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I know very well how to do these type of questions . First I redefined f in different intervals to so that there is no modulus left. Then I used second derivative test But I am not getting any of the option am i right? As my book says that (A) is correct.

Please guide me

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    Include your working, and then we can see what may or may not be going wrong. – 2017-01-20
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    Okay I am uploading my work. Editing the question – 2017-01-20
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    The only you need is to check the solutions of the derivative, where it exists, and critical points as the extremes or if there is some point where the derivative is not defined. – 2017-01-20

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For $x \in [-1,0] $, we have, $$f (x) =\frac {1}{1-x} +\frac {1}{2-x}$$ and for $x\in [0,1] $, we have $$f (x)=\frac {1}{1+x} +\frac {1}{2-x} $$ We can see $f (x) $ is continuous at $x=0$.

For $x\in [-1,0] $, we have $$f'(x) =\frac {1}{(1-x)^2} +\frac {1}{(2-x)^2} $$ and when $f'(x)=0$ no solution is possible. We also check the value of $f (x)$ at $-1$ and $0$ also. We thus get $$\operatorname {max}_{f (x)} =f (0)=\frac {3}{2} $$

For $x\in [0,1] $, we have, $$f'(x) =\frac {-1}{(1+x)^2} +\frac {1}{(2-x)^2} $$ and $f'(x)=0$ gives us $$2-x =\pm (1+x) $$ $$\Rightarrow x=0.5 \text { as an acceptable solution} $$ But however we need to check the value of $f (x) $ at $0$ and $1$ also. We thus have that $$\operatorname {max}_{f (x)} =f (1) =\frac {3}{2} $$

The result thus follows. Hope it helps.

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    I am doing the same way. But I am taking the LCM of terms . Am I not allowed to do that or I am making calculation mistakes? – 2017-01-20
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    @AbhishekChandra The function f (x) for x>1 is wrong. The function is not defined in that domain. Taking LCM is not advised as it makes calculations seemingly difficult. – 2017-01-20
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$\bf{For\; Maximum}$

Using $$1+\frac{1}{1+|1|}-f(x)=1+\frac{1}{1+|1|}-\frac{1}{1+|x|}-\frac{1}{1+|x-1|}$$

$$ = \frac{|x|}{1+|x|}+\frac{|x-1|-|1|}{\left(1+|1|\right)(1+|x-1|)}\geq \frac{|x|}{1+|x|}+\frac{|1|-|x|-|1|}{\left(1+|1|\right)(1+|x-1|)}$$

$$ =|x|\bigg[\frac{1+|1|+|x-1|+|1||x-1|-1-|x|}{(1+|x|)\left(1+|1|\right)(1+|x-1|)}\bigg]\geq |x|\bigg(\frac{|1||x-1|}{(1+|x|)\left(1+|1|\right)(1+|x-1|)}\bigg)\geq 0$$

So $$f(x)\leq 1+\frac{1}{1+|1|} = \frac{3}{2}$$

and equality hold when $x=0$ and $x=1$

so function $f(x)$ is maximum at $x=0,1$ i e $\displaystyle f(0)=f(1) = \frac{3}{2}.$

$\bf{For\; Minimum}$

Function $$f(x) = \frac{1}{1+|x|}+\frac{1}{1+|x-1|}$$ is symmetrical about $\displaystyle x = \frac{1}{2}$

Bcz $$f\left(\frac{1}{2}-x\right)=f\left(\frac{1}{2}+x\right)$$

So $\displaystyle x = \frac{1}{2}$ is a point of minimum i. e $\displaystyle f\left(\frac{1}{2}\right) = \frac{4}{3}.$