If $27 a+9 b+3 c+d=0$, then the equation $4 a x^3+3 b x^2+2 c x+d=0$ has at least one real root lying between |
0 and 1 1 and 3 0 and 3 none of these |
0 and 3 |
Consider the polynomial $f(x)$ given by $f(x)=a x^4+b x^3+c x^2+d x$ $\Rightarrow f^{\prime}(x)=4 a x^3+3 b x^2+2 c d+d$ We have, $f(0)=0$ and, $f(3)=81 a+27 b+9 c+3 d=3(27 a+9 b+3 c+d)=0$ [Given] Therefore, 0 and 3 are roots of $f(x)=0$. Consequently, by Rolle's theorem $f^{\prime}(x)=0$ i.e. $4 a x^3+3 b x^2+2 c x+d=0$ has at least one real root between 0 and 3. |