If the function f-1- - -1- - is defined by fx -2xx-1- then f-1x is

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+4 log_2 y)/2 ∴ f^ 1(x)=1/2(1+√(1+4 log_2 x)) ∴ ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

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)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N o n e o f t h e s e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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).

Suggest Corrections

Similar questions

Q. If the function

f : [1, \infty) \to [1, \infty)

is defined by

f (x) = 2^{x (x - 1)}

, then

f^{- 1} (x)

Q. If the function

f : [1, \infty) \to [1, \infty)

is defined by

f (x) = 2^{x (x - 1)}

, then

f^{- 1} (x)

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

Q. If the function

f : [1, \infty) \to [1, \infty)

is defined by

f (x) = 2^{x (x - 1)}

, then

f^{- 1} (x)

Q. The function

f : (0, \infty) \to (- \infty, \infty)

is defined by

f (x) = {log}_{3} x

then

f^{- 1} (x) =

Join BYJU'S Learning Program

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