A) \[(0,\,\,\pi /2)\]
B) (0, 1)
C) \[(\pi /2,\,\,\pi )\]
D) None of these
Correct Answer: D
Solution :
\[\operatorname{f}(x)\,={{x}^{100}}\,+sin\,x\,-1\,\,\Rightarrow \,\,f'(x)=100{{x}^{99}}+cos\,x\] If \[0<x< \pi /2, then f(x)>0.\] Therefore f(x) is increasing on \[(0,\,\,\pi /2)\]. If \[0<x<1\], then \[100{{x}^{99}}>0 and cos\,\,x>0\] [\[\because \] x lies between 0 and 1 radian] \[\Rightarrow \,\,\,f(x)\,\,=\,\,100{{x}^{99}}''+cos\,\,x>0\] f(x) is increasing on (0, 1). If \[\pi /2<x<\pi \], then \[100 {{x}^{99}}> 100\] \[[\because \,\,x>1,\,\,\,\therefore \,\,{{x}^{99}}>1]\] \[\Rightarrow \,\,\,100{{x}^{99}}\,+\,\,cos\,x>0\] \[\left[ \because cos x \ge -1, \,\therefore \,100{{x}^{99}}+ cos x > 99 \right]\] \[\Rightarrow \,\,\,\,\,f'\,(x) > 0 \Rightarrow \, f(x)\] is increasing on \[(\pi /2,\,\pi ).\]
You need to login to perform this action.
You will be redirected in
3 sec
