Functions 2 Question 7
8.
Let $f(x)=\frac{\alpha x}{x+1}, x \neq-1$. Then, for what value of $\alpha$ is $f[f(x)]=x ?$
(a) $\sqrt{2}$
(b) $-\sqrt{2}$
(c) 1
(d) -1
(2001, 1M)
Show Answer
Answer:
Correct Answer: 8. (d)
Solution:
- Given, $f(x)=\frac{\alpha x}{x+1}$
$ f[f(x)]=f \frac{\alpha x}{x+1}=\frac{\alpha \frac{\alpha x}{x+1}}{\frac{\alpha x}{x+1}+1} $
$ =\frac{\frac{\alpha^{2} x}{x+1}}{\frac{\alpha x+(x+1)}{x+1}}=\frac{\alpha^{2} x}{(\alpha+1) x+1}=x \text { [given] …(i) } $
$\Rightarrow \quad \alpha^{2} x=(\alpha+1) x^{2}+x $
$\Rightarrow \quad x\left[\alpha^{2}-(\alpha+1) x-1\right]=0 $
$\Rightarrow \quad x(\alpha+1)(\alpha-1-x) =0 $
$\Rightarrow \quad \text { and } \quad \alpha+1 =0 $
$\Rightarrow \quad \alpha =-1$
But $\alpha=1$ does not satisfy the Eq. (i).
