Let $f(x)=(x-4)(x-5)(x-6)(x-7)$ then
(a) $f^{\prime}(x)=0$ has four real roots
(b) three roots of $f^{\prime}(x)=0$ lie in $(4,5) \cup(5,6)$ $\cup(6,7)$
(c) the equation $f^{\prime}(x)=$ has only two roots
(d) three roots of $f^{\prime}(x)=0$ lie in $(3,4) \cup(4,5)$ $\cup(5,6)$