Question

Assertion :Consider the function
$f\left(x\right)=|x-2|+|x-5|,xϵR$
$f\left(4\right)=0$ Reason: $f$ is continuous in $[2,5]$
differential in $(2,5)$ and $f\left(2\right)=f\left(5\right)$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is incorrect but Reason is correct
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is incorrect but Reason is correct

We have,

$f\left(x\right)=|x-2|+|x-5|,x\in R$

Now,

put $x=4$ then,

$f\left(4\right)=|4-2|+|4-5|$

$f\left(4\right)=2+1=3$

$f\left(4\right)=3$

Now,

$f\left(x\right)=|x-2|+|x-5|$

Put $x=2$then,

$f\left(2\right)=|2-2|+|5-2|$

$f\left(2\right)=0+3$

$f\left(2\right)=3$

Put $x=5$

$f\left(5\right)=|5-2|+|5-5|$

$f\left(5\right)=3$

So,

$f\left(2\right)=f\left(5\right)$

So,

Assertion is incorrect but reason is correct.

Hence, this is the answer.