Question
Let \(f:x \mapsto {\sin ^3}x\) .
(i) Write down the range of the function f .
(ii) Consider \(f(x) = 1\) , \(0 \le x \le 2\pi \) . Write down the number of solutions to this equation. Justify your answer.
Find \(f'(x)\) , giving your answer in the form \(a{\sin ^p}x{\cos ^q}x\) where \(a{\text{, }}p{\text{, }}q \in \mathbb{Z}\) .
Let \(g(x) = \sqrt 3 \sin x{(\cos x)^{\frac{1}{2}}}\) for \(0 \le x \le \frac{\pi }{2}\) . Find the volume generated when the curve of g is revolved through \(2\pi \) about the x-axis.
Answer/Explanation
Markscheme
(i) range of f is \([ – 1{\text{, }}1]\) , \(( – 1 \le f(x) \le 1)\) A2 N2
(ii) \({\sin ^3}x \Rightarrow 1 \Rightarrow \sin x = 1\) A1
justification for one solution on \([0{\text{, }}2\pi ]\) R1
e.g. \(x = \frac{\pi }{2}\) , unit circle, sketch of \(\sin x\)
1 solution (seen anywhere) A1 N1
[5 marks]
\(f'(x) = 3{\sin ^2}x\cos x\) A2 N2
[2 marks]
using \(V = \int_a^b {\pi {y^2}{\rm{d}}x} \) (M1)
\(V = \int_0^{\frac{\pi }{2}} {\pi (\sqrt 3 } \sin x{\cos ^{\frac{1}{2}}}x{)^2}{\rm{d}}x\) (A1)
\( = \pi \int_0^{\frac{\pi }{2}} {3{{\sin }^2}x\cos x{\rm{d}}x} \) A1
\(V = \pi \left[ {{{\sin }^3}x} \right]_0^{\frac{\pi }{2}}\) \(\left( { = \pi \left( {{{\sin }^3}\left( {\frac{\pi }{2}} \right) – {{\sin }^3}0} \right)} \right)\) A2
evidence of using \(\sin \frac{\pi }{2} = 1\) and \(\sin 0 = 0\) (A1)
e.g. \(\pi \left( {1 – 0} \right)\)
\(V = \pi \) A1 N1
[7 marks]
Question
Let \(f(x) = \frac{{2x}}{{{x^2} + 5}}\).
Use the quotient rule to show that \(f'(x) = \frac{{10 – 2{x^2}}}{{{{({x^2} + 5)}^2}}}\).
Find \(\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x} \).
The following diagram shows part of the graph of \(f\).
The shaded region is enclosed by the graph of \(f\), the \(x\)-axis, and the lines \(x = \sqrt 5 \) and \(x = q\). This region has an area of \(\ln 7\). Find the value of \(q\).
Answer/Explanation
Markscheme
derivative of \(2x\) is \(2\) (must be seen in quotient rule) (A1)
derivative of \({x^2} + 5\) is \(2x\) (must be seen in quotient rule) (A1)
correct substitution into quotient rule A1
eg \(\frac{{({x^2} + 5)(2) – (2x)(2x)}}{{{{({x^2} + 5)}^2}}},{\text{ }}\frac{{2({x^2} + 5) – 4{x^2}}}{{{{({x^2} + 5)}^2}}}\)
correct working which clearly leads to given answer A1
eg \(\frac{{2{x^2} + 10 – 4{x^2}}}{{{{({x^2} + 5)}^2}}},{\text{ }}\frac{{2{x^2} + 10 – 4{x^2}}}{{{x^4} + 10{x^2} + 25}}\)
\(f'(x) = \frac{{10 – 2{x^2}}}{{{{({x^2} + 5)}^2}}}\) AG N0
[4 marks]
valid approach using substitution or inspection (M1)
eg \(u = {x^2} + 5,{\text{ d}}u = 2x{\text{d}}x,{\text{ }}\frac{1}{2}\ln ({x^2} + 5)\)
\(\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x = \int {\frac{1}{u}{\text{d}}u} } \) (A1)
\(\int {\frac{1}{u}{\text{d}}u = \ln u + c} \) (A1)
\(\ln ({x^2} + 5) + c\) A1 N4
[4 marks]
correct expression for area (A1)
eg \(\left[ {\ln \left( {{x^2} + 5} \right)} \right]_{\sqrt 5 }^q,{\text{ }}\int\limits_{\sqrt 5 }^q {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x} \)
substituting limits into their integrated function and subtracting (in either order) (M1)
eg \(\ln ({q^2} + 5) – \ln \left( {{{\sqrt 5 }^2} + 5} \right)\)
correct working (A1)
eg \(\ln \left( {{q^2} + 5} \right) – \ln 10,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}}\)
equating their expression to \(\ln 7\) (seen anywhere) (M1)
eg \(\ln \left( {{q^2} + 5} \right) – \ln 10 = \ln 7,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}} = \ln 7,{\text{ }}\ln ({q^2} + 5) = \ln 7 + \ln 10\)
correct equation without logs (A1)
eg \(\frac{{{q^2} + 5}}{{10}} = 7,{\text{ }}{q^2} + 5 = 70\)
\({q^2} = 65\) (A1)
\(q = \sqrt {65} \) A1 N3
Note: Award A0 for \(q = \pm \sqrt {65} \).
[7 marks]
Question
The following diagram shows the graph of \(f(x) = \frac{x}{{{x^2} + 1}}\), for \(0 \le x \le 4\), and the line \(x = 4\).
Let \(R\) be the region enclosed by the graph of \(f\) , the \(x\)-axis and the line \(x = 4\).
Find the area of \(R\).
Answer/Explanation
Markscheme
substitution of limits or function (A1)
eg\(\;\;\;A = \int_0^4 {f(x),{\text{ }}\int {\frac{x}{{{x^2} + 1}}{\text{d}}x} } \)
correct integration by substitution/inspection A2
\(\frac{1}{2}\ln ({x^2} + 1)\)
substituting limits into their integrated function and subtracting (in any order) (M1)
eg\(\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) – \ln ({0^2} + 1)} \right)\)
correct working A1
eg\(\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) – \ln ({0^2} + 1)} \right),{\text{ }}\frac{1}{2}\left( {\ln (17) – \ln (1)} \right),{\text{ }}\frac{1}{2}\ln 17 – 0\)
\(A = \frac{1}{2}\ln (17)\) A1 N3
Note: Exception to FT rule. Allow full FT on incorrect integration involving a \(\ln \) function.
[6 marks]
Question
Let \(f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}\), 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
Find \(\int {x{{\text{e}}^{{x^2} – 1}}{\text{d}}x} \).
Find \(f(x)\), given that \(f’(x) = x{{\text{e}}^{{x^2} – 1}}\) and \(f( – 1) = 3\).
Answer/Explanation
Markscheme
valid approach to set up integration by substitution/inspection (M1)
eg\(\,\,\,\,\,\)\(u = {x^2} – 1,{\text{ d}}u = 2x,{\text{ }}\int {2x{{\text{e}}^{{x^2} – 1}}{\text{d}}x} \)
correct expression (A1)
eg\(\,\,\,\,\,\)\(\frac{1}{2}\int {2x{{\text{e}}^{{x^2} – 1}}{\text{d}}x,{\text{ }}\frac{1}{2}\int {{{\text{e}}^u}{\text{d}}u} } \)
\(\frac{1}{2}{{\text{e}}^{{x^2} – 1}} + c\) A2 N4
Notes: Award A1 if missing “\( + c\)”.
[4 marks]
substituting \(x = – 1\) into their answer from (a) (M1)
eg\(\,\,\,\,\,\)\(\frac{1}{2}{{\text{e}}^0},{\text{ }}\frac{1}{2}{{\text{e}}^{1 – 1}} = 3\)
correct working (A1)
eg\(\,\,\,\,\,\)\(\frac{1}{2} + c = 3,{\text{ }}c = 2.5\)
\(f(x) = \frac{1}{2}{{\text{e}}^{{x^2} – 1}} + 2.5\) A1 N2
[3 marks]
Question
Let \(f’(x) = \frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}\). Given that \(f(0) = 1\), find \(f(x)\).
Answer/Explanation
Markscheme
valid approach (M1)
eg\(\,\,\,\,\,\)\(\int {f'{\text{d}}x,{\text{ }}\int {\frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}{\text{d}}x} } \)
correct integration by substitution/inspection A2
eg\(\,\,\,\,\,\)\(f(x) = – \frac{1}{4}{({x^3} + 1)^{ – 4}} + c,{\text{ }}\frac{{ – 1}}{{4{{({x^3} + 1)}^4}}}\)
correct substitution into their integrated function (must include \(c\)) M1
eg\(\,\,\,\,\,\)\(1 = \frac{{ – 1}}{{4{{({0^3} + 1)}^4}}} + c,{\text{ }} – \frac{1}{4} + c = 1\)
Note: Award M0 if candidates substitute into \(f’\) or \(f’’\).
\(c = \frac{5}{4}\) (A1)
\(f(x) = – \frac{1}{4}{({x^3} + 1)^{ – 4}} + \frac{5}{4}{\text{ }}\left( { = \frac{{ – 1}}{{4{{({x^3} + 1)}^4}}} + \frac{5}{4},{\text{ }}\frac{{5{{({x^3} + 1)}^4} – 1}}{{4{{({x^3} + 1)}^4}}}} \right)\) A1 N4
[6 marks]
Question
Let \(f\left( x \right) = \frac{1}{{\sqrt {2x – 1} }}\), for \(x > \frac{1}{2}\).
Find \(\int {{{\left( {f\left( x \right)} \right)}^2}{\text{d}}x} \).
Part of the graph of f is shown in the following diagram.
The shaded region R is enclosed by the graph of f, the x-axis, and the lines x = 1 and x = 9 . Find the volume of the solid formed when R is revolved 360° about the x-axis.
Answer/Explanation
Markscheme
correct working (A1)
eg \(\int {\frac{1}{{2x – 1}}{\text{d}}x,\,\,\int {{{\left( {2x – 1} \right)}^{ – 1}},\,\,\frac{1}{{2x – 1}},\,\,\int {{{\left( {\frac{1}{{\sqrt u }}} \right)}^2}\frac{{{\text{d}}u}}{2}} } } \)
\({\int {\left( {f\left( x \right)} \right)} ^2}{\text{d}}x = \frac{1}{2}{\text{ln}}\left( {2x – 1} \right) + c\) A2 N3
Note: Award A1 for \(\frac{1}{2}{\text{ln}}\left( {2x – 1} \right)\).
[3 marks]
attempt to substitute either limits or the function into formula involving f 2 (accept absence of \(\pi \) / dx) (M1)
eg \(\int_1^9 {{y^2}{\text{d}}x,\,\,} \pi {\int {\left( {\frac{1}{{\sqrt {2x – 1} }}} \right)} ^2}{\text{d}}x,\,\,\left[ {\frac{1}{2}{\text{ln}}\left( {2x – 1} \right)} \right]_1^9\)
substituting limits into their integral and subtracting (in any order) (M1)
eg \(\frac{\pi }{2}\left( {{\text{ln}}\left( {17} \right) – {\text{ln}}\left( 1 \right)} \right),\,\,\pi \left( {0 – \frac{1}{2}{\text{ln}}\left( {2 \times 9 – 1} \right)} \right)\)
correct working involving calculating a log value or using log law (A1)
eg \({\text{ln}}\left( 1 \right) = 0,\,\,{\text{ln}}\left( {\frac{{17}}{1}} \right)\)
\(\frac{\pi }{2}{\text{ln}}17\,\,\,\,\left( {{\text{accept }}\pi {\text{ln}}\sqrt {17} } \right)\) A1 N3
Note: Full FT may be awarded as normal, from their incorrect answer in part (a), however, do not award the final two A marks unless they involve logarithms.
[4 marks]
Question
Find \(\int {\frac{{{{\rm{e}}^x}}}{{1 + {{\rm{e}}^x}}}} {\rm{d}}x\) .
Find \(\int {\sin 3x\cos 3x{\rm{d}}x} \) .
Answer/Explanation
Markscheme
attempt to use substitution or inspection M1
e.g. \(u = 1 + {{\rm{e}}^x}\) so \(\frac{{{\rm{d}}u}}{{{\rm{d}}x}} = {{\rm{e}}^x}\)
correct working A1
e.g. \(\int {\frac{{{\rm{d}}u}}{u}} = \ln u\)
\(\ln (1 + {{\rm{e}}^x}) + C\) A1 N3
[3 marks]
METHOD 1
attempt to use substitution or inspection M1
e.g. let \(u = \sin 3x\)
\(\frac{{{\rm{d}}u}}{{{\rm{d}}x}} = 3\cos 3x\) A1
\(\frac{1}{3}\int {u{\rm{d}}u = } \frac{1}{3} \times \frac{{{u^2}}}{2} + C\) A1
\(\int {\sin 3x\cos 3x{\rm{d}}x = } \frac{{{{\sin }^2}3x}}{6} + C\) A1 N2
METHOD 2
attempt to use substitution or inspection M1
e.g. let \(u = \cos 3x\)
\(\frac{{{\rm{d}}u}}{{{\rm{d}}x}} = – 3\sin 3x\) A1
\( – \frac{1}{3}\int {u{\rm{d}}u = } – \frac{1}{3} \times \frac{{{u^2}}}{2} + C\) A1
\(\int {\sin 3x\cos 3x{\rm{d}}x = } \frac{{{{\cos }^2}3x}}{6} + C\) A1 N2
METHOD 3
recognizing double angle M1
correct working A1
e.g. \(\frac{1}{2}\sin 6x\)
\(\int {\sin 6x{\rm{d}}x = } \frac{{ – \cos 6x}}{6} + C\) A1
\(\int {\frac{1}{2}\sin 6x{\rm{d}}x = – \frac{{\cos 6x}}{{12}}} + C\) A1 N2
[4 marks]