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Question: If f(x) = \(\frac{x}{\sin x}\)and g(x) = \(\frac{x}{\tan x}\)where 0 \< x £ 1, then in this interval...

If f(x) = xsin⁡x\frac{x}{\sin x}and g(x) = xtan⁡x\frac{x}{\tan x}where 0 < x £ 1, then in this interval

A

Both f(x) and g(x) are increasing functions

B

Both f(x) and g(x) are decreasing functions

C

f(x) is an increasing function

D

g(x) is an increasing function

Answer

f(x) is an increasing function

Explanation

Solution

f¢(x) =sin⁡x−xcos⁡xsin⁡2x\frac{\sin x - x\cos x}{\sin^{2}x}, g¢(x) = tan⁡x−xsec⁡2xtan⁡2x\frac{\tan x - x\sec^{2}x}{\tan^{2}x}

Let u(x) = sin x – x cos x, so that u¢(x) = x sin x > 0 for 0 < x £ 1. So u(x) > u(0) = 0. So f¢(x) > 0 for 0 < x £ 1. Hence f increasing on (0, 1]. Let v(x) = tan x –x sec2 x, so that

v¢(x) = –2x sec2 x tan x < 0 for 0 < x £ 1. Thus v(x) < v (0), i.e., g¢(x) < 0 for 0 < x £ 1. So g decreases on (0, 1]