Let \(x = \sec \theta - \cos \theta \text{ and } y = \sec^4 \theta - \cos^4 \theta \\ \) What is \(\left[ \frac{x^2+4}{y^2+4} \frac{dy}{dx} \left( x^2+4 \frac{d^2y}{dx^2} - 16y \right) \right] \) equal to?

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NDA-I (Mathematics) Official Paper (Held On: 13 Apr, 2025)
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  1. \(16x\)
  2. \(16y\)
  3. \(-16x \)
  4. \(-16y\)

Answer (Detailed Solution Below)

Option 3 : \(-16x \)
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Detailed Solution

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Calculation:

Given,

\(x = \sec\theta - \cos\theta\)

\(y = \sec^4\theta - \cos^4\theta\)

We know from before:

\(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = \frac{16\,(y^{2}+4)}{\,x^{2}+4\,} \)

Differentiate both sides w.r.t. x:

\(\displaystyle \frac{d}{dx}\Bigl[(x^{2}+4)\bigl(\tfrac{dy}{dx}\bigr)^{2}\Bigr] = \frac{d}{dx}\bigl[16\,(y^{2}+4)\bigr] \)

This yields the differential relation:

\(\displaystyle (x^{2}+4)\,\frac{d^{2}y}{dx^{2}} \;+\;x\,\frac{dy}{dx} \;-\;16\,y \;=\;0\)

The required expression is

\(\displaystyle \frac{x^{2}+4}{y^{2}+4}\,\frac{dy}{dx} \Bigl[(x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y\Bigr] \)

From the differential relation,

\(\displaystyle (x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y = -\,x\,\frac{dy}{dx} \)

So the expression becomes

\(\displaystyle \frac{x^{2}+4}{y^{2}+4}\,\frac{dy}{dx}\, \bigl(-x\,\tfrac{dy}{dx}\bigr) = -\,x\,\frac{x^{2}+4}{y^{2}+4} \Bigl(\frac{dy}{dx}\Bigr)^{2} \)

Using \(\bigl(\tfrac{dy}{dx}\bigr)^{2} = \tfrac{16\,(y^{2}+4)}{x^{2}+4}\), all factors cancel except -16x.

∴  The value of the given expression is  \(-16x\).

Hence, the correct answer is Option 3. 

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