A) 4
B) 3
C) 2
D) 1
Correct Answer: B
Solution :
\[\mu (x)=\frac{1}{x-1},\]which is discontinous at\[x=1\] \[f(u)=\frac{1}{{{u}^{2}}+u-2}=\frac{1}{(u+2)(u-1)},\] which is discontinous at \[u=-2,1\] when \[u=-2,\]then\[\frac{1}{x-1}=-2\Rightarrow x=\frac{1}{2}\] when \[u=1,\]then\[\frac{1}{x-1}=1\Rightarrow x=2\] Hence given composite function is discontinous at three points, \[x=1,\frac{1}{2}\]and 2.You need to login to perform this action.
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