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NDA 1 2025 Maths, question 71

Topic: Differential calculus

Passage · Q71 to Q72

Let x=sec⁡θ−cos⁡θx = \sec\theta - \cos\theta and y=sec⁡4θ−cos⁡4θy = \sec^{4}\theta - \cos^{4}\theta.

Q71

What is (dydx)2\left( \frac{dy}{dx} \right)^{2} equal to?

  1. (a) 4(y2+4)x2+4\frac{4\left( y^{2} + 4 \right)}{x^{2} + 4}
  2. (b) 4(y2−4)x2−4\frac{4\left( y^{2} - 4 \right)}{x^{2} - 4}
  3. (c) 16(y2+4)x2+4\frac{16\left( y^{2} + 4 \right)}{x^{2} + 4}Correct answer
  4. (d) 16(y2−4)x2−4\frac{16\left( y^{2} - 4 \right)}{x^{2} - 4}

Solution.

Answer: (c) 16(y2+4)x2+4\frac{16\left( y^{2} + 4 \right)}{x^{2} + 4}

  1. Let t=sec⁡θt = \sec\theta, so x=t−t−1x = t - t^{- 1} and y=t4−t−4y = t^{4} - t^{- 4}.
  2. Differentiating with respect to tt gives dy/dx=4(t4+t−4)/(t+t−1)dy/dx = 4\left( t^{4} + t^{- 4} \right)/\left( t + t^{- 1} \right).
  3. Now (t4+t−4)2=y2+4\left( t^{4} + t^{- 4} \right)^{2} = y^{2} + 4 and (t+t−1)2=x2+4\left( t + t^{- 1} \right)^{2} = x^{2} + 4.
  4. Squaring the derivative yields 16(y2+4)/(x2+4)16\left( y^{2} + 4 \right)/\left( x^{2} + 4 \right).