CBSE 2026 · Region 4 · Set 1 · Q22 · 2 marks
Show that the function $\displaystyle \mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\frac{\cos \mathrm{x}}{-\mathrm{x}+\dfrac{\pi}{2}}, & \mathrm{x} \neq \frac{\pi}{2} \\ 1, & \mathrm{x}=\frac{\pi}{2}\end{array}\right.$ is continuous at $\displaystyle \mathrm{x}=\frac{\pi}{2}$.Find whether the function $\displaystyle \mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\mathrm{x}-1, & \mathrm{x}<2 \\ 2 \mathrm{x}-3, & \mathrm{x} \geq 2\end{array}\right.$ at $\displaystyle \mathrm{x}=2$ is differentiable or not.
Show that the function $\displaystyle \mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\frac{\cos \mathrm{x}}{-\mathrm{x}+\dfrac{\pi}{2}}, & \mathrm{x} \neq \frac{\pi}{2} \\ 1, & \mathrm{x}=\frac{\pi}{2}\end{array}\right.$ is continuous at $\displaystyle \mathrm{x}=\frac{\pi}{2}$.
Find whether the function $\displaystyle \mathrm{f}(\mathrm{x})=\left\{\begin{array}{cc}\mathrm{x}-1, & \mathrm{x}<2 \\ 2 \mathrm{x}-3, & \mathrm{x} \geq 2\end{array}\right.$ at $\displaystyle \mathrm{x}=2$ is differentiable or not.
Marking-scheme solution
For showing: $\displaystyle \lim_{x \rightarrow \frac{\pi}{2}} f(x)=1$
For showing: $\displaystyle f\left(\dfrac{\pi}{2}\right)=1$
$\displaystyle \therefore f(x)$ is continuous at $\displaystyle x=\dfrac{\pi}{2}$.
Getting $\displaystyle L H D=1$
Getting $\displaystyle R H D=2$
$\displaystyle \therefore L H D \neq R H D$, so $\displaystyle f(x)$ is not differentiable at $\displaystyle x=2$.
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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.