Let f(x) = (sin x + 3 3x + 5 5x + 3 7x)/( 2x + 2 4x + 3 6x) , wherever defined. If x _ 1 + x _ 2 = (π)/(2), where f (x) is defined at x _ 1 and x _ 2 , then f^ 2 ( x _ 1 ) +f^ 2 ( x _ 2 ) is

Correct answer: C. 4 Numerator of f(x) = (sin x + 3x) + 2 ( 3x + 5x) + 3 ( 5x + 7x) = 2cos x ( 2x + 2 4x + 3 6x) Hence, f(x) =2cos x, if f( x _ 1 ) =2cos x _ 1 & f( x _ 2 ) =2sin x _ 1 i.e. f^ 2 ( x _ 1 )+f^ 2 ( x _ 2 ) =4

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

Let f(x) = (sin x + 3 3x + 5 5x + 3 7x)/( 2x + 2 4x + 3 6x) ,...

Let

f\left(x\right) = \dfrac{sin x + 3\sin 3x + 5\sin 5x + 3\sin 7x}{\sin 2x + 2\sin 4x + 3\sin 6x}

, wherever defined. If

{x}_{1} + {x}_{2} = \dfrac{\pi}{2},

where

f \left(x\right)

is defined at

{x}_{1} and {x}_{2}, then

f^{2} \left({x}_{1}\right)

+f^{2} \left({x}_{2}\right)

Options

A $\cos^{2} x$
B $\sin^{2} x$
C $4$
D $1$

Correct answer

C. $4$

Step-by-step solution

Numerator of $f\left(x\right) = \left(sin x + \sin 3x\right) + 2 \left(\sin 3x + \sin 5x\right) + 3 \left(\sin 5x + \sin 7x\right)<br />$ = 2cos x \left(\sin 2x + 2\sin 4x + 3\sin 6x\right) $<br />Hence,$ f\left(x\right) =2cos x, $if$ f\left({x}_{1}\right) =2cos {x}_{1} & f\left({x}_{2}\right) =2sin {x}_{1} $</p><p>i.e.$ f^{2}\left({x}_{1}\right)+f^{2}\left({x}_{2}\right) =4

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