If $f(\theta)=|\sin \theta|+|\cos \theta|, \theta \in R$, then

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

f(\theta)=|\sin \theta|+|\cos \theta|, \theta \in R

, then

View answer and explanation

Correct option: D

f(\theta) \in[1, \sqrt{2}]

Option D (

f(\theta) \in[1, \sqrt{2}]

) is correct because the function's range is determined by the minimum and maximum values of

|\sin \theta| + |\cos \theta|

Practice All Chapter MCQs →

\sin \,\,x\,\,\;cos\,\,3x = \sin \,\,3x\,\,\cos \,\,5x

\left[ {0,\;\,\frac{\pi }{2}} \right]

\left( {x,\,y} \right)

x,\,\,y \in \left[ {0,10} \right]

\left( {\sqrt {{{\sin }^2} - \sin \,\,x + \frac{1}{2}} } \right){2^{{{\sec }^2}y}}\,\, \cdot \,\, \le 1

\mathrm{X}

\theta \rightarrow \frac{\pi}{2}

\sin x+\cos x=\frac{\sqrt{7}}{2}

x \in\left[0, \frac{\pi}{4}\right]

\tan \frac{x}{2}

\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}

0 < x < \pi

\cos x+\sin x=\frac{1}{2}

\tan x