| Let \[\frac{dF(x)}{dx}={{x}^{\sqrt{1-{{x}^{2}}}}},0<x\le 1\] and \[I=\int\limits_{\pi /6}^{\pi /2}{{{\left( \sin x \right)}^{\cos x}}\cos x\,dx=F\left( n \right)-F\left( \frac{1}{2} \right)}\] |
| Given the following statements |
| (i) domain of \[F'(x)\]is \[(0,1]\] |
| (ii) Possible value of n is \[\frac{{{\pi }^{2}}}{4}\] |
| (iii) Possible value of n is 1 |
| (iv) The value of I can be evaluated by substitution. |
| The correct statement are |
A) (i), (ii)
B) (ii), (iii)
C) (i), (iii)
D) (i), (iv)
Correct Answer: C
Solution :
| v\[\frac{dF(x)}{dx}={{x}^{\sqrt{1-{{x}^{2}}}}}\] |
| Domain of \[F'(x)\]is \[(0,1]\] |
| \[I=\int\limits_{\pi /16}^{\pi /2}{{{(\sin \,x)}^{\cos \,x}}\cos x\,dx}\] \[=F(n)-F(1/2)\] |
| \[=\int\limits_{0}^{1}{{{t}^{\sqrt{1-{{t}^{2}}}}}dt=F\left( 1 \right)-F\left( 1/2 \right)\Rightarrow n=1}\] |
You need to login to perform this action.
You will be redirected in
3 sec
