Question

If ƒ(x) = $\frac{x}{\sin x}$ and g(x) = $\frac{x}{\tan x}$, where 0 < x £ 1,

then in this interval –

• A. ƒ(x) and g(x) are increasing functions
• B. Both ƒ(x) and g(x) are decreasing functions
• C. ƒ(x) is an increasing function
• D. g(x) is an increasing function

Answer 1

Answer: (C)

ƒ(x) is an increasing function

Answer 2

Let ƒ(x) = $\frac{x}{\sin x}$ Ž ƒ¢(x) = $\frac{\sin x - x\cos x}{\sin^{2}x}$

Let u(x) = sin x – x cos x so u¢(x) = cos x – cos x + x sin x = x sin x > 0 for 0 < x £ 1. Hence u is an increasing so u(x) > u(0) = 0. Thus ƒ¢(x) > 0 for 0 < x £ 1. i.e. ƒ is an increasing function. Now

g¢(x) = $\frac{\tan x - x\sec^{2}x}{\tan^{2}x}$. Let v(x) = tan x – x sec²x.

Since v¢(x) = sec²x

– sec²x – 2x sec² x tan x = – 2x sec² x tan x < 0 for 0 < x £ 1. Hence v(x) < v(0) = 0 so g¢(x)< 0 for 0 < x £ 1. Thus g is a decreasing function.