Question

For a suitably chosen real constant a, let a function, \(f:\mathbb{R}-\{-a\}\to\mathbb{R}\) be defined by \(f(x)=\frac{a-x}{a+x}\). Further suppose that for any real number \(x\ne -a\) and \(f(x)\ne -a\), \((f\circ f)(x)=x\). Then \(f\left(-\frac12\right)\) is equal to :

• A. \(\frac13\)
• B. \(-3\)
• C. \(-\frac13\)
• D. \(3\)

Hint:

Compute \(f(f(x))\) and demand that it equals x for all admissible x; this fixes a.

Answer 1

The Correct Option is D

\(f(f(x))=\dfrac{a-\frac{a-x}{a+x}}{a+\frac{a-x}{a+x}}=\dfrac{a(a+x)-(a-x)}{a(a+x)+(a-x)}=\dfrac{a^2-a+(a+1)x}{a^2+a+(a-1)x}\).
Setting this equal to x for all x: \(a^2-a+(a+1)x = (a^2+a)x+(a-1)x^2\).
Comparing coefficients: \(x^2\): \(a-1=0 \Rightarrow a=1\); constant: \(a^2-a=0\) (true); x: \(a+1=a^2+a\Rightarrow a^2=1\) (true).
So \(a=1\), \(f(x)=\frac{1-x}{1+x}\).
\(f\left(-\frac12\right)=\dfrac{1+\frac12}{1-\frac12}=\dfrac{3/2}{1/2}=3\).
Answer: (4) 3