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  • If <img alt="\mathrm{f(x)=\left\{\begin{array}{lll} =0 & \text { at } & x=0 \\ =\frac{1}{2}-x+[x] & \text { if } & 0<x<\frac{1}{2} \\ =\frac{1}{2} & \text { if } &amp

If \mathrm{f(x)=\left\{\begin{array}{lll} =0 & \text { at } & x=0 \\ =\frac{1}{2}-x+[x] & \text { if } & 0<x<\frac{1}{2} \\ =\frac{1}{2} & \text { if } & x=\frac{1}{2} \\ =\frac{2}{3}-x & \text { if } & \frac{1}{2}<x<1 \\ =1 & \text { if } & x=1 \end{array}\right.} then \mathrm{f(x)} is

 

Option: 1

Continuous at \mathrm{x=\frac{1}{2}}


Option: 2

continuous at x = 1

 


Option: 3

continuous at x = 0


Option: 4

discontinuous at x = 0


Answers (1)

best_answer

\mathrm{\begin{aligned} \lim _{x \rightarrow 0^{+}} f(x) & =\lim _{x \rightarrow 0} \frac{1}{2}-x+[x]=\frac{1}{2}-0+[0+]=\frac{1}{2} \\ f(0) & =0 \end{aligned}}

\mathrm{\therefore f(x)} is discontinuous at x = 0

Since, \mathrm{\lim _{x \rightarrow \frac{1}{2}+} f(x)=\lim _{x \rightarrow \frac{1}{2}-}\left(\frac{2}{3}-x\right)=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}}

\mathrm{\therefore f(x)} is not continuous at x = 1/2

\mathrm{\lim _{x \rightarrow 1} f(x) \lim _{x \rightarrow 1}\left(\frac{2}{3}-x\right)=-\frac{1}{3} \neq f(1)}

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

 

Posted by

jitender.kumar

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