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  • <img alt="\text{Let } \mathrm{f(x)}= \left\{\begin{matrix} \mathrm{\sqrt{1+x^{2}},x<\sqrt{3}}\\ \mathrm{\sqrt{3}x-1,\sqrt{3}\leq x< 5}\\ \mathrm{[x],4\leq x< 5}\\ \mathrm{|1-x|,x\geq 5} \e

\text{Let } \mathrm{f(x)}= \left\{\begin{matrix} \mathrm{\sqrt{1+x^{2}},x<\sqrt{3}}\\ \mathrm{\sqrt{3}x-1,\sqrt{3}\leq x< 5}\\ \mathrm{[x],4\leq x< 5}\\ \mathrm{|1-x|,x\geq 5} \end{matrix}\right. , where \mathrm{[x]}  is the greatest integer \mathrm{\leq x}

The number of points of discontinuity of \mathrm{f(x)} in \mathrm{R} is

Option: 1

1


Option: 2

2


Option: 3

0


Option: 4

none of these


Answers (1)

best_answer

11. We know that \mathrm{|1-x|}  is continuous everywhere, \mathrm{[x]} is continuous everywhere except at integers, \mathrm{\sqrt{1+x^{2}}} and \mathrm{\sqrt{3} x-1} are continuous in their respective intervals of definition. So, the only doubtful points of continuity are \mathrm{x=\sqrt{3},4,5}

\begin{aligned} & \mathrm{f(\sqrt{3}-0)=\lim _{h \rightarrow 0} \sqrt{1+(\sqrt{3}-h)^2}=2 ; }\\ & \mathrm{f(\sqrt{3}+0)=\lim _{h \rightarrow 0}\{\sqrt{ } 3(\sqrt{3}+h)-1\}=2 ; \quad f(\sqrt{3})=2 .} \\ & \mathrm{f(4-0)=\lim _{h \rightarrow 0}\{\sqrt{3} 3(4-h)-1\}=4 \sqrt{3}-1 ; \quad f(4+0)=\lim _{h \rightarrow 0}[4+h]=4 .} \\ &\mathrm{ f(5-0)=\lim _{h \rightarrow 0}[5-h]=\lim _{h \rightarrow 0} 4=4 ; }\\ &\mathrm{ f(5+0)=\lim _{h \rightarrow 0}|1-(5+h)|=4 ; \quad f(5)=4 .} \\ & \therefore \quad \mathrm{f(x)} \text { is discontinuous at }\mathrm{ x=4} \text { only. } \end{aligned}

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