A) e
B) 1/e
C) 1
D) 0
Correct Answer: C
Solution :
Given \[y={{e}^{(2{{x}^{2}}-2x+1){{\sin }^{2}}x}}\] For minima or maxima, \[\frac{dy}{dx}=0\] \[\therefore {{e}^{(2{{x}^{2}}-2x+1){{\sin }^{2}}x}}[(4x-2){{\sin }^{2}}x+2(2{{x}^{2}}-2x+1)\sin x\cos x]=0\] Þ \[[(4x-2){{\sin }^{2}}x+2(2{{x}^{2}}-2x+1)\sin x\cos x]=0\] Þ \[2\sin x[(2x-1)\sin x+(2{{x}^{2}}-2x+1)\cos x]=0\] Þ \[\sin x=0\] \[\therefore y\]is minimum for \[\sin x=0\] Thus minimum value of \[y={{e}^{(2{{x}^{2}}-2x+1)(0)}}={{e}^{0}}=1\].
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