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MHTCET 16 Apr 2026 Shift 2 Paper

Question : 122 of 150

Marks: +1, -0

\cot [f(x)]=\frac{3 x-x^{3}}{1-3 x^{2}}

\sin [g(x)]=\frac{1-x^{2}}{1+x^{2}}

\lim _{x \rightarrow t} \frac{f(x)-f(t)}{g(x)-g(t)}=\ldots

Solution:

Concept:

Use trigonometric substitutions and the derivative ratio for the limit of an indeterminate form.

Explanation:

Put

x=\tan \theta

, so

\theta=\tan^{-1}x

Since

\tan 3\theta = \frac{3x-x^3}{1-3x^2}

, we have

\cot[f(x)] = \tan 3\theta = \cot\!\left(\frac{\pi}{2}-3\theta\right)

Hence we may take

f(x)=\frac{\pi}{2}-3\tan^{-1}x

Also,

\frac{1-x^2}{1+x^2}=\cos 2\theta = \sin\!\left(\frac{\pi}{2}-2\theta\right)

, so

g(x)=\frac{\pi}{2}-2\tan^{-1}x

The required limit equals

\frac{f'(t)}{g'(t)}

Now

f'(x)=\frac{-3}{1+x^2}

and

g'(x)=\frac{-2}{1+x^2}

Thus

\frac{f'(t)}{g'(t)}=\frac{\frac{-3}{1+t^2}}{\frac{-2}{1+t^2}}=\frac{3}{2}

Answer:

\frac{3}{2}

, i.e. option B.

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