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Mathematics · 2025 · 3 marks
CBSE 2025 · Region 4 · Set 1 · Q27
(a) Find k so that f ( x )= {(x^2-2 x -3)/(x +1), x ≠-1; k, x =-1} is continuous at x =-1.
Find k so that \[\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc} \frac{\mathrm{x}^{2}-2 \mathrm{x}-3}{\mathrm{x}+1}, & \mathrm{x} \neq-1 \\ k, & \mathrm{x}=-1 \end{array}\right. \] is continuous at $\displaystyle \mathrm{x}=-1$.
Check the differentiability of function $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{x}|\mathrm{x}|$ at $\displaystyle \mathrm{x}=0$.
Marking-scheme solution
(a)
For continuity at $\displaystyle \mathrm{x}=-1$, $\displaystyle k=\lim_{\mathrm{x}\to-1}\frac{\mathrm{x}^2-2\mathrm{x}-3}{\mathrm{x}+1}=\lim_{\mathrm{x}\to-1}\frac{(\mathrm{x}-3)(\mathrm{x}+1)}{\mathrm{x}+1}=\lim_{\mathrm{x}\to-1}(\mathrm{x}-3)=-4$. Hence $\displaystyle \boxed{k=-4}$.
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CBSE Class 12 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.