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Date	November 2018	Marks available	5	Reference code	18N.1.AHL.TZ0.H_10
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Find	Question number	H_10	Adapted from	N/A

Question

The function $f$ is defined by $f\left( x \right) = {{\text{e}}^x}\,{\text{cos}}{\,^2}x$ , where 0 ≤ $x$ ≤ 5. The curve $y = f\left( x \right)$ is shown on the following graph which has local maximum points at A and C and touches the $x$ -axis at B and D.

Use integration by parts to show that $\int {{{\text{e}}^x}\,{\text{cos}}\,2x{\text{d}}x = } \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x + c$ , $c \in \mathbb{R}$ .

[5]

Hence, show that $\int {{{\text{e}}^x}\,{\text{cos}}{\,^2}x{\text{d}}x = } \frac{{{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{{10}}{\text{cos}}\,2x + \frac{{{{\text{e}}^x}}}{2} + c$ , $c \in \mathbb{R}$ .

[3]

Find the $x$ -coordinates of A and of C , giving your answers in the form $a + {\text{arctan}}\,b$ , where $a$ , $b \in \mathbb{R}$ .

[6]

Find the area enclosed by the curve and the $x$ -axis between B and D, as shaded on the diagram.

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

attempt at integration by parts with $u = {{\text{e}}^x}$ , $\frac{{{\text{d}}v}}{{{\text{d}}x}} = {\text{cos}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x = } \frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x\,{\text{d}}x - \int {\frac{{{{\text{e}}^x}}}{2}} {\text{sin}}\,2x\,{\text{d}}x$ A1

= $\frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x - \frac{1}{2}\left( { - \frac{{{{\text{e}}^x}}}{2}{\text{cos}}\,2x + \int {\frac{{{{\text{e}}^x}}}{2}} {\text{cos}}\,2x} \right)$ M1A1

= $\frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{4}{\text{cos}}\,2x - \frac{1}{4}\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x}$

$\therefore \frac{5}{4}\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{{{\text{e}}^x}}}{2}{\text{sin}}\,2x\, + \frac{{{{\text{e}}^x}}}{4}{\text{cos}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x\left( { + c} \right)$ AG

METHOD 2

attempt at integration by parts with $u = {\text{cos}}\,2x$ , $\frac{{{\text{d}}v}}{{{\text{d}}x}} = {{\text{e}}^x}$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x = } {{\text{e}}^x}\,{\text{cos}}\,2x + 2\int {{{\text{e}}^x}\,{\text{sin}}\,2x\,{\text{d}}x}$ A1

$= {{\text{e}}^x}\,{\text{cos}}\,2x + 2\left( {{{\text{e}}^x}\,{\text{sin}}\,2x - 2\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} } \right)$ M1A1

$= {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x - 4\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x}$

$\therefore 5\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x\left( { + c} \right)$ AG

METHOD 3

attempt at use of table M1

A1A1

Note: A1 for first 2 lines correct, A1 for third line correct.

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x = \,} {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x - 4\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x}$ M1

$\therefore 5\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = {{\text{e}}^x}\,{\text{cos}}\,2x + 2{{\text{e}}^x}\,{\text{sin}}\,2x$ M1

$\int {{{\text{e}}^x}\,{\text{cos}}\,2x\,{\text{d}}x} = \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x\left( { + c} \right)$ AG

[5 marks]

$\int {{{\text{e}}^x}\,{\text{co}}{{\text{s}}^2}\,x{\text{d}}x = } \int {\frac{{{{\text{e}}^x}}}{2}} \left( {{\text{cos}}\,2x + 1} \right){\text{d}}x$ M1A1

$= \frac{1}{2}\left( {\frac{{{\text{2}}{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x} \right) + \frac{{{{\text{e}}^x}}}{2}$ A1

$= \frac{{{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{{10}}{\text{cos}}\,2x + \frac{{{{\text{e}}^x}}}{2}\left( { + c} \right)$ AG

Note: Do not accept solutions where the RHS is differentiated.

[3 marks]

$f'\left( x \right) = {{\text{e}}^x}\,{\text{co}}{{\text{s}}^{\text{2}}}\,x - 2{{\text{e}}^x}\,{\text{sin}}\,x\,{\text{cos}}\,x$ M1A1

Note: Award M1 for an attempt at both the product rule and the chain rule.

${{\text{e}}^x}\,{\text{cos}}\,x\left( {{\text{cos}}\,x - 2\,{\text{sin}}\,x} \right) = 0$ (M1)

Note: Award M1 for an attempt to factorise ${{\text{cos}}\,x}$ or divide by ${\text{cos}}\,x\left( {{\text{cos}}\,x \ne 0} \right)$ .

discount ${\text{cos}}\,x = 0$ (as this would also be a zero of the function)

$\Rightarrow {\text{cos}}\,x - 2\,{\text{sin}}\,x = 0$

$\Rightarrow {\text{tan}}\,x = \frac{1}{2}$ (M1)

$\Rightarrow x = {\text{arctan}}\left( {\frac{1}{2}} \right)$ (at A) and $x = \pi + {\text{arctan}}\left( {\frac{1}{2}} \right)$ (at C) A1A1

Note: Award A1 for each correct answer. If extra values are seen award A1A0.

[6 marks]

${\text{cos}}\,x = 0 \Rightarrow x = \frac{\pi }{2}$ or $\frac{{3\pi }}{2}$ A1

Note: The A1may be awarded for work seen in part (c).

$\int_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} {\left( {{{\text{e}}^x}\,{\text{co}}{{\text{s}}^{\text{2}}}\,x} \right)} \,{\text{d}}x = \left[ {\frac{{{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{{10}}{\text{cos}}\,2x + \frac{{{{\text{e}}^x}}}{2}} \right]_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}}$ M1

$= \left( { - \frac{{{{\text{e}}^{\frac{{3\pi }}{2}}}}}{{10}} + \frac{{{{\text{e}}^{\frac{{3\pi }}{2}}}}}{2}} \right) - \left( { - \frac{{{{\text{e}}^{\frac{\pi }{2}}}}}{{10}} + \frac{{{{\text{e}}^{\frac{\pi }{2}}}}}{2}} \right)\left( { = \frac{{{\text{2}}{{\text{e}}^{\frac{{3\pi }}{2}}}}}{5} - \frac{{{\text{2}}{{\text{e}}^{\frac{\pi }{2}}}}}{5}} \right)$ M1(A1)A1

Note: Award M1 for substitution of the end points and subtracting, (A1) for ${\text{sin}}\,3\pi = {\text{sin}}\,\pi = 0$ and ${\text{cos}}\,3\pi = {\text{cos}}\,\pi = - 1$ and A1 for a completely correct answer.

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.7—The second derivative

Show 36 related questions

22M.1.SL.TZ1.7e.ii:
Find the values of $x$ for which the graph of $f$ is concave-down. Justify your answer.
22M.1.SL.TZ1.7e.i:
Write down the value of $a$ .
18M.1.AHL.TZ1.H_9c:
Sketch the graph of $y = f\left( x \right)$ showing clearly the position of the points A and B.
19M.1.AHL.TZ1.H_8d:
Sketch the curve $y = f\left( x \right)$ , 0 ≤ $x$ ≤ 5 indicating clearly the coordinates of the maximum and minimum points and any intercepts with the coordinate axes.
21N.3.AHL.TZ0.1f:
Sketch the graph of $x^{2} - y^{2} = 1$ , stating the coordinates of any axis intercepts and the equation of each asymptote.
19N.1.SL.TZ0.S_8a:
Write down an expression for $y$ in terms of $x$ .
18N.1.AHL.TZ0.H_10b:
Hence, show that $\int {{{\text{e}}^x}\,{\text{cos}}{\,^2}x{\text{d}}x = } \frac{{{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{{10}}{\text{cos}}\,2x + \frac{{{{\text{e}}^x}}}{2} + c$ , $c \in \mathbb{R}$ .
21N.1.SL.TZ0.9c.ii:
Find the values of $x$ where the graph of $f$ has points of inflexion. Justify your answer.
19N.1.SL.TZ0.S_8d.i:
Find the value of $x$ for which $V$ is a maximum.
18M.1.AHL.TZ1.H_9b.ii:
The coordinates of B can be expressed in the form B $\left( {{2^a},\,b \times {2^{ - 3a}}} \right)$ where a, b $\in \mathbb{Q}$ . Find the value of a and the value of b.
19N.1.SL.TZ0.S_8b:
Find an expression for $V$ in terms of $x$ .
18N.1.AHL.TZ0.H_10a:
Use integration by parts to show that $\int {{{\text{e}}^x}\,{\text{cos}}\,2x{\text{d}}x = } \frac{{2{{\text{e}}^x}}}{5}{\text{sin}}\,2x + \frac{{{{\text{e}}^x}}}{5}{\text{cos}}\,2x + c$ , $c \in \mathbb{R}$ .
22M.1.AHL.TZ1.12c.i:
Show that $g (x)$ satisfies the equation $g'' (x) = 2 (g' (x) - g (x))$ .
22M.2.AHL.TZ2.12d:
Hence determine the maximum value of $\frac{d P}{d t}$ in terms of $k$ and $N$ .
19M.1.AHL.TZ1.H_8b:
find the value of $f\left( 1 \right)$ .
19M.1.AHL.TZ2.H_8a:
Show that the volume of the cone may be expressed by $V = \frac{\pi }{3}\left( {2R{h^2} - {h^3}} \right)$ .
19N.1.SL.TZ0.S_8e:
Find the maximum volume.
19N.1.SL.TZ0.S_8d.ii:
Justify your answer.
19N.1.SL.TZ0.S_8c:
Find $\frac{{{\text{d}}V}}{{{\text{d}}x}}$ .
19M.1.AHL.TZ1.H_8c:
find the value of $f\left( 4 \right)$ .
19M.1.AHL.TZ2.H_8b:
Given that there is one inscribed cone having a maximum volume, show that the volume of this cone is $\frac{{32\pi {R^3}}}{{81}}$ .
21N.1.SL.TZ0.9a:
Find all the values of $x$ where the graph of $f$ is increasing. Justify your answer.
21N.1.SL.TZ0.9c.i:
Find the value of $x$ where the graph of $f$ has a local minimum. Justify your answer.
21N.1.SL.TZ0.9d:
The total area of the region enclosed by the graph of $f'$ , the derivative of $f$ , and the $x$ -axis is $20$ .

Given that $f (p) + f (t) = 4$ , find the value of $f (0)$ .
21N.3.AHL.TZ0.1b:
Show that ${(f (t))}^{2} + {(g (t))}^{2} = f (2 t)$ .
21N.3.AHL.TZ0.1c.i:
$f (i u)$ .
21N.3.AHL.TZ0.1g:
The hyperbola with equation $x^{2} - y^{2} = 1$ can be rotated to coincide with the curve defined by $x y = k, k \in ℝ$ .

Find the possible values of $k$ .
21N.1.SL.TZ0.9b:
Find the value of $x$ where the graph of $f$ has a local maximum.
21N.3.AHL.TZ0.1a:
Verify that $u = f (t)$ satisfies the differential equation $\frac{d^{2} u}{d t^{2}} = u$ .
21N.3.AHL.TZ0.1e:
Show that ${(f (t))}^{2} - {(g (t))}^{2} = {(f (i u))}^{2} - {(g (i u))}^{2}$ .
19M.1.AHL.TZ1.H_8a:
Write down the $x$ -coordinate of the point of inflexion on the graph of $y = f\left( x \right)$ .
17M.1.AHL.TZ1.H_12e.i:
Show that the graph of $y = q(x)$ is concave up for $x > 1$ .
17M.1.AHL.TZ1.H_12e.ii:
Sketch the graph of $y = q(x)$ showing clearly any intercepts with the axes.
18N.1.AHL.TZ0.H_10c:
Find the $x$ -coordinates of A and of C , giving your answers in the form $a + {\text{arctan}}\,b$ , where $a$ , $b \in \mathbb{R}$ .
21N.3.AHL.TZ0.1c.ii:
$g (i u)$ .
21N.3.AHL.TZ0.1d:
Hence find, and simplify, an expression for ${(f (i u))}^{2} + {(g (i u))}^{2}$ .

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