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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 1 · Set 1 · Q21
(a) Check whether function f (x) defined as f (x)={ (|x-3|)/(2(x-3)), x<3; (x-6)/6, x ≥ 3 is continuous at x=3. or not?
Check whether function $\displaystyle \mathrm{f}(x)$ defined as
\[\mathrm{f}(x)=\left\{\begin{array}{ll}
\frac{|x-3|}{2(x-3)}, & x<3 \\
\frac{x-6}{6}, & x \geq 3
\end{array} \text { is continuous at } x=3\right. \text { or not? }
\]
If $\displaystyle \sqrt{3}\left(x^{2}+\mathrm{y}^{2}\right)=4 x \mathrm{y}$, then find $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}$ at $\displaystyle \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.
Official answer
The function is not continuous at x = $\displaystyle 3$
Marking-scheme solution
$\displaystyle \mathrm{f}(x)=\begin{cases}\dfrac{-(x-3)}{2(x-3)}, & x<3 \\[6pt] \dfrac{x-6}{6}, & x \geq 3\end{cases} \Rightarrow \mathrm{f}(x)=\begin{cases}\dfrac{-1}{2}, & x<3 \\[6pt] \dfrac{x-6}{6}, & x \geq 3\end{cases}$
$\displaystyle \mathrm{f}(3) = \dfrac{-1}{2}$
$\displaystyle \mathrm{LHL} = \lim_{x \to 3^-} \mathrm{f}(x) = \lim_{x \to 3^-}\left(\dfrac{-1}{2}\right) = \dfrac{-1}{2}$
$\displaystyle \mathrm{RHL} = \lim_{x \to 3^+} \mathrm{f}(x) = \lim_{x \to 3^+} \dfrac{x-6}{6} = \dfrac{-1}{2}$
LHL $\displaystyle =$ RHL $\displaystyle = \mathrm{f}(3)$, hence $\displaystyle \mathrm{f}(x)$ is continuous at $\displaystyle x=3$.
Differentiating $\displaystyle \sqrt3(x^2+\mathrm{y}^2)=4xy$ w.r.t. $\displaystyle x$: $\displaystyle \sqrt3\left(2x+2\mathrm{y}\dfrac{dy}{dx}\right) = 4\left(x\dfrac{dy}{dx}+\mathrm{y}\right)$
$\displaystyle \dfrac{dy}{dx} = \dfrac{2\mathrm{y}-\sqrt3x}{\sqrt3\mathrm{y}-2x}$
At $\displaystyle \left(\dfrac12,\dfrac{\sqrt3}{2}\right)$: $\displaystyle \dfrac{dy}{dx} = \sqrt3$
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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.