SATHEE: Functions 2 Question 7

Functions 2 Question 7

8.

Let $f (x) = \frac{α x}{x + 1}, x \neq - 1$ . Then, for what value of $α$ is $f [f (x)] = x ?$

(a) $\sqrt{2}$

(b) $- \sqrt{2}$

(d) -1

(2001, 1M)

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Answer:

Correct Answer: 8. (d)

Solution:

Given, $f (x) = \frac{α x}{x + 1}$

$f [f (x)] = f \frac{α x}{x + 1} = \frac{α \frac{α x}{x + 1}}{\frac{α x}{x + 1} + 1}$

$= \frac{\frac{α^{2} x}{x + 1}}{\frac{α x + (x + 1)}{x + 1}} = \frac{α^{2} x}{(α + 1) x + 1} = x [given] …(i)$

$\Rightarrow α^{2} x = (α + 1) x^{2} + x$

$\Rightarrow x [α^{2} - (α + 1) x - 1] = 0$

$\Rightarrow x (α + 1) (α - 1 - x) = 0$

$\Rightarrow and α + 1 = 0$

$\Rightarrow α = - 1$

But $α = 1$ does not satisfy the Eq. (i).

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Functions 2 Question 7

8.

Answer:

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