One variable Maxima minima problem 
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

 A: 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. 
A: $\bf{For\; Maximum}$
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 $\displaystyle f\left(\frac{1}{2}\right) = \frac{4}{3}.$
