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The function  \mathrm{f(x)=|x|+|x-1|}   is

Option: 1

continuous at x=1, but not differentiable


Option: 2

both continuous and differentiable at x=1


Option: 3

not continuous at x=1


Option: 4

none of these.


Answers (1)

best_answer

\mathrm{ f(x)=|x|+|x-1|=\left\{\begin{array}{rr} -2 x+1, & x<0 \\ 1 & 0 \leq x<1 \\ 2 x-1 & 1 \leq x \end{array}\right.}

Since,   \mathrm{ \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} 1=1, \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}(2 x-1)=1}
and,   \mathrm{ f(1)=2 \times 1-1=1}
\mathrm{ \therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)}

So, f(x) is continuous at x=1.
Now,  \mathrm{ \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{1-1}{-h}=0} 
and,  \mathrm{ \lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}} 
\mathrm{ =\lim _{h \rightarrow 0} \frac{2(1+h)-1-1}{h}=2 }
\mathrm{ \therefore \quad( LHD ~at~ x=1) \neq( RHD } 

at x=1). So, f(x) is not differentiable at x=1.

Posted by

Deependra Verma

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