Differentiation of trigonometric functions and differentiation rules — Question 160

Original QCE Vault practice · 5 marks

Q160 · Practice questionTechnology-freeComplex unfamiliar5 marks

QUESTION 160 (5 marks)

Let

f(x)=\dfrac{\sin x}{2+\cos x}

for

0\le x\le2\pi

. Determine the global maximum and global minimum values of

f

on this interval, and the

x

-values at which they occur.

WORKED SOLUTION

5 marks

ANSWER

Global maximum

\dfrac{\sqrt3}{3}

x=\dfrac{2\pi}{3}

; global minimum

-\dfrac{\sqrt3}{3}

x=\dfrac{4\pi}{3}

Worked solution

f'(x)=\dfrac{2\cos x+1}{(2+\cos x)^2}

. Stationary points occur when

\cos x=-1/2

, so

x=2\pi/3,4\pi/3

. Evaluate these and the endpoints:

f(0)=f(2\pi)=0

f(2\pi/3)=\sqrt3/3

f(4\pi/3)=-\sqrt3/3

Differentiates using the quotient rule.

[1 mark]

Solves the stationary-point equation.

[1 mark]

Evaluates the first stationary value.

[1 mark]

Evaluates the second stationary value and endpoints.

[1 mark]

States the global maximum and minimum correctly.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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