If f(x) = frac{{α ,x}}{{x + 1}},x ne - 1 . Then, for what value of α is f(f(x)) = x

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1.Relation and Function

medium

If $f(x) = \frac{{\alpha \,x}}{{x + 1}},\;x \ne - 1$. Then, for what value of $\alpha $ is $f(f(x)) = x$

$\sqrt 2 $

$ - \sqrt 2 $

$1$

$-1$

(IIT-2001)

Solution

(d) $f(f(x)) = \frac{{\alpha \,f(x)}}{{f(x) + 1}} $

$= \frac{{\alpha \left( {\frac{{\alpha x}}{{x + 1}}} \right)}}{{\left( {\frac{{\alpha x}}{{x + 1}} + 1} \right)}} = \frac{{{\alpha ^2}.x}}{{\alpha x + x + 1}}$

$\therefore$ $x = \frac{{{\alpha ^2}.x}}{{(\alpha + 1)x + 1}}$ or $x((\alpha + 1)x + 1 – {\alpha ^2}) = 0$

==> $(\alpha + 1){x^2} + (1 – {\alpha ^2})x = 0$. This should hold for all $x.$

==> $\alpha + 1 = 0,\,\,1 – {\alpha ^2} = 0$,

$\because \alpha = – 1$.

Standard 12

Mathematics

Suppose $f$ is a function satisfying $f ( x + y )= f ( x )+ f ( y )$ for all $x , y \in N$ and $f (1)=\frac{1}{5}$. If $\sum \limits_{n=1}^m \frac{f(n)}{n(n+1)(n+2)}=\frac{1}{12}$, then $m$ is equal to $……………$.

hard

(JEE MAIN-2023)

Which of the following function is even function

medium

Which of the following is true

normal

Let $R$ be the set of real numbers and $f: R \rightarrow R$ be defined by $f(x)=\frac{\{x\}}{1+[x]^2}$, where $[x]$ is the greatest integer less than or equal to $x$, and $\left\{x{\}}=x-[x]\right.$. Which of the following statements are true?

$I.$ The range of $f$ is a closed interval.

$II.$ $f$ is continuous on $R$.

$III.$ $f$ is one-one on $R$

normal

(KVPY-2017)

The domain of the function $f(x) = \frac{{{{\sin }^{ – 1}}(3 – x)}}{{\ln (|x|\; – 2)}}$ is

medium

If $f(x) = \frac{{\alpha \,x}}{{x + 1}},\;x \ne - 1$. Then, for what value of $\alpha $ is $f(f(x)) = x$

Solution

Similar Questions

Suppose $f$ is a function satisfying $f ( x + y )= f ( x )+ f ( y )$ for all $x , y \in N$ and $f (1)=\frac{1}{5}$. If $\sum \limits_{n=1}^m \frac{f(n)}{n(n+1)(n+2)}=\frac{1}{12}$, then $m$ is equal to $……………$.

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