Consider the function f ( x ) = | x - 2 | + | x - 5 |,x R Stateme… - Vidyadip

MCQ

Consider the function $f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right|,x \in R$

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

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

A
Statement $-1$ is false, Statement $-2$ is true;
B
Statement $-1$ is true, Statement $-2$ is false
C
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is not a correct explanation for Statement $-1$
✓
Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$

✓

Answer

Correct option: D.

Statement $-1$ is true, Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$

d
${f_1}\left( x \right) = \left| {x - 2} \right| = \left\{ {\begin{array}{*{20}{c}}
{x - 2}&{x - 2 \ge 0}\\
{2 - x}&{x - 2 \le 0}
\end{array}} \right.$

$ \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{x - 2}&{x \ge 2}\\
{2 - x}&{x \le 2}
\end{array}} \right.$

similarly ${f_2}\left( x \right) = \left| {x - 5} \right| = \left\{ {\begin{array}{*{20}{c}}
{x - 5}&{x \ge 5}\\
{5 - x}&{x \le 5}
\end{array}} \right.$

$f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right| = \left\{ {\begin{array}{*{20}{c}}
{ - 2x + 7,}&{when\,x < 2}\\
{x - 2 + 5 - x = 3,}&{when\,2 \le x < 5}\\
{2x - 7}&{when\,x \ge 5}
\end{array}} \right.$

Thus $f(x)=3, \quad$ when $2 \leq x \leq 5$

$\therefore f^{\prime}(x)=0, \quad$ when $2 < x < 5$

$\therefore f^{\prime}(4)=0 \Rightarrow$ 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 $2.$

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