For x, t in R,let p_t(x) = (sin t) x^2 - (2 c | vedclass

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(B) Given $p_t(x) = (\sin t) x^2 - (2 \cos t) x + \sin t$.$A(t) = \int_0^1 ((\sin t) x^2 - (2 \cos t) x + \sin t) dx$$A(t) = [\frac{x^3}{3} \sin t - x

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$II$ and $III$ only

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(B) Given $p_t(x) = (\sin t) x^2 - (2 \cos t) x + \sin t$.$A(t) = \int_0^1 ((\sin t) x^2 - (2 \cos t) x + \sin t) dx$$A(t) = [\frac{x^3}{3} \sin t - x^2 \cos t + x \sin t]_0^1$$A(t) = \frac{1}{3} \sin t - \cos t + \sin t = \frac{4}{3} \sin t - \cos t$.$A'(t) = \frac{4}{3} \cos t + \sin t$.$I$. $A(t) = \frac{4}{3} \sin t - \cos t$. This can be positive or negative depending on $t$,so $I$ is false.$II$. $A'(t) = 0 \implies \sin t = -\frac{4}{3} \cos t \implies 	an t = -\frac{4}{3}$. This equation has infinitely many solutions for $t$,so $A(t)$ has infinitely many critical points. $II$ is true.$III$. $A(t) = 0 \implies \frac{4}{3} \sin t = \cos t \implies 	an t = \frac{3}{4}$. This equation has infinitely many solutions for $t$,so $III$ is true.$IV$. $A'(t) = \frac{4}{3} \cos t + \sin t$. This expression changes sign as $t$ varies,so $IV$ is false.Thus,statements $II$ and $III$ are true.

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