CBSE 2026 · Region 3 · Set 1 · Q21 · 2 marks
If $\displaystyle \mathrm{x}=\mathrm{t}+\frac{1}{\mathrm{t}}$ and $\displaystyle \mathrm{y}=\mathrm{t}-\frac{1}{\mathrm{t}}$, find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$ at $\displaystyle \mathrm{t}=2$.
Marking-scheme solution
$\displaystyle x=t+\dfrac{1}{t} \Rightarrow \dfrac{dx}{dt}=1-\dfrac{1}{t^{2}}$
$\displaystyle \mathrm{y}=t-\dfrac{1}{t} \Rightarrow \dfrac{dy}{dt}=1+\dfrac{1}{t^{2}}$
$\displaystyle \dfrac{dy}{dx}=\dfrac{t^{2}+1}{t^{2}-1}$
$\displaystyle \left(\dfrac{dy}{dx}\right)_{(t=2)}=\dfrac{5}{3}$
Continuity and DifferentiabilityDerivatives of Functions in Parametric FormsApplyvery_short_answereasy
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.