First slide

Monotonicity

Question

$L e t A = \sin x + \tan x a n d B = 2 x i n t h e i n t e r v a l 0 < x < \frac{π}{2}$ then

Moderate

A

$A > B$
B

$A = B$
C

$A < B$
D

None of these

Solution

$\begin{array}{l} W e h a v e A = \sin x + \tan x, B = 2 x \\ A - B = \sin x + \tan x - 2 x \\ = \sin x - x + \tan x - x \\ L e t f (x) = \sin x - x + \tan x - x \\ ⇒ f^{1} (x) = \cos x - 1 + \sec^{2} x - 1 \\ = \cos x - 1 + \tan^{2} x \\ \Rightarrow f^{11} (x) = - \sin x + 2 \tan x \sec^{2} x \\ = \tan x (2 \sec^{2} x - \cos x) > 0 i n (0, \frac{π}{2}) \\ \Rightarrow f^{1} (x) i s i n c r e a s i n g i n (0, \frac{π}{2}) \\ ∴ x > 0 \Rightarrow f^{1} (x) > f^{1} (0) \\ \Rightarrow f^{1} (x) > 0 \\ \Rightarrow f (x) i s a n i n c r e a s i n g f u n c t i o n i n (0, \frac{π}{2}) \\ ∴ x > 0 \Rightarrow f (x) > f (0) \\ ⇒ f (x) > 0 \\ ⇒ A - B > 0 \Rightarrow A > B \end{array}$

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