The maximum and minimum values of the express | vedclass

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Question

(A) Let $f(	heta) = 2\sin^2	heta - 3\sin	heta$. Let $x = \sin	heta$,where $x \in [-1, 1]$.Then the expression becomes $g(x) = 2x^2 - 3x$.This is a

Vedclass · Accepted Answer

$5, -\frac{9}{8}$

Vedclass · Answer

(A) Let $f(	heta) = 2\sin^2	heta - 3\sin	heta$. Let $x = \sin	heta$,where $x \in [-1, 1]$.Then the expression becomes $g(x) = 2x^2 - 3x$.This is a downward-opening parabola if the coefficient of $x^2$ were negative,but here it is upward-opening with vertex at $x = -b/(2a) = -(-3)/(2 	imes 2) = 3/4$.Since $3/4 \in [-1, 1]$,the minimum value occurs at $x = 3/4$:$g(3/4) = 2(3/4)^2 - 3(3/4) = 2(9/16) - 9/4 = 9/8 - 18/8 = -9/8$.The maximum value occurs at the boundaries of the interval $x \in [-1, 1]$.At $x = -1$: $g(-1) = 2(-1)^2 - 3(-1) = 2 + 3 = 5$.At $x = 1$: $g(1) = 2(1)^2 - 3(1) = 2 - 3 = -1$.Comparing $5$ and $-1$,the maximum value is $5$.Thus,the maximum and minimum values are $5$ and $-9/8$ respectively.

The maximum and minimum values of the expression $2\sin^2\theta - 3\sin\theta$ are respectively:

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