A) \[x=1\] only
B) \[x=1\] and \[x=-1\] only
C) \[x=1,x=-1,x=-3\] only
D) \[x=1,x=-1,x=-3\] and other values of x
Correct Answer: C
Solution :
\[f(x)=\frac{2{{x}^{2}}+7}{{{x}^{2}}(x+3)-1(x+3)}\frac{9{{x}^{2}}+7}{({{x}^{2}}-1)(x+3)}\] \[=\frac{2{{x}^{2}}+7}{(x-1)(x+1)(x+3)}\] Hence points of discontinuity are \[x=1\], \[x=-1\] and \[x=-3\] only.You need to login to perform this action.
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