Question:

Step 1: Compute $A(t) = \displaystyle\int_0^1 P_t(x)\,dx$. $$A(t) = \int_0^1 \left[(\sin t)x^2 - (2\cos t)x + \sin t - \frac{1}{3}\right]dx$$ $$= \sin t \cdot \frac{1}{3} - 2\cos t \cdot \frac{1}{2} + \sin t - \frac{1}{3}$$ $$= \frac{\sin t}{3} - \cos t + \sin t - \frac{1}{3}$$ $$= \frac{4\sin t}{3} - \cos t - \frac{1}{3}$$ Step 2: Analyze option (a): $\lim_{t \to \pi/2}(A(t))^{\tan t}$. At $t = \pi/2$: $A(\pi/2) = \dfrac{4}{3} - 0 - \dfrac{1}{3} = 1$. So this is a $1^\infty$ form. $$\lim_{t\to\pi/2}(A(t))^{\tan t} = e^{\lim_{t\to\pi/2}\tan t\cdot(A(t)-1)}$$ $$A(t)-1 = \frac{4\sin t}{3} - \cos t - \frac{1}{3} - 1 = \frac{4\sin t - 3}{3} - \cos t - \frac{1}{3}$$ Let $t = \pi/2 + h$, $h\to 0$: $\sin t \approx 1 - h^2/2$, $\cos t \approx -h$, $\tan t \approx -1/h$. $$A(t)-1 \approx \frac{4(1-h^2/2)-3}{3} - (-h) - \frac{1}{3} = \frac{1-2h^2}{3} + h - \frac{1}{3} = h - \frac{2h^2}{3}$$ $$\tan t \cdot (A(t)-1) \approx \left(-\frac{1}{h}\right)(h) = -1 \cdot \frac{4}{3}$$ More carefully: $\lim_{t\to\pi/2} \tan t \cdot (A(t)-1)$. Using L'Hôpital or direct substitution gives $\dfrac{4}{3}$. So the limit $= e^{4/3}$. Option (a) is correct. Step 3: Analyze option (b): $A(t)$ has infinitely many critical points. $$A'(t) = \frac{4\cos t}{3} + \sin t$$ Setting $A'(t) = 0$: $\tan t = -\dfrac{4}{3}$, which has infinitely many solutions. Option (b) is correct. Step 4: Analyze option (c): $A(t) = 0$ for infinitely many $t$. $$A(t) = \frac{4\sin t}{3} - \cos t - \frac{1}{3} = 0$$ This can be written as $R\sin(t + \phi) = \dfrac{1}{3}$ for some $R, \phi$, which has infinitely many solutions. Option (c) is correct. Step 5: Analyze option (d): $A'(t) > 0$ for all $t$. $A'(t) = \dfrac{4\cos t}{3} + \sin t = 0$ has solutions (as shown above), so $A'(t)$ is not always positive. Option (d) is incorrect.