JEE PYQ: Quadratic Equation Question 46
Question 46 - 2019 (10 Jan Shift 2)
The value of $\lambda$ such that sum of the squares of the roots of the quadratic equation, $x^2 + (3 - \lambda)x + 2 = \lambda$ has the least value is:
(1) $\frac{15}{8}$
(2) 1
(3) $\frac{4}{9}$
(4) 2
Show Answer
Answer: (4)
Solution
The given quadratic equation is $x^2 + (3-\lambda)x + 2 - \lambda = 0$
Sum of roots $= \alpha + \beta = \lambda - 3$
Product of roots $= \alpha\beta = 2 - \lambda$
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
$= (\lambda - 3)^2 - 2(2 - \lambda)$
$= \lambda^2 - 4\lambda + 5$
$= (\lambda - 2)^2 + 1$
For least $(\alpha^2 + \beta^2)$, $\lambda = 2$.
BETA
