A) 1
B) 2
C) 0
D) None of these
Correct Answer: A
Solution :
[a] Let \[y{{=}_{e}}(2{{x}^{2}}-2x-1){{\sin }^{2}}x\] and \[u=(2{{x}^{2}}-2x-1){{\sin }^{2}}x\] Now \[\frac{du}{dx}=(2{{x}^{2}}-2x-1)2\sin \,x\,\,\cos \,x+(4x-2){{\sin }^{2}}\] \[x=\sin \,\,x[2(2{{x}^{2}}-2x)\cos \,\,x+(4x-2)\sin \,\,x]\] \[\frac{du}{dx}=0\Rightarrow \sin x=0\Rightarrow x=n\pi \] \[\frac{{{d}^{2}}u}{d{{x}^{2}}}=\sin x\frac{d}{dx}[2(2{{x}^{2}}-2x-1)\cos x+\] \[(4x-2)\sin x]\] At \[x=n\pi ,\frac{{{d}^{2}}u}{d{{x}^{2}}}=0+2{{\cos }^{2}}n\pi (2{{n}^{2}}{{\pi }^{2}}-1)>0\] Hence at \[x=n\pi \], the value of u and so its corresponding the value of y is minima and minimum value \[={{e}^{0}}=1\]
You need to login to perform this action.
You will be redirected in
3 sec
