If ^{-1}≤ft( π/4)- ^{-1}≤ft({3/x})=π/6 then x is a root of the equation

Correct Answer is : x2−x−12=0x²-x-12=0x2−x−12=0

Correct Answer is : x2−x−12=0x²-x-12=0x2−x−12=0

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MHTCET 19 Apr 2026 Shift 1 Paper

Question : 119 of 150

Marks: +1, -0

\sin ^{-1}\left(\tan \frac{\pi}{4}\right)-\sin ^{-1}\left(\sqrt{\frac{3}{x}}\right)=\frac{\pi}{6}

x

is a root of the equation

Solution:

Concept:

Use principal values of inverse trigonometric functions and simplify to find

x

Explanation:

Since

\tan \frac{\pi}{4}=1

, we get

\sin^{-1}(1)=\frac{\pi}{2}

The given equation becomes:

\frac{\pi}{2}-\sin^{-1}\left(\sqrt{\frac{3}{x}}\right)=\frac{\pi}{6}

So,

\sin^{-1}\left(\sqrt{\frac{3}{x}}\right)=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}

Taking sine on both sides:

\sqrt{\frac{3}{x}}=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}

Squaring both sides:

\frac{3}{x}=\frac{3}{4} \Rightarrow x=4

Now check which quadratic equation has

x=4

as a root.

For

x=4

x^{2}-x-12=16-4-12=0

Hence,

x=4

satisfies

x^{2}-x-12=0

Answer:

Option B:

x^{2}-x-12=0

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