If $0 < \alpha < \beta < \frac{\pi}{2}$, then

$\frac{\tan \alpha}{\tan \beta} < \frac{a}{\beta}$

$\frac{\tan \alpha}{\tan \beta} > \frac{\alpha}{\beta}$

JEE MATHEMATICS MCQ: INCREASING & DECREASING

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Analyzing +4 | -1

0 < \alpha < \beta < \frac{\pi}{2}

, then

View answer and explanation

Correct option: B

\frac{\tan \beta}{\tan \alpha} > \frac{\alpha}{\beta}

The correct answer is B (

\frac{{\tan \beta}}{{\tan \alpha}} > \frac{{\alpha}}{{\beta}}

) because the function

f(x) = x \tan x

is increasing on

(0, \pi/2)

Practice All Chapter MCQs →

f\left( x \right) = \frac{{K\sin x + 2\cos x}}{{\sin x + \cos x}}

x

f\left( x \right) = {x^2} + ax + 1

a

f(x)=

\sin x-\cos x-a x+b

f(x)=x e^{x(1-x)}, x \in R

f

g

fog

f\left( x \right) = {x^3} + b{x^2} + cx + d

0 < {b^2} < c

\left( { - \infty ,\;\infty } \right)