The number of roots of the equation sin^ - 1 x - cos^ - 1 x = sin^ - 1 (5x - 3) is/are

Correct answer: B. 1 sin^ - 1 x - cos^ - 1 x = sin^ - 1 (5x - 3) ⇒ (π)/(2) - cos^ - 1 x - cos^ - 1 x = (π)/(2) - cos^ - 1 (5x - 3) ⇒ 2cos^ - 1 x = cos^ - 1 (5x - 3). Alsox ∈ [-1,1] … (1) ⇒ cos^ - 1 (2x^ 2 - 1) = cos^ - 1 (5x - 3) and (5x - 3)∈ [-1,1] , i.e., - 1 ≤ 5x - 3 ≤ 1 ⇒ 2x^ 2 - 1 = 5x - 3, hence, x ∈ [(2)/(5), (4)/(5)] ⇒ 2x^ 2 - 5x + 2 = 0 ⇒ x = 2 or (1)/(2) but, x = 2 does not satisfy the equation (1) Hence, the given equation has only one root

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

The number of roots of the equation sin^ - 1 x - cos^ - 1 x = sin^...

The number of roots of the equation

sin^{- 1}x - cos^{- 1}x = sin^{- 1} \left(5x - 3\right) is/are

Options

A $3$
B $1$
C $2$
D $0$

Correct answer

B. $1$

Step-by-step solution

$sin^{- 1}x - cos^{- 1}x = sin^{- 1} \left(5x - 3\right)$
$⇒ \dfrac{\pi}{2} -$ cos^{- 1}x - cos^{- 1}x = $\dfrac{\pi}{2} - cos^{- 1} \left(5x - 3\right)$
$⇒ 2cos^{- 1}x = cos^{- 1} \left(5x - 3\right). Alsox \in \left[-1,1\right]$ … $\left(1\right)$
$⇒ cos^{- 1} \left(2x^{2} - 1\right) = cos^{- 1}$ \left(5x - 3\right) $and \left(5x - 3\right)\in \left[-1,1\right]$ , $i.e.,$ - 1 ≤ 5x - 3 ≤ 1 $<br />$ ⇒ 2x^{2} - 1 = 5x - 3, hence, x \in \left[\dfrac{2}{5}, \dfrac{4}{5}\right]< $br />$ ⇒ 2x^{2} - 5x + 2 = 0 ⇒ x = 2 $or$ \dfrac{1}{2} $<br />but,$ x = 2 $does not satisfy the equation \left(1\right)$

Hence, the given equation has only one root

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