A) \[0<x<3\]
B) \[-3<x<0\]
C) \[0<x<9\]
D) \[-3<x<3\]
Correct Answer: B
Solution :
[b] \[f(x)=\sqrt{9-{{x}^{2}}}\] \[f'(x)=\frac{1}{2\sqrt{9-{{x}^{2}}}}\times (-2x)=-\frac{x}{\sqrt{9-{{x}^{2}}}}\] For function to be increasing \[-\frac{x}{\sqrt{9-{{x}^{2}}}}>0\] or \[-x>0\] or \[x<0\] but \[\sqrt{9-{{x}^{2}}}\] is defined only when \[9-{{x}^{2}}>0\] or \[{{x}^{2}}-9<0\] \[(x+3)(x-3)<0\] i.e., \[-3<x<3\] \[-3<x<3\cap x<0\Rightarrow -3<x<0\]
You need to login to perform this action.
You will be redirected in
3 sec
