Let f(x) = (x-4)(x-5)(x-6)(x-7),then - | vedclass

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(B) Given $f(x) = (x-4)(x-5)(x-6)(x-7)$.Since $f(x)$ is a polynomial of degree $4$,$f'(x)$ is a polynomial of degree $3$. Thus,$f'(x) = 0$ has $3$ roo

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Three roots of $f'(x) = 0$ lie in $(4, 5) \cup (5, 6) \cup (6, 7)$

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(B) Given $f(x) = (x-4)(x-5)(x-6)(x-7)$.Since $f(x)$ is a polynomial of degree $4$,$f'(x)$ is a polynomial of degree $3$. Thus,$f'(x) = 0$ has $3$ roots.According to Rolle's Theorem,if $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$ with $f(a) = f(b)$,then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.Here,$f(4) = f(5) = 0$,so there exists a root $c_1 \in (4, 5)$ such that $f'(c_1) = 0$.$f(5) = f(6) = 0$,so there exists a root $c_2 \in (5, 6)$ such that $f'(c_2) = 0$.$f(6) = f(7) = 0$,so there exists a root $c_3 \in (6, 7)$ such that $f'(c_3) = 0$.Thus,the three roots of $f'(x) = 0$ lie in the intervals $(4, 5)$,$(5, 6)$,and $(6, 7)$ respectively.Therefore,the three roots lie in $(4, 5) \cup (5, 6) \cup (6, 7)$.

Let $f(x) = (x-4)(x-5)(x-6)(x-7)$,then -

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