The number of elements in the set S = {x : x | vedclass

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(B) Let $I = \int_{0}^{x} t^{2} \sin(x-t) dt$. Using the property $\int_{0}^{x} f(t) dt = \int_{0}^{x} f(x-t) dt$, we have $I = \int_{0}^{x} (x-t)^{2}

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$16$

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(B) Let $I = \int_{0}^{x} t^{2} \sin(x-t) dt$. Using the property $\int_{0}^{x} f(t) dt = \int_{0}^{x} f(x-t) dt$, we have $I = \int_{0}^{x} (x-t)^{2} \sin(t) dt$. Expanding this, $I = \int_{0}^{x} (x^{2} - 2xt + t^{2}) \sin(t) dt = x^{2} \int_{0}^{x} \sin(t) dt - 2x \int_{0}^{x} t \sin(t) dt + \int_{0}^{x} t^{2} \sin(t) dt$. Evaluating the integrals: $1. \int_{0}^{x} \sin(t) dt = [-\cos(t)]_{0}^{x} = 1 - \cos(x)$. $2. \int_{0}^{x} t \sin(t) dt = [-t \cos(t)]_{0}^{x} + \int_{0}^{x} \cos(t) dt = -x \cos(x) + \sin(x)$. $3. \int_{0}^{x} t^{2} \sin(t) dt = [-t^{2} \cos(t)]_{0}^{x} + 2 \int_{0}^{x} t \cos(t) dt = -x^{2} \cos(x) + 2(t \sin(t) + \cos(t))_{0}^{x} = -x^{2} \cos(x) + 2x \sin(x) + 2 \cos(x) - 2$. Substituting these back: $I = x^{2}(1 - \cos(x)) - 2x(-x \cos(x) + \sin(x)) + (-x^{2} \cos(x) + 2x \sin(x) + 2 \cos(x) - 2) = x^{2} - x^{2} \cos(x) + 2x^{2} \cos(x) - 2x \sin(x) - x^{2} \cos(x) + 2x \sin(x) + 2 \cos(x) - 2 = x^{2} + 2 \cos(x) - 2$. Given $I = x^{2}$, we have $x^{2} + 2 \cos(x) - 2 = x^{2}$, which simplifies to $2 \cos(x) = 2$, or $\cos(x) = 1$. For $x \in [0, 100]$, $\cos(x) = 1$ implies $x = 2n\pi$ for $n = 0, 1, 2, \dots$. Since $2n\pi \le 100$, $n \le \frac{100}{2\pi} \approx \frac{100}{6.28} \approx 15.92$. Thus, $n$ can be $0, 1, 2, \dots, 15$, which gives $16$ values.

The number of elements in the set $S = \{x : x \in [0, 100] \text{ and } \int_{0}^{x} t^{2} \sin(x-t) dt = x^{2}\}$ is:

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