Applications Of Derivatives Question 305
Question: The minimum value of $ {e^{(2x^{2}-2x+1){{\sin }^{2}}x}} $ is
[Roorkee Qualifying 1998]
Options:
A) e
B) 1/e
C) 1
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
Given $ y={e^{(2x^{2}-2x+1){{\sin }^{2}}x}} $ For minima or maxima, $ \frac{dy}{dx}=0 $
$ \therefore {e^{(2x^{2}-2x+1){{\sin }^{2}}x}}[(4x-2){{\sin }^{2}}x+2(2x^{2}-2x+1)\sin x\cos x]=0 $
therefore $ [(4x-2){{\sin }^{2}}x+2(2x^{2}-2x+1)\sin x\cos x]=0 $
therefore $ 2\sin x[(2x-1)\sin x+(2x^{2}-2x+1)\cos x]=0 $
therefore $ \sin x=0 $
$ \therefore y $ is minimum for $ \sin x=0 $ Thus minimum value of $ y={e^{(2x^{2}-2x+1)(0)}}=e^{0}=1 $ .
BETA
