Verified Solution • 2025–26 Session moderate

QJEEM21Ma7dbc: JEE Main 2021 Mathematics - Class 11 JEEM21Ma7dbc Solution: Let a, b, c be in arithmetic progression. Let the centroid of…

Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be (103,73)\left( {{{10} \over 3},{7 \over 3}} \right) . If α\alpha , β\beta are the roots of the equation ax2+bx+1=0a{x^2} + bx + 1 = 0 , then the value of α2+β2−αβ{\alpha ^2} + {\beta ^2} - \alpha \beta is :

Multiple Choice Options

  • A 69256{{69} \over {256}}
  • B 71256{{71} \over {256}}
  • C −71256- {{71} \over {256}}
  • D −69256- {{69} \over {256}}

Step-by-Step Verified Solution

2b = a + c

2a+23=103{{2a + 2} \over 3} = {{10} \over 3} and 2b+c3=73{{2b + c} \over 3} = {7 \over 3}

a = 4,

{2b+c=72b−c=4}, solving

b=114b = {{11} \over 4}

c=32c = {3 \over 2}

∴\therefore Quadratic Equation is 4x2+114x+1=04{x^2} + {{11} \over 4}x + 1 = 0

∴\therefore The value of

α2+β2−αβ{\alpha ^2} + {\beta ^2} - \alpha \beta

= α2+β2+2αβ−3αβ{\alpha ^2} + {\beta ^2} + 2\alpha \beta - 3\alpha \beta

= (α+β)2−3αβ{(\alpha + \beta )^2} - 3\alpha \beta

=121256−34=−71256= {{121} \over {256}} - {3 \over 4} = - {{71} \over {256}}

Related Learning Resources & Exam Preparation