If x ∈ (0, (π)/(2)) then the value of ^ -1 ((7)/(2)(1+ 2 x)+ ( ^2x-48 ^2 x) x ) is equal to

Correct answer: A. x- ^ -1 (7 x) y= ^ -1 ((7)/(2)(1+ 2 x)+ ( ^2 x-48 ^2 x) x)= ^ -1 ((7 x)( x)+ 1-49 ^2 x 1- ^2 x )= ^ -1 ( x)- ^ -1 (7 x) [ x < 7 x]=x- ^ -1 (7 x)

BITSAT 2022

Exam: BITSAT
Year: 2022
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

BITSAT 2022 · Mathematics · Inverse Trigonometric Functions

If x ∈ (0, (π)/(2)) then the value of ^ -1 ((7)/(2)(1+ 2 x)+ (...

x \in \left(0, \frac{\pi}{2}\right)

then the value of

\cos ^{-1} \left(\frac{7}{2}(1+\cos 2 x)+\sqrt{ \left(\sin ^2x-48 \cos ^2 x\right) \sin x}\right)

is equal to

Options

A $x-\cos ^{-1}(7 \cos x)$
B $x+\sin ^{-1}(7 \cos x)$
C $x+\cos ^{-1}(6 \cos x)$
D $x+\cos ^{-1}(7 \cos x)$

Correct answer

A. $x-\cos ^{-1}(7 \cos x)$

Step-by-step solution

y=\cos ^{-1}\left(\frac{7}{2}(1+\cos 2 x)+\sqrt{\left(\sin ^2 x-48 \cos ^2 x\right)} \sin x\right)=\cos ^{-1}\left((7 \cos x)(\cos x)+\sqrt{1-49 \cos ^2 x} \sqrt{1-\cos ^2 x}\right)=\cos ^{-1}(\cos x)-\cos ^{-1}(7 \cos x) \quad[\because \cos x < 7 \cos x]=x-\cos ^{-1}(7 \cos x)

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