The function S(x) = intlimits_0^x {sin left( | vedclass

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(C) Given $S(x) = \int\limits_0^x {\sin \left( {\frac{{\pi {t^2}}}{2}} \right)\,dt} $.By the Fundamental Theorem of Calculus,$S'(x) = \sin \left( {\fr

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(C) Given $S(x) = \int\limits_0^x {\sin \left( {\frac{{\pi {t^2}}}{2}} \right)\,dt} $.By the Fundamental Theorem of Calculus,$S'(x) = \sin \left( {\frac{{\pi {x^2}}}{2}} \right)$.Critical points occur where $S'(x) = 0$,so $\sin \left( {\frac{{\pi {x^2}}}{2}} \right) = 0$.This implies $\frac{{\pi {x^2}}}{2} = n\pi$ for some integer $n$,so $x^2 = 2n$.Given the interval $[1, 2.4]$,we have $1 \le x^2 \le 5.76$.For $n=1$,$x^2 = 2 \implies x = \sqrt{2} \approx 1.414$.For $n=2$,$x^2 = 4 \implies x = 2$.Now,find the second derivative: $S''(x) = \cos \left( {\frac{{\pi {x^2}}}{2}} \right) \cdot \pi x$.At $x = \sqrt{2}$,$S''(\sqrt{2}) = \cos(\pi) \cdot \pi \sqrt{2} = -\pi \sqrt{2} At $x = 2$,$S''(2) = \cos(2\pi) \cdot 2\pi = 2\pi > 0$,so $x = 2$ is a local minimum.Thus,the local minimum occurs at $x = 2$.

The function $S(x) = \int\limits_0^x {\sin \left( {\frac{{\pi {t^2}}}{2}} \right)\,dt} $ has two critical points in the interval $[1, 2.4]$. One of the critical points is a local minimum and the other is a local maximum. The local minimum occurs at $x =$

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