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(C) Let $f(x) = x^2 - x \sin x - \cos x$.We want to find the number of roots of $f(x) = 0$.First,calculate the derivative: $f'(x) = 2x - (\sin x + x \

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$2$

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(C) Let $f(x) = x^2 - x \sin x - \cos x$.We want to find the number of roots of $f(x) = 0$.First,calculate the derivative: $f'(x) = 2x - (\sin x + x \cos x) + \sin x = 2x - x \cos x = x(2 - \cos x)$.Since $-1 \leq \cos x \leq 1$,we have $2 - \cos x \geq 1 > 0$ for all $x$.Thus,$f'(x) > 0$ for $x > 0$,$f'(x) This means $f(x)$ is strictly decreasing on $(-\infty, 0]$ and strictly increasing on $[0, \infty)$.The global minimum occurs at $x = 0$,where $f(0) = 0^2 - 0 \sin(0) - \cos(0) = -1$.Since the minimum value is $-1$ and the function is continuous,we check the limits as $x 	o \pm \infty$.As $x 	o \pm \infty$,$f(x) \approx x^2 	o \infty$.Since $f(0) = -1 Similarly,since $f(0) = -1 Therefore,there are exactly $2$ points where $f(x) = 0$.

The number of points in $(-\infty, \infty)$ for which $x^2 - x \sin x - \cos x = 0$ is

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