QJEEAdv24Ma6ce2: JEE Advanced 2024 Mathematics - Class 11 JEEAdv24Ma6ce2 Solution: Let be a polynomial with real coefficients such that . Suppose…

$f(1)=1+a+b+c=-9 \quad \Rightarrow \quad a+b+c=-10\quad \text{.... (i)}$

$\begin{aligned}& 4 x^3+3 a x^2+2 b x=0 \text { roots are } \sqrt{3} i,-\sqrt{3} i, 0 \\& \Rightarrow \quad 4 \mathrm{x}^2+3 \mathrm{ax}+2 \mathrm{~b}=0<\begin{array}{l}\sqrt{3} i \\-\sqrt{3} i\end{array} \\& \Rightarrow \quad \mathrm{a}=0 \& \frac{2 \mathrm{~b}}{4}=(\sqrt{3} \mathrm{i})(-\sqrt{3} \mathrm{i}) \\& \mathrm{b}=6 \text { use } \mathrm{a}, \mathrm{b} \text { in (1) } \Rightarrow \mathrm{c}=-16 \\& \Rightarrow \quad \mathrm{f}(\mathrm{x})=\mathrm{x}^4+6 \mathrm{x}^2 \quad 16=0 \\& \left(x^2+8\right)\left(x^2-2\right)=0 \\& \Rightarrow \quad \mathrm{x}= \pm \sqrt{8} \mathrm{i}, \pm \sqrt{2} \quad \Rightarrow \quad\left|\alpha_1\right|^2+\left|\alpha_2\right|^2+\left|\alpha_3\right|^2+\left|\alpha_4\right|^2=20 \\&\end{aligned}$