NIMCET 2021 — Mathematics PYQ

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Question

NIMCET 2021 — Mathematics PYQ

NIMCET | Mathematics | 2021

If $y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right), \; -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}, \; \text{then } \frac{dy}{dx} \text{ is}$

Choose the correct answer:

Explanation

1. Trigonometric Substitution

To simplify the expression inside the inverse tangent, let's substitute:

x = \tan \theta \implies \theta = \tan^{-1} x

The expression becomes:

y = \tan^{-1}\left(\frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}\right)

2. Apply the Identity

We recognize the formula for $\tan 3\theta$ :

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

Substituting this back into the equation:

y = \tan^{-1}(\tan 3\theta)

3. Evaluate the Interval

The given condition is $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$ .

Substituting $x = \tan \theta$ :

-\frac{1}{\sqrt{3}} < \tan \theta < \frac{1}{\sqrt{3}}

-\frac{\pi}{6} < \theta < \frac{\pi}{6}

Multiplying by $3$ :

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

Since $3\theta$ falls within the principal value branch of $\tan^{-1}$ , we can simplify:

y = 3\theta

4. Substitute back for x

Replace $\theta$ with $\tan^{-1} x$ :

y = 3 \tan^{-1} x

5. Find the Derivative

Now, differentiate $y$ with respect to $x$ :

\frac{dy}{dx} = \frac{d}{dx} (3 \tan^{-1} x)

\frac{dy}{dx} = 3 \cdot \frac{1}{1 + x^2}

\frac{dy}{dx} = \frac{3}{1 + x^2}

Correct Option: B) $\frac{3}{1+x^2}$

Choose the correct answer:

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