If the function f -1- - -1- - is defined by f x -2 x x -1- then f -1 x isA- 1-2xx-1- 1-21-1-4 log 2 x C- 1-21-1-4 log 2 xD- None of these

The correct option is B 1/2(1+ √(1+4 log_2 x))f(x)=y⇒ 2^x(x 1)=y ⇒ x(x 1) log_2 2=log_2y ⇒ x(x 1)=log_2 y ⇒ x^2 x log_2 y=0 ⇒ x=1 ±√(1+4 log_2 y)/2 ∴ x=1+ √(1 ...

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Byju's Answer

Standard XII

Mathematics

Inverse of a Function

If the functi...

Question

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

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

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

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

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N o n e o f t h e s e

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Solution

The correct option is B $\frac{1}{2} (1 + \sqrt{1 + 4 l o g_{2} x})$
$f (x) = y \Rightarrow 2^{x (x - 1)} = y \Rightarrow x (x - 1) l o g_{2} 2 = l o g_{2} y$
$\Rightarrow x (x - 1) = l o g_{2} y \Rightarrow x^{2} - x - l o g_{2} y = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{1 + 4 l o g_{2} y}}{2}$
$∴ x = \frac{1 + \sqrt{1 + 4 l o g_{2} y}}{2}$
$∴ f^{- 1} (x) = \frac{1}{2} (1 + \sqrt{1 + 4 l o g_{2} x})$
$∴$ The correct answer is (b).

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

The correct option is B 12(1+√1+4 log2 x)f(x)=y⇒2x(x−1)=y⇒x(x−1) log2 2=log2y ⇒x(x−1)=log2 y⇒x2−x−log2 y=0 ⇒x=1±√1+4 log2 y2 ∴x=1+√1+4 log2 y2 ∴f−1(x)=12(1+√1+4 log2 x) ∴ The correct answer is (b).

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