Question

Let f(x) =$\frac{\alpha x}{x + 1}$, x ≠ 1. Then, for what value of α is f(f(x))

= x ?

• A. $\sqrt{2}$
• B. −$\sqrt{2}$
• C. 1
• D. −1

Answer 1

Answer: (D)

−1

Answer 2

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

f(f(x)) = x ⇒ $\frac{\alpha\left( \frac{\alpha x}{x + 1} \right)}{\frac{\alpha x}{x + 1} + 1}$= x

⇒ $\frac{\alpha^{2}x}{(\alpha + 1)x + 1}$= x ⇒ (α + 1)x² + (1 − α²) x = 0 ……(1)

⇒ α² − α − 2 = 0

⇒ (α − 2) (α + 1) = 0 ⇒ α = 2, − 1

Putting x = 2 in (1)

α² - 2α − 3 = 0

⇒ (α + 1) (α −3) = 0 ⇒ α = −1, 3

The common solution is α = −1

∴ (4) is the correct alternative