Let f(x) = (x - 1)^2 + 1 for x ge 1.Statement | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x) = (x - 1)^2 + 1$ for $x \ge 1$.To find $f^{-1}(x)$,set $y = (x - 1)^2 + 1$.Then $y - 1 = (x - 1)^2$. Since $x \ge 1$,we have $x - 1 =

Vedclass · Accepted Answer

Statement-$1$ is true,Statement-$2$ is true; Statement-$2$ is a correct explanation for Statement-$1$.

Vedclass · Answer

(D) Given $f(x) = (x - 1)^2 + 1$ for $x \ge 1$.To find $f^{-1}(x)$,set $y = (x - 1)^2 + 1$.Then $y - 1 = (x - 1)^2$. Since $x \ge 1$,we have $x - 1 = \sqrt{y - 1}$,so $x = 1 + \sqrt{y - 1}$.Thus,$f^{-1}(x) = 1 + \sqrt{x - 1}$ for $x \ge 1$. So,Statement-$2$ is true.To find $S = \{x : f(x) = f^{-1}(x)\}$,we solve $f(x) = x$.$(x - 1)^2 + 1 = x \implies x^2 - 2x + 1 + 1 = x \implies x^2 - 3x + 2 = 0$.$(x - 1)(x - 2) = 0$,so $x = 1$ or $x = 2$.Since both values satisfy $x \ge 1$,$S = \{1, 2\}$. So,Statement-$1$ is true.Since $f(x) = f^{-1}(x)$ is equivalent to $f(x) = x$ for increasing functions,Statement-$2$ explains Statement-$1$.

Let $f(x) = (x - 1)^2 + 1$ for $x \ge 1$.
Statement-$1$: $S = \{x : f(x) = f^{-1}(x)\} = \{1, 2\}$.
Statement-$2$: $f$ is a bijection and $f^{-1}(x) = 1 + \sqrt{x - 1}$ for $x \ge 1$.

Similar Questions

If the functions $f$ and $g$ are defined by $f(x) = 3x - 4$ and $g(x) = 2 + 3x$ for $x \in R$,then $g^{-1}(f^{-1}(5))$ is equal to

Let the function $f$ be defined by $f(x) = \frac{2x + 1}{1 - 3x}$,then $f^{-1}(x)$ is

If $y = f(x) = \frac{x + 2}{x - 1}$,then $x = $

If $g$ is the inverse of the function $f(x)$ and $g(x) = x + \tan x$,then $f^{\prime}(x) = $

If $f(x) = (2x - 3\pi)^5 + \frac{4}{3}x + \cos x$ and $g$ is the inverse of $f$,then $g'(2\pi) = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $f(x) = (x - 1)^2 + 1$ for $x \ge 1$.Statement-$1$: $S = \{x : f(x) = f^{-1}(x)\} = \{1, 2\}$.Statement-$2$: $f$ is a bijection and $f^{-1}(x) = 1 + \sqrt{x - 1}$ for $x \ge 1$.