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Solution:

(c) (0, 2)

Let

$f(x)=a x^{3}+b x^{2}+c x+d$             ......(1)

$f(0)=d$

$f(2)=8 a+4 b+2 c+d$

$=2(4 a+2 b+c)+d$

$=d$              $(\because(4 a+2 b+c)=0)$

f is continuous in the closed interval [0, 2] and f is derivable in the open interval (0, 2).

Also, f(0) = f(2)

By Rolle's Theorem,

$f^{\prime}(\alpha)=0 \quad$ for $0<\alpha<2$

Now, $f^{\prime}(x)=3 a x^{2}+2 b x+c$

$\Rightarrow f^{\prime}(\alpha)=3 a \alpha^{2}+2 b \alpha+c=0$

Equation (1) has atleast one root in the interval $(0,2)$.

Thus, $f^{\prime}(\mathrm{x})$ must have root in the interval $(0,2)$.