The range of values of the function f(x) = fr | vedclass

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

Question

(C) We know that for any real $x$,the range of $\sin x$ is $[-1, 1]$.So,$-1 \leq \sin x \leq 1$.Multiplying by $-3$,we get $-3 \leq -3 \sin x \leq 3$.

Vedclass · Accepted Answer

$( -\infty, -1 ] \cup [ 1/5, \infty )$

Vedclass · Answer

(C) We know that for any real $x$,the range of $\sin x$ is $[-1, 1]$.So,$-1 \leq \sin x \leq 1$.Multiplying by $-3$,we get $-3 \leq -3 \sin x \leq 3$.Adding $2$ to all parts,we get $2 - 3 \leq 2 - 3 \sin x \leq 2 + 3$,which simplifies to $-1 \leq 2 - 3 \sin x \leq 5$.Now,we consider the function $f(x) = \frac{1}{2 - 3 \sin x}$.Since $2 - 3 \sin x$ takes values in $[-1, 5]$ excluding $0$ (where the function is undefined),we analyze the intervals.When $2 - 3 \sin x \in [-1, 0)$,the value of $\frac{1}{2 - 3 \sin x}$ lies in $(-\infty, -1]$.When $2 - 3 \sin x \in (0, 5]$,the value of $\frac{1}{2 - 3 \sin x}$ lies in $[1/5, \infty)$.Thus,the range is $(-\infty, -1] \cup [1/5, \infty)$.

The range of values of the function $f(x) = \frac{1}{2 - 3\sin x}$ is

Similar Questions

The domain of $f(x) = \sqrt{\log_2\left(\frac{10x - 4}{4 - x^2}\right) - 1}$ is

If $f: R \to R$,then the range of the function $f(x) = \frac{x^2}{x^2 + 1}$ is

If $f: R \rightarrow R$ is defined as $f(x) = \frac{x^6}{x^6+2020}$,for all $x \in R$,then the range of $f$ is .......

The domain of the function $f(x) = \sqrt{2 - 2x - x^2}$ is

The domain of the real-valued function $f(x) = \sqrt{\frac{2-|x|}{3-|x|}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module