Practicing Success
Let $A = \{x∈R:x ≥1\}$. The inverse of the function f:A→A given by $f(x) = 2^{x(x-1)}$, is |
$(\frac{1}{2})^{x(x-1)}$ $\frac{1}{2}\left\{1+\sqrt{1+4\log_2x}\right\}$ $\frac{1}{2}\left\{1-\sqrt{1+4\log_2x}\right\}$ not defined |
$\frac{1}{2}\left\{1+\sqrt{1+4\log_2x}\right\}$ |
The correct answer is Option (2) → $\frac{1}{2}\left\{1+\sqrt{1+4\log_2x}\right\}$ It can be easily verified that f: A → A is a bijection. Let $f(x) = y$. Then, $f(x)=y$ $⇒2^{x (x − 1)} = y$ $⇒x (x-1) = \log_2 y$ $⇒x^2-x-\log_2 y = 0$ $x=\frac{1±\sqrt{1 + 4 \log_2 y}}{2}$ $⇒x=\frac{1}{2}\{1+\sqrt{1 + 4 \log_2 y}\}$ $[∵x>1]$ $⇒f^{-1}(y)=\frac{1}{2}\left\{1+\sqrt{1 + 4 \log_2 y}\right\}$ Hence $f^{-1}: A → A$ is given by $f^{-1}(x)=\frac{1}{2}\left\{1+\sqrt{1 + 4 \log_2 y}\right\}$ |