The equation, x + 2 2x + 3 3x = 8/πhas at least one root in

Correct Answer is : [0,π2]≤ft[0, π/2][0,2π]

Correct Answer is : [0,π2]≤ft[0, π/2][0,2π]

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CGPET 2017 Solved Paper

Section: Mathematics

Question : 125 of 150

Marks: +1, -0

The equation,

\sin x + 2 \sin 2x + 3 \sin 3x = \frac{8}{\pi}

has at least one root in

Solution:

We have,

\sin x + 2 \sin 2x + 3 \sin 3x = \frac{8}{\pi}

\Rightarrow \frac{8}{\pi} - \sin x - 2 \sin 2x - 3 \sin 3x = 0

Let,

f(x) = \frac{8}{\pi} x + \cos x + \cos 2x + \cos 3x

f(0) = 0 + 1 + 1 + 1 = 3

f\left(\frac{\pi}{2}\right) = 4 + 0 - 1 + 0 = 3

∴

f(0) = f\left(\frac{\pi}{2}\right)

Now,

f'(x) = \frac{8}{\pi} - \sin x - 2 \sin 2x - 3 \sin 3x

By Rolles theorem,

at least one root of

f(x)

lies between

\left(0, \frac{\pi}{2}\right) \leq \left[0, \frac{\pi}{2}\right]

i.e.

\frac{8}{\pi} - \sin x - 2 \sin 2x - 3 \sin 3x = 0

for some

x \in \left[0, \frac{\pi}{2}\right]

\Rightarrow \sin x + 2 \sin 2x + 3 \sin 3x = \frac{8}{\pi}

for some

x \in \left[0, \frac{\pi}{2}\right]

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