A) \[1-\frac{1}{a}\]
B) \[\frac{1}{a}\]
C) \[1+\frac{1}{a}\]
D) \[\frac{1}{a}-1\]
E) \[-\frac{1}{a}\]
Correct Answer: C
Solution :
We have\[f(x)=\frac{\alpha {{x}^{2}}}{x+1}x\ne -1\] According to question \[f(a)=a\] \[\Rightarrow \] \[\frac{\alpha {{a}^{2}}}{a+1}=a\] \[\Rightarrow \] \[\alpha {{a}^{2}}={{a}^{2}}+a\] \[\Rightarrow \] \[\alpha =\frac{{{a}^{2}}+a}{{{a}^{2}}}=1+\frac{1}{a}\]
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