Question:medium

A function satisfying the conditions of Lagrange's mean value theorem on the interval $[1, 5]$ is:

Show Hint

Always remember that LMVT requires differentiability on the open interval $(a, b)$. Modulus functions are non-differentiable only where their inner expression becomes zero. For $|x-1| + |x-5|$, the non-differentiable points are $x=1$ and $x=5$, which lie exactly on the boundaries, keeping the open interval $(1, 5)$ perfectly differentiable!
Updated On: Jul 9, 2026
  • $f(x) = |x - 1| + |x - 5|$
  • $g(x) = x - [x]$ ($[x]$ represents integral part of $x$)
  • $h(x) = \begin{cases} 3x - 1, & 0 \le x \le 1 \\ x^2 + 1, & 1 \le x \le 4 \\ 17, & 4 \le x \le 6 \end{cases}$
  • $k(x) = \begin{cases} x^2, & -\infty < x \le 1
    x, & 1 \le x \le \infty \end{cases}$
Show Solution

The Correct Option is A

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