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
Correct Answer: C
Solution :
\[f'(x)=\frac{\sin x-x\cos x}{{{\sin }^{2}}x}=\frac{\cos x(\tan x-x)}{{{\sin }^{2}}x}\] \[0<x\le 1\Rightarrow x\in {{Q}_{1}}\Rightarrow \tan x>x,\,\,\cos x>0\] \[\therefore f'(x)>0\]for \[0<x\le 1\] \[\therefore \]\[f(x)\] is an increasing function. \[g'(x)=\frac{\tan x-x{{\sec }^{2}}x}{{{\tan }^{2}}x}=\frac{\sin x\cos x-x}{{{\sin }^{2}}x}=\frac{\sin 2x-2x}{2{{\sin }^{2}}x}\] \[(\sin 2x-2x)'=2\cos 2x-2=2[\cos 2x-1]<0\] Þ \[\sin 2x-2x\] is decreasing Þ\[\sin 2x-2x<0\] \[\therefore \] \[g'(x)<0\Rightarrow g(x)\]is decreasing.
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