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 \(\theta \) .
Explain why the area of triangle OPA is the same as the area triangle OPB.
Show that \(S = 2(\pi – 2\sin \theta )\) .
Find the value of \(\theta \) when S is a local minimum, justifying that it is a minimum.
Find a value of \(\theta \) for which S has its greatest value.
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]
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]
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]
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]
S is greatest when \(4\sin \theta \) is smallest (or equivalent) (R1)
\(\theta = 0\) (or \(\pi \) ) A1 N2
[2 marks]
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.
Which of the points A, B, D corresponds to a maximum on the graph of f ?
Show that B corresponds to a point of inflexion on the graph of f .
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 A1 N1
[1 mark]
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]
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.
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.
Describe the behaviour of the graph of \(f\) for large \(|x|\) .
Write down the range of \(f\) .
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]
(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]
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]
correct inequalities, \(y \le – 2\) , \(y > 3\) , FT from (a)(i) and (c) A1A1 N2
[2 marks]
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.
Given that \(f'(x) = \frac{3}{2}{x^2} – x + p\) , show that \(p = – 4\) .
Find \(f(x)\) .
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]
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]
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]
Question
A function \(f\) has its derivative given by \(f'(x) = 3{x^2} – 2kx – 9\), where \(k\) is a constant.
Find \(f”(x)\).
The graph of \(f\) has a point of inflexion when \(x = 1\).
Show that \(k = 3\).
Find \(f'( – 2)\).
Find the equation of the tangent to the curve of \(f\) at \(( – 2,{\text{ }}1)\), giving your answer in the form \(y = ax + b\).
Given that \(f'( – 1) = 0\), explain why the graph of \(f\) has a local maximum when \(x = – 1\).
Answer/Explanation
Markscheme
\(f”(x) = 6x – 2k\) A1A1 N2
[2 marks]
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]
correct substitution into \(f'(x)\) (A1)
eg\(\;\;\;3{( – 2)^2} – 6( – 2) – 9\)
\(f'( – 2) = 15\) A1 N2
[2 marks]
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]
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]
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 \(x = 5\).
Find the set of values of \(x\) for which the graph of \(f\) is concave down.
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 \(f(6)\).
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 \(x = 6\).
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]
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]
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]
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]
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.
Find \(f(x)\), expressing your answer as a single logarithm.
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 \(a,{\text{ }}b \in \mathbb{N}\).
Answer/Explanation
Markscheme
recognizing \(f'(x) = 0\) (M1)
correct working (A1)
eg\(\,\,\,\,\,\)\(6 – 2x = 0\)
\(x = 3\) A1 N2
[3 marks]
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]
\(a = 3\) 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]
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).
The graph of f has a point of inflexion at x = p. Find p.
Find the values of x for which the graph of f is concave-down.
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 f = 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]
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)
6p + 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]
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 x ≤ −3) A1 N2
[3 marks]
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 \(f'(x)\) .
Hence
(i) show that \(q = – 2\) ;
(ii) verify that A is a minimum point.
Find the maximum value of \(f(x)\) .
The function \(f(x)\) can be written in the form \(r\cos (x – a)\) .
Write down the value of r and of a .
Answer/Explanation
Markscheme
\(f'(x) = – \sin x + \sqrt 3 \cos x\) A1A1 N2
[2 marks]
(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]
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]
\(r = 2\) , \(a = \frac{\pi }{3}\) A1A1 N2
[2 marks]