Q67 / 150 · MHT CET 2026 (19 April Shift 1) · Mathematics
If $\sin^{-1}\left(\tan\dfrac{\pi}{4}\right) - \sin^{-1}\left(\sqrt{\dfrac{3}{x}}\right) = \dfrac{\pi}{6}$ then $x$ is a root of the equation
- A$x^2 - x - 6 = 0$
- Correct answerB$x^2 - x - 12 = 0$
- C$x^2 + x - 12 = 0$
- D$x^2 + x - 6 = 0$
Solution
Given equation is $\sin^{-1}\left(\tan\dfrac{\pi}{4}\right) - \sin^{-1}\left(\sqrt{\dfrac{3}{x}}\right) = \dfrac{\pi}{6}$
Since $\tan\dfrac{\pi}{4} = 1$, we have:
$\sin^{-1}(1) - \sin^{-1}\left(\sqrt{\dfrac{3}{x}}\right) = \dfrac{\pi}{6}$
$\dfrac{\pi}{2} - \sin^{-1}\left(\sqrt{\dfrac{3}{x}}\right) = \dfrac{\pi}{6}$
$\sin^{-1}\left(\sqrt{\dfrac{3}{x}}\right) = \dfrac{\pi}{2} - \dfrac{\pi}{6} = \dfrac{\pi}{3}$
$\sqrt{\dfrac{3}{x}} = \sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2}$
Squaring both sides, we get:
$\dfrac{3}{x} = \dfrac{3}{4}$
$x = 4$
Substituting $x = 4$ into the given options, we find that it satisfies the equation $x^2 - x - 12 = 0$ since $4^2 - 4 - 12 = 16 - 16 = 0$.
Answer: $x^2 - x - 12 = 0$
Since $\tan\dfrac{\pi}{4} = 1$, we have:
$\sin^{-1}(1) - \sin^{-1}\left(\sqrt{\dfrac{3}{x}}\right) = \dfrac{\pi}{6}$
$\dfrac{\pi}{2} - \sin^{-1}\left(\sqrt{\dfrac{3}{x}}\right) = \dfrac{\pi}{6}$
$\sin^{-1}\left(\sqrt{\dfrac{3}{x}}\right) = \dfrac{\pi}{2} - \dfrac{\pi}{6} = \dfrac{\pi}{3}$
$\sqrt{\dfrac{3}{x}} = \sin\left(\dfrac{\pi}{3}\right) = \dfrac{\sqrt{3}}{2}$
Squaring both sides, we get:
$\dfrac{3}{x} = \dfrac{3}{4}$
$x = 4$
Substituting $x = 4$ into the given options, we find that it satisfies the equation $x^2 - x - 12 = 0$ since $4^2 - 4 - 12 = 16 - 16 = 0$.
Answer: $x^2 - x - 12 = 0$
Hence, the correct answer is option: 'B'.