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If \mathrm{f(x)= \begin{cases}{[x]+\sqrt{\{x\}},} & x<1 \\ \frac{1}{[x]+\{x\}^2}, & x \geq 1\end{cases}}

[where [.] and {.} represent the greatest integer and fractional part functions respectively]

Option: 1

f(x) is continuous at x=1 but not differentiable


Option: 2

f(x) is not continuous at x=1


Option: 3

f(x) is differentiable at x=1


Option: 4

\lim _{x \rightarrow 1} f(x) does not exist


Answers (1)

best_answer

\mathrm{f(x)= \begin{cases}{[x]+\sqrt{\langle x|}} & x<1 \\ \frac{1}{[x]+\{x\}^2} & x \geq 1\end{cases}}

consider the function \mathrm{f(x)} in the interval (0,2).

\mathrm{\begin{aligned} & f(x)= \begin{cases}\sqrt{x} & 0<x<1 \\ \frac{1}{1+(x-1)^2} & 1 \leq x<2\end{cases} \\ & f(1)=1 \text { and } \lim _{x \rightarrow 1} f(x)=1 \end{aligned}}

\mathrm{\therefore f(x)} is continuous at \mathrm{x=1}

\mathrm{\begin{aligned} f^{\prime}\left(1^{+}\right)=\lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h} & =\lim _{h \rightarrow 0^{+}} \frac{\frac{1}{1+h^2}-1}{h} \\ & =\lim _{h \rightarrow 0^{+}} \frac{-h}{1+h^2}=0 \\ f^{\prime}\left(1^{-}\right)=\lim _{h \rightarrow 0^{+}} \frac{f(1-h)-f(1)}{-h} & =\lim _{h \rightarrow 0^{+}} \frac{\sqrt{1-h}-1}{-h} \\ & =\lim _{h \rightarrow 0^{+}} \frac{1}{\sqrt{1-h}+1}=\frac{1}{2} \end{aligned}}

\mathrm{\therefore f(x)} is not differentiable at \mathrm{x=1}

Posted by

Ajit Kumar Dubey

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