GATE Electrical Engineering (EE) 2021 Solved Papers

Number of real $\text{(+ve)}$ roots $\leq$ Number of sign changes in $p(x)=2$ Number of real $\text{(-ve)}$ rots $\leq$ Number of sign changes in $p(-x)=0$ So, $p(x)=0$ has no $\text{(-ve)}$ real roots and a maximum of $\text{2(+ve)}$ real roots $\therefore$ Number of real roots of ' $P(x)=0' \leq 2$ $\begin{array}{l} P(0)=2 \\ P(t)=-1 \end{array}$One real root lies between 0 & 1 $\begin{array}{l} P(3)=81-108+2 < 0 \\ P(4)=256-256+2 > 0 \end{array}$One real root lies between 3 & 4 P(x) has only two real roots. One real root lies in (0, 1) and One real root lies is (3, 4)