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
The graph of \(y = \sqrt x \) between \(x = 0\) and \(x = a\) is rotated \(360^\circ \) about the x-axis. The volume of the solid formed is \(32\pi \) . Find the value of a.
Answer/Explanation
Markscheme
attempt to substitute into formula \(V = \int {\pi {y^2}{\rm{d}}x} \) (M1)
integral expression A1
e.g. \(\pi \int_0^a {(\sqrt x } {)^2}{\rm{d}}x\) , \(\pi \int x \)
correct integration (A1)
e.g. \(\int x{{\rm{d}}x = \frac{1}{2}{x^2}}\)
correct substitution \(V = \pi \left[ {\frac{1}{2}{a^2}} \right]\) (A1)
equating their expression to \(32\pi \) M1
e.g. \(\pi \left[ {\frac{1}{2}{a^2}} \right] = 32\pi \)
\({a^2} = 64\)
\(a = 8\) A2 N2
[7 marks]
Question
Let \(f(x) = \sqrt x \) . Line L is the normal to the graph of f at the point (4, 2) .
In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the line L .
Show that the equation of L is \(y = – 4x + 18\) .
Point A is the x-intercept of L . Find the x-coordinate of A.
Find an expression for the area of R .
The region R is rotated \(360^\circ \) about the x-axis. Find the volume of the solid formed, giving your answer in terms of \(\pi \) .
Answer/Explanation
Markscheme
finding derivative (A1)
e.g. \(f'(x) = \frac{1}{2}{x^{\frac{1}{2}}},\frac{{1}}{{2\sqrt x }}\)
correct value of derivative or its negative reciprocal (seen anywhere) A1
e.g. \(\frac{1}{{2\sqrt 4 }}\) , \(\frac{1}{4}\)
gradient of normal = \(\frac{1}{{{\text{gradient of tangent}}}}\) (seen anywhere) A1
e.g. \( – \frac{1}{{f'(4)}} = – 4\) , \( – 2\sqrt x \)
substituting into equation of line (for normal) M1
e.g. \(y – 2 = – 4(x – 4)\)
\(y = – 4x + 18\) AG N0
[4 marks]
recognition that \(y = 0\) at A (M1)
e.g. \( – 4x + 18 = 0\)
\(x = \frac{{18}}{4}\) \(\left( { = \frac{9}{2}} \right)\) A1 N2
[2 marks]
splitting into two appropriate parts (areas and/or integrals) (M1)
correct expression for area of R A2 N3
e.g. area of R = \(\int_0^4 {\sqrt x } {\rm{d}}x + \int_4^{4.5} {( – 4x + 18){\rm{d}}x} \) , \(\int_0^4 {\sqrt x } {\rm{d}}x + \frac{1}{2} \times 0.5 \times 2\) (triangle)
Note: Award A1 if dx is missing.
[3 marks]
correct expression for the volume from \(x = 0\) to \(x = 4\) (A1)
e.g. \(V = \int_0^4 {\pi \left[ {f{{(x)}^2}} \right]} {\rm{d}}x\) , \({\int_0^4 {\pi \sqrt x } ^2}{\rm{d}}x\) , \(\int_0^4 {\pi x{\rm{d}}x} \)
\(V = \left[ {\frac{1}{2}\pi {x^2}} \right]_0^4\) A1
\(V = \pi \left( {\frac{1}{2} \times 16 – \frac{1}{2} \times 0} \right)\) (A1)
\(V = 8\pi \) A1
finding the volume from \(x = 4\) to \(x = 4.5\)
EITHER
recognizing a cone (M1)
e.g. \(V = \frac{1}{3}\pi {r^2}h\)
\(V = \frac{1}{3}\pi {(2)^2} \times \frac{1}{2}\) (A1)
\( = \frac{{2\pi }}{3}\) A1
total volume is \(8\pi + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\) A1 N4
OR
\(V = \pi \int_4^{4.5} {{{( – 4x + 18)}^2}{\rm{d}}x} \) (M1)
\( = \int_4^{4.5} {\pi (16{x^2} – 144x + 324){\rm{d}}x} \)
\( = \pi \left[ {\frac{{16}}{3}{x^3} – 72{x^2} + 324x} \right]_4^{4.5}\) A1
\( = \frac{{2\pi }}{3}\) A1
total volume is \(8\pi + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\) A1 N4
[8 marks]
Question
The graph of \(f(x) = \sqrt {16 – 4{x^2}} \) , for \( – 2 \le x \le 2\) , is shown below.
The region enclosed by the curve of f and the x-axis is rotated \(360^\circ \) about the x-axis.
Find the volume of the solid formed.
Answer/Explanation
Markscheme
attempt to set up integral expression M1
e.g. \(\pi {\int {\sqrt {16 – 4{x^2}} } ^2}{\rm{d}}x\) , \(2\pi \int_0^2 {(16 – 4{x^2})} \) , \({\int {\sqrt {16 – 4{x^2}} } ^2}{\rm{d}}x\)
\(\int {16} {\rm{d}}x = 16x\) , \(\int {4{x^2}{\rm{d}}x = } \frac{{4{x^3}}}{3}\) (seen anywhere) A1A1
evidence of substituting limits into the integrand (M1)
e.g. \(\left( {32 – \frac{{32}}{3}} \right) – \left( { – 32 + \frac{{32}}{3}} \right)\) , \(64 – \frac{{64}}{3}\)
volume \(= \frac{{128\pi }}{3}\) A2 N3
[6 marks]
Question
Find \(\int_4^{10} {(x – 4){\rm{d}}x} \) .
Part of the graph of \(f(x) = \sqrt {{x^{}} – 4} \) , for \(x \ge 4\) , is shown below. The shaded region R is enclosed by the graph of \(f\) , the line \(x = 10\) , and the x-axis.
The region R is rotated \({360^ \circ }\) about the x-axis. Find the volume of the solid formed.
Answer/Explanation
Markscheme
correct integration A1A1
e.g. \(\frac{{{x^2}}}{2} – 4x\), \(\left[ {\frac{{{x^2}}}{2} – 4x} \right]_4^{10}\), \(\frac{{{{(x – 4)}^2}}}{2}\)
Notes: In the first 2 examples, award A1 for each correct term.
In the third example, award A1 for \(\frac{1}{2}\) and A1 for \({(x – 4)^2}\).
substituting limits into their integrated function and subtracting (in any order) (M1)
e.g. \(\left( {\frac{{{{10}^2}}}{2} – 4(10)} \right) – \left( {\frac{{{4^2}}}{2} – 4(4)} \right),10 – ( – 8),\frac{1}{2}({6^2} – 0)\)
\(\int_4^{10} {(x – 4){\rm{d}}x = 18} \) A1 N2
attempt to substitute either limits or the function into volume formula (M1)
e.g. \(\pi \int_4^{10} {{f^2}} {\rm{d}}x{\text{, }}{\int_a^b {(\sqrt {x – 4} )} ^2}{\text{, }}\pi \int_4^{10} {\sqrt {x – 4} } \)
Note: Do not penalise for missing \(\pi \) or dx.
correct substitution (accept absence of dx and \(\pi \)) (A1)
e.g. \(\pi {\int_4^{10} {(\sqrt {x – 4} )} ^2}{\text{, }}\pi \int_4^{10} {(x – 4){\rm{d}}x} {\text{, }}\int_4^{10} {(x – 4){\rm{d}}x} \)
volume = \(18\pi \) A1 N2
[3 marks]
Question
Let \(f(x) = {x^2}\).
Find \(\int_1^2 {{{\left( {f(x)} \right)}^2}{\text{d}}x} \).
The following diagram shows part of the graph of \(f\).
The shaded region \(R\) is enclosed by the graph of \(f\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\).
Find the volume of the solid formed when \(R\) is revolved \({360^ \circ }\) about the \(x\)-axis.
Answer/Explanation
Markscheme
substituting for \({\left( {f(x)} \right)^2}\) (may be seen in integral) A1
eg \({\left( {{x^2}} \right)^2}{\text{, }}{x^4}\)
correct integration, \(\int {{x^4}{\text{d}}x = \frac{1}{5}{x^5}} \) (A1)
substituting limits into their integrated function and subtracting (in any order) (M1)
eg \(\frac{{{2^5}}}{5} – \frac{1}{5}{\text{, }}\frac{1}{5}(1 – 4)\)
\(\int_1^2 {{{\left( {f(x)} \right)}^2}{\text{d}}x} = \frac{{31}}{5}( = 6.2) \) A1 N2
[4 marks]
attempt to substitute limits or function into formula involving \({f^2}\) (M1)
eg \(\int_1^2 {{{\left( {f(x)} \right)}^2}{\text{d}}x{\text{, }}\pi \int {{x^4}{\text{d}}x} } \)
\(\frac{{31}}{5}\pi {\text{ }}( = 6.2\pi )\) A1 N2
[2 marks]
Question
The following table shows the probability distribution of a discrete random variable \(A\), in terms of an angle \(\theta \).
Show that \(\cos \theta = \frac{3}{4}\).
Given that \(\tan \theta > 0\), find \(\tan \theta \).
Let \(y = \frac{1}{{\cos x}}\), for \(0 < x < \frac{\pi }{2}\). The graph of \(y\)between \(x = \theta \) and \(x = \frac{\pi }{4}\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.
Answer/Explanation
Markscheme
evidence of summing to 1 (M1)
eg\(\,\,\,\,\,\)\(\sum {p = 1} \)
correct equation A1
eg\(\,\,\,\,\,\)\(\cos \theta + 2\cos 2\theta = 1\)
correct equation in \(\cos \theta \) A1
eg\(\,\,\,\,\,\)\(\cos \theta + 2(2{\cos ^2}\theta – 1) = 1,{\text{ }}4{\cos ^2}\theta + \cos \theta – 3 = 0\)
evidence of valid approach to solve quadratic (M1)
eg\(\,\,\,\,\,\)factorizing equation set equal to \(0,{\text{ }}\frac{{ – 1 \pm \sqrt {1 – 4 \times 4 \times ( – 3)} }}{8}\)
correct working, clearly leading to required answer A1
eg\(\,\,\,\,\,\)\((4\cos \theta – 3)(\cos \theta + 1),{\text{ }}\frac{{ – 1 \pm 7}}{8}\)
correct reason for rejecting \(\cos \theta \ne – 1\) R1
eg\(\,\,\,\,\,\)\(\cos \theta \) is a probability (value must lie between 0 and 1), \(\cos \theta > 0\)
Note: Award R0 for \(\cos \theta \ne – 1\) without a reason.
\(\cos \theta = \frac{3}{4}\) AG N0
valid approach (M1)
eg\(\,\,\,\,\,\)sketch of right triangle with sides 3 and 4, \({\sin ^2}x + {\cos ^2}x = 1\)
correct working
(A1)
eg\(\,\,\,\,\,\)missing side \( = \sqrt 7 ,{\text{ }}\frac{{\frac{{\sqrt 7 }}{4}}}{{\frac{3}{4}}}\)
\(\tan \theta = \frac{{\sqrt 7 }}{3}\) A1 N2
[3 marks]
attempt to substitute either limits or the function into formula involving \({f^2}\) (M1)
eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{f^2},{\text{ }}\int {{{\left( {\frac{1}{{\cos x}}} \right)}^2}} } \)
correct substitution of both limits and function (A1)
eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{{\left( {\frac{1}{{\cos x}}} \right)}^2}{\text{d}}x} \)
correct integration (A1)
eg\(\,\,\,\,\,\)\(\tan x\)
substituting their limits into their integrated function and subtracting (M1)
eg\(\,\,\,\,\,\)\(\tan \frac{\pi }{4} – \tan \theta \)
Note: Award M0 if they substitute into original or differentiated function.
\(\tan \frac{\pi }{4} = 1\) (A1)
eg\(\,\,\,\,\,\)\(1 – \tan \theta \)
\(V = \pi – \frac{{\pi \sqrt 7 }}{3}\) A1 N3
[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]