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Date November 2017 Marks available 5 Reference code 17N.2.hl.TZ0.10
Level HL only Paper 2 Time zone TZ0
Command term Show that Question number 10 Adapted from N/A

Question

Consider the function \(f(x) = \frac{{\sqrt x }}{{\sin x}},{\text{ }}0 < x < \pi \).

Consider the region bounded by the curve \(y = f(x)\), the \(x\)-axis and the lines \(x = \frac{\pi }{6},{\text{ }}x = \frac{\pi }{3}\).

Show that the \(x\)-coordinate of the minimum point on the curve \(y = f(x)\) satisfies the equation \(\tan x = 2x\).

[5]
a.i.

Determine the values of \(x\) for which \(f(x)\) is a decreasing function.

[2]
a.ii.

Sketch the graph of \(y = f(x)\) showing clearly the minimum point and any asymptotic behaviour.

[3]
b.

Find the coordinates of the point on the graph of \(f\) where the normal to the graph is parallel to the line \(y =  - x\).

[4]
c.

This region is now rotated through \(2\pi \) radians about the \(x\)-axis. Find the volume of revolution.

[3]
d.

Markscheme

attempt to use quotient rule or product rule     M1

\(f’(x) = \frac{{\sin x\left( {\frac{1}{2}{x^{ - \frac{1}{2}}}} \right) - \sqrt x \cos x}}{{{{\sin }^2}x}}{\text{ }}\left( { = \frac{1}{{2\sqrt x \sin x}} - \frac{{\sqrt x \cos x}}{{{{\sin }^2}x}}} \right)\)     A1A1

 

Note:     Award A1 for \(\frac{1}{{2\sqrt x \sin x}}\) or equivalent and A1 for \( - \frac{{\sqrt x \cos x}}{{{{\sin }^2}x}}\) or equivalent.

 

setting \(f’(x) = 0\)     M1

\(\frac{{\sin x}}{{2\sqrt x }} - \sqrt x \cos x = 0\)

\(\frac{{\sin x}}{{2\sqrt x }} = \sqrt x \cos x\) or equivalent     A1

\(\tan x = 2x\)     AG

[5 marks]

a.i.

\(x = 1.17\)

\(0 < x \leqslant 1.17\)     A1A1

 

Note:     Award A1 for \(0 < x\) and A1 for \(x \leqslant 1.17\). Accept \(x < 1.17\).

 

[2 marks]

a.ii.

N17/5/MATHL/HP2/ENG/TZ0/10.b/M

concave up curve over correct domain with one minimum point above the \(x\)-axis.     A1

approaches \(x = 0\) asymptotically     A1

approaches \(x = \pi \) asymptotically     A1

 

Note:     For the final A1 an asymptote must be seen, and \(\pi \) must be seen on the \(x\)-axis or in an equation.

 

[3 marks]

b.

\(f’(x){\text{ }}\left( { = \frac{{\sin x\left( {\frac{1}{2}{x^{ - \frac{1}{2}}}} \right) - \sqrt x \cos x}}{{{{\sin }^2}x}}} \right) = 1\)     (A1)

attempt to solve for \(x\)     (M1)

\(x = 1.96\)     A1

\(y = f(1.96 \ldots )\)

\( = 1.51\)     A1

[4 marks]

c.

\(V = \pi \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{x{\text{d}}x}}{{{{\sin }^2}x}}} \)     (M1)(A1)

 

Note:     M1 is for an integral of the correct squared function (with or without limits and/or \(\pi \)).

 

\( = 2.68{\text{ }}( = 0.852\pi )\)     A1

[3 marks]

d.

Examiners report

[N/A]
a.i.
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a.ii.
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b.
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c.
[N/A]
d.

Syllabus sections

Topic 6 - Core: Calculus » 6.1 » Informal ideas of limit, continuity and convergence.

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