A) \[2,\,\,\frac{2}{3},\,\,\frac{7}{3}\]
B) \[2,\,\,\frac{3}{2},\,\,\frac{7}{3}\]
C) \[2,\,\,\frac{3}{2},\,\,\frac{3}{7}\]
D) None of the above
Correct Answer: B
Solution :
Clearly; the function\[t=\frac{1}{x-2}\]is discontinuous at \[x=2\] and the function\[y=\frac{1}{{{t}^{2}}-t-6}\]is discontinuous at the points, where \[{{t}^{2}}-t-6=0\] \[\Rightarrow \] \[(t+2)(t-3)=0\] \[\Rightarrow \] \[t=-2,\,\,3\] But \[t=-2\] \[\Rightarrow \] \[\frac{1}{x-2}=-2\] \[\Rightarrow \] \[x-2=-\frac{1}{2}\] \[\Rightarrow \] \[x=\frac{3}{2}\] and \[t=3\] \[\Rightarrow \] \[\frac{1}{x-2}=3\] \[\Rightarrow \] \[x-2=\frac{1}{3}\] \[\Rightarrow \] \[x=\frac{7}{3}\] Therefore, the values of \[x\] which make the function \[y\] discontinuous are\[x=2,\,\,\,\frac{3}{2}\]and\[\frac{7}{3}\].
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