Limits Continuity And Differentiability Question 39
Question: 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 points of discontinuity of $ | f(x) | $ is
Options:
A) 1
B) 2
C) 3
D) none of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)= \begin{cases} x+2, \\ -x^{2}-2, \\ x, \\ \end{cases} .\begin{matrix} x<0 \\ 0\le x<1 \\ x\ge 1 \\ \end{matrix} $
$ \therefore | f(x) |= \begin{cases} -x-2, \\ x+2, \\ x^{2}+2, \\ x, \\ \end{cases} .\begin{matrix} x<-2 \\ -2\le x<0 \\ 0\le x<1 \\ x\ge 1 \\ \end{matrix} $
It is discontinuous at x=1.
Therefore, number of discontinuity is 1.