widehat {\rm{O}}{\rm{B}} = \theta$$ radians.

Let S be the total area of the two segments shaded in the diagram below.

Find the area of the triangle OPB, in terms of $$\theta$$ .

[2]
a.

Explain why the area of triangle OPA is the same as the area triangle OPB.

[3]
b.

Show that $$S = 2(\pi – 2\sin \theta )$$ .

[3]
c.

Find the value of $$\theta$$ when S is a local minimum, justifying that it is a minimum.

[8]
d.

Find a value of $$\theta$$ for which S has its greatest value.

[2]
e.

## Markscheme

evidence of using area of a triangle     (M1)

e.g. $$A = \frac{1}{2} \times 2 \times 2 \times \sin \theta$$

$$A = 2\sin \theta$$     A1     N2

[2 marks]

a.

METHOD 1

$${\rm{P}}\widehat {\rm{O}}{\rm{A = }}\pi – \theta$$     (A1)

$${\text{area }}\Delta {\rm{OPA}} = \frac{1}{2}2 \times 2 \times \sin (\pi – \theta )$$ $$( = 2\sin (\pi – \theta ))$$     A1

since $$\sin (\pi – \theta ) = \sin \theta$$     R1

then both triangles have the same area     AG     N0

METHOD 2

triangle OPA has the same height and the same base as triangle OPB     R3

then both triangles have the same area     AG     N0

[3 marks]

b.

area semicircle $$= \frac{1}{2} \times \pi {(2)^2}$$ $$( = 2\pi )$$     A1

$${\text{area }}\Delta {\rm{APB}} = 2\sin \theta + 2\sin \theta$$ $$( = 4\sin \theta )$$     A1

$$S{\text{ = area of semicircle}} – {\text{area }}\Delta {\rm{APB}}$$ $$( = 2\pi – 4\sin \theta )$$     M1

$$S = 2(\pi – 2\sin \theta )$$     AG     N0

[3 marks]

c.

METHOD 1

attempt to differentiate     (M1)

e.g. $$\frac{{{\rm{d}}S}}{{{\rm{d}}\theta }} = – 4\cos \theta$$

setting derivative equal to 0     (M1)

correct equation     A1

e.g. $$– 4\cos \theta = 0$$ , $$\cos \theta = 0$$ , $$4\cos \theta = 0$$

$$\theta = \frac{\pi }{2}$$     A1     N3

EITHER

evidence of using second derivative     (M1)

$$S”(\theta ) = 4\sin \theta$$   A1

$$S”\left( {\frac{\pi }{2}} \right) = 4$$     A1

it is a minimum because $$S”\left( {\frac{\pi }{2}} \right) > 0$$     R1     N0

OR

evidence of using first derivative      (M1)

for $$\theta < \frac{\pi }{2},S'(\theta ) < 0$$ (may use diagram)     A1

for $$\theta > \frac{\pi }{2},S'(\theta ) > 0$$ (may use diagram)    A1

it is a minimum since the derivative goes from negative to positive     R1     N0

METHOD 2

$$2\pi – 4\sin \theta$$ is minimum when $$4\sin \theta$$ is a maximum     R3

$$4\sin \theta$$ is a maximum when $$\sin \theta = 1$$     (A2)

$$\theta = \frac{\pi }{2}$$     A3     N3

[8 marks]

d.

S is greatest when $$4\sin \theta$$ is smallest (or equivalent)     (R1)

$$\theta = 0$$ (or $$\pi$$ )     A1     N2

[2 marks]

e.

## Question

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

(i)     Write down the value of $$f'(x)$$ at C.

(ii)    Hence, show that C corresponds to a minimum on the graph of f , i.e. it has the same x-coordinate.

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

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

[1]
b.

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

[3]
c.

## Markscheme

(i) $$f'(x) = 0$$     A1     N1

(ii) METHOD 1

$$f'(x) < 0$$ to the left of C, $$f'(x) > 0$$ to the right of C     R1R1     N2

METHOD 2

$$f”(x) > 0$$     R2     N2

[3 marks]

a(i) and (ii).

A     A1     N1

[1 mark]

b.

METHOD 1

$$f”(x) = 0$$     R2

discussion of sign change of $$f”(x)$$     R1

e.g. $$f”(x) < 0$$ to the left of B and $$f”(x) > 0$$ to the right of B; $$f”(x)$$ changes sign either side of B

B is a point of inflexion     AG     N0

METHOD 2

B is a minimum on the graph of the derivative $${f’}$$     R2

discussion of sign change of $$f”(x)$$     R1

e.g. $$f”(x) < 0$$ to the left of B and $$f”(x) > 0$$ to the right of B; $$f”(x)$$ changes sign either side of B

B is a point of inflexion     AG     N0

[3 marks]

c.

## Question

Consider $$f(x) = {x^2} + \frac{p}{x}$$ , $$x \ne 0$$ , where p is a constant.

Find $$f'(x)$$ .

[2]
a.

There is a minimum value of $$f(x)$$ when $$x = – 2$$ . Find the value of $$p$$ .

[4]
b.

## Markscheme

$$f'(x) = 2x – \frac{p}{{{x^2}}}$$     A1A1     N2

Note: Award A1 for $$2x$$ , A1 for $$– \frac{p}{{{x^2}}}$$ .

[2 marks]

a.

evidence of equating derivative to 0 (seen anywhere)     (M1)

evidence of finding $$f'( – 2)$$ (seen anywhere)     (M1)

correct equation     A1

e.g. $$– 4 – \frac{p}{4} = 0$$ , $$– 16 – p = 0$$

$$p = – 16$$     A1     N3

[4 marks]

b.

## Question

Let $$f(x) = \frac{1}{2}{x^3} – {x^2} – 3x$$ . Part of the graph of f is shown below.

There is a maximum point at A and a minimum point at B(3, − 9) .

Find the coordinates of A.

[8]
a.

Write down the coordinates of

(i)     the image of B after reflection in the y-axis;

(ii)    the image of B after translation by the vector $$\left( {\begin{array}{*{20}{c}} { – 2}\\ 5 \end{array}} \right)$$ ;

(iii)   the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor $$\frac{1}{2}$$ .

[6]
b(i), (ii) and (iii).

## Markscheme

$$f(x) = {x^2} – 2x – 3$$     A1A1A1

evidence of solving $$f'(x) = 0$$     (M1)

e.g. $${x^2} – 2x – 3 = 0$$

evidence of correct working     A1

e.g. $$(x + 1)(x – 3)$$ ,  $$\frac{{2 \pm \sqrt {16} }}{2}$$

$$x = – 1$$ (ignore $$x = 3$$ )     (A1)

evidence of substituting their negative x-value into $$f(x)$$     (M1)

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

$$y = \frac{5}{3}$$     A1

coordinates are $$\left( { – 1,\frac{5}{3}} \right)$$     N3

[8 marks]

a.

(i) $$( – 3{\text{, }} – 9)$$     A1     N1

(ii) $$(1{\text{, }} – 4)$$     A1A1    N2

(iii) reflection gives $$(3{\text{, }}9)$$     (A1)

stretch gives $$\left( {\frac{3}{2}{\text{, }}9} \right)$$     A1A1     N3

[6 marks]

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

## Question

Consider the function f with second derivative $$f”(x) = 3x – 1$$ . The graph of f has a minimum point at A(2, 4) and a maximum point at $${\rm{B}}\left( { – \frac{4}{3},\frac{{358}}{{27}}} \right)$$ .

Use the second derivative to justify that B is a maximum.

[3]
a.

Given that $$f'(x) = \frac{3}{2}{x^2} – x + p$$ , show that $$p = – 4$$ .

[4]
b.

Find $$f(x)$$ .

[7]
c.

## Markscheme

substituting into the second derivative     M1

e.g. $$3 \times \left( { – \frac{4}{3}} \right) – 1$$

$$f”\left( { – \frac{4}{3}} \right) = – 5$$     A1

since the second derivative is negative, B is a maximum     R1     N0

[3 marks]

a.

setting $$f'(x)$$ equal to zero     (M1)

evidence of substituting $$x = 2$$ (or $$x = – \frac{4}{3}$$ )     (M1)

e.g. $$f'(2)$$

correct substitution     A1

e.g. $$\frac{3}{2}{(2)^2} – 2 + p$$ , $$\frac{3}{2}{\left( { – \frac{4}{3}} \right)^2} – \left( { – \frac{4}{3}} \right) + p$$

correct simplification

e.g. $$6 – 2 + p = 0$$ , $$\frac{8}{3} + \frac{4}{3} + p = 0$$ , $$4 + p = 0$$     A1

$$p = – 4$$     AG     N0

[4 marks]

b.

evidence of integration     (M1)

$$f(x) = \frac{1}{2}{x^3} – \frac{1}{2}{x^2} – 4x + c$$     A1A1A1

substituting (2, 4) or $$\left( { – \frac{4}{3},\frac{{358}}{{27}}} \right)$$ into their expression     (M1)

correct equation     A1

e.g. $$\frac{1}{2} \times {2^3} – \frac{1}{2} \times {2^2} – 4 \times 2 + c = 4$$ , $$\frac{1}{2} \times 8 – \frac{1}{2} \times 4 – 4 \times 2 + c = 4$$ , $$4 – 2 – 8 + c = 4$$

$$f(x) = \frac{1}{2}{x^3} – \frac{1}{2}{x^2} – 4x + 10$$     A1     N4

[7 marks]

c.

## 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  $$f(x) = \frac{x}{{ – 2{x^2} + 5x – 2}}$$ 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. $$\frac{{ – 2{x^2} + 5x – 2 – ( – 4{x^2} + 5x)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}$$

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

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.

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

[2]
a.

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

[3]
b.

Write down the value of $$p$$.

[2]
c.

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

Find the value of $$q$$.

[3]
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 $$\frac{1}{2}$$.

[5]
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

The following diagram shows the graph of a function $$f$$. There is a local minimum point at $$A$$, where $$x > 0$$.

The derivative of $$f$$ is given by $$f'(x) = 3{x^2} – 8x – 3$$.

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

[5]
a.

The $$y$$-intercept of the graph is at ($$0,6$$). Find an expression for $$f(x)$$.

The graph of a function $$g$$ is obtained by reflecting the graph of $$f$$ in the $$y$$-axis, followed by a translation of $$\left({\begin{array}{*{20}{c}}m\\n\end{array}}\right)$$.

[6]
b.

Find the $$x$$-coordinate of the local minimum point on the graph of $$g$$.

[3]
c.

## Markscheme

recognizing that the local minimum occurs when $$f'(x) = 0$$     (M1)

valid attempt to solve $$3{x^2} – 8x – 3 = 0$$     (M1)

eg$$\;\;\;$$factorization, formula

correct working     A1

$$(3x + 1)(x – 3),{\text{ }}x = \frac{{8 \pm \sqrt {64 + 36} }}{6}$$

$$x = 3$$     A2     N3

Note:     Award A1 if both values $$x = \frac{{ – 1}}{3},{\text{ }}x = 3$$ are given.

[5 marks]

a.

valid approach     (M1)

$$f(x) = \int {f'(x){\text{d}}x}$$

$$f(x) = {x^3} – 4{x^2} – 3x + c\;\;\;$$(do not penalize for missing “$$+ c$$”)     A1A1A1

$$c = 6$$     (A1)

$$f(x) = {x^3} – 4{x^2} – 3x + 6$$     A1     N6

[6 marks]

b.

applying reflection     (A1)

eg$$\;\;\;f( – x)$$

recognizing that the minimum is the image of $$A$$     (M1)

eg$$\;\;\;x = – 3$$

correct expression for $$x$$     A1     N3

eg$$\;\;\; – 3 + m,{\text{ }}\left( {\begin{array}{*{20}{c}} { – 3 + m} \\ { – 12 + n} \end{array}} \right),{\text{ }}(m – 3,{\text{ }}n – 12)$$

[3 marks]

Total [14 marks]

c.

## Question

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

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

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

[2]
a.

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

[2]
b.

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

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

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

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

[5]
c.

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

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

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

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

[6]
d.

## Markscheme

METHOD 1

$$f'(5) = 0$$     (A1)

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

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

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

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

METHOD 2

$$f'(5) = 0$$     A1

valid reasoning referring to second derivative     R1

eg$$\;\;\;f”(5) > 0$$

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

[2 marks]

a.

attempt to find relevant interval     (M1)

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

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

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

[2 marks]

b.

METHOD 1 (one integral)

correct application of Fundamental Theorem of Calculus     (A1)

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

attempt to link definite integral with areas     (M1)

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

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

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

correct working     A1

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

$$f(6) = 2$$     A1     N3

METHOD 2 (more than one integral)

correct application of Fundamental Theorem of Calculus     (A1)

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

attempt to link definite integrals with areas     (M1)

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

correct values for integrals     (A1)

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

one correct intermediate value     A1

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

$$f(6) = 2$$     A1     N3

[5 marks]

c.

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

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

choosing chain rule or product rule     (M1)

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

correct derivative     (A1)

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

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

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

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

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

correct equation in any form     A1     N2

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

[6 marks]

[Total 15 marks]

d.

## Question

Let $$f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}$$, for $$0 < x < 6$$.

The graph of $$f$$ has a maximum point at P.

The $$y$$-coordinate of P is $$\ln 27$$.

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

[3]
a.

Find $$f(x)$$, expressing your answer as a single logarithm.

[8]
b.

The graph of $$f$$ is transformed by a vertical stretch with scale factor $$\frac{1}{{\ln 3}}$$. The image of P under this transformation has coordinates $$(a,{\text{ }}b)$$.

Find the value of $$a$$ and of $$b$$, where $$a,{\text{ }}b \in \mathbb{N}$$.

[[N/A]]
c.

## Markscheme

recognizing $$f'(x) = 0$$     (M1)

correct working     (A1)

eg$$\,\,\,\,\,$$$$6 – 2x = 0$$

$$x = 3$$    A1     N2

[3 marks]

a.

evidence of integration     (M1)

eg$$\,\,\,\,\,$$$$\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} }$$

using substitution     (A1)

eg$$\,\,\,\,\,$$$$\int {\frac{1}{u}{\text{d}}u}$$ where $$u = 6x – {x^2}$$

correct integral     A1

eg$$\,\,\,\,\,$$$$\ln (u) + c,{\text{ }}\ln (6x – {x^2})$$

substituting $$(3,{\text{ }}\ln 27)$$ into their integrated expression (must have $$c$$)     (M1)

eg$$\,\,\,\,\,$$$$\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$c = \ln 27 – \ln 9$$

EITHER

$$c = \ln 3$$    (A1)

attempt to substitute their value of $$c$$ into $$f(x)$$     (M1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln 3$$     A1     N4

OR

attempt to substitute their value of $$c$$ into $$f(x)$$     (M1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9$$

correct use of a log law     (A1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9$$

$$f(x) = \ln \left( {3(6x – {x^2})} \right)$$    A1     N4

[8 marks]

b.

$$a = 3$$    A1     N1

correct working     A1

eg$$\,\,\,\,\,$$$$\frac{{\ln 27}}{{\ln 3}}$$

correct use of log law     (A1)

eg$$\,\,\,\,\,$$$$\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27$$

$$b = 3$$    A1     N2

[4 marks]

c.

## Question

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

Find f (x).

[6]
a.

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

[4]
b.

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

[3]
c.

## Markscheme

evidence of integration       (M1)

eg  $$\int {f’\left( x \right)}$$

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

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

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

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

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

correct working       (A1)

eg  $$8 + C = 0,\,\,\,C = – 8$$

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

[6 marks]

a.

METHOD 1 (using 2nd derivative)

recognizing that f” = 0 (seen anywhere)      M1

correct expression for f”      (A1)

eg   6x + 18, 6p + 18

correct working      (A1)

6+ 18 = 0

p = −3       A1 N3

METHOD 1 (using 1st derivative)

recognizing the vertex of f′ is needed       (M2)

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

correct substitution      (A1)

eg   $$\frac{{ – 18}}{{2 \times 3}}$$

p = −3       A1 N3

[4 marks]

b.

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

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

correct working       (A1)

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

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

[3 marks]

c.

## Question

The following diagram shows a semicircle centre O, diameter [AB], with radius 2.

Let P be a point on the circumference, with $${\rm{P}}\widehat {\rm{O}}{\rm{B}} = \theta$$ radians.

Let S be the total area of the two segments shaded in the diagram below.

Find the area of the triangle OPB, in terms of $$\theta$$ .

[2]
a.

Explain why the area of triangle OPA is the same as the area triangle OPB.

[3]
b.

Show that $$S = 2(\pi – 2\sin \theta )$$ .

[3]
c.

Find the value of $$\theta$$ when S is a local minimum, justifying that it is a minimum.

[8]
d.

Find a value of $$\theta$$ for which S has its greatest value.

[2]
e.

## Markscheme

evidence of using area of a triangle     (M1)

e.g. $$A = \frac{1}{2} \times 2 \times 2 \times \sin \theta$$

$$A = 2\sin \theta$$     A1     N2

[2 marks]

a.

METHOD 1

$${\rm{P}}\widehat {\rm{O}}{\rm{A = }}\pi – \theta$$     (A1)

$${\text{area }}\Delta {\rm{OPA}} = \frac{1}{2}2 \times 2 \times \sin (\pi – \theta )$$ $$( = 2\sin (\pi – \theta ))$$     A1

since $$\sin (\pi – \theta ) = \sin \theta$$     R1

then both triangles have the same area     AG     N0

METHOD 2

triangle OPA has the same height and the same base as triangle OPB     R3

then both triangles have the same area     AG     N0

[3 marks]

b.

area semicircle $$= \frac{1}{2} \times \pi {(2)^2}$$ $$( = 2\pi )$$     A1

$${\text{area }}\Delta {\rm{APB}} = 2\sin \theta + 2\sin \theta$$ $$( = 4\sin \theta )$$     A1

$$S{\text{ = area of semicircle}} – {\text{area }}\Delta {\rm{APB}}$$ $$( = 2\pi – 4\sin \theta )$$     M1

$$S = 2(\pi – 2\sin \theta )$$     AG     N0

[3 marks]

c.

METHOD 1

attempt to differentiate     (M1)

e.g. $$\frac{{{\rm{d}}S}}{{{\rm{d}}\theta }} = – 4\cos \theta$$

setting derivative equal to 0     (M1)

correct equation     A1

e.g. $$– 4\cos \theta = 0$$ , $$\cos \theta = 0$$ , $$4\cos \theta = 0$$

$$\theta = \frac{\pi }{2}$$     A1     N3

EITHER

evidence of using second derivative     (M1)

$$S”(\theta ) = 4\sin \theta$$   A1

$$S”\left( {\frac{\pi }{2}} \right) = 4$$     A1

it is a minimum because $$S”\left( {\frac{\pi }{2}} \right) > 0$$     R1     N0

OR

evidence of using first derivative      (M1)

for $$\theta < \frac{\pi }{2},S'(\theta ) < 0$$ (may use diagram)     A1

for $$\theta > \frac{\pi }{2},S'(\theta ) > 0$$ (may use diagram)    A1

it is a minimum since the derivative goes from negative to positive     R1     N0

METHOD 2

$$2\pi – 4\sin \theta$$ is minimum when $$4\sin \theta$$ is a maximum     R3

$$4\sin \theta$$ is a maximum when $$\sin \theta = 1$$     (A2)

$$\theta = \frac{\pi }{2}$$     A3     N3

[8 marks]

d.

S is greatest when $$4\sin \theta$$ is smallest (or equivalent)     (R1)

$$\theta = 0$$ (or $$\pi$$ )     A1     N2

[2 marks]

e.

## Question

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

(i)     Write down the value of $$f'(x)$$ at C.

(ii)    Hence, show that C corresponds to a minimum on the graph of f , i.e. it has the same x-coordinate.

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

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

[1]
b.

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

[3]
c.

## Markscheme

(i) $$f'(x) = 0$$     A1     N1

(ii) METHOD 1

$$f'(x) < 0$$ to the left of C, $$f'(x) > 0$$ to the right of C     R1R1     N2

METHOD 2

$$f”(x) > 0$$     R2     N2

[3 marks]

a(i) and (ii).

A     A1     N1

[1 mark]

b.

METHOD 1

$$f”(x) = 0$$     R2

discussion of sign change of $$f”(x)$$     R1

e.g. $$f”(x) < 0$$ to the left of B and $$f”(x) > 0$$ to the right of B; $$f”(x)$$ changes sign either side of B

B is a point of inflexion     AG     N0

METHOD 2

B is a minimum on the graph of the derivative $${f’}$$     R2

discussion of sign change of $$f”(x)$$     R1

e.g. $$f”(x) < 0$$ to the left of B and $$f”(x) > 0$$ to the right of B; $$f”(x)$$ changes sign either side of B

B is a point of inflexion     AG     N0

[3 marks]

c.

## Question

Let $$f(x) = 3 + \frac{{20}}{{{x^2} – 4}}$$ , for $$x \ne \pm 2$$ . The graph of f is given below.

The y-intercept is at the point A.

(i)     Find the coordinates of A.

(ii)    Show that $$f'(x) = 0$$ at A.

[7]
a.

The second derivative $$f”(x) = \frac{{40(3{x^2} + 4)}}{{{{({x^2} – 4)}^3}}}$$ . Use this to

(i)     justify that the graph of f has a local maximum at A;

(ii)    explain why the graph of f does not have a point of inflexion.

[6]
b.

Describe the behaviour of the graph of $$f$$ for large $$|x|$$ .

[1]
c.

Write down the range of $$f$$ .

[2]
d.

## Markscheme

(i) coordinates of A are $$(0{\text{, }} – 2)$$     A1A1     N2

(ii) derivative of $${x^2} – 4 = 2x$$ (seen anywhere)     (A1)

evidence of correct approach     (M1)

e.g. quotient rule, chain rule

finding $$f'(x)$$     A2

e.g. $$f'(x) = 20 \times ( – 1) \times {({x^2} – 4)^{ – 2}} \times (2x)$$ , $$\frac{{({x^2} – 4)(0) – (20)(2x)}}{{{{({x^2} – 4)}^2}}}$$

substituting $$x = 0$$ into $$f'(x)$$ (do not accept solving $$f'(x) = 0$$ )     M1

at A $$f'(x) = 0$$     AG     N0

[7 marks]

a.

(i) reference to $$f'(x) = 0$$ (seen anywhere)     (R1)

reference to $$f”(0)$$ is negative (seen anywhere)     R1

evidence of substituting $$x = 0$$ into $$f”(x)$$     M1

finding $$f”(0) = \frac{{40 \times 4}}{{{{( – 4)}^3}}}$$ $$\left( { = – \frac{5}{2}} \right)$$     A1

then the graph must have a local maximum     AG

(ii) reference to $$f”(x) = 0$$ at point of inflexion     (R1)

recognizing that the second derivative is never 0     A1     N2

e.g. $$40(3{x^2} + 4) \ne 0$$ , $$3{x^2} + 4 \ne 0$$ , $${x^2} \ne – \frac{4}{3}$$ , the numerator is always positive

Note: Do not accept the use of the first derivative in part (b).

[6 marks]

b.

correct (informal) statement, including reference to approaching $$y = 3$$     A1     N1

e.g. getting closer to the line $$y = 3$$ , horizontal asymptote at $$y = 3$$

[1 mark]

c.

correct inequalities, $$y \le – 2$$ , $$y > 3$$ , FT from (a)(i) and (c)     A1A1     N2

[2 marks]

d.

## Question

Consider the function f with second derivative $$f”(x) = 3x – 1$$ . The graph of f has a minimum point at A(2, 4) and a maximum point at $${\rm{B}}\left( { – \frac{4}{3},\frac{{358}}{{27}}} \right)$$ .

Use the second derivative to justify that B is a maximum.

[3]
a.

Given that $$f'(x) = \frac{3}{2}{x^2} – x + p$$ , show that $$p = – 4$$ .

[4]
b.

Find $$f(x)$$ .

[7]
c.

## Markscheme

substituting into the second derivative     M1

e.g. $$3 \times \left( { – \frac{4}{3}} \right) – 1$$

$$f”\left( { – \frac{4}{3}} \right) = – 5$$     A1

since the second derivative is negative, B is a maximum     R1     N0

[3 marks]

a.

setting $$f'(x)$$ equal to zero     (M1)

evidence of substituting $$x = 2$$ (or $$x = – \frac{4}{3}$$ )     (M1)

e.g. $$f'(2)$$

correct substitution     A1

e.g. $$\frac{3}{2}{(2)^2} – 2 + p$$ , $$\frac{3}{2}{\left( { – \frac{4}{3}} \right)^2} – \left( { – \frac{4}{3}} \right) + p$$

correct simplification

e.g. $$6 – 2 + p = 0$$ , $$\frac{8}{3} + \frac{4}{3} + p = 0$$ , $$4 + p = 0$$     A1

$$p = – 4$$     AG     N0

[4 marks]

b.

evidence of integration     (M1)

$$f(x) = \frac{1}{2}{x^3} – \frac{1}{2}{x^2} – 4x + c$$     A1A1A1

substituting (2, 4) or $$\left( { – \frac{4}{3},\frac{{358}}{{27}}} \right)$$ into their expression     (M1)

correct equation     A1

e.g. $$\frac{1}{2} \times {2^3} – \frac{1}{2} \times {2^2} – 4 \times 2 + c = 4$$ , $$\frac{1}{2} \times 8 – \frac{1}{2} \times 4 – 4 \times 2 + c = 4$$ , $$4 – 2 – 8 + c = 4$$

$$f(x) = \frac{1}{2}{x^3} – \frac{1}{2}{x^2} – 4x + 10$$     A1     N4

[7 marks]

c.

## Question

A function $$f$$ has its derivative given by $$f'(x) = 3{x^2} – 2kx – 9$$, where $$k$$ is a constant.

Find $$f”(x)$$.

[2]
a.

The graph of $$f$$ has a point of inflexion when $$x = 1$$.

Show that $$k = 3$$.

[3]
b.

Find $$f'( – 2)$$.

[2]
c.

Find the equation of the tangent to the curve of $$f$$ at $$( – 2,{\text{ }}1)$$, giving your answer in the form $$y = ax + b$$.

[4]
d.

Given that $$f'( – 1) = 0$$, explain why the graph of $$f$$ has a local maximum when $$x = – 1$$.

[3]
e.

## Markscheme

$$f”(x) = 6x – 2k$$     A1A1     N2

[2 marks]

a.

substituting $$x = 1$$ into $$f”$$     (M1)

eg$$\;\;\;f”(1),{\text{ }}6(1) – 2k$$

recognizing $$f”(x) = 0\;\;\;$$(seen anywhere)     M1

correct equation     A1

eg$$\;\;\;6 – 2k = 0$$

$$k = 3$$     AG     N0

[3 marks]

b.

correct substitution into $$f'(x)$$     (A1)

eg$$\;\;\;3{( – 2)^2} – 6( – 2) – 9$$

$$f'( – 2) = 15$$     A1     N2

[2 marks]

c.

recognizing gradient value (may be seen in equation)     M1

eg$$\;\;\;a = 15,{\text{ }}y = 15x + b$$

attempt to substitute $$( – 2,{\text{ }}1)$$ into equation of a straight line     M1

eg$$\;\;\;1 = 15( – 2) + b,{\text{ }}(y – 1) = m(x + 2),{\text{ }}(y + 2) = 15(x – 1)$$

correct working     (A1)

eg$$\;\;\;31 = b,{\text{ }}y = 15x + 30 + 1$$

$$y = 15x + 31$$     A1     N2

[4 marks]

d.

METHOD 1 ($${{\text{2}}^{{\text{nd}}}}$$ derivative)

recognizing $$f” < 0\;\;\;$$(seen anywhere)     R1

substituting $$x = – 1$$ into $$f”$$     (M1)

eg$$\;\;\;f”( – 1),{\text{ }}6( – 1) – 6$$

$$f”( – 1) = – 12$$     A1

therefore the graph of $$f$$ has a local maximum when $$x = – 1$$     AG     N0

METHOD 2 ($${{\text{1}}^{{\text{st}}}}$$ derivative)

recognizing change of sign of $$f'(x)\;\;\;$$(seen anywhere)     R1

eg$$\;\;\;$$sign chart$$\;\;\;$$

correct value of $$f’$$ for $$– 1 < x < 3$$     A1

eg$$\;\;\;f'(0) = – 9$$

correct value of $$f’$$ for $$x$$ value to the left of $$– 1$$     A1

eg$$\;\;\;f'( – 2) = 15$$

therefore the graph of $$f$$ has a local maximum when $$x = – 1$$     AG     N0

[3 marks]

Total [14 marks]

e.

## Question

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

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

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

[2]
a.

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

[2]
b.

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

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

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

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

[5]
c.

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

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

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

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

[6]
d.

## Markscheme

METHOD 1

$$f'(5) = 0$$     (A1)

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

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

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

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

METHOD 2

$$f'(5) = 0$$     A1

valid reasoning referring to second derivative     R1

eg$$\;\;\;f”(5) > 0$$

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

[2 marks]

a.

attempt to find relevant interval     (M1)

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

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

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

[2 marks]

b.

METHOD 1 (one integral)

correct application of Fundamental Theorem of Calculus     (A1)

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

attempt to link definite integral with areas     (M1)

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

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

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

correct working     A1

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

$$f(6) = 2$$     A1     N3

METHOD 2 (more than one integral)

correct application of Fundamental Theorem of Calculus     (A1)

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

attempt to link definite integrals with areas     (M1)

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

correct values for integrals     (A1)

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

one correct intermediate value     A1

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

$$f(6) = 2$$     A1     N3

[5 marks]

c.

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

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

choosing chain rule or product rule     (M1)

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

correct derivative     (A1)

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

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

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

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

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

correct equation in any form     A1     N2

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

[6 marks]

[Total 15 marks]

d.

## Question

Let $$f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}$$, for $$0 < x < 6$$.

The graph of $$f$$ has a maximum point at P.

The $$y$$-coordinate of P is $$\ln 27$$.

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

[3]
a.

Find $$f(x)$$, expressing your answer as a single logarithm.

[8]
b.

The graph of $$f$$ is transformed by a vertical stretch with scale factor $$\frac{1}{{\ln 3}}$$. The image of P under this transformation has coordinates $$(a,{\text{ }}b)$$.

Find the value of $$a$$ and of $$b$$, where $$a,{\text{ }}b \in \mathbb{N}$$.

[[N/A]]
c.

## Markscheme

recognizing $$f'(x) = 0$$     (M1)

correct working     (A1)

eg$$\,\,\,\,\,$$$$6 – 2x = 0$$

$$x = 3$$    A1     N2

[3 marks]

a.

evidence of integration     (M1)

eg$$\,\,\,\,\,$$$$\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} }$$

using substitution     (A1)

eg$$\,\,\,\,\,$$$$\int {\frac{1}{u}{\text{d}}u}$$ where $$u = 6x – {x^2}$$

correct integral     A1

eg$$\,\,\,\,\,$$$$\ln (u) + c,{\text{ }}\ln (6x – {x^2})$$

substituting $$(3,{\text{ }}\ln 27)$$ into their integrated expression (must have $$c$$)     (M1)

eg$$\,\,\,\,\,$$$$\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27$$

correct working     (A1)

eg$$\,\,\,\,\,$$$$c = \ln 27 – \ln 9$$

EITHER

$$c = \ln 3$$    (A1)

attempt to substitute their value of $$c$$ into $$f(x)$$     (M1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln 3$$     A1     N4

OR

attempt to substitute their value of $$c$$ into $$f(x)$$     (M1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9$$

correct use of a log law     (A1)

eg$$\,\,\,\,\,$$$$f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9$$

$$f(x) = \ln \left( {3(6x – {x^2})} \right)$$    A1     N4

[8 marks]

b.

$$a = 3$$    A1     N1

correct working     A1

eg$$\,\,\,\,\,$$$$\frac{{\ln 27}}{{\ln 3}}$$

correct use of log law     (A1)

eg$$\,\,\,\,\,$$$$\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27$$

$$b = 3$$    A1     N2

[4 marks]

c.

## Question

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

Find f (x).

[6]
a.

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

[4]
b.

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

[3]
c.

## Markscheme

evidence of integration       (M1)

eg  $$\int {f’\left( x \right)}$$

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

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

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

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

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

correct working       (A1)

eg  $$8 + C = 0,\,\,\,C = – 8$$

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

[6 marks]

a.

METHOD 1 (using 2nd derivative)

recognizing that f” = 0 (seen anywhere)      M1

correct expression for f”      (A1)

eg   6x + 18, 6p + 18

correct working      (A1)

6+ 18 = 0

p = −3       A1 N3

METHOD 1 (using 1st derivative)

recognizing the vertex of f′ is needed       (M2)

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

correct substitution      (A1)

eg   $$\frac{{ – 18}}{{2 \times 3}}$$

p = −3       A1 N3

[4 marks]

b.

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

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

correct working       (A1)

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

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

[3 marks]

c.

## Question

Let $$f(x) = \cos x + \sqrt 3 \sin x$$ , $$0 \le x \le 2\pi$$ . The following diagram shows the graph of $$f$$ .

The $$y$$-intercept is at ($$0$$, $$1$$) , there is a minimum point at A ($$p$$, $$q$$) and a maximum point at B.

Find $$f'(x)$$ .

[2]
a.

Hence

(i)     show that $$q = – 2$$ ;

(ii)    verify that A is a minimum point.

[10]
b(i) and (ii).

Find the maximum value of $$f(x)$$ .

[3]
c.

The function $$f(x)$$ can be written in the form $$r\cos (x – a)$$ .

Write down the value of r and of a .

[2]
d.

## Markscheme

$$f'(x) = – \sin x + \sqrt 3 \cos x$$     A1A1     N2

[2 marks]

a.

(i) at A, $$f'(x) = 0$$     R1

correct working     A1

e.g. $$\sin x = \sqrt 3 \cos x$$

$$\tan x = \sqrt 3$$     A1

$$x = \frac{\pi }{3}$$ , $$\frac{{4\pi }}{3}$$     A1

attempt to substitute their x into $$f(x)$$     M1

e.g. $$\cos \left( {\frac{{4\pi }}{3}} \right) + \sqrt 3 \sin \left( {\frac{{4\pi }}{3}} \right)$$

correct substitution     A1

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

correct working that clearly leads to $$– 2$$     A1

e.g. $$– \frac{1}{2} – \frac{3}{2}$$

 $$q = – 2$$     AG     N0

(ii) correct calculations to find $$f'(x)$$ either side of $$x = \frac{{4\pi }}{3}$$     A1A1

e.g. $$f'(\pi ) = 0 – \sqrt 3$$ ,  $$f'(2\pi ) = 0 + \sqrt 3$$   

$$f'(x)$$ changes sign from negative to positive     R1

so A is a minimum     AG     N0

[10 marks]

b(i) and (ii).

max when $$x = \frac{\pi }{3}$$     R1

correctly substituting $$x = \frac{\pi }{3}$$ into $$f(x)$$     A1

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

max value is 2     A1     N1

[3 marks]

c.

$$r = 2$$ , $$a = \frac{\pi }{3}$$     A1A1     N2

[2 marks]

d.