If `f(x)=x/(sinx)a n dg(x)=x/(tanx),w h e r e0

Options

A. both `f(x) and g(x)` are increasing functions
B. both `f(x) and g(x)` are decreasing functions
C. `f(x)` is and increasing function
D. `g(x)` is an increasing fuction

Answer

B

Explanation

Given, `g(x) = (x) /( tan x ), ` where ` 0 lt x le 1 `
Now, ` g(x)` is continuous in ` [ 0, 1 ]` and differentiable in `]0, 1]`.
For `" " 0 lt x lt 1 `
`" " g' (x) = (tanx - x sec ^(2) x )/( tan ^(2) x)`
Again, ` H (x) = tan x - x sec ^(2) x, 0 le x le 1 `
Now, `H (x) ` is continuous in `[0, 1]` and differentiable in `]0, 1[`.
For `0 lt x lt 1, H (x) = tan x - x sec^(2) x, 0 le x le 1 `
`rArr " "H ' (x) = sec^(2) x - sec^(2) x - 2 x sec ^(2)x tan x `
`" " = - 2x sec ^(2)x tanx lt 0`
Hence, ` H (x) ` is decreasing function in `[ 0, 1]`.
Thus, `H (x) lt H (0) ` for ` 0 lt x lt 1 `
`rArr " " H (x) lt 0 ` for ` 0 lt x lt 1 `
`rArr " " g ' (x) lt 0 ` for ` 0 lt x lt 1 `
`rArr g(x)` is decreasing function in (0, 1].
Therefore, `g(x) = (x)/( tan x ) ` is a decreasing function in `0 lt x le 1 `.
Also, `" " g(x) lt g(0)` for ` 0 lt x le 1 `
`rArr " " (x) /(tan x ) lt 1 ` for ` 0 lt x le 1 `
`rArr " " x lt tan x ` for ` 0 lt x le 1 `
Now, let ` f (x) = {{:(x//sin x ,,"for",,0 lt x le 1 ),(1,,"for",, x=0):}`
Now, `f ` is continouous in `[0, 1]` differentiating in `]0, 1[`.
For `0 lt x lt1`,
`f' (x) = ( sin x - x cos x ) /( sin ^(2) x ) = ((tan x - x) cos x ) /( sin ^(2) x) gt 0 "for " 0 lt x lt 1 `
`rArr f(x) `increases in ` [0, 1]`
Thus, `f (x) = (x)/(sin x )` increases in `0 lt x le 1 `.
Therefore, option (c) is the answer.