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Choose the correct option.
If f(x) = {x+2, x < 0; -x²-2, 0 ≤ x < 1; x, x ≥ 1} then the number of point(s) of discontinuity of…
If $f(x) = \begin{cases} x+2 & x < 0 \\ -x^2-2 & 0 \le x < 1 \\ x & x \ge 1 \end{cases}$ then the number of point(s) of discontinuity of $f(x)$, is / are:
- (a)$1$
- (b)$3$
- (c)$2$
- (d)$0$
Answer
Answer
Correct option: (c)
Answer: (c) $2$
At $x=0$: left limit $2$, $f(0)=-2$, discontinuous. At $x=1$: left limit $-3$, $f(1)=1$, discontinuous.