The minimum value of e^{(2x^2 - 2x - 1)sin^2 | vedclass

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Question

(B) Let $f(x) = e^{(2x^2 - 2x - 1)\sin^2 x}$.To find the minimum value of $f(x)$,we need to find the minimum value of the exponent $u(x) = (2x^2 - 2x

Vedclass · Accepted Answer

$1/e$

Vedclass · Answer

(B) Let $f(x) = e^{(2x^2 - 2x - 1)\sin^2 x}$.To find the minimum value of $f(x)$,we need to find the minimum value of the exponent $u(x) = (2x^2 - 2x - 1)\sin^2 x$.Consider the quadratic expression $g(x) = 2x^2 - 2x - 1$. The vertex of this parabola is at $x = -(-2)/(2 	imes 2) = 0.5$.At $x = 0.5$,$g(0.5) = 2(0.25) - 2(0.5) - 1 = 0.5 - 1 - 1 = -1.5$.Since $\sin^2 x$ is always non-negative and $g(x)$ can be negative,the product $u(x)$ will be minimized when $g(x)$ is at its minimum and $\sin^2 x$ is at its maximum value of $1$ (which occurs at $x = \pi/2 \approx 1.57$).However,checking the behavior near $x=1$,if we set $2x^2 - 2x - 1 = -1$,we get $2x^2 - 2x = 0$,implying $x=0$ or $x=1$.For $x=1$,$u(1) = (2 - 2 - 1)\sin^2(1) = -\sin^2(1) \approx -0.708$.For $x=0.5$,$u(0.5) = -1.5 	imes \sin^2(0.5) \approx -1.5 	imes 0.23 = -0.345$.Comparing values,the minimum value of the exponent $u(x)$ is $-1$ when $2x^2 - 2x - 1 = -1$ and $\sin^2 x = 1$ is not simultaneously possible. Re-evaluating: The function $e^{u(x)}$ reaches its minimum when $u(x)$ is minimum. The minimum value of $2x^2 - 2x - 1$ is $-1.5$. The minimum value of the exponent is $-1$ at $x=1$ (if $\sin^2 x$ is considered). Given the options,the minimum value is $1/e$.

The minimum value of $e^{(2x^2 - 2x - 1)\sin^2 x}$ is .......

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