Verified Solution • 2025–26 Session moderate

QJEEM19Mac29d: JEE Main 2019 Mathematics - Class 11 JEEM19Mac29d Solution: If m is chosen in the quadratic equation (m 2 + 1) x 2 – 3x +…

If m is chosen in the quadratic equation

(m2 + 1) x2 – 3x + (m2 + 1)2 = 0

such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is :-

Multiple Choice Options

  • A 434\sqrt 3
  • B 838\sqrt 3
  • C 858\sqrt 5
  • D 10510\sqrt 5

Step-by-Step Verified Solution

Given quadratic equation

(m2 + 1) x2 – 3x + (m2 + 1)2 = 0

Let roots of the equation α\alpha and β\beta .

∴\therefore Sum of roots = α\alpha + β\beta = 3m2+1{3 \over {{m^2} + 1}}

Product of roots = α\alpha β\beta = m2 + 1

3m2+1{3 \over {{m^2} + 1}} is maximum when m = 0

Hence equation becomes x2 – 3x + 1 = 0

α+β=3\alpha + \beta = 3 ,    αβ=1\alpha \beta = 1   

∣α−β∣\left| {\alpha - \beta } \right| = (α+β)2−4αβ\sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } = (3)2−4.1\sqrt {{{\left( 3 \right)}^2} - 4.1} = 5\sqrt 5

∣α3−β3∣=∣(α−β)(α2+β2+αβ)∣\left| {{\alpha ^3} - {\beta ^3}} \right| = \left| {(\alpha - \beta )({\alpha ^2} + {\beta ^2} + \alpha \beta )} \right|

= 5∣(α+β)2−αβ∣\sqrt 5 \left| {{{\left( {\alpha + \beta } \right)}^2} - \alpha \beta } \right|

=5(9−1)=85= \sqrt 5 (9 - 1) = 8\sqrt 5

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