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\s

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 $f: R \rightarrow R$ be defined by $f(x)=3 x^2-5$ and $g: R \rightarrow R$ by $g(x)=\frac{x}{x^2+1}$,then $g \circ f$ is

If $f : R \rightarrow R$ is defined by $f(x) = x^{2} - 3x + 2$,find $f(f(x))$.

For a suitably chosen real constant $a$,let the function $f: R-\{-a\} \rightarrow R$ be defined by $f(x)=\frac{a-x}{a+x}$. Further,suppose that for any real number $x \neq-a$ and $f(x) \neq-a$,$(f \circ f)(x)=x$. Then,$f\left(-\frac{1}{2}\right)$ is equal to

If $g(x)=x^2+x-1$ and $(g \circ f)(x)=4 x^2-10 x+5$,then $f(2)$ is equal to

If $f(x) = \log \left(\frac{1+x}{1-x}\right)$ and $g(x) = \frac{3x+x^3}{1+3x^2}$,then $(fog)(x) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module