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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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).
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

Markscheme

correct derivatives applied in quotient rule     (A1)A1A1

\(1\), \( – 4x + 5\)

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

correct substitution into quotient rule     A1

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

correct working     (A1)

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

expression clearly leading to the answer     A1

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

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

[6 marks]

a.

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

e.g. \(2{x^2} – 2 = 0\)

evidence of correct working     A1

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

correct solution to quadratic     (A1)

e.g. \(x = \pm 1\)

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

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

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

correct working

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

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

[7 marks]

b.

recognizing values between max and min     (R1)

\(\frac{1}{9} < k < 1\)     A2     N3

[3 marks]

c.

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.
Answer/Explanation

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.

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