JEE Main 2018 — Mathematics Algebra

Let, the roots of the equation, $x^2+(2-\lambda) x+(10-\lambda)=0$x2+(2−λ)x+(10−λ)=0 are $\alpha$α and $\beta$β. Also roots of the given equation are $$\frac{\lambda-2 \pm \sqrt{4-4 \lambda+\lambda^2-40+4 \lambda}}{2}=\frac{\lambda-2 \pm \sqrt{\lambda^2-36}}{2}$$2λ−2±4−4λ+λ2−40+4λ​​=2λ−2±λ2−36​​ The magnitude of the difference of the roots is $\left|\sqrt{\lambda^2-36}\right|$​λ2−36​​ So, $\alpha^3+\beta^3=\frac{(\lambda-2)^3}{4}+\frac{3(\lambda-2)\left(\lambda^2-36\right)}{4}$α3+β3=4(λ−2)3​+43(λ−2)(λ2−36)​ $$\begin{aligned} =\frac{(\lambda-2)\left(4 \lambda^2-4 \lambda-104\right)}{4} \\ =&(\lambda-2)\left(\lambda^2-\lambda-26\right)=f(\lambda) \end{aligned}$$=4(λ−2)(4λ2−4λ−104)​=​(λ−2)(λ2−λ−26)=f(λ)​ As $f(\lambda)$f(λ) attains its minimum value at $\lambda=4$λ=4. Therefore, the magintude of the difference of the