Let f(x)=sqrt{x^{2}-3x+2} and g(x)=sqrt{x} be | vedclass

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

Question

(A) For $f \circ g(x) = f(g(x)) = \sqrt{(\sqrt{x})^2 - 3\sqrt{x} + 2} = \sqrt{x - 3\sqrt{x} + 2}$.For the domain $S$, we require $x \ge 0$ and $x - 3\

Vedclass · Accepted Answer

$S=T$

Vedclass · Answer

(A) For $f \circ g(x) = f(g(x)) = \sqrt{(\sqrt{x})^2 - 3\sqrt{x} + 2} = \sqrt{x - 3\sqrt{x} + 2}$.For the domain $S$, we require $x \ge 0$ and $x - 3\sqrt{x} + 2 \ge 0$.Let $u = \sqrt{x}$, then $u^2 - 3u + 2 \ge 0 \implies (u-1)(u-2) \ge 0$.This gives $u \le 1$ or $u \ge 2$, so $\sqrt{x} \le 1$ or $\sqrt{x} \ge 2$.Thus, $x \in [0, 1] \cup [4, \infty)$, so $S = [0, 1] \cup [4, \infty)$.For $g \circ f(x) = g(f(x)) = \sqrt{\sqrt{x^2 - 3x + 2}}$.For the domain $T$, we require $x^2 - 3x + 2 \ge 0 \implies (x-1)(x-2) \ge 0$.This gives $x \le 1$ or $x \ge 2$, so $T = (-\infty, 1] \cup [2, \infty)$.Now, $S \cap T = ([0, 1] \cup [4, \infty)) \cap ((-\infty, 1] \cup [2, \infty)) = [0, 1] \cup [4, \infty)$.Since $S \cap T = S$, and $S$ is a union of two disjoint intervals, the intersection is not a single interval.

Let $f(x)=\sqrt{x^{2}-3x+2}$ and $g(x)=\sqrt{x}$ be two given functions. If $S$ is the domain of $f \circ g$ and $T$ is the domain of $g \circ f$, then:

Similar Questions

Let $f(x) = \begin{cases} x, & x < 0 \\ 1 + x^2, & x \geq 0 \end{cases}$ and $g(x) = 1 + x - [x]$,then the range of $f(g(x))$ is (where $[.]$ denotes the greatest integer function).

If $f(x)=-|x|$, then $(f \circ f \circ f)(x) + (f \circ f \circ f)(-x) =$

If $f(x) = \begin{cases} \sin x, & x \neq n\pi, n \in \mathbb{Z} \\ 0, & \text{otherwise} \end{cases}$ and $g(x) = \begin{cases} x^2 + 1, & x \neq 0, 2 \\ 4, & x = 0 \\ 5, & x = 2 \end{cases}$,then $\lim_{x \to 0} g(f(x)) = $

If $f(x) = \sin^2 x$ and the composite function $g(f(x)) = |\sin x|$,then the function $g(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module