# Consider the function $f(x)=|x-2|+|x-5|\;x\in R$.

Statement- 1: $f'(4)=0$

Statement- 2: $f$ is continuous in [2,5] differentiable in (2,5) and $f(2)=f(5)$

$\begin{array}{1 1}(a)\;\text{Statement-1 is false,Statement-2 is true}\\(b)\;\text{Statement-1 is true,Statement-2 is true,Statement-2 is a correct explanation for Statement-1}\\(c)\;\text{Statement-1 is true,Statement-2 is true,Statement-2 is not a correct explanation for Statement-1}\\(d)\;\text{Statement-1 is true,Statement-2 is false}\end{array}$

$f_1(x)=|x-2|=\left\{\begin{array}{1 1}x-2&x-2\geq 0\\2-x&x-2\leq 0\end{array}\right.$
$\Rightarrow \left\{\begin{array}{1 1}x-2&x\geq 2\\2-x&x\leq 2\end{array}\right.$
Similarly $f_2(x)=|x-5|=\left\{\begin{array}{1 1}x-5&x\geq 5\\5-x&x\leq 5\end{array}\right.$
$f(x)=|x-2|+|x-5|$$=\left\{\begin{array}{1 1 1} -2x+7,\:& when\:\: x < 2 \\ x-2+5-x=3, \:\: & when \:2 \leq x <5 \\ 2x-7\:\: & when\: x\geq 5 \end{array}\right. Thus f(x)=3,\:\:\:when\:2\leq x\leq 5 \therefore\:f'(x)=0,\:\:\:when\:2 < x < 5 \therefore\:f'(4)=0$$\Rightarrow\:$ Statement 1 is true.
$f(2)=0+|2-5|=3$
$F(5)=|5-2|+0=3$
$\therefore$ Statement 2 is also true
But statement 1 is not the explanation for statement 2.
Hence (c) is the correct answer.
edited Mar 25, 2014