Range of the function $ \frac{1}{2-\sin 3 x} $ is

JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

Range of the function

\frac{1}{2-\sin 3 x}

View answer and explanation

Correct option: B

\left[\frac{1}{3}, 1\right]

Option B (

[\frac{1}{3}, 1]

) is correct because the minimum value of the function is achieved when sin(3x) = 1 and the maximum value when sin(3x) = -1.

Practice All Chapter MCQs →

f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

g(f(x))=|\sin x|

f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}