If a is the minimum value of sin^2 theta - si | vedclass

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(D) Let $f(	heta) = \sin^2 	heta - \sin 	heta + \frac{1}{2}$.We can rewrite this as $f(	heta) = (\sin 	heta - \frac{1}{2})^2 + \frac{1}{4}$.The m

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$2$

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(D) Let $f(	heta) = \sin^2 	heta - \sin 	heta + \frac{1}{2}$.We can rewrite this as $f(	heta) = (\sin 	heta - \frac{1}{2})^2 + \frac{1}{4}$.The minimum value $a$ occurs when $\sin 	heta = \frac{1}{2}$,so $a = \frac{1}{4}$.Now,calculate $b = \lim_{x 	o \infty} (\sqrt{(x + 1)(x + 2)} - x)$.$b = \lim_{x 	o \infty} \frac{(\sqrt{x^2 + 3x + 2} - x)(\sqrt{x^2 + 3x + 2} + x)}{\sqrt{x^2 + 3x + 2} + x} = \lim_{x 	o \infty} \frac{x^2 + 3x + 2 - x^2}{\sqrt{x^2 + 3x + 2} + x} = \lim_{x 	o \infty} \frac{3x + 2}{\sqrt{x^2 + 3x + 2} + x}$.Dividing numerator and denominator by $x$,we get $b = \lim_{x 	o \infty} \frac{3 + 2/x}{\sqrt{1 + 3/x + 2/x^2} + 1} = \frac{3}{1 + 1} = \frac{3}{2}$.Finally,$|2a + b| = |2(\frac{1}{4}) + \frac{3}{2}| = |\frac{1}{2} + \frac{3}{2}| = |2| = 2$.

If $a$ is the minimum value of $\sin^2 \theta - \sin \theta + \frac{1}{2}$ and $b = \lim_{x \to \infty} (\sqrt{(x + 1)(x + 2)} - x)$,then $|2a + b| = $

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