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CUET · MATHS · PYQ PAPER 2025

If \(x=\frac{1-t}{1+t}\) and \(y=\frac{2 t}{1+t}\), then \(\frac{d^2 y}{d x^2}\) is equal to :

  1. A \(\frac{1}{1+t^2}\)
  2. B 0
  3. C \(-1\)
  4. D \(-\frac{1}{1+t^2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) 0

Step-by-step Solution

Detailed explanation

\(\frac{dx}{dt} = \frac{d}{dt}\left(\frac{1-t}{1+t}\right) = \frac{-1(1+t)-(1-t)(1)}{(1+t)^2} = \frac{-2}{(1+t)^2}\) \(\frac{dy}{dt} = \frac{d}{dt}\left(\frac{2t}{1+t}\right) = \frac{2(1+t)-2t(1)}{(1+t)^2} = \frac{2}{(1+t)^2}\)…
From CUET
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