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JEE MAIN 2019
08-04-2019 S2
Question
Let $f:[-1,3] \rightarrow R$ be defined as $$ f(x)=\left\{\begin{array}{cc} |x|+[x], & -1 \leq x<1 \\ x+|x|, & 1 \leq x<2 \\ x+[x], & 2 \leq x \leq 3 \end{array}\right. $$ where $[t]$ denotes the greatest integer less than or equal to $t$. Then, $f$ is discontinuous at :
Select the correct option:
A
Only one point
B
Only two points
C
Only three points
D
Four or more points
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & f(x)= \begin{cases}|x|+[x], & -1 \leq x<1 \\ x+|x|, & 1 \leq x<2 \\ x+[x], & 2 \leq x \leq 3\end{cases} \\ & =\left\{\begin{array}{cc}-x-1, & -1 \leq x<0 \\ x, & 0 \leq x<1 \\ 2 x, & 1 \leq x<2 \\ x+2, & 2 \leq x<3 \\ 6, & x=3\end{array}\right. \\ & \Rightarrow \quad \begin{array}{l}f(-1)=0, f\left(-1^{+}\right)=0 \\ f(0-)=-1, f(0)=0, f\left(0^{+}\right)=0 \\ f(1-)=1, f(1)=2, f\left(1^{+}\right)=2 \\ f(2-)=4, f(2)=4, f\left(2^{+}\right)=4 \\ f(3-)=5, f(3)=6\end{array} \\ & f(x) \text { is discontinuous at } x=\{0,1,3\}\end{aligned}$
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