If the function f : [ 1 , ∞ ) → [ 1 , ∞ ) is defined by f ( x ) = 2 x ( x − 1 ) , then f − 1 ( x ) is Options: A . &nbsp ( 1 2 ) x ( x − 1 ) B . &nbsp 1 2 ( 1 + √ 1 + 4 l o g 2 x ) C . &nbsp ( 1 2 ) ( 1 − √ 1 + 4 l o g 2 x ) D . &nbsp N o n e o f t h e s e

Answer: Option B : B f ( x ) = y ⇒ 2 x ( x − 1 ) = y ⇒ x ( x − 1 ) l o g 2 2 = l o g 2 y ⇒ x ( x − 1 ) = l o g 2 y ⇒ x 2 − x − l o g 2 y = 0 ⇒ x = 1 ± √ 1 + 4 l o g 2 y 2 ∴ x = 1 + √ 1 + 4 l o g 2 y 2 ∴ f − 1 ( x ) = 1 2 ( 1 + √ 1 + 4 l o g 2 x ) ∴ The correct answer is (b).

If the function f:[1,∞)→[1,∞) is defined by f(x)=2x(x

Discussion Forum : Relations And Functions Ii

Question -

If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ , then $f^{- 1} (x)$ is

Options:

A .

{(\frac{1}{2})}^{x (x - 1)}

B .

\frac{1}{2} (1 + \sqrt{1 + 4 l o g_{2} x})

C .

(\frac{1}{2}) (1 - \sqrt{1 + 4 l o g_{2} x})

D .

N o n e o f t h e s e

Answer: Option B
:
B

f (x) = y \Rightarrow 2^{x (x - 1)} = y \Rightarrow x (x - 1) l o g_{2} 2 = l o g_{2} y

\Rightarrow x (x - 1) = l o g_{2} y \Rightarrow x^{2} - x - l o g_{2} y = 0

\Rightarrow x = \frac{1 \pm \sqrt{1 + 4 l o g_{2} y}}{2}

∴ x = \frac{1 + \sqrt{1 + 4 l o g_{2} y}}{2}

∴ f^{- 1} (x) = \frac{1}{2} (1 + \sqrt{1 + 4 l o g_{2} x})

∴

The correct answer is (b).

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