If f(x)= ^{-1}≤ft({3 x-x³}{1-3 x²}) and g(x)= ^{-1}≤ft({1-x²}{1+x²}), then ₓ arrow a} f(x)-f(a)/g(x)-g(a),≤ft(0

Correct Answer is : −32-3/2−23

Correct Answer is : −32-3/2−23

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MHTCET 17 Apr 2026 Shift 2 Paper

Question : 123 of 150

Marks: +1, -0

f(x)=\cot ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)

g(x)=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)

\lim _{x \rightarrow a} \frac{f(x)-f(a)}{g(x)-g(a)},\left(0 < a < \frac{1}{2}\right)

Solution:

Concept:

Use the substitution

x = \tan\theta

to simplify both inverse trigonometric functions, then find the limit as the ratio of derivatives at

x = a

Explanation:

Let

x = \tan\theta

Since

0 < a < \frac{1}{2}

, we have

0 < \theta = \tan^{-1}x < \tan^{-1}\left(\frac{1}{2}\right) < \frac{\pi}{6}

Using the identity

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

we get

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

Because

0 < 3\theta < \frac{\pi}{2}

, we can write:

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

Differentiating,

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

Also,

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

Since

0 < 2\theta < \pi

g(x) = \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1}x

Differentiating,

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

Now, the given limit is the ratio of derivatives at

x = a

\lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(a)}{g'(a)}

= \frac{-\frac{3}{1 + a^2}}{\frac{2}{1 + a^2}} = -\frac{3}{2}

Answer:

Option C:

-\frac{3}{2}

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