If f: R rightarrow R is defined by f(x)=e^{x} | vedclass

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

Question

(C) Given $f: R \rightarrow R$ where $f(x) = e^{x}$ and $g: R \rightarrow R$ where $g(x) = x^{2}$.We calculate the composite function $(g \circ f)(x)

Vedclass · Accepted Answer

$g \circ f$ is injective but $g$ is not bijective

Vedclass · Answer

(C) Given $f: R \rightarrow R$ where $f(x) = e^{x}$ and $g: R \rightarrow R$ where $g(x) = x^{2}$.We calculate the composite function $(g \circ f)(x) = g(f(x)) = g(e^{x}) = (e^{x})^{2} = e^{2x}$.For $(g \circ f)(x) = e^{2x}$, if $(g \circ f)(x_{1}) = (g \circ f)(x_{2})$, then $e^{2x_{1}} = e^{2x_{2}}$, which implies $2x_{1} = 2x_{2}$, so $x_{1} = x_{2}$. Thus, $g \circ f$ is injective.However, the range of $g \circ f$ is $(0, \infty)$, which is not equal to the codomain $R$, so $g \circ f$ is not surjective.For $g(x) = x^{2}$, $g(-1) = 1$ and $g(1) = 1$, so $g$ is not injective. Also, the range of $g$ is $[0, \infty)$, so $g$ is not surjective.Therefore, $g \circ f$ is injective but $g$ is not bijective.

If $f: R \rightarrow R$ is defined by $f(x)=e^{x}$ and $g: R \rightarrow R$ is defined by $g(x)=x^{2}$, then the mapping $(g \circ f): R \rightarrow R$ is defined by $(g \circ f)(x) = g(f(x))$ for all $x \in R$. Which of the following is true?

Similar Questions

Let $f^1(x) = \frac{3x + 2}{2x + 3}$,$x \in R - \left\{-\frac{3}{2}\right\}$. For $n \geq 2$,define $f^n(x) = f^1 \circ f^{n-1}(x)$. If $f^5(x) = \frac{ax + b}{bx + a}$ and $\gcd(a, b) = 1$,then $a + b$ is equal to $............$.

If $g(f(x)) = |\sin x|$ and $f(g(x)) = (\sin \sqrt{x})^2$,then

Consider functions $f: A \rightarrow B$ and $g: B \rightarrow C$ $(A, B, C \subseteq \mathbb{R})$ such that $(g \circ f)^{-1}$ exists. Then:

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

Let $f: R - \{-\frac{1}{2}\} \rightarrow R$ and $g: R - \{-\frac{5}{2}\} \rightarrow R$ be defined as $f(x) = \frac{2x+3}{2x+1}$ and $g(x) = \frac{|x|+1}{2x+5}$. Then the domain of the function $f \circ g$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module