If `f(x)=(x^2)/(2-2cosx);g(x)=(x^2)/(6x-6sinx)` where `0 < x < 1,` then (A) both 'f' and 'g' are increasing functions

Options

A. f is increasing function
B. g is increasing function
C. f is decreasing function
D. g is decreasing function

Answer

1,4

Explanation

`f(X)=(x^(2))/(2-2 cos x)=(x^(2))/(4)cosec^(2)(x)/(2)`
`rarr f(x)=(x)/(2) cosec ^(2)(x)/(2)cot(x)/(2) tan(x)/(2)-(x)/(2)`
for `xltxlt1 tan (x)/(2)gt(x)/(2)`
`rarr f(X)gt0`
So f(x) is increasing
`g(x)=(x^(2))/(6x-6 sin x)`
`rarr g(x)=1/6(2x(x-sinx)-x^(2)(1-cosx))/(x-sinx)^(2)`
`=(x)/(6)(x-2sinx+xcosx)/(x-sinx)^(2)`
Let h(x) =x-2 sinx+cosx
h(x) =1 -2 cosx +cosx-sinx
=1-cosx - x sin x
`=2sinx^(2)(x)/(2)cos (x)/(2)-2 x sin(x)/(2)cos (x)/(2)`
`=2sin(x)/(2)cos(x)/(2)-2xsin(x)/(2)cos(x)/(2)`
`=2sin(x)/(2)cos(x)/(2)tan(x)/(2)-x`
for `0ltxlt1,tan (x)/(2) lt x`
`therefore h(x)lt0`
So h(x) is decreasing
For `xgt0,h(x) lth(x)`
`therefore h(x)lt0`
`therefore g(x) lt0`
so g(x) is decreasing