Let f(x) = intlimits_0^{x^2} {(t - 1)(t - 4)( | vedclass

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(B) Given $f(x) = \int\limits_0^{x^2} {(t - 1)(t - 4)(t - 9)} dt$.Using the Leibniz rule for differentiation,$f'(x) = (x^2 - 1)(x^2 - 4)(x^2 - 9) \cdo

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$f'''(x) = 0$ has $2$ distinct positive solutions.

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(B) Given $f(x) = \int\limits_0^{x^2} {(t - 1)(t - 4)(t - 9)} dt$.Using the Leibniz rule for differentiation,$f'(x) = (x^2 - 1)(x^2 - 4)(x^2 - 9) \cdot \frac{d}{dx}(x^2) = 2x(x^2 - 1)(x^2 - 4)(x^2 - 9)$.$f'(x) = 2x(x - 1)(x + 1)(x - 2)(x + 2)(x - 3)(x + 3)$.The critical points are $x = 0, \pm 1, \pm 2, \pm 3$,which are $7$ distinct points.$f'(x)$ is an odd function,so $f''(x)$ is an even function and $f'''(x)$ is an odd function.Since $f'(x)$ is a polynomial of degree $7$,$f'''(x)$ is a polynomial of degree $5$.As $f'''(x)$ is an odd function,$x = 0$ is a root of $f'''(x) = 0$.By Rolle's Theorem,between any two roots of $f'(x)$,there is at least one root of $f''(x)$. Since $f'(x)$ has $7$ roots,$f''(x)$ has $6$ roots.Similarly,$f'''(x)$ has $5$ roots. Since $f'''(x)$ is an odd function and $x=0$ is a root,the remaining $4$ roots must be symmetric about the origin,i.e.,$2$ positive and $2$ negative roots.Thus,$f'''(x) = 0$ has $2$ distinct positive solutions.

Let $f(x) = \int\limits_0^{x^2} {(t - 1)(t - 4)(t - 9)} dt$,then:

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