A) \[1-\frac{1}{a}\]
B) \[\frac{1}{a}\]
C) \[1+\frac{1}{a}\]
D) \[\frac{1}{a}-1\]
Correct Answer: C
Solution :
[c] Given \[f(x)=\frac{\alpha {{x}^{2}}}{x+1},x\ne -1;f(a)=a\] \[\frac{\alpha {{a}^{2}}}{a+1}=a\Rightarrow \alpha =1+\frac{1}{a}.\]
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