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  • If <img alt="\mathrm{f(x)=\left\{\begin{array}{ll}{[x]+\sqrt{\{x\}},} & x<1 \\ \frac{1}{[x]+\{x\}^{2}}, & x \geq 1\end{array}\right.}" src="https://entrancecorner.oncodecogs.com/gif.late

If \mathrm{f(x)=\left\{\begin{array}{ll}{[x]+\sqrt{\{x\}},} & x<1 \\ \frac{1}{[x]+\{x\}^{2}}, & x \geq 1\end{array}\right.}, then

[where \mathrm{[\cdot]} and \mathrm{\left \{ \cdot \right \}} 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{\{x \mid}} & x<1 \\ \frac{1}{[x]+\{x\}^{2}} & x \geq 1\end{cases}}

Consider the function \mathrm{f(x)} in the interval \mathrm{(0,2)}.
\mathrm{f(x)= \begin{cases}\sqrt{x} & 0<x<1 \\ \frac{1}{1+(x-1)^{2}} & 1 \leq x<2\end{cases}}
\mathrm{f(1)=1\, and \, \lim _{x \rightarrow 1} f(x)=1}

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

\mathrm{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}}
                                                               \mathrm{ =\lim _{h \rightarrow 0^{+}} \frac{-h}{1+h^{2}}=0 }

\mathrm{ 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} }
                                                                \mathrm{ =\lim _{h \rightarrow 0^{+}} \frac{1}{\sqrt{1-h}+1}=\frac{1}{2}}
\mathrm{ \therefore f\left ( x \right )} is not differentiable at \mathrm{ x= 1}



                                                                 

 

Posted by

Divya Prakash Singh

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