
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
163.5k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions
