If f: R rightarrow R is defined by f(x) = fra | vedclass

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

Question

(B) Given, $f(x) = \frac{1}{2 - \cos 3x}$.We know that for any $x \in R$, $-1 \leq \cos 3x \leq 1$.Multiplying by $-1$, we get $-1 \leq -\cos 3x \leq

Vedclass · Accepted Answer

$[1/3, 1]$

Vedclass · Answer

(B) Given, $f(x) = \frac{1}{2 - \cos 3x}$.We know that for any $x \in R$, $-1 \leq \cos 3x \leq 1$.Multiplying by $-1$, we get $-1 \leq -\cos 3x \leq 1$.Adding $2$ to all sides, we get $2 - 1 \leq 2 - \cos 3x \leq 2 + 1$, which simplifies to $1 \leq 2 - \cos 3x \leq 3$.Taking the reciprocal, the inequality sign reverses: $\frac{1}{3} \leq \frac{1}{2 - \cos 3x} \leq \frac{1}{1}$.Thus, $\frac{1}{3} \leq f(x) \leq 1$.Therefore, the range of $f$ is $[1/3, 1]$.

If $f: R \rightarrow R$ is defined by $f(x) = \frac{1}{2 - \cos 3x}$ for each $x \in R$, then the range of $f$ is

Similar Questions

The domain of definition of the function $y(x)$ given by $2^x + 2^y = 2$ is

What is the range of values for $\frac{x}{x^2 + 4}$ for all real values of $x$?

Range of the function $f(x) = 3 + 2^x + 4^x$ is

Let $\rho$ be a relation defined on the set of natural numbers $N$, as $\rho = \{(x, y) \in N \times N: 2x + y = 41\}$. Then the domain $A$ and range $B$ are:

The range of the function $f(x) = \sqrt{9 - x^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module