Question

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

• A. $f^{\prime }\left(x\right)=0$ has four real roots
• B. three roots of $f^{\prime }\left(x\right)=0$ lie in $(4,5)\cup (5,6)\cup (6,7)$
• C. the euation $f^{\prime }\left(x\right)=$ has only two roots>d. three roots of $f^{\prime }\left(x\right)=0(3,4)\cup (4,5)\cup (5,6)$
• D. three roots of $f^{\prime }\left(x\right)=0$ $(3,4)\cup (4,5)\cup (5,6)$

Answer 1

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

Since $f\left(4\right)=f\left(5\right)=f\left(6\right)=f\left(7\right)=0$
So by Rolle's theorem applied on the interval $[4,5],[5,6],[6,7]$
There are $x_1\in (4,5),x_2\in (5,6),x_3\in (6,7)$
Such that $f^{\prime }\left(x_1\right)=f^{\prime }\left(x_2\right)=f^{\prime }\left(x_3\right)=0$.
Since $f^{\prime }$ is polynomial of degree $3$ so cannot have four roots

Hence $f\left(x\right)$ has 3 roots in $(4,5)\cup (5,6)\cup (6,7)$