JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Medium

1 of 1

Analyzing +4 | -1

In a

\triangle A B C

, the maximum value of

\frac{a \cos ^2 \frac{A}{2}+b \cos ^2 \frac{B}{2}+c \cos ^2 \frac{C}{2}}{a+b+c}

View answer and explanation

Correct option: C

\frac{3}{4}

Option C (

\frac{3}{4}

) is correct because the maximum value of the expression is

\frac{3}{4}

when using trigonometric identities and the inequality

R \geq 2r

More Questions from TRIGONOMETRIC EQUATIONS

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