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

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

The value(s) of x satisfying the equation ^ -1 (1 - x) - 2 ^ -1 x =...

The value(s)ofxof x satisfying the equation sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{-1} \left(1 - x\right) - 2\sin^{-1} x = \dfrac{\pi}{2}& is/are

Options

  1. A00
  2. B12\dfrac{1}{2}
  3. C0,120, \dfrac{1}{2}
  4. D−12- \dfrac{1}{2}

Correct answer

A. 00

Step-by-step solution

We have,
sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{- 1} \left(1 - x\right) - 2\sin^{- 1} x = \dfrac{\pi}{2}
⇒sin⁡−1(1−x)=π2+2sin⁡−1x⇒ \sin^{- 1} \left(1 - x\right) = \dfrac{\pi}{2} + 2\sin^{- 1} x
⇒1−x=sin⁡(π2+2sin⁡−1x)⇒ 1 - x = \sin \left(\dfrac{\pi}{2} + 2\sin^{- 1} x\right)
⇒1−x=cos⁡(2sin⁡−1x)⇒ 1 - x = \cos \left(2\sin^{- 1} x\right)
⇒1−x=cos⁡{cos⁡−1(1−2x2)}[∵2sin⁡−1x=cos⁡−1(1−2x2)]⇒ 1 - x = \cos \left\{\cos^{- 1} \left(1 - 2{x}^{2}\right)\right\}\left[∵ 2\sin^{- 1} x = \cos^{- 1} \left(1 - 2{x}^{2}\right)\right]
⇒1−x=(1−2x2)⇒ 1 - x = \left(1 - 2{x}^{2}\right)
⇒x=2x2⇒x(2x−1)=0⇒x=0,12⇒ x = 2{x}^{2} ⇒ x\left(2x - 1\right) = 0 ⇒ x = 0, \dfrac{1}{2}
For, x=12,x = \dfrac{1}{2}, we have
LHS =sin⁡−1(1−x)−2sin⁡−1x= \sin^{- 1} \left(1 - x\right) - 2\sin^{- 1} x
=sin⁡−112−2sin⁡−112=−sin⁡−112=−π6≠R.H.S.= \sin^{- 1} \dfrac{1}{2} - 2\sin^{- 1} \dfrac{1}{2} = - \sin^{- 1} \dfrac{1}{2} = \dfrac{- \pi}{6} \ne R . H . S .
So, x=12x = \dfrac{1}{2} is not a root of the given equation.
Clearly, x=0x = 0 satisfies the equation
Here, x=0x = 0 is the root of the given equation.
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