For the function $f(x) = |x|$ defined on $[-1, 1]$, with $f'$ denoting the derivative, Rolle's / Lagrange's Mean Value Theorem is not applicable because:
Options:
(A) $f(x)$ is not continuous at $x = 0$
(B) $f'(x) = 0$ at $x = 0$
(C) $df/dx$ is not defined at $x = 0$
(D) $df/dx = 0$ at $x = 0$
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2020 Q.11)**
**1. Intuitive Physical Setup & Given Parameters**:
We are given $f(x) = |x|$ on the symmetric interval $[-1, 1]$ and asked why the Mean Value Theorem (MVT) cannot be applied.
**2. Step-by-Step Mathematical Derivation**:
1. **Hypotheses of Lagrange's Mean Value Theorem**:
For a function $f(x)$ to satisfy the MVT on $[a, b]$, two necessary conditions must hold:
- Condition 1: $f(x)$ is continuous on the closed interval $[a, b]$.
- Condition 2: $f(x)$ is differentiable on the open interval $(a, b)$.
2. **Analysis of $f(x) = |x|$ on $[-1, 1]$**:
- Continuity: $|x|$ is continuous everywhere on $\mathbb{R}$, including at $x = 0$ ($\lim_{x \to 0} |x| = 0 = f(0)$).
- Differentiability at the origin $x = 0 \in (-1, 1)$:
$\lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h - 0}{h} = +1$
$\lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h - 0}{h} = -1$
Because the left-hand derivative ($-1$) does not equal the right-hand derivative ($+1$), the derivative $f'(0)$ does not exist.
3. **Conclusion**:
Condition 2 is violated because $df/dx$ is undefined at $x = 0$.
**3. Comprehensive Options & Distractor Analysis**:
- **(A) f(x) is not continuous at x = 0**: ❌ **INCORRECT**. $|x|$ is continuous everywhere.
- **(B) f'(x) = 0 at x = 0**: ❌ **INCORRECT**. The derivative does not exist at $x = 0$, so it cannot equal 0.
- **(C) df/dx is not defined at x = 0**: ✅ **CORRECT**. The sharp corner (cusp) at $x = 0$ has unequal left and right derivatives, violating differentiability on $(-1, 1)$.
- **(D) df/dx = 0 at x = 0**: ❌ **INCORRECT**. Contradicts non-differentiability.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **C**
- **Exam Takeaway**: Lagrange's MVT requires BOTH continuity on $[a, b]$ and differentiability on $(a, b)$. A single point of non-differentiability inside the interval disqualifies the theorem.
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