If $f(x)=\frac{x^{2}}{2-2 \cos x}$; $g(x)=\frac{x^{2}}{6x-6\sin x}$ where $0<x<1$ then are $f$, $g$ increasing/decreasing?

Question

Question If $$f(x)=\frac{x^{2}}{2-2\cos x}$$ $$g(x)=\frac{x^{2}}{6x-6\sin x}$$ where $0< x <1$, then

---

(a) both $f$ and $g$ $\uparrow$

(b) both $f$ and $g$ $\downarrow$

(c) $f$ $\uparrow$, $g$$\downarrow$

(d) $f$$\downarrow$, $g$$\uparrow$

My Approach Only method I know is derivative test but that didn't work

Answer 1

I may contribute for one function$$ f(x) = \frac{x^{2}}{2 - 2 \cos(x)}, \:\: 0 < x < 1 $$ The 1st derivative is $$ f'(x) = \frac{2x (1-\cos(x))- \sin(x)x^{2}}{2 ( 1 - \cos(x) )^{2}} = \frac{x}{ 1 - \cos(x) } - \frac{ 0.5 \sin(x)x^{2}}{ ( 1 - \cos(x) )^{2}} $$ $$ = \frac{x}{ 1 - \cos(x) } \left( 1 - \frac{ 0.5 \sin(x)x}{ 1 - \cos(x) } \right) $$ Since $0 < x < 1$, we would have $x/(1 - \cos(x)) $ positive. Now we may check the other term : we can prove by contradiction, that $$ 1 - \frac{ 0.5 \sin(x)x}{ 1 - \cos(x) } > 0 $$

Assume the opposite is true, which means that $$ 1 - \frac{ 0.5 \sin(x)x}{ 1 - \cos(x) } < 0 $$ $$ 0.5 \sin(x)x + \cos(x)> 1 $$ by squaring.. we may get $$ [ 0.25 \sin(x) ] x^{2} + [ \cos(x) ] x - \sin(x) > 0 $$ By viewing it as a quadratic polynomial wrt to $x$, it should be that $D < 0$, but instead we get $$ D = \cos^{2}(x) + \sin^{2}(x) = 1 >0 $$ a contradiction. So it must be that $$ 1 - \frac{ 0.5 \sin(x)x}{ 1 - \cos(x) } > 0 $$ also positive.

This means $f'(x)>0$ for $0<x<1$.

Hope this helps.

Answer 2

HINT: $$f'(x)=-1/2\,{\frac {x \left( x\sin \left( x \right) +2\,\cos \left( x \right) -2 \right) }{ \left( -1+\cos \left( x \right) \right) ^{2}}} $$ $$g'(x)=1/6\,{\frac {x \left( x\cos \left( x \right) -2\,\sin \left( x \right) +x \right) }{ \left( -x+\sin \left( x \right) \right) ^{2}}} $$