Integral Calculus
Discontinuity of Functions
MMTS_Full_Test_08
Grade 12
Question:
Let $f(x)=\begin{cases}|1-2x^2|, & 0\le x<1 \\ [x^2-2x], & 1\le x<2\end{cases}$. If $m,n$ are number of points of discontinuity and non-differentiability of $f(x)$ in $(0,2)$, then
m=1, n=2
m=1, n=1
m=2, n=2
m=3, n=2
Step-by-Step Solution
Key Concept: Check at $x=1$ and within each piece
Discontinuity: at $x=1$ (from GIF). Non-differentiability: at $x=1/\sqrt{2}$ (absolute value) and $x=1$. $m=1,n=2$.
Correct Answer: 1