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The function  \mathrm{ f(x)=\left\{\begin{array}{cc} |x-3|, & x \geq 1 \\ \frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4}, & x<1 \end{array}\right. \text { is }}

Option: 1

continuous at x=1


Option: 2

derivable at x=4


Option: 3

continuous at x= -1


Option: 4

derivable at x=3


Answers (1)

best_answer

\mathrm{ \text { We have, } \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}\left(\frac{x^2}{4}-\frac{3 x}{4}+\frac{13}{4}\right) }
\mathrm{ =\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=2 \\ }

\mathrm{ \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}|x-3|=2 \\ }
and, f(1)=|1-3|=2.
\mathrm{ \therefore f(x) }   is continuous at x=1.
Since

\mathrm{ \lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3}|x-3|=0, \\ }
and, f(3)=0. Therefore, f(x) is continuous at x=3.

Now,
\mathrm{ \text { Now, } (\text { LHD at } x=1)=\left[\frac{d}{d x}\left(\frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4}\right)\right]_{x=1} }
\mathrm{ \qquad=\left[\frac{x}{2}-\frac{3}{2}\right]_{x=1}=\frac{1}{2}-\frac{3}{2}=-1 \\ }
\mathrm{ \text { (RHD at } x=1)=\left(\frac{d}{d x}(-(x-3))\right)_{x=1}=-1 }

So, f(x) is differentiable at x=1.

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Shailly goel

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