If f:[1, +infty) to [2, +infty) is given by f | vedclass

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

Question

(A) Given $y = f(x) = x + \frac{1}{x}$ where $x \ge 1$ and $y \ge 2$. Multiplying by $x$,we get $yx = x^2 + 1$,which rearranges to the quadratic equat

Vedclass · Accepted Answer

$\frac{x + \sqrt{x^2 - 4}}{2}$

Vedclass · Answer

(A) Given $y = f(x) = x + \frac{1}{x}$ where $x \ge 1$ and $y \ge 2$. Multiplying by $x$,we get $yx = x^2 + 1$,which rearranges to the quadratic equation $x^2 - yx + 1 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{y \pm \sqrt{y^2 - 4}}{2}$. Since the domain of $f$ is $[1, +\infty)$,we must have $x \ge 1$. If we take the negative sign,$x = \frac{y - \sqrt{y^2 - 4}}{2}$. For $y \ge 2$,this value is $\le 1$. Specifically,as $y 	o \infty$,this expression approaches $0$. Thus,to satisfy $x \ge 1$,we must take the positive sign: $x = \frac{y + \sqrt{y^2 - 4}}{2}$. Replacing $y$ with $x$,we obtain ${f^{-1}}(x) = \frac{x + \sqrt{x^2 - 4}}{2}$.

If $f:[1, +\infty) \to [2, +\infty)$ is given by $f(x) = x + \frac{1}{x}$,then ${f^{-1}}(x)$ equals:

Similar Questions

Let $f(x)=(x+1)^2-1$ for $x \geq -1$.
Statement-$1$: $S=\{x:f(x)=f^{-1}(x)\}=\{0, -1\}$
Statement-$2$: $f$ is a bijection.

If $g(x)$ is the inverse of the function $f(x)$ and $f^{\prime}(x) = \frac{1}{h(x)}$,then $g^{\prime}(x) = $

The number of integral values of $a$ for which the function $f: R \to R, f(x) = 2x^3 - 3(a + 2)x^2 + 12ax - 7$ where $a \in [-4, 6]$ is invertible,is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module