Let f(x)=(x−4)(x−5)(x−6)(x−7) then

# Let $f\left(x\right)=\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)$ then

1. A

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

2. B

three roots of ${f}^{\mathrm{\prime }}\left(x\right)=0$ lie in $\left(4,5\right)\cup \left(5,6\right)\cup \left(6,7\right)$

3. C

the equation ${f}^{\mathrm{\prime }}\left(x\right)=0$ has only one root

4. D

three roots of ${f}^{\mathrm{\prime }}\left(x\right)=0$ lie in $\left(3,4\right)\cup \left(4,5\right)\cup \left(5,6\right)$

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### Solution:

We have,

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

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

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

${\alpha }_{2}\in \left(6.7\right)$ such that ${f}^{\mathrm{\prime }}\left({\alpha }_{i}\right)=0,i=1,2,3.$

Since   is a cubic polynomial. Therefore, ${\alpha }_{1},{\alpha }_{2},{\alpha }_{3}$ are the only roots of ${f}^{\mathrm{\prime }}\left(x\right)=0.$

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