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Question

Consider the following in respect of the function $$f(x) = \left\{\begin{matrix}2+ x, & x \geq 0\\ 2 - x, & x < 0\end{matrix}\right.$$
1. $$\displaystyle \lim_{x \rightarrow 1} f(x)$$ does not exist.
2. f(x) is differentiable at x = 0.
3. f(x) is continuous at x = 0.
Which of the above statements is/are correct?


A
1 only
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B
3 only
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C
2 and 3 only
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D
1 and 3 only
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Solution

The correct option is B $$3$$ only
Left hand limit$$ = \lim_{x\to 1^-}{f(x)}$$
$$= \displaystyle \lim_{h\to 0}{f(1-h)}$$
$$=\displaystyle \lim_{h\to 0}2+(1-h)=2+1=3$$
Right hand limit$$ = \displaystyle \lim_{x\to 1^+}{f(x)}$$
$$= \displaystyle \lim_{h\to 0}{f(1+h)}$$
$$=\displaystyle \lim_{h\to 0}2+(1+h)=2+1=3$$
So, the limit exists at x=1

Left hand limit$$ =\displaystyle  \lim_{x\to 0^-}{f(x)}$$
$$= \displaystyle \lim_{h\to 0}{f(0-h)}$$
$$=\displaystyle \lim_{h\to 0} 2-(0-h)=2-0=2$$
Right hand limit$$ = \displaystyle \lim_{x\to 0^+}{f(x)}$$
$$= \displaystyle \lim_{h\to 0}{f(0+h)}$$
$$=\displaystyle \lim_{h\to 0}2+(0+h)=2+0=2$$
So, LHL=RHL=f(0)
So, f(x) is continuous at x=0

Right hand limit$$ = \displaystyle \lim_{h\to 0}\dfrac{f(0+h)-f(0)}{h}$$
$$= \displaystyle \lim_{h\to 0}\dfrac{(2+(0+h))-2}{h}$$
$$=\displaystyle \lim_{h\to 0}\dfrac{h}{h}=1$$
Left hand limit$$ = \displaystyle \lim_{h\to 0}\dfrac{f(0-h)-f(0)}{-h}$$
$$= \displaystyle \lim_{h\to 0}\dfrac{(2-(0-h))-(2)}{-h}$$
$$=\displaystyle \lim_{h\to 0}\dfrac{h}{-h}=-1$$
LHL$$\neq$$RHL
So, f(x) is not differentiable at x=0

The answer is option (B)

Mathematics

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