Consider the function f(x) = |x - 2| + |x - 5 | vedclass

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(C) For $x \in [2, 5]$,the function is defined as $f(x) = (x - 2) - (x - 5) = 3$.Since $f(x) = 3$ is a constant function on the interval $[2, 5]$,its

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

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(C) For $x \in [2, 5]$,the function is defined as $f(x) = (x - 2) - (x - 5) = 3$.Since $f(x) = 3$ is a constant function on the interval $[2, 5]$,its derivative $f'(x) = 0$ for all $x \in (2, 5)$.Thus,$f'(4) = 0$,so Statement-$1$ is true.For Statement-$2$,$f(x) = |x - 2| + |x - 5|$ is a sum of continuous functions,hence continuous on $[2, 5]$.On the interval $(2, 5)$,$f(x) = 3$,which is a constant function and therefore differentiable.Also,$f(2) = |2 - 2| + |2 - 5| = 3$ and $f(5) = |5 - 2| + |5 - 5| = 3$,so $f(2) = f(5)$.Statement-$2$ is true.Since $f(x)$ is constant on $[2, 5]$,its derivative is zero everywhere in $(2, 5)$,which directly explains why $f'(4) = 0$. Thus,Statement-$2$ is the correct explanation for Statement-$1$.

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 $f(2) = f(5)$.

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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 $f(2) = f(5)$.