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

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(D) The function is defined as $f(x) = |x - 2| + |x - 5|$.We can express this piecewise as:$f(x) = \begin{cases} -(x-2) - (x-5) = -2x + 7 & 	ext{if }

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

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(D) The function is defined as $f(x) = |x - 2| + |x - 5|$.We can express this piecewise as:$f(x) = \begin{cases} -(x-2) - (x-5) = -2x + 7 & 	ext{if } x For Statement-$1$:In the interval $(2, 5)$,$f(x) = 3$. Therefore,$f'(x) = 0$ for all $x \in (2, 5)$.Since $4 \in (2, 5)$,$f'(4) = 0$. Thus,Statement-$1$ is true.For Statement-$2$:$f(x)$ is a sum of continuous functions,so it is continuous everywhere. In $[2, 5]$,$f(x) = 3$,which is a constant function,so it is differentiable in $(2, 5)$.Also,$f(2) = |2-2| + |2-5| = 3$ and $f(5) = |5-2| + |5-5| = 3$. Thus,$f(2) = f(5)$.Statement-$2$ is true.Statement-$2$ describes the conditions of Rolle's Theorem,which implies the existence of a point $c \in (2, 5)$ such that $f'(c) = 0$. Since $f(x)$ is constant on $[2, 5]$,$f'(x) = 0$ for all $x \in (2, 5)$,which confirms Statement-$1$. Thus,Statement-$2$ is a 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 $f(2) = f(5)$.