NIMCET 2021 — Mathematics PYQ

Q: How do I solve NIMCET 2021 Mathematics questions?

Practice NIMCET 2021 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Mathematics questions?

NIMCET Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2021 — Mathematics PYQ

NIMCET | Mathematics | 2021

If $y = sin^{- 1} (\frac{x ^{2} - 1}{1 + 3 x ^{2} + x ^{4}})$ for $x > 0$ , then find $\frac{d y}{d x}$ .

Choose the correct answer:

Explanation

Solution

To simplify the expression inside the inverse sine, we divide both the numerator and the denominator by $x$ :

y = \sin^{-1} \left( \frac{\frac{x^2 - 1}{x}}{\frac{\sqrt{1 + 3x^2 + x^4}}{x}} \right)

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

Notice that $x^2 + \frac{1}{x^2}$ can be written as $(x - \frac{1}{x})^2 + 2$ . Substituting this into the denominator:

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

y = \sin^{-1} \left( \frac{x - \frac{1}{x}}{\sqrt{(x - \frac{1}{x})^2 + 5}} \right)

Substitution:

Let $x - \frac{1}{x} = \sqrt{5} \tan \theta$ .

Then the expression becomes:

y = \sin^{-1} \left( \frac{\sqrt{5} \tan \theta}{\sqrt{5 \tan^2 \theta + 5}} \right)

y = \sin^{-1} \left( \frac{\sqrt{5} \tan \theta}{\sqrt{5} \sec \theta} \right)

y = \sin^{-1} (\sin \theta) = \theta

From our substitution:

\tan \theta = \frac{x - \frac{1}{x}}{\sqrt{5}} \implies \theta = \tan^{-1} \left( \frac{x^2 - 1}{\sqrt{5}x} \right)

So,

y = \tan^{-1} \left( \frac{x^2 - 1}{\sqrt{5}x} \right)

Differentiating with respect to $x$ :

Using the formula $\frac{d}{dx} \tan^{-1} u = \frac{1}{1 + u^2} \frac{du}{dx}$ :

\frac{dy}{dx} = \frac{1}{1 + \left( \frac{x^2 - 1}{\sqrt{5}x} \right)^2} \cdot \frac{d}{dx} \left( \frac{x}{\sqrt{5}} - \frac{1}{\sqrt{5}x} \right)

\frac{dy}{dx} = \frac{5x^2}{5x^2 + (x^2 - 1)^2} \cdot \frac{1}{\sqrt{5}} \left( 1 + \frac{1}{x^2} \right)

\frac{dy}{dx} = \frac{5x^2}{x^4 + 3x^2 + 1} \cdot \frac{1}{\sqrt{5}} \left( \frac{x^2 + 1}{x^2} \right)

\frac{dy}{dx} = \frac{\sqrt{5}(x^2 + 1)}{x^4 + 3x^2 + 1}