IBDP Maths analysis and approaches Topic :SL 5.5 Area of the region enclosed by a curve and x axis HL Paper 1

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Question

Consider the function f defined by f (x) = 6 + 6 cos x , for 0 ≤ x ≤ 4π .
The following diagram shows the graph of y = f (x) .

The graph of f touches the x-axis at points A and B, as shown. The shaded region is enclosed by the graph of y = f (x) and the x-axis, between the points A and B.

(a) Find the x-coordinates of A and B. [3]

(b) Show that the area of the shaded region is 12π . [5]

The right cone in the following diagram has a total surface area of 12π , equal to the shaded area in the previous diagram.
The cone has a base radius of 2, height h , and slant height l .

(d) Hence, find the volume of the cone. [4]

Answer/Explanation

Ans:

(a) When $f\left(x\right)=0$,
$$\begin{eqnarray}
6\left(1+\cos x\right) = 0 \nonumber \\
\cos x = -1.
\end{eqnarray}$$
Thus, $x=\pi$ or $x=3\pi$.<br>
(b)
$$\begin{eqnarray}
\text{req’d area} &=& \int_{\pi}^{3\pi} f\left(x\right) \text{d}x \nonumber \\
&=& 6\int_{\pi}^{3\pi} 1+\cos x \text{d}x \nonumber \\
&=& 6\left[ 1+\sin x \right]_{\pi}^{3\pi} \nonumber \\
&=& 6\left(3\pi-\pi\right) \nonumber \\
&=& 12\pi.
\end{eqnarray}$$
(c)
$$\begin{eqnarray}
\text{total surface area} &=& \pi\left(2\right)^2+\pi\left(2\right)l \nonumber \\
12\pi &=& 4\pi+2\pi l \nonumber \\
2\pi l &=& 8\pi \nonumber \\
l &=& 4.
\end{eqnarray}$$
(d) By Pythagoras’ Theorem, we have $h=\sqrt{4^2-2^2}$, i.e., $h=\sqrt{12}$. Thus,
$$\begin{eqnarray}
\text{vol. of cone} &=& \frac{1}{3}\pi\left(2\right)^2\left(h\right) \nonumber \\
&=& \frac{4\pi\sqrt{12}}{3} \nonumber \\
&=& \frac{8\pi\sqrt{3}}{3}.
\end{eqnarray}$$

Question

Find the area between the curves $y = 2 + x – {x^2}{\text{ and }}y = 2 – 3x + {x^2}$ .

Answer/Explanation

Markscheme

$2 + x – {x^2} = 2 – 3x + {x^2}$ M1

$ \Rightarrow 2{x^2} – 4x = 0$

$ \Rightarrow 2x(x – 2) = 0$

$ \Rightarrow x = 0,{\text{ }}x = 2$ A1A1

Note: Accept graphical solution.

Award M1 for correct graph and A1A1 for correctly labelled roots.

$\therefore {\text{A}} = \int_0^2 {\left( {(2 + x – {x^2}) – (2 – 3x + {x^2})} \right){\text{d}}x} $ (M1)

$ = \int_0^2 {(4x – 2{x^2}){\text{d}}x\,\,\,\,\,{\text{or equivalent}}} $ A1

$ = \left[ {2{x^2} – \frac{{2{x^3}}}{3}} \right]_0^2$ A1

$ = \frac{8}{3}\left( { = 2\frac{2}{3}} \right)$ A1

[7 marks]

Question

A function is defined as $f(x) = k\sqrt x $, with $k > 0$ and $x \geqslant 0$ .

(a) Sketch the graph of $y = f(x)$ .

(b) Show that f is a one-to-one function.

(d) If the graphs of $y = f(x)$ and $y = {f^{ – 1}}(x)$ intersect at the point (4, 4) find the value of k .

(e) Consider the graphs of $y = f(x)$ and $y = {f^{ – 1}}(x)$ using the value of k found in part (d).

(i) Find the area enclosed by the two graphs.

(ii) The line x = c cuts the graphs of $y = f(x)$ and $y = {f^{ – 1}}(x)$ at the points P and Q respectively. Given that the tangent to $y = f(x)$ at point P is parallel to the tangent to $y = {f^{ – 1}}(x)$ at point Q find the value of c .

Answer/Explanation

Markscheme

(a)

Note: Award A1 for correct concavity, passing through (0, 0) and increasing.

Scales need not be there.

[1 mark]

(b) a statement involving the application of the Horizontal Line Test or equivalent A1

[1 mark]

for either $x = k\sqrt y $ or $x = \frac{{{y^2}}}{{{k^2}}}$ A1

${f^{ – 1}}(x) = \frac{{{x^2}}}{{{k^2}}}$ A1

${\text{dom}}\left( {{f^{ – 1}}(x)} \right) = \left[ {0,\infty } \right[$ A1

[3 marks]

(d) $\frac{{{x^2}}}{{{k^2}}} = k\sqrt x \,\,\,\,\,$or equivalent method M1

$k = \sqrt x $

$k = 2$ A1

[2 marks]

(e) (i) $A = \int_a^b {({y_1} – {y_2}){\text{d}}x} $ (M1)

$A = \int_0^4 {\left( {2{x^{\frac{1}{2}}} – \frac{1}{4}{x^2}} \right){\text{d}}x} $ A1

$ = \left[ {\frac{4}{3}{x^{\frac{3}{2}}} – \frac{1}{{12}}{x^3}} \right]_0^4$ A1

$ = \frac{{16}}{3}$ A1

(ii) attempt to find either $f'(x)$ or $({f^{ – 1}})'(x)$ M1

$f'(x) = \frac{1}{{\sqrt x }},{\text{ }}\left( {({f^{ – 1}})'(x) = \frac{x}{2}} \right)$ A1A1

$\frac{1}{{\sqrt c }} = \frac{c}{2}$ M1

$c = {2^{\frac{2}{3}}}$ A1

[9 marks]

Total [16 marks]

Question

Consider the curve $y = x{{\text{e}}^x}$ and the line $y = kx,{\text{ }}k \in \mathbb{R}$.

(a) Let k = 0.

(i) Show that the curve and the line intersect once.

(ii) Find the angle between the tangent to the curve and the line at the point of intersection.

(b) Let k =1. Show that the line is a tangent to the curve.

(c) (i) Find the values of k for which the curve $y = x{{\text{e}}^x}$ and the line $y = kx$ meet in two distinct points.

(ii) Write down the coordinates of the points of intersection.

(iii) Write down an integral representing the area of the region A enclosed by the curve and the line.

(iv) Hence, given that $0 < k < 1$, show that $A < 1$.

Answer/Explanation

Markscheme

(a) (i) $x{{\text{e}}^x} = 0 \Rightarrow x = 0$ A1

so, they intersect only once at (0, 0)

(ii) $y’ = {{\text{e}}^x} + x{{\text{e}}^x} = (1 + x){{\text{e}}^x}$ M1A1

$y'(0) = 1$ A1

$\theta = \arctan 1 = \frac{\pi }{4}{\text{ }}(\theta = 45^\circ )$ A1

[5 marks]

(b) when $k = 1,{\text{ }}y = x$

$x{{\text{e}}^x} = x \Rightarrow x({{\text{e}}^x} – 1) = 0$ M1

$ \Rightarrow x = 0$ A1

$y'(0) = 1$ which equals the gradient of the line $y = x$ R1

so, the line is tangent to the curve at origin AG

Note: Award full credit to candidates who note that the equation $x({{\text{e}}^x} – 1) = 0$ has a double root x = 0 so y = x is a tangent.

[3 marks]

$ \Rightarrow x = 0{\text{ or }}x = \ln k$ A1

$k > 0{\text{ and }}k \ne 1$ A1

(ii) (0, 0) and $(\ln k,{\text{ }}k\ln k)$ A1A1

(iii) $A = \left| {\int_0^{\ln k} {kx – x{{\text{e}}^x}{\text{d}}x} } \right|$ M1A1

Note: Do not penalize the omission of absolute value.

(iv) attempt at integration by parts to find $\int {x{{\text{e}}^x}{\text{d}}x} $ M1

$\int {x{{\text{e}}^x}{\text{d}}x} = x{{\text{e}}^x} – \int {{{\text{e}}^x}{\text{d}}x = {{\text{e}}^x}(x – 1)} $ A1

as $0 < k < 1 \Rightarrow \ln k < 0$ R1

$A = \int_{\ln k}^0 {kx – x{{\text{e}}^x}{\text{d}}x = \left[ {\frac{k}{2}{x^2} – (x – 1){{\text{e}}^x}} \right]_{\ln k}^0} $ A1

$ = 1 – \left( {\frac{k}{2}{{(\ln k)}^2} – (\ln k – 1)k} \right)$ A1

$ = 1 – \frac{k}{2}\left( {{{(\ln k)}^2} – 2\ln k + 2} \right)$

$ = 1 – \frac{k}{2}\left( {{{(\ln k – 1)}^2} + 1} \right)$ M1A1

since $\frac{k}{2}\left( {{{(\ln k – 1)}^2} + 1} \right) > 0$ R1

$A < 1$ AG

[15 marks]

Total [23 marks]

Question

Find the area enclosed by the curve $y = \arctan x$ , the x-axis and the line $x = \sqrt 3 $ .

Answer/Explanation

Markscheme

METHOD 1

${\text{area}} = \mathop \smallint \limits_0^{\sqrt 3 } \arctan x{\text{d}}x$ A1

attempting to integrate by parts M1

$ = \left[ {x\arctan x} \right]_0^{\sqrt 3 } – \mathop \smallint \limits_0^{\sqrt 3 } x\frac{1}{{1 + {x^2}}}{\text{d}}x$ A1A1

$ = \left[ {x\arctan x} \right]_0^{\sqrt 3 } – \left[ {\frac{1}{2}\ln \left( {1 + {x^2}} \right)} \right]_0^{\sqrt 3 }$ A1

Note: Award A1 even if limits are absent.

$ = \frac{\pi }{{\sqrt 3 }} – \frac{1}{2}\ln 4$ A1

$\left( { = \frac{{\pi \sqrt 3 }}{3} – \ln 2} \right)$

METHOD 2

${\text{area}} = \frac{{\pi \sqrt 3 }}{3} – \mathop \smallint \limits_0^{\frac{\pi }{3}} \tan y{\text{d}}y$ M1A1A1

$\ { = \frac{{\pi \sqrt 3 }}{3} + \left[ {\ln \left| {\cos y} \right|} \right]_0^{\frac{\pi }{3}}} $ M1A1

$ = \frac{{\pi \sqrt 3 }}{3} + \ln \frac{1}{2}$ $\left( { = \frac{{\pi \sqrt 3 }}{3} – \ln 2} \right)$ A1

[6 marks]

Question

(i) Sketch the graphs of $y = \sin x$ and $y = \sin 2x$ , on the same set of axes, for $0 \leqslant x \leqslant \frac{\pi }{2}$ .

(ii) Find the x-coordinates of the points of intersection of the graphs in the domain $0 \leqslant x \leqslant \frac{\pi }{2}$ .

(iii) Find the area enclosed by the graphs.[9]

Find the value of $\int_0^1 {\sqrt {\frac{x}{{4 – x}}} }{{\text{d}}x} $ using the substitution $x = 4{\sin ^2}\theta $ .[8]

The increasing function f satisfies $f(0) = 0$ and $f(a) = b$ , where $a > 0$ and $b > 0$ .

(i) By reference to a sketch, show that $\int_0^a {f(x){\text{d}}x = ab – \int_0^b {{f^{ – 1}}(x){\text{d}}x} } $ .

(ii) Hence find the value of $\int_0^2 {\arcsin \left( {\frac{x}{4}} \right){\text{d}}x} $ .[8]

Answer/Explanation

Markscheme

(i)

Note: Award A1 for correct $\sin x$ , A1 for correct $\sin 2x$ .

Note: Award A1A0 for two correct shapes with $\frac{\pi }{2}$ and/or 1 missing.

Note: Condone graph outside the domain.

(ii) $\sin 2x = \sin x$ , $0 \leqslant x \leqslant \frac{\pi }{2}$

$2\sin x\cos x – \sin x = 0$ M1

$\sin x(2\cos x – 1) = 0$

$x = 0,\frac{\pi }{3}$ A1A1 N1N1

(iii) area $ = \int_0^{\frac{\pi }{3}} {(\sin 2x – \sin x){\text{d}}x} $ M1

Note: Award M1 for an integral that contains limits, not necessarily correct, with $\sin x$ and $\sin 2x$ subtracted in either order.

$ = \left[ { – \frac{1}{2}\cos 2x + \cos x} \right]_0^{\frac{\pi }{3}}$ A1

$ = \left( { – \frac{1}{2}\cos \frac{{2\pi }}{3} + \cos \frac{\pi }{3}} \right) – \left( { – \frac{1}{2}\cos 0 + \cos 0} \right)$ (M1)

$ = \frac{3}{4} – \frac{1}{2}$

$ = \frac{1}{4}$ A1

[9 marks]

$\int_0^1 {\sqrt {\frac{x}{{4 – x}}} } {\text{d}}x = \int_0^{\frac{\pi }{6}} {\sqrt {\frac{{4{{\sin }^2}\theta }}{{4 – 4{{\sin }^2}\theta }}} \times 8\sin \theta \cos \theta {\text{d}}\theta } $ M1A1A1

Note: Award M1 for substitution and reasonable attempt at finding expression for dx in terms of ${\text{d}}\theta $ , first A1 for correct limits, second A1 for correct substitution for dx .

$\int_0^{\frac{\pi }{6}} {8{{\sin }^2}\theta {\text{d}}\theta } $ A1

$\int_0^{\frac{\pi }{6}} {4 – 4\cos 2\theta {\text{d}}\theta } $ M1

$ = [4\theta – 2\sin 2\theta ]_0^{\frac{\pi }{6}}$ A1

$ = \left( {\frac{{2\pi }}{3} – 2\sin \frac{\pi }{3}} \right) – 0$ (M1)

$ = \frac{{2\pi }}{3} – \sqrt 3 $ A1

[8 marks]

(i)

from the diagram above

the shaded area $ = \int_0^a {f(x){\text{d}}x = ab – \int_0^b {{f^{ – 1}}(y){\text{d}}y} } $ R1

${ = ab – \int_0^b {{f^{ – 1}}(x){\text{d}}x} }$ AG

(ii) $f(x) = \arcsin \frac{x}{4} \Rightarrow {f^{ – 1}}(x) = 4\sin x$ A1

$\int_0^2 {\arcsin \left( {\frac{x}{4}} \right){\text{d}}x = \frac{\pi }{3} – \int_0^{\frac{\pi }{6}} {4\sin x{\text{d}}x} } $ M1A1A1

Note: Award A1 for the limit $\frac{\pi }{6}$ seen anywhere, A1 for all else correct.

$ = \frac{\pi }{3} – [ – 4\cos x]_0^{\frac{\pi }{6}}$ A1

$ = \frac{\pi }{3} – 4 + 2\sqrt 3 $ A1

Note: Award no marks for methods using integration by parts.

[8 marks]

Question

The graphs of $f(x) = – {x^2} + 2$ and $g(x) = {x^3} – {x^2} – bx + 2,{\text{ }}b > 0$, intersect and create two closed regions. Show that these two regions have equal areas.

Answer/Explanation

Markscheme

to find the points of intersection of the two curves

$ – {x^2} + 2 = {x^3} – {x^2} – bx + 2$ M1

${x^3} – bx = x({x^2} – b) = 0$

$ \Rightarrow x = 0;{\text{ }}x = \pm \sqrt b $ A1A1

${A_1} = \int_{ – \sqrt b }^0 {\left[ {({x^3} – {x^2} – bx + 2) – ( – {x^2} + 2)} \right]} {\text{d}}x\left( { = \int_{ – \sqrt b }^0 {({x^3} – bx){\text{d}}x} } \right)$ M1

$ = \left[ {\frac{{{x^4}}}{4} – \frac{{b{x^2}}}{2}} \right]_{ – \sqrt b }^0$

$ = – \left( {\frac{{{{( – \sqrt b )}^4}}}{4} – \frac{{b{{( – \sqrt b )}^2}}}{2}} \right) = – \frac{{{b^2}}}{4} + \frac{{{b^2}}}{2} = \frac{{{b^2}}}{4}$ A1

${A_2} = \int_0^{\sqrt b } {\left[ {( – {x^2} + 2) – ({x^3} – {x^2} – bx + 2)} \right]{\text{d}}x} $ M1

$ = \int_0^{\sqrt b } {( – {x^3} + bx){\text{d}}x} $

$ = \left[ { – \frac{{{x^4}}}{4} + \frac{{b{x^2}}}{2}} \right]_0^{\sqrt b } = \frac{{{b^2}}}{4}$ A1

therefore ${A_1} = {A_2} = \frac{{{b^2}}}{4}$ AG

[7 marks]

Question

The graphs of $y = \left| {x + 1} \right|$ and $y = \left| {x – 3} \right|$ are shown below.

Let f (x) = $\left| {\,x + 1\,} \right| – \left| {\,x – 3\,} \right|$.

Draw the graph of y = f (x) on the blank grid below.

[4]

Hence state the value of

(i) $f'( – 3)$;

(ii) $f'(2.7)$;

(iii) $\int_{ – 3}^{ – 2} {f(x)dx} $.[4]

Answer/Explanation

Markscheme

M1A1A1A1

Note: Award M1 for any of the three sections completely correct, A1 for each correct segment of the graph.

[4 marks]

(i) 0 A1

(ii) 2 A1

(iii) finding area of rectangle (M1)

$ – 4$ A1

Note: Award M1A0 for the answer 4.

[4 marks]

Question

The graph below shows the two curves $y = \frac{1}{x}$ and $y = \frac{k}{x}$, where $k > 1$.

Find the area of region A in terms of k .[3]

Find the area of region B in terms of k .[2]

Find the ratio of the area of region A to the area of region B .[3]

Answer/Explanation

Markscheme

$\int_{\frac{1}{6}}^1 {\frac{k}{x} – \frac{1}{x}{\text{d}}x = (k – 1} )[\ln x]_{\frac{1}{6}}^1$ M1 A1

Note: Award M1 for $\int {\frac{k}{x} – \frac{1}{x}{\text{d}}x{\text{ or }}\int {\frac{1}{x} – \frac{k}{x}{\text{d}}x} } $ and A1 for $(k – 1)\ln x$ seen in part (a) or later in part (b).

$ = (1 – k)\ln \frac{1}{6}$ A1

[3 marks]

$\int_1^{\sqrt 6 } {\frac{k}{x} – \frac{1}{x}{\text{d}}x = (k – 1} )[\ln x]_1^{\sqrt 6 }$ (A1)

Note: Award A1 for correct change of limits.

$ = (k – 1)\ln \sqrt 6 $ A1

[2 marks]

$(1 – k)\ln \frac{1}{6} = (k – 1)\ln 6$ A1

$(k – 1)\ln \sqrt 6 = \frac{1}{2}(k – 1)\ln 6$ A1

Note: This simplification could have occurred earlier, and marks should still be awarded.

ratio is 2 (or 2:1) A1

[3 marks]

Question

The function f is defined on the domain $\left[ {0,\,\frac{{3\pi }}{2}} \right]$ by $f(x) = {e^{ – x}}\cos x$ .

State the two zeros of f .[1]

Sketch the graph of f .[1]

The region bounded by the graph, the x-axis and the y-axis is denoted by A and the region bounded by the graph and the x-axis is denoted by B . Show that the ratio of the area of A to the area of B is

\[\frac{{{e^\pi }\left( {{e^{\frac{\pi }{2}}} + 1} \right)}}{{{e^\pi } + 1}}.\][7]

Answer/Explanation

Markscheme

${e^{ – x}}\cos x = 0$

$ \Rightarrow x = \frac{\pi }{2},{\text{ }}\frac{{3\pi }}{2}$ A1

[1 mark]

Note: Accept any form of concavity for $x \in \left[ {0,\frac{\pi }{2}} \right]$.

Note: Do not penalize unmarked zeros if given in part (a).

Note: Zeros written on diagram can be used to allow the mark in part (a) to be awarded retrospectively.

[1 mark]

attempt at integration by parts M1

EITHER

$I = \int {{{\text{e}}^{ – x}}\cos x{\text{d}}x = – {{\text{e}}^{ – x}}\cos x{\text{d}}x – \int {{{\text{e}}^{ – x}}\sin x{\text{d}}x} } $ A1

$ \Rightarrow I = – {{\text{e}}^{ – x}}\cos x{\text{d}}x – \left[ { – {{\text{e}}^{ – x}}\sin x + \int {{{\text{e}}^{ – x}}\cos x{\text{d}}x} } \right]$ A1

$ \Rightarrow I = \frac{{{{\text{e}}^{ – x}}}}{2}(\sin x – \cos x) + C$ A1

Note: Do not penalize absence of C.

$I = \int {{{\text{e}}^{ – x}}\cos x{\text{d}}x = {{\text{e}}^{ – x}}\sin x + \int {{{\text{e}}^{ – x}}\sin x{\text{d}}x} } $ A1

$I = {{\text{e}}^{ – x}}\sin x – {{\text{e}}^{ – x}}\cos x – \int {{{\text{e}}^{ – x}}\cos x{\text{d}}x} $ A1

$ \Rightarrow I = \frac{{{{\text{e}}^{ – x}}}}{2}(\sin x – \cos x) + C$ A1

Note: Do not penalize absence of C.

THEN

$\int_0^{\frac{\pi }{2}} {{{\text{e}}^{ – x}}\cos x{\text{d}}x = \left[ {\frac{{{{\text{e}}^{ – x}}}}{2}(\sin x – \cos x)} \right]} _0^{\frac{\pi }{2}} = \frac{{{{\text{e}}^{ – \frac{\pi }{2}}}}}{2} + \frac{1}{2}$ A1

$\int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {{{\text{e}}^{ – x}}\cos x{\text{d}}x = \left[ {\frac{{{{\text{e}}^{ – x}}}}{2}(\sin x – \cos x)} \right]_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} = – \frac{{{{\text{e}}^{ – \frac{{3\pi }}{2}}}}}{2} – \frac{{{{\text{e}}^{ – \frac{\pi }{2}}}}}{2}} $ A1

ratio of A:B is $\frac{{\frac{{{{\text{e}}^{ – \frac{\pi }{2}}}}}{2} + \frac{1}{2}}}{{\frac{{{{\text{e}}^{ – \frac{{3\pi }}{2}}}}}{2} + \frac{{{{\text{e}}^{ – \frac{\pi }{2}}}}}{2}}}$

$ = \frac{{{{\text{e}}^{\frac{{3\pi }}{2}}}\left( {{{\text{e}}^{ – \frac{\pi }{2}}} + 1} \right)}}{{{{\text{e}}^{\frac{{3\pi }}{2}}}\left( {{{\text{e}}^{ – \frac{{3\pi }}{2}}} + {{\text{e}}^{ – \frac{\pi }{2}}}} \right)}}$ M1

$ = \frac{{{{\text{e}}^\pi }\left( {{{\text{e}}^{\frac{\pi }{2}}} + 1} \right)}}{{{{\text{e}}^\pi } + 1}}$ AG

[7 marks]

Question

The function $f$ is given by $f(x) = x{{\text{e}}^{ – x}}{\text{ }}(x \geqslant 0)$.

(i) Find an expression for $f'(x)$.

(ii) Hence determine the coordinates of the point A, where $f'(x) = 0$.[3]

a(i)(ii).

Find an expression for $f”(x)$ and hence show the point A is a maximum.[3]

Find the coordinates of B, the point of inflexion.[2]

The graph of the function $g$ is obtained from the graph of $f$ by stretching it in the x-direction by a scale factor 2.

(i) Write down an expression for $g(x)$.

(ii) State the coordinates of the maximum C of $g$.

(iii) Determine the x-coordinates of D and E, the two points where $f(x) = g(x)$.[5]

Sketch the graphs of $y = f(x)$ and $y = g(x)$ on the same axes, showing clearly the points A, B, C, D and E.[4]

Find an exact value for the area of the region bounded by the curve $y = g(x)$, the x-axis and the line $x = 1$.[3]

Answer/Explanation

Markscheme

(i) $f'(x) = {{\text{e}}^{ – x}} – x{{\text{e}}^{ – x}}$ M1A1

(ii) $f'(x) = 0 \Rightarrow x = 1$

coordinates $\left( {1,{\text{ }}{{\text{e}}^{ – 1}}} \right)$ A1

[3 marks]

a(i)(ii).

$f”(x) = – {{\text{e}}^{ – x}} – {{\text{e}}^{ – x}} + x{{\text{e}}^{ – x}}{\text{ }}\left( { = – {{\text{e}}^{ – x}}(2 – x)} \right)$ A1

substituting $x = 1$ into $f”(x)$ M1

$f”(1){\text{ }}\left( { = – {{\text{e}}^{ – 1}}} \right) < 0$ hence maximum R1AG

[3 marks]

$f”(x) = 0{\text{ (}} \Rightarrow x = 2)$ M1

coordinates $\left( {2,{\text{ 2}}{{\text{e}}^{ – 2}}} \right)$ A1

[2 marks]

(i) $g(x) = \frac{x}{2}{{\text{e}}^{ – \frac{x}{2}}}$ A1

(ii) coordinates of maximum $\left( {2,{\text{ }}{{\text{e}}^{ – 1}}} \right)$ A1

(iii) equating $f(x) = g(x)$ and attempting to solve $x{{\text{e}}^{ – x}} = \frac{x}{2}{{\text{e}}^{ – \frac{x}{2}}}$

$ \Rightarrow x\left( {2{{\text{e}}^{\frac{x}{2}}} – {{\text{e}}^x}} \right) = 0$ (A1)

$ \Rightarrow x = 0$ A1

or $2{{\text{e}}^{\frac{x}{2}}} = {{\text{e}}^x}$

$ \Rightarrow {{\text{e}}^{\frac{x}{2}}} = 2$

$ \Rightarrow x = 2\ln 2$ $(\ln 4)$ A1

Note: Award first (A1) only if factorisation seen or if two correct

solutions are seen.

Note: Award A1 for shape of $f$, including domain extending beyond $x = 2$.

Ignore any graph shown for $x < 0$.

Award A1 for A and B correctly identified.

Award A1 for shape of $g$, including domain extending beyond $x = 2$.

Ignore any graph shown for $x < 0$. Allow follow through from $f$.

Award A1 for C, D and E correctly identified (D and E are interchangeable).

[4 marks]

$A = \int_0^1 {\frac{x}{2}{{\text{e}}^{ – \frac{x}{2}}}{\text{d}}x} $ M1

$ = \left[ { – x{{\text{e}}^{ – \frac{x}{2}}}} \right]_0^1 – \int_0^1 { – {{\text{e}}^{ – \frac{x}{2}}}{\text{d}}x} $ A1

Note: Condone absence of limits or incorrect limits.

$ = – {{\text{e}}^{ – \frac{1}{2}}} – \left[ {2{{\text{e}}^{ – \frac{x}{2}}}} \right]_0^1$

$ = – {{\text{e}}^{ – \frac{1}{2}}} – \left( {2{{\text{e}}^{ – \frac{1}{2}}} – 2} \right) = 2 – 3{{\text{e}}^{ – \frac{1}{2}}}$ A1

[3 marks]

Question

The first set of axes below shows the graph of $y = {\text{ }}f(x)$ for $ – 4 \leqslant x \leqslant 4$.

Let $g(x) = \int_{ – 4}^x {f(t){\text{d}}t} $ for $ – 4 \leqslant x \leqslant 4$.

(a) State the value of x at which $g(x)$ is a minimum.

(b) On the second set of axes, sketch the graph of $y = g(x)$.

Answer/Explanation

Markscheme

(a) $x = 1$ A1

[1 mark]

(b) A1 for point (–4, 0)

A1 for (0, − 4)

A1 for min at $x = 1$ in approximately the correct place

A1 for (4, 0)

A1 for shape including continuity at $x = 0$

[5 marks]

Total [6 marks]

Question

Consider the function $f(x) = \frac{{\ln x}}{x},{\text{ }}x > 0$.

The sketch below shows the graph of $y = {\text{ }}f(x)$ and its tangent at a point A.

Show that $f'(x) = \frac{{1 – \ln x}}{{{x^2}}}$.[2]

Find the coordinates of B, at which the curve reaches its maximum value.[3]

Find the coordinates of C, the point of inflexion on the curve.[5]

The graph of $y = {\text{ }}f(x)$ crosses the $x$-axis at the point A.

Find the equation of the tangent to the graph of $f$ at the point A.

[4]d.

The graph of $y = {\text{ }}f(x)$ crosses the $x$-axis at the point A.

Find the area enclosed by the curve $y = f(x)$, the tangent at A, and the line $x = {\text{e}}$.[7]

Answer/Explanation

Markscheme

$f'(x) = \frac{{x \times \frac{1}{x} – \ln x}}{{{x^2}}}$ M1A1

$ = \frac{{1 – \ln x}}{{{x^2}}}$ AG

[2 marks]

$\frac{{1 – \ln x}}{{{x^2}}} = 0$ has solution $x = {\text{e}}$ M1A1

$y = \frac{1}{{\text{e}}}$ A1

hence maximum at the point $\left( {{\text{e, }}\frac{1}{{\text{e}}}} \right)$

[3 marks]

$f”(x) = \frac{{{x^2}\left( { – \frac{1}{x}} \right) – 2x(1 – \ln x)}}{{{x^4}}}$ M1A1

$ = \frac{{2\ln x – 3}}{{{x^3}}}$

Note: The M1A1 should be awarded if the correct working appears in part (b).

point of inflexion where $f”(x) = 0$ M1

so $x = {{\text{e}}^{\frac{3}{2}}},{\text{ }}y = \frac{3}{2}{{\text{e}}^{\frac{{ – 3}}{2}}}$ A1A1

C has coordinates $\left( {{{\text{e}}^{\frac{3}{2}}},{\text{ }}\frac{3}{2}{{\text{e}}^{\frac{{ – 3}}{2}}}} \right)$

[5 marks]

$f(1) = 0$ A1

$f'(1) = 1$ (A1)

$y = x + c$ (M1)

through (1, 0)

equation is $y = x – 1$ A1

[4 marks]

METHOD 1

area $ = \int_1^{\text{e}} {x – 1 – \frac{{\ln x}}{x}{\text{d}}x} $ M1A1A1

Note: Award M1 for integration of difference between line and curve, A1 for correct limits, A1 for correct expressions in either order.

$\int {\frac{{\ln x}}{x}{\text{d}}x = \frac{{{{(\ln x)}^2}}}{2}} ( + c)$ (M1)A1

$\int {(x – 1){\text{d}}x = \frac{{{x^2}}}{2} – x( + c)} $ A1

$ = \left[ {\frac{1}{2}{x^2} – x – \frac{1}{2}{{(\ln x)}^2}} \right]_1^{\text{e}}$

$ = \left( {\frac{1}{2}{{\text{e}}^2} – {\text{e}} – \frac{1}{2}} \right) – \left( {\frac{1}{2} – 1} \right)$

$ = \frac{1}{2}{{\text{e}}^2} – {\text{e}}$ A1

METHOD 2

area = area of triangle $ – \int_1^e {\frac{{\ln x}}{x}{\text{d}}x} $ M1A1

Note: A1 is for correct integral with limits and is dependent on the M1.

$\int {\frac{{\ln x}}{x}{\text{d}}x = \frac{{{{(\ln x)}^2}}}{2}( + c)} $ (M1)A1

area of triangle $ = \frac{1}{2}(e – 1)(e – 1)$ M1A1

$\frac{1}{2}(e – 1)(e – 1) – \left( {\frac{1}{2}} \right) = \frac{1}{2}{{\text{e}}^2} – {\text{e}}$ A1

[7 marks]

Question

The graph of the function $f(x) = \frac{{x + 1}}{{{x^2} + 1}}$ is shown below.

The point (1, 1) is a point of inflexion. There are two other points of inflexion.

Find $f'(x)$.[2]

Hence find the $x$-coordinates of the points where the gradient of the graph of $f$ is zero.[1]

Find $f”(x)$ expressing your answer in the form $\frac{{p(x)}}{{{{({x^2} + 1)}^3}}}$, where $p(x)$ is a polynomial of degree 3.[3]

Find the $x$-coordinates of the other two points of inflexion.[4]

Find the area of the shaded region. Express your answer in the form $\frac{\pi }{a} – \ln \sqrt b $, where $a$ and $b$ are integers.[6]

Answer/Explanation

Markscheme

(a) $f'(x) = \frac{{\left( {{x^2} + 1} \right) – 2x(x + 1)}}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{ }}\left( { = \frac{{ – {x^2} – 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right)$ M1A1

[2 marks]

$\frac{{ – {x^2} – 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}} = 0$

$x = – 1 \pm \sqrt 2 $ A1

[1 mark]

$f”(x) = \frac{{( – 2x – 2){{\left( {{x^2} + 1} \right)}^2} – 2(2x)\left( {{x^2} + 1} \right)\left( { – {x^2} – 2x + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}}$ A1A1

Note: Award A1 for $( – 2x – 2){\left( {{x^2} + 1} \right)^2}$ or equivalent.

Note: Award A1 for $ – 2(2x)\left( {{x^2} + 1} \right)\left( { – {x^2} – 2x + 1} \right)$ or equivalent.

$ = \frac{{( – 2x – 2)\left( {{x^2} + 1} \right) – 4x\left( { – {x^2} – 2x + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}}$

$ = \frac{{2{x^3} + 6{x^2} – 6x – 2}}{{{{\left( {{x^2} + 1} \right)}^3}}}$ A1

$\left( { = \frac{{2\left( {{x^3} + 3{x^2} – 3x – 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}}} \right)$

[3 marks]

recognition that $(x – 1)$ is a factor (R1)

$(x – 1)\left( {{x^2} + bx + c} \right) = \left( {{x^3} + 3{x^2} – 3x – 1} \right)$ M1

$ \Rightarrow {x^2} + 4x + 1 = 0$ A1

$x = – 2 \pm \sqrt 3 $ A1

Note: Allow long division / synthetic division.

[4 marks]

$\int_{ – 1}^0 {\frac{{x + 1}}{{{x^2} + 1}}{\text{d}}x} $ M1

$\int {\frac{{x + 1}}{{{x^2} + 1}}{\text{d}}x = \int {\frac{x}{{{x^2} + 1}}{\text{d}}x + \int {\frac{1}{{{x^2} + 1}}{\text{d}}x} } } $ M1

$ = \frac{1}{2}\ln \left( {{x^2} + 1} \right) + \arctan (x)$ A1A1

$ = \left[ {\frac{1}{2}\ln \left( {{x^2} + 1} \right) + \arctan (x)} \right]_{ – 1}^0 = \frac{1}{2}\ln 1 + \arctan 0 – \frac{1}{2}\ln 2 – \arctan ( – 1)$ M1

$ = \frac{\pi }{4} – \ln \sqrt 2 $ A1

[6 marks]

Question

The function $f$ is defined as $f(x) = {{\text{e}}^{3x + 1}},{\text{ }}x \in \mathbb{R}$.

(i) Find ${f^{ – 1}}(x)$.

(ii) State the domain of ${f^{ – 1}}$.[4]

The function $g$ is defined as $g(x) = \ln x,{\text{ }}x \in {\mathbb{R}^ + }$.

The graph of $y = g(x)$ and the graph of $y = {f^{ – 1}}(x)$ intersect at the point $P$.

Find the coordinates of $P$.[5]

The graph of $y = g(x)$ intersects the $x$-axis at the point $Q$.

Show that the equation of the tangent $T$ to the graph of $y = g(x)$ at the point $Q$ is $y = x – 1$.[3]

A region $R$ is bounded by the graphs of $y = g(x)$, the tangent $T$ and the line $x = {\text{e}}$.

Find the area of the region $R$.[5]

A region $R$ is bounded by the graphs of $y = g(x)$, the tangent $T$ and the line $x = {\text{e}}$.

(i) Show that $g(x) \le x – 1,{\text{ }}x \in {\mathbb{R}^ + }$.

(ii) By replacing $x$ with $\frac{1}{x}$ in part (e)(i), show that $\frac{{x – 1}}{x} \le g(x),{\text{ }}x \in {\mathbb{R}^ + }$.[6]

Answer/Explanation

Markscheme

(i) $x = {{\text{e}}^{3y + 1}}$ M1

Note: The M1 is for switching variables and can be awarded at any stage.

Further marks do not rely on this mark being awarded.

taking the natural logarithm of both sides and attempting to transpose M1

$\left( {{f^{ – 1}}(x)} \right) = \frac{1}{3}(\ln x – 1)$ A1

(ii) $x \in {\mathbb{R}^ + }$ or equivalent, for example $x > 0$. A1

[4 marks]

$\ln x = \frac{1}{3}(\ln x – 1) \Rightarrow \ln x – \frac{1}{3}\ln x = – \frac{1}{3}$ (or equivalent) M1A1

$\ln x = – \frac{1}{2}$ (or equivalent) A1

$x = {{\text{e}}^{ – \frac{1}{2}}}$ A1

coordinates of $P$ are $\left( {{{\text{e}}^{ – \frac{1}{2}}},{\text{ }} – \frac{1}{2}} \right)$ A1

[5 marks]

coordinates of $Q$ are ($1,{\rm{ }}0$) seen anywhere A1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{1}{x}$ M1

at ${\text{Q, }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1$ A1

$y = x – 1$ AG

[3 marks]

let the required area be $A$

$A = \int_1^e {x – 1{\text{d}}x – \int_1^e {\ln x{\text{d}}x} } $ M1

Note: The M1 is for a difference of integrals. Condone absence of limits here.

attempting to use integration by parts to find $\int {\ln x{\text{d}}x} $ (M1)

$ = \left[ {\frac{{{x^2}}}{2} – x} \right]_1^{\text{e}} – [x\ln x – x]_1^{\text{e}}$ A1A1

Note: Award A1 for $\frac{{{x^2}}}{2} – x$ and A1 for $x\ln x – x$.

Note: The second M1 and second A1 are independent of the first M1 and the first A1.

$ = \frac{{{{\text{e}}^2}}}{2} – {\text{e}} – \frac{1}{2}\left( { = \frac{{{{\text{e}}^2} – 2{\text{e}} – 1}}{2}} \right)$ A1

[5 marks]

(i) METHOD 1

consider for example $h(x) = x – 1 – \ln x$

$h(1) = 0\;\;\;{\text{and}}\;\;\;h'(x) = 1 – \frac{1}{x}$ (A1)

as $h'(x) \ge 0\;\;\;{\text{for}}\;\;\;x \ge 1,\;\;\;{\text{then}}\;\;\;h(x) \ge 0\;\;\;{\text{for}}\;\;\;x \ge 1$ R1

as $h'(x) \le 0\;\;\;{\text{for}}\;\;\;0 < x \le 1,\;\;\;{\text{then}}\;\;\;h(x) \ge 0\;\;\;{\text{for}}\;\;\;0 < x \le 1$ R1

so $g(x) \le x – 1,{\text{ }}x \in {\mathbb{R}^ + }$ AG

METHOD 2

$g”(x) = – \frac{1}{{{x^2}}}$ A1

$g”(x) < 0\;\;\;$(concave down) for$\;\;\;x \in {\mathbb{R}^ + }$ R1

the graph of $y = g(x)$ is below its tangent $(y = x – 1\;\;\;{\text{at}}\;\;\;x = 1)$ R1

so $g(x) \le x – 1,{\text{ }}x \in {\mathbb{R}^ + }$ AG

Note: The reasoning may be supported by drawn graphical arguments.

METHOD 3

clear correct graphs of $y = x – 1\;\;\;{\text{and}}\;\;\;\ln x\;\;\;{\text{for}}\;\;\;x > 0$ A1A1

statement to the effect that the graph of $\ln x$ is below the graph of its tangent at $x = 1$ R1AG

(ii) replacing $x$ by $\frac{1}{x}$ to obtain $\ln \left( {\frac{1}{x}} \right) \le \frac{1}{x} – 1\left( { = \frac{{1 – x}}{x}} \right)$ M1

$ – \ln x \le \frac{1}{x} – 1\left( { = \frac{{1 – x}}{x}} \right)$ (A1)

$\ln x \ge 1 – \frac{1}{x}\left( { = \frac{{x – 1}}{x}} \right)$ A1

so $\frac{{x – 1}}{x} \le g(x),{\text{ }}x \in {\mathbb{R}^ + }$ AG

[6 marks]

Total [23 marks]

Question

Consider the functions $f(x) = \tan x,{\text{ }}0 \le \ x < \frac{\pi }{2}$ and $g(x) = \frac{{x + 1}}{{x – 1}},{\text{ }}x \in \mathbb{R},{\text{ }}x \ne 1$.

Find an expression for $g \circ f(x)$, stating its domain.[2]

Hence show that $g \circ f(x) = \frac{{\sin x + \cos x}}{{\sin x – \cos x}}$.[2]

Let $y = g \circ f(x)$, find an exact value for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ at the point on the graph of $y = g \circ f(x)$ where $x = \frac{\pi }{6}$, expressing your answer in the form $a + b\sqrt 3 ,{\text{ }}a,{\text{ }}b \in \mathbb{Z}$.[6]

Show that the area bounded by the graph of $y = g \circ f(x)$, the $x$-axis and the lines $x = 0$ and $x = \frac{\pi }{6}$ is $\ln \left( {1 + \sqrt 3 } \right)$.[6]

Answer/Explanation

Markscheme

$g \circ f(x) = \frac{{\tan x + 1}}{{\tan x – 1}}$ A1

$x \ne \frac{\pi }{4},{\text{ }}0 \le x < \frac{\pi }{2}$ A1

[2 marks]

$\frac{{\tan x + 1}}{{\tan x – 1}} = \frac{{\frac{{\sin x}}{{\cos x}} + 1}}{{\frac{{\sin x}}{{\cos x}} – 1}}$ M1A1

$ = \frac{{\sin x + \cos x}}{{\sin x – \cos x}}$ AG

[2 marks]

METHOD 1

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(\sin x – \cos x)(\cos x – \sin x) – (\sin x + \cos x)(\cos x + \sin x)}}{{{{(\sin x – \cos x)}^2}}}$ M1(A1)

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(2\sin x\cos x – {{\cos }^2}x – {{\sin }^2}x) – (2\sin x\cos x + {{\cos }^2}x + {{\sin }^2}x)}}{{{{\cos }^2}x + {{\sin }^2}x – 2\sin x\cos x}}$

$ = \frac{{ – 2}}{{1 – \sin 2x}}$

Substitute $\frac{\pi }{6}$ into any formula for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ M1

$\frac{{ – 2}}{{1 – \sin \frac{\pi }{3}}}$

$ = \frac{{ – 2}}{{1 – \frac{{\sqrt 3 }}{2}}}$ A1

$ = \frac{{ – 4}}{{2 – \sqrt 3 }}$

$ = \frac{{ – 4}}{{2 – \sqrt 3 }}\left( {\frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }}} \right)$ M1

$ = \frac{{ – 8 – 4\sqrt 3 }}{1} = – 8 – 4\sqrt 3 $ A1

METHOD 2

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(\tan x – 1){{\sec }^2}x – (\tan x + 1){{\sec }^2}x}}{{{{(\tan x – 1)}^2}}}$ M1A1

$ = \frac{{ – 2{{\sec }^2}x}}{{{{(\tan x – 1)}^2}}}$ A1

$ = \frac{{ – 2{{\sec }^2}\frac{\pi }{6}}}{{{{\left( {\tan \frac{\pi }{6} – 1} \right)}^2}}} = \frac{{ – 2\left( {\frac{4}{3}} \right)}}{{{{\left( {\frac{1}{{\sqrt 3 }} – 1} \right)}^2}}} = \frac{{ – 8}}{{{{\left( {1 – \sqrt 3 } \right)}^2}}}$ M1

Note: Award M1 for substitution $\frac{\pi }{6}$.

$\frac{{ – 8}}{{{{\left( {1 – \sqrt 3 } \right)}^2}}} = \frac{{ – 8}}{{\left( {4 – 2\sqrt 3 } \right)}}\frac{{\left( {4 + 2\sqrt 3 } \right)}}{{\left( {4 + 2\sqrt 3 } \right)}} = – 8 – 4\sqrt 3 $ M1A1

[6 marks]

Area $\left| {\int_0^{\frac{\pi }{6}} {\frac{{\sin x + \cos x}}{{\sin x – \cos x}}{\text{d}}x} } \right|$ M1

$ = \left| {\left[ {\ln \left| {\sin x – \cos x} \right|} \right]_0^{\frac{\pi }{6}}} \right|$ A1

Note: Condone absence of limits and absence of modulus signs at this stage.

$ = \left| {\ln \left| {\sin \frac{\pi }{6} – \cos \frac{\pi }{6}} \right| – \ln \left| {\sin 0 – \cos 0} \right|} \right|$ M1

$ = \left| {\ln \left| {\frac{1}{2} – \frac{{\sqrt 3 }}{2}} \right| – 0} \right|$

$ = \left| {\ln \left( {\frac{{\sqrt 3 – 1}}{2}} \right)} \right|$ A1

$ = – \ln \left( {\frac{{\sqrt 3 – 1}}{2}} \right) = \ln \left( {\frac{2}{{\sqrt 3 – 1}}} \right)$ A1

$ = \ln \left( {\frac{2}{{\sqrt 3 – 1}} \times \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}}} \right)$ M1

$ = \ln \left( {\sqrt 3 + 1} \right)$ AG

[6 marks]

Total [16 marks]

Question

Consider the function defined by $f(x) = x\sqrt {1 – {x^2}} $ on the domain $ – 1 \le x \le 1$.

Show that $f$ is an odd function.[2]

Find $f'(x)$.[3]

Hence find the $x$-coordinates of any local maximum or minimum points.[3]

Find the range of $f$.[3]

Sketch the graph of $y = f(x)$ indicating clearly the coordinates of the $x$-intercepts and any local maximum or minimum points.[3]

Find the area of the region enclosed by the graph of $y = f(x)$ and the $x$-axis for $x \ge 0$.[4]

Show that $\int_{ – 1}^1 {\left| {x\sqrt {1 – {x^2}} } \right|{\text{d}}x > \left| {\int_{ – 1}^1 {x\sqrt {1 – {x^2}} {\text{d}}x} } \right|} $.[2]

Answer/Explanation

Markscheme

$f( – x) = ( – x)\sqrt {1 – {{( – x)}^2}} $ M1

$ = – x\sqrt {1 – {x^2}} $

$ = – f(x)$ R1

hence $f$ is odd AG

[2 marks]

$f'(x) = x \bullet \frac{1}{2}{(1 – {x^2})^{ – \frac{1}{2}}} \bullet – 2x + {(1 – {x^2})^{\frac{1}{2}}}$ M1A1A1

[3 marks]

$f'(x) = \sqrt {1 – {x^2}} – \frac{{{x^2}}}{{\sqrt {1 – {x^2}} }}\;\;\;\left( { = \frac{{1 – 2{x^2}}}{{\sqrt {1 – {x^2}} }}} \right)$ A1

Note: This may be seen in part (b).

Note: Do not allow FT from part (b).

$f'(x) = 0 \Rightarrow 1 – 2{x^2} = 0$ M1

$x = \pm \frac{1}{{\sqrt 2 }}$ A1

[3 marks]

$y$-coordinates of the Max Min Points are $y = \pm \frac{1}{2}$ M1A1

so range of $f(x)$ is $\left[ { – \frac{1}{2},{\text{ }}\frac{1}{2}} \right]$ A1

Note: Allow FT from (c) if values of $x$, within the domain, are used.

[3 marks]

Shape: The graph of an odd function, on the given domain, s-shaped,

where the max(min) is the right(left) of $0.5{\text{ }}( – 0.5)$ A1

$x$-intercepts A1

turning points A1

[3 marks]

${\text{area}} = \int_0^1 {x\sqrt {1 – {x^2}} {\text{d}}x} $ (M1)

attempt at “backwards chain rule” or substitution M1

$ = – \frac{1}{2}\int_0^1 {( – 2x)\sqrt {1 – {x^2}} {\text{d}}x} $

Note: Condone absence of limits for first two marks.

$ = \left[ {\frac{2}{3}{{(1 – {x^2})}^{\frac{3}{2}}} \bullet – \frac{1}{2}} \right]_0^1$ A1

$ = \left[ { – \frac{1}{3}{{(1 – {x^2})}^{\frac{3}{2}}}} \right]_0^1$

$ = 0 – \left( { – \frac{1}{3}} \right) = \frac{1}{3}$ A1

[4 marks]

$\int_{ – 1}^1 {\left| {x\sqrt {1 – {x^2}} } \right|{\text{d}}x > 0} $ R1

$\left| {\int_{ – 1}^1 {x\sqrt {1 – {x^2}} {\text{d}}x} } \right| = 0$ R1

so $\int_{ – 1}^1 {\left| {x\sqrt {1 – {x^2}} } \right|{\text{d}}x > \left| {\int_{ – 1}^1 {x\sqrt {1 – {x^2}} {\text{d}}x} } \right|} $ AG

[2 marks]

Total [20 marks]

Question

The following graph shows the relation $x = 3\cos 2y + 4,{\text{ }}0 \leqslant y \leqslant \pi $.

The curve is rotated 360° about the $y$-axis to form a volume of revolution.

A container with this shape is made with a solid base of diameter 14 cm . The container is filled with water at a rate of ${\text{2 c}}{{\text{m}}^{\text{3}}}\,{\text{mi}}{{\text{n}}^{ – 1}}$. At time $t$ minutes, the water depth is $h{\text{ cm, }}0 \leqslant h \leqslant \pi $ and the volume of water in the container is $V{\text{ c}}{{\text{m}}^{\text{3}}}$.

Calculate the value of the volume generated.[8]

(i) Given that $\frac{{{\text{d}}V}}{{{\text{d}}h}} = \pi {(3\cos 2h + 4)^2}$, find an expression for $\frac{{{\text{d}}h}}{{{\text{d}}t}}$.

(ii) Find the value of $\frac{{{\text{d}}h}}{{{\text{d}}t}}$ when $h = \frac{\pi }{4}$.[4]

(i) Find $\frac{{{{\text{d}}^2}h}}{{{\text{d}}{t^2}}}$.

(ii) Find the values of $h$ for which $\frac{{{{\text{d}}^2}h}}{{{\text{d}}{t^2}}} = 0$.

(iii) By making specific reference to the shape of the container, interpret $\frac{{{\text{d}}h}}{{{\text{d}}t}}$ at the values of $h$ found in part (c)(ii).[7]

Answer/Explanation

Markscheme

use of $\pi \int_a^b {{x^2}{\text{d}}y} $ (M1)

Note: Condone any or missing limits.

$V = \pi \int_0^\pi {{{(3\cos 2y + 4)}^2}{\text{d}}y} $ (A1)

$ = \pi \int_0^\pi {(9{{\cos }^2}2y + 24\cos 2y + 16){\text{d}}y} $ A1

$9{\cos ^2}2y = \frac{9}{2}(1 + \cos 4y)$ (M1)

$ = \pi \left[ {\frac{{9y}}{2} + \frac{9}{8}\sin 4y + 12\sin 2y + 16y} \right]_0^\pi $ M1A1

$ = \pi \left( {\frac{{9\pi }}{2} + 16\pi } \right)$ (A1)

$ = \frac{{41{\pi ^2}}}{2}{\text{ (c}}{{\text{m}}^3})$ A1

Note: If the coefficient “$\pi $” is absent, or eg, “$2\pi $” is used, only M marks are available.

[8 marks]

(i) attempting to use $\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{{{\text{d}}V}}{{{\text{d}}t}} \times \frac{{{\text{d}}h}}{{{\text{d}}V}}$ with $\frac{{{\text{d}}V}}{{{\text{d}}t}} = 2$ M1

$\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{2}{{\pi {{(3\cos 2h + 4)}^2}}}$ A1

(ii) substituting $h = \frac{\pi }{4}$ into $\frac{{{\text{d}}h}}{{{\text{d}}t}}$ (M1)

$\frac{{{\text{d}}h}}{{{\text{d}}t}} = \frac{1}{{8\pi }}{\text{ (cm}}\,{\text{mi}}{{\text{n}}^{ – 1}})$ A1

Note: Do not allow FT marks for (b)(ii).

[4 marks]

(i) $\frac{{{{\text{d}}^2}h}}{{{\text{d}}{t^2}}} = \frac{{\text{d}}}{{{\text{d}}t}}\left( {\frac{{{\text{d}}h}}{{{\text{d}}t}}} \right) = \frac{{{\text{d}}h}}{{{\text{d}}t}} \times \frac{{\text{d}}}{{{\text{d}}h}}\left( {\frac{{{\text{d}}h}}{{{\text{d}}t}}} \right)$ (M1)

$ = \frac{2}{{\pi {{(3\cos 2h + 4)}^2}}} \times \frac{{24\sin 2h}}{{\pi {{(3\cos 2h + 4)}^3}}}$ M1A1

Note: Award M1 for attempting to find $\frac{{\text{d}}}{{{\text{d}}h}}\left( {\frac{{{\text{d}}h}}{{{\text{d}}t}}} \right)$.

$ = \frac{{48\sin 2h}}{{{\pi ^2}{{(3\cos 2h + 4)}^5}}}$ A1

(ii) $\sin 2h = 0 \Rightarrow h = 0,{\text{ }}\frac{\pi }{2},{\text{ }}\pi $ A1

Note: Award A1 for $\sin 2h = 0 \Rightarrow h = 0,{\text{ }}\frac{\pi }{2},{\text{ }}\pi $ from an incorrect $\frac{{{{\text{d}}^2}h}}{{{\text{d}}{t^2}}}$.

(iii) METHOD 1

$\frac{{{\text{d}}h}}{{{\text{d}}t}}$ is a minimum at $h = 0,{\text{ }}\pi $ and the container is widest at these values R1

$\frac{{{\text{d}}h}}{{{\text{d}}t}}$ is a maximum at $h = \frac{\pi }{2}$ and the container is narrowest at this value R1

[7 marks]

Question

Given that $\int_{ – 2}^2 {f\left( x \right){\text{d}}x = 10} $ and $\int_0^2 {f\left( x \right){\text{d}}x = 12} $, find

$\int_{ – 2}^0 {\left( {f\left( x \right){\text{ + 2}}} \right){\text{d}}x} $.[4]

$\int_{ – 2}^0 {f\left( {x{\text{ + 2}}} \right){\text{d}}x} $.[2]

Answer/Explanation

Markscheme

$\int_{ – 2}^0 {f\left( x \right){\text{d}}x = 10} – 12 = – 2$ (M1)(A1)

$\int_{ – 2}^0 {2{\text{d}}x = \left[ {2x} \right]} _{ – 2}^0 = 4$ A1

$\int_{ – 2}^0 {\left( {f\left( x \right){\text{ + 2}}} \right){\text{d}}x} = 2$ A1

[4 marks]

$\int_{ – 2}^0 {f\left( {x{\text{ + 2}}} \right){\text{d}}x} = \int_0^2 {f\left( x \right){\text{d}}x} $ (M1)

= 12 A1

[2 marks]

Question

Let $y = {\text{arccos}}\left( {\frac{x}{2}} \right)$

Find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$.[2]

Find $\int_0^1 {{\text{arccos}}\left( {\frac{x}{2}} \right){\text{d}}x} $.[7]

Answer/Explanation

Markscheme

$y = {\text{arccos}}\left( {\frac{x}{2}} \right) \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{1}{{2\sqrt {1 – {{\left( {\frac{x}{2}} \right)}^2}} }}\left( { = – \frac{1}{{\sqrt {4 – {x^2}} }}} \right)$ M1A1

Note: M1 is for use of the chain rule.

[2 marks]

attempt at integration by parts M1

$u = {\text{arccos}}\left( {\frac{x}{2}} \right) \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} = – \frac{1}{{\sqrt {4 – {x^2}} }}$

$\frac{{{\text{d}}v}}{{{\text{d}}x}} = 1 \Rightarrow v = x$ (A1)

$\int_0^1 {{\text{arccos}}\left( {\frac{x}{2}} \right){\text{d}}x} = \left[ {x\,\,{\text{arccos}}\left( {\frac{x}{2}} \right)} \right]_0^1 + \int_0^1 {\frac{1}{{\sqrt {4 – {x^2}} }}} dx$ A1

using integration by substitution or inspection (M1)

$\left[ {x\,\,{\text{arccos}}\left( {\frac{x}{2}} \right)} \right]_0^1 + \left[ { – {{\left( {4 – {x^2}} \right)}^{\frac{1}{2}}}} \right]_0^1$ A1

Note: Award A1 for ${ – {{\left( {4 – {x^2}} \right)}^{\frac{1}{2}}}}$ or equivalent.

Note: Condone lack of limits to this point.

attempt to substitute limits into their integral M1

$ = \frac{\pi }{3} – \sqrt 3 + 2$ A1

[7 marks]

Question

It is given that ${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$.

Show that ${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{2}{\text{lo}}{{\text{g}}_r}\,x$ where $r,\,x \in {\mathbb{R}^ + }$.[2]

Express $y$ in terms of $x$. Give your answer in the form $y = p{x^q}$, where p , q are constants.[5]

The region R, is bounded by the graph of the function found in part (b), the x-axis, and the lines $x = 1$ and $x = \alpha $ where $\alpha > 1$. The area of R is $\sqrt 2 $.

Find the value of $\alpha $.[5]

Answer/Explanation

Markscheme

METHOD 1

${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{lo}}{{\text{g}}_r}\,{r^2}}}\left( { = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{2}}\,{\text{lo}}{{\text{g}}_r}\,r}}} \right)$ M1A1

$ = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}$ AG

[2 marks]

METHOD 2

${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{{{\text{lo}}{{\text{g}}_x}\,{r^2}}}$ M1

$ = \frac{1}{{2\,{\text{lo}}{{\text{g}}_x}\,r}}$ A1

$ = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}$ AG

[2 marks]

METHOD 1

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,2{x^2} = 0$ M1

${\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2} = 0$ M1

${\text{lo}}{{\text{g}}_2}\,y = – \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2}$

${\text{lo}}{{\text{g}}_2}\,y = {\text{lo}}{{\text{g}}_2}\left( {\frac{1}{{\sqrt {2x} }}} \right)$ M1A1

$y = \frac{1}{{\sqrt 2 }}{x^{ – 1}}$ A1

Note: For the final A mark, $y$ must be expressed in the form $p{x^q}$.

[5 marks]

METHOD 2

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$

${\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,x + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2x = 0$ M1

${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_2}\,{x^{\frac{1}{2}}} + {\text{lo}}{{\text{g}}_2}\,{\left( {2x} \right)^{\frac{1}{2}}} = 0$ M1

${\text{lo}}{{\text{g}}_2}\,\left( {\sqrt 2 xy} \right) = 0$ M1

$\sqrt 2 xy = 1$ A1

$y = \frac{1}{{\sqrt 2 }}{x^{ – 1}}$ A1

Note: For the final A mark, $y$ must be expressed in the form $p{x^q}$.

[5 marks]

the area of R is $\int\limits_1^\alpha {\frac{1}{{\sqrt 2 }}} {x^{ – 1}}{\text{d}}x$ M1

$ = \left[ {\frac{1}{{\sqrt 2 }}{\text{ln}}\,x} \right]_1^\alpha $ A1

$ = \frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha $ A1

$\frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha = \sqrt 2 $ M1

$\alpha = {{\text{e}}^2}$ A1

Note: Only follow through from part (b) if $y$ is in the form $y = p{x^q}$

[5 marks]

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