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

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f^{'} (x) = 0

has four roots

three roots of

f^{'} (x) = 0

lie in

(4, 5) \cup (5, 6) \cup (6, 7)

the equation

f^{'} (x) = 0

has only one root

three roots of

f^{'} (x) = 0

lie in

(3, 4) \cup (4, 5) \cup (5, 6)

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detailed solution

Correct option is B

We have,

$f (x) = (x - 4) (x - 5) (x - 6) (x - 7) .$

Clearly, $f (4) = f (5) = f (6) = f (7) = 0$

By Rolle's theorem, there exist $α_{1} \in (4, 5), α_{2} \in (5, 6),$

$α_{2} \in (6.7)$ such that $f^{'} (α_{i}) = 0, i = 1, 2, 3 .$

Since $f' (x)$ is a cubic polynomial. Therefore, $α_{1}, α_{2}, α_{3}$ are the only roots of $f^{'} (x) = 0 .$

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Let f(x)=(x−4)(x−5)(x−6)(x−7) then