CBSE 2024 · Region 5 · Set 1 · Q22 · 2 marks
Verify whether the function f defined by \[\mathrm{f}(x)=\left\{\begin{array}{cc} x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{array}\right. \] is continuous at $\displaystyle x=0$ or not.Check for differentiability of the function f defined by $\displaystyle \mathrm{f}(x)=|x-5|$, at the point $\displaystyle x=5$.
Verify whether the function f defined by \[\mathrm{f}(x)=\left\{\begin{array}{cc} x \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{array}\right. \] is continuous at $\displaystyle x=0$ or not.
Check for differentiability of the function f defined by $\displaystyle \mathrm{f}(x)=|x-5|$, at the point $\displaystyle x=5$.
Marking-scheme solution
(a)
$\displaystyle \lim _{x \rightarrow 0}(x)=\lim _{x \rightarrow 0} x . \sin \frac{1}{x}=0 \times$ Finite value in $\displaystyle [-1,1]=0=\mathrm{f}(0)$
$\displaystyle \therefore \mathbf{f}$ is a continuous function.
Or
(b) LHD $\displaystyle =\lim _{x \rightarrow 5^{-}} \frac{|x-5|-0}{x-5}=\lim _{x \rightarrow 5^{-}} \frac{-(x-5)}{x-5}=-1$
$$R H D=\lim _{x \rightarrow $\displaystyle 5$^{+}} \frac{|x-5|-0}{x-5}=\lim _{x \rightarrow $\displaystyle 5$^{+}} \frac{(x-5)}{x-5}=$\displaystyle 1$
$$LHD $\displaystyle \boldsymbol{F} \mathbf{R H D ,} \therefore \mathbf{f}$ is not differentiable at $\displaystyle \mathbf{x}=\mathbf{5}$
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.