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If $f (x) = (x - 2) (x - 3) (x - 4) (x - 5) (x - 6)$ then

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f^{'} (x) = 0

has five real roots

four roots of

f^{'} (x) = 0

lie in

(2, 3) \cup (3, 4) \cup

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

the equation

f^{'} (x)

has only three roots.

four roots of

f^{'} (x)

= 0 lie in

(1, 2) \cup (2, 3) \cup (3, 4)

\cup (4, 5)

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detailed solution

Correct option is B

If $f (x) = (x - 2) (x - 3) (x - 4) (x - 5) (x - 6)$

so, by Rolle's theorem applied on [2, 3], [3, 4], [4,

5], [5, 6] there are $x_{1} \in (2, 3), x_{2} \in (3, 4), x_{3} \in (4$

$5), x_{4} \in (5, 6)$ such that $f^{'} (x_{i}) = 0, i = 1, 2, 3, 4 .$

Since $f'$ is polynomial of degree 4 so cannot have five roots

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If f(x)=(x−2)(x−3)(x−4)(x−5)(x−6) then