Question

Let $f\left(x\right)=\frac{\alpha x^2}{\left(x+1\right)}$,$x\ne -1$. The value of $\alpha$ for which $f\left(a\right)=a,(a\ne 0)$ is

• A. $1-\frac{1}{a}$
• B. $\frac{1}{a}$
• C. $1+\frac{1}{a}$
• D. $\frac{1}{a}-1$

Answer 1

The correct option is C

$1+\frac{1}{a}$

Explanation for Correct Answer:

Given that,.

$f\left(x\right)=\frac{\alpha x^2}{\left(x+1\right)}$,$x\ne -1$

$f\left(a\right)=a,$$(a\ne 0)$

Calculating $f\left(a\right)$:

$f\left(a\right)=\frac{\alpha a^2}{a+1}$

Now,$a=\frac{\alpha a^2}{a+1}$

$$\begin{gathered} \Rightarrow \alpha .a^2=a(a+1) \\ \Rightarrow \alpha .a=(a+1) \\ \Rightarrow \alpha =\frac{a+1}{a} \end{gathered}$$

Therefore,$\alpha =1+\frac{1}{a}$

Hence, option (C) is the correct answer