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

# If $f\left(x\right)=\left(x-2\right)\left(x-3\right)\left(x-4\right)\left(x-5\right)\left(x-6\right)$ then

1. A

${f}^{\mathrm{\prime }}\left(x\right)=0$has five real roots

2. B

four roots of  lie in $\left(4,5\right)\cup \left(5,6\right)$

3. C

the equation ${f}^{\mathrm{\prime }}\left(x\right)$ has only three roots.

4. D

four roots of ${f}^{\mathrm{\prime }}\left(x\right)$ = 0 lie in $\left(1,2\right)\cup \left(2,3\right)\cup \left(3,4\right)$$\cup \left(4,5\right)$

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### Solution:

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

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

5], [5, 6] there are ${x}_{1}\in \left(2,3\right),{x}_{2}\in \left(3,4\right),{x}_{3}\in \left(4$

such that

Since $f\text{'}$ is polynomial of degree 4 so cannot have five roots  