Verified Solution • 2025–26 Session moderate

QJEEM20Mabdd0: JEE Main 2020 Mathematics - Class 11 JEEM20Mabdd0 Solution

If α\alpha and β\beta are the roots of the equation,
7x2 – 3x – 2 = 0, then the value of
α1−α2+β1−β2{\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}} is equal to :

Multiple Choice Options

  • A 124{1 \over {24}}
  • B 2732{{27} \over {32}}
  • C 2716{{27} \over {16}}
  • D 38{3 \over 8}

Step-by-Step Verified Solution

Given, 7x2 – 3x – 2 = 0

∴\therefore α\alpha + β\beta = 37{3 \over 7}

α\alpha β\beta = - 27{2 \over 7}

α1−α2+β1−β2{\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}

= α+β−αβ(α+β)1−α2−β2+α2β2{{\alpha + \beta - \alpha \beta \left( {\alpha + \beta } \right)} \over {1 - {\alpha ^2} - {\beta ^2} + {\alpha ^2}{\beta ^2}}}

= 37+27(37)1−(α+β)2+2αβ+(−27)2{{{3 \over 7} + {2 \over 7}\left( {{3 \over 7}} \right)} \over {1 - {{\left( {\alpha + \beta } \right)}^2} + 2\alpha \beta + {{\left( { - {2 \over 7}} \right)}^2}}}

= 37+27(37)1−(37)2+2(−27)+(−27)2{{{3 \over 7} + {2 \over 7}\left( {{3 \over 7}} \right)} \over {1 - {{\left( {{3 \over 7}} \right)}^2} + 2\left( { - {2 \over 7}} \right) + {{\left( { - {2 \over 7}} \right)}^2}}}

= 2716{{27} \over {16}}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring