Let f(x) = sin x and g(x) = ln |x|. If the ra | vedclass

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

Question

(D) Given $f(x) = \sin x$ and $g(x) = \ln |x|$.For $fog(x) = f(g(x)) = \sin(\ln |x|)$.Since the range of $\ln |x|$ is $(-\infty, \infty)$,the range of

Vedclass · Accepted Answer

$R_1 = \{ u: - 1 \le u \le 1\} , R_2 = \{ v: - \infty < v \le 0\}$

Vedclass · Answer

(D) Given $f(x) = \sin x$ and $g(x) = \ln |x|$.For $fog(x) = f(g(x)) = \sin(\ln |x|)$.Since the range of $\ln |x|$ is $(-\infty, \infty)$,the range of $\sin(\ln |x|)$ is $[-1, 1]$. Thus,$R_1 = [-1, 1]$.For $gof(x) = g(f(x)) = \ln |\sin x|$.Since the range of $|\sin x|$ is $(0, 1]$,the range of $\ln |\sin x|$ is $(-\infty, 0]$. Thus,$R_2 = (-\infty, 0]$.Therefore,$R_1 = \{u: -1 \le u \le 1\}$ and $R_2 = \{v: -\infty < v \le 0\}$.

Let $f(x) = \sin x$ and $g(x) = \ln |x|$. If the ranges of the composite functions $fog$ and $gof$ are $R_1$ and $R_2$ respectively,then:

Similar Questions

Define $f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$. If $f \circ f(x)$ is discontinuous at $a$ and $b$ in $[0, 3]$ and $a < b$, then $2 a + 3 b = $

Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on the set $A = \{1, 2, 3\}$. Then $R \circ S = $

If $f(x) = (a - x^n)^{1/n},$ where $a > 0$ and $n$ is a positive integer,then $f[f(x)] = $

$f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{\sqrt{x}}$,then:

If $f(x) = 3x + 10$ and $g(x) = x^2 - 1$,then $(fog)^{-1}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module