NDA 1 2025 Maths, question 72

Topic: Differential calculus

Passage · Q71 to Q72

Let $x = \sec\theta - \cos\theta$ and $y = \sec^{4}\theta - \cos^{4}\theta$ .

Q72

What is $\left( \frac{x^{2} + 4}{y^{2} + 4} \right)\frac{dy}{dx}\left\lbrack \left( x^{2} + 4 \right)\frac{d^{2}y}{dx^{2}} - 16y \right\rbrack$ equal to?

(a) $16x$
(b) $16y$
(c) $- 16x$ Correct answer
(d) $- 16y$

Solution.

Answer: (c) $- 16x$

The derivative identity is $\left( x^{2} + 4 \right)(y\prime)^{2} = 16\left( y^{2} + 4 \right)$ .
Differentiating gives $2x(y\prime)^{2} + 2\left( x^{2} + 4 \right)y\prime y\prime\prime = 32yy\prime$ .
Here $y\prime \neq 0$ , so $\left( x^{2} + 4 \right)y\prime\prime - 16y = - xy\prime$ .
The required expression becomes $- x\frac{x^{2} + 4}{y^{2} + 4}(y\prime)^{2} = - 16x$ .