CUET · MATHS · PYQ PAPER 2025
If \(y=t-\frac{1}{t}\) and \(x=t+\frac{1}{t}\), then \(\frac{d y}{d x}\) is equal to
- A \(\frac{t^2-1}{t^2+1}\)
- B \(\frac{t^2+1}{t^2-1}\)
- C \(\frac{t-1}{t+1}\)
- D \(\frac{t+1}{t-1}\)
Answer & Solution
Correct Answer
(B) \(\frac{t^2+1}{t^2-1}\)
Step-by-step Solution
Detailed explanation
\( \frac{dy}{dt} = 1 - (-\frac{1}{t^2}) = 1 + \frac{1}{t^2} = \frac{t^2+1}{t^2} \) \( \frac{dx}{dt} = 1 + (-\frac{1}{t^2}) = 1 - \frac{1}{t^2} = \frac{t^2-1}{t^2} \) \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{(t^2+1)/t^2}{(t^2-1)/t^2} = \frac{t^2+1}{t^2-1} \)
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