Applications-Of-Derivatives Question 439
Question: Find the minimum value of $ {e^{(2x^{2}-2x-1){{\sin }^{2}}x}} $ .
Options:
1
2
0
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ y{=_{e}}(2x^{2}-2x-1){{\sin }^{2}}x $ and $ u=(2x^{2}-2x-1){{\sin }^{2}}x $ Now $ \frac{du}{dx}=(2x^{2}-2x-1)2\sin ,x\cos ,x+(4x-2){{\sin }^{2}}x $ $ =\sin x[2(2x^{2}-2x-1)\cos x+(4x-2)\sin x] $ $ \frac{du}{dx}=0\Rightarrow \sin x=0\Rightarrow x=n\pi $ $ \frac{d^{2}u}{dx^{2}}=\sin x\frac{d}{dx}[2(2x^{2}-2x-1)\cos x+ $ $ (4x-2)\sin x] $ At $ x=n\pi ,\frac{d^{2}u}{dx^{2}}=0+2{{\cos }^{2}}n\pi (2n^{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 $
BETA
