A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
[c] \[t=\frac{1}{x-1}\] is discontinuous at \[x=1.\] Also \[y=\frac{1}{{{t}^{2}}+t-2}\] discontinuous at \[t=-2\] and \[t=1\] When \[t=-2,\frac{1}{x-1}=-2\Rightarrow x=\frac{1}{2}\] When \[t=1,\,\,\frac{1}{x-1}\,\,\,\,\Rightarrow x=2\] So, \[y=f(x)\] is discontinuous at three points \[x=1,\,\,\frac{1}{2},\,\,2\]You need to login to perform this action.
You will be redirected in
3 sec