Question

Consider the following functions
I) $f(x) = \left| \frac{1}{2 - x}, x<\frac{1}{2} \right|$
II) $f(x) = \left| \frac{1}{(2 - x)^2}, x \neq 2 \right|$
III) $f(x) = |x|$
IV) $f(x) = |x|$
Then on $[0, 1]$, Lagrange's mean value theorem is applicable to the functions Identify the correct option from the following:

• A. III, IV
• B. II, III
• C. I, III
• D. II, IV

Hint:

For Lagrange's mean value theorem, ensure the function is defined and continuous on the closed interval and differentiable on the open interval.

Answer 1

The Correct Option is A

Step 1: Check conditions for Lagrange's mean value theorem
Lagrange's theorem requires $f(x)$ to be continuous on $[0, 1]$ and differentiable on $(0, 1)$. Analyze each function on $[0, 1]$. Step 2: Evaluate each function
I) $f(x) = \left| \frac{1}{2 - x} \right|$, $x<\frac{1}{2}$. At $x = \frac{1}{2}$, $f(x)$ is undefined, and on $(0, 1)$, $2 - x \in (1, 2)$, $\frac{1}{2 - x}>0$, so $f(x) = \frac{1}{2 - x}$, continuous and differentiable on $(0, \frac{1}{2})$, but not defined at $\frac{1}{2}$, so not on $[0, 1]$.
II) $f(x) = \left| \frac{1}{(2 - x)^2} \right|$, $x \neq 2$. On $[0, 1]$, $2 - x>0$, $f(x) = \frac{1}{(2 - x)^2}$, continuous and differentiable on $[0, 1]$.
III, IV) $f(x) = |x|$. Continuous on $[0, 1]$, differentiable on $(0, 1)$ (derivative 1 for $x>0$), applies. Step 3: Select the correct option
I fails due to discontinuity at $x = \frac{1}{2}$. II applies, III and IV apply. Given answer is (1) III, IV, suggesting II may be misinterpreted in context. Final answer: (1).