Question

Let $\alpha ,\beta $ be the roots of ${x^2} + (3 - \lambda )x - \lambda = 0$. The value of $\lambda $ for which ${\alpha ^2} + {\beta ^2}$ is minimum, is
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: Find \[\alpha + \beta \] and $\alpha \beta $ using the fact that the sum of roots of a quadratic polynomial in the standard form is $ - \dfrac{b}{a}$ and the product of the roots is $\dfrac{c}{a}$. After getting a quadratic equation in the variable $\lambda $, find its minimum value using the first order and second order derivative tests.

Formula used: ${a^2} + {b^2} = {(a + b)^2} - 2ab$, $\dfrac{{d({x^n} + l{x^m})}}{{dx}} = n{x^{n - 1}} + ml{x^{m - 1}}$

Complete step-by-step solution:
We know that for a quadratic equation in its standard form, $a{x^2} + bx + c = 0$, the sum of the roots is $ - \dfrac{b}{a}$ and the product of the roots is $\dfrac{c}{a}$.
The sum and product of the roots of the equation ${x^2} + (3 - \lambda )x - \lambda = 0$ are,
$\alpha + \beta = - \dfrac{{3 - \lambda }}{1} = \lambda - 3$
$\alpha \beta = \dfrac{{ - \lambda }}{1} = - \lambda $
We can write ${\alpha ^2} + {\beta ^2}$ as ${(\alpha + \beta )^2} - 2\alpha \beta $
Therefore, ${\alpha ^2} + {\beta ^2} = {(\lambda - 3)^2} - 2( - \lambda )$
${\alpha ^2} + {\beta ^2} = {\lambda ^2} - 6\lambda + 9 + 2\lambda $
${\alpha ^2} + {\beta ^2} = {\lambda ^2} - 4\lambda + 9$
We now need to find the minimum value of ${\lambda ^2} - 4\lambda + 9$.
Let ${\lambda ^2} - 4\lambda + 9$ be y
Find the first derivative of y and equate it to 0 to get the critical point.
$\dfrac{{dy}}{{d\lambda }} = 2\lambda - 4$
$2\lambda - 4 = 0$
$\lambda = 2$ is the critical point.
Find the second derivative at $\lambda = 2$ to check whether we get the minimum value or maximum value.
$\dfrac{{{d^2}y}}{{d{\lambda ^2}}} = 2$ which is greater than 0. Therefore, we get the minimum value at the critical point.
The value of ${\lambda ^2} - 4\lambda + 9$ at $\lambda = 2$ is the minimum value.
Therefore, the correct answer is option C. 2.

Note: Once we get ${\alpha ^2} + {\beta ^2} = {\lambda ^2} - 4\lambda + 9$, we can substitute all the four values of $\lambda $ given in the options. When $\lambda = 0$, ${\alpha ^2} + {\beta ^2} = {\lambda ^2} - 4\lambda + 9 = 9$; when $\lambda = 1$, ${\alpha ^2} + {\beta ^2} = {\lambda ^2} - 4\lambda + 9 = 6$; when $\lambda = 2$, ${\alpha ^2} + {\beta ^2} = {\lambda ^2} - 4\lambda + 9 = 5$; when $\lambda = 3$, ${\alpha ^2} + {\beta ^2} = {\lambda ^2} - 4\lambda + 9 = 6$. As we can see at $\lambda = 2$, we get the minimum value.