Verified Solution • 2025–26 Session moderate

QJEEM19Ma2482: JEE Main 2019 Mathematics - Class 11 JEEM19Ma2482 Solution: If be the ratio of the roots of the quadratic equation in x, 3m …

If λ\lambda be the ratio of the roots of the quadratic equation in x, 3m2x2 + m(m – 4)x + 2 = 0, then the least value of m for which λ+1λ=1,\lambda + {1 \over \lambda } = 1, is

Multiple Choice Options

  • A −2+2- 2 + \sqrt 2
  • B 4 −- 3 2\sqrt 2
  • C 2 −- 3\sqrt 3
  • D 4 −- 2 3\sqrt 3

Step-by-Step Verified Solution

3m2x2 + m(m −- 4) x + 2 = 0

λ+1λ=1,αβ+βα=1,α2+β2=αβ\lambda + {1 \over \lambda } = 1,{\alpha \over \beta } + {\beta \over \alpha } = 1,{\alpha ^2} + {\beta ^2} = \alpha \beta

( α\alpha + β\beta )2 = 3 α\alpha β\beta

(−m(m−4)3m2)2=3(2)3m2,(m−4)29m2=63m{\left( { - {{m\left( {m - 4} \right)} \over {3{m^2}}}} \right)^2} = {{3\left( 2 \right)} \over {3{m^2}}},{{{{\left( {m - 4} \right)}^2}} \over {9{m^2}}} = {6 \over {3m}}

(m−4)2=18,m=4±18,  4±32{\left( {m - 4} \right)^2} = 18,m = 4 \pm \sqrt {18,} \,\,4 \pm 3\sqrt 2

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