Practicing Success
Which of the following is the solution set of the equation $sin^{-1}x = cos^{-1}x + sin^{-1}(3x-1)$? |
$\left[0,\frac{1}{3}\right]$ $\left[\frac{1}{3},\frac{2}{3}\right]$ $\left[0,\frac{2}{3}\right]$ none of these |
none of these |
For the existence of the given equation, we must have -1 ≤ x ≤ 1 and -1 ≤ 3x -1 ≤ 1 ⇒ 0 ≤ x ≤ $\frac{2}{3}$ Now, $sin^{-1}x = cos^{-1}x +sin^{-1}(3x-1)$ $⇒sin^{-1}x - cos^{-1}x =sin^{-1}(3x-1)$ $⇒ 2sin^{-1}x-\frac{\pi}{2}=sin^{-1}(3x-1)$ $⇒ sin\left(2sin^{-1}x-\frac{\pi}{2}\right) = sin \left(sin^{-1}(3x-1)\right)$ $⇒ -cos (2sin^{-1}x) = 3x -1 $ $⇒ - \begin{Bmatrix}1- 2 sin^2 (sin^{-1}x)\end{Bmatrix}= 3x - 1$ $⇒ -(1 -2x^2) = 3x - 1$ $⇒ 2x^2 -3x = 0 ⇒x = 0, \frac{3}{2} ⇒ x = 0 $ [∵ 0 ≤ x ≤ 2/3] |