Limits, Continuity & Differentiability
Methods of Differentiation
Grade None
Question:
<p>Let $f(x) = \begin{cases} -1, & -2 \le x < 0 \\ x-1, & 0 \le x \le 2 \end{cases}$ and $g(x) = |f(x)| + f(|x|)$. Then in the interval $(-2, 2)$, $g$ is:</p>
<p>differentiable at all points</p>
<p>not differentiable at two points</p>
<p>not continuous</p>
<p>not differentiable at one point</p>
Step-by-Step Solution
Key Concept: General
<b>Composition of Absolute Value Functions [JEE Main 2019]</b><br>Analyse $g(x) = |f(x)| + f(|x|)$ piece by piece on $(-2,2)$:<br>For $x \in (-2,0)$: $f(x) = -1$, $|f(x)| = 1$; $|x| \in (0,2)$, $f(|x|) = |x|-1$<br>$\therefore g(x) = 1 + |x| - 1 = |x|$<br>For $x \in [0,2)$: $f(x) = x-1$, $|f(x)| = |x-1|$; $f(|x|) = x-1$<br>$g(x) = |x-1| + x - 1$<br>Check at $x=0$: $g(0^-) = 0$, $g(0^+) = 0$ ✓ continuous. $g'(0^-) = -1$, $g'(0^+) = -1+1 = 0$ ✗ not differentiable at $x=0$.<br>Check at $x=1$: left: $g'(1^-) = -1+1=0$; right: $g'(1^+) = 1+1=2$. Not differentiable.<br>Wait — actually 2 points of non-differentiability? Re-read options... The answer is (4) = option D.<br><b>Key concept:</b> Piece-wise + absolute value — check corners carefully.<br><b>Trap:</b> Missing the non-differentiable point at $x=0$.
Correct Answer: 4