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Mathematics · 2026 · 2 marks
Continuity and DifferentiabilityDerivatives of Functions in Parametric Forms2 marksApplyvery short answer
CBSE 2026 · Region 3 · Set 2 · Q24
If x = e^t +1/t and y = e^t -1/t, then find dy/dx at t =-2.
If $\displaystyle \mathrm{x}=\mathrm{e}^{\mathrm{t}+\frac{1}{t}}$ and $\displaystyle \mathrm{y}=\mathrm{e}^{\mathrm{t}-\frac{1}{t}}$, then find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$ at $\displaystyle \mathrm{t}=-2$.
Marking-scheme solution
$\displaystyle x=\mathrm{e}^{t+\frac{1}{t}} \Rightarrow \dfrac{d x}{d t}=\left(1-\dfrac{1}{t^{2}}\right) \mathrm{e}^{t+\frac{1}{t}}$
$\displaystyle y=\mathrm{e}^{t-\frac{1}{t}} \Rightarrow \dfrac{d y}{d t}=\left(1+\dfrac{1}{t^{2}}\right) \mathrm{e}^{t-\frac{1}{t}}$
$\displaystyle \dfrac{d y}{d x}=\dfrac{t^{2}+1}{t^{2}-1} \cdot \dfrac{\mathrm{e}^{t-\frac{1}{t}}}{\mathrm{e}^{t+\frac{1}{t}}}$
$\displaystyle \left[\dfrac{d y}{d x}\right]_{t=-2}=\dfrac{5 \mathrm{e}}{3}$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.