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.