Limits, Continuity & Differentiability
Continuity and Differentiability of Absolute Value Functions
nta_pyq_2024_jan
Grade 12
Question:
Let $f(x)=|2x^2+5|x|-3|$, $x\in\mathbb{R}$. If $m$ and $n$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $m+n$ is equal to:
Step-by-Step Solution
Key Concept: First analyze $g(x)=2x^2+5|x|-3$. For $x>0$: $g(x)=2x^2+5x-3=(2x-1)(x+3)$, roots at $x=1/2$ and $x=-3$ (rejected). For $x<0$: $g(x)=2x^2-5x-3=(2x+1)(x-3)$, roots at $x=-1/2$. Then $f(x)=|g(x)|$ — count points where inner expression changes sign (non-differentiable corners).
$f(x)=|2x^2+5|x|-3|$. Polynomial with absolute values — continuous everywhere, $m=0$.
Non-differentiable points: where the expression inside the outer absolute value is 0 ($x=\pm1/2$) and where $|x|$ is non-differentiable ($x=0$, check if $g(0)=|-3|=3\ne0$ — but $|x|$ causes corner in $g$, which carries through to $f$).
Total: $n=3$ ($x=-1/2, 0, 1/2$). $m+n=0+3=3$.
Correct Answer: 4