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Quadratic Equation and Inequalities question

2019 · 12 Jan · Shift 1 · Q34
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  5. /2019 · 12 Jan · Shift 1 · Q34

Quadratic Equation and Inequalities question

2019 · 12 Jan · Shift 1 · Q34

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If λ\lambdaλ 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,λ+λ1​=1, is
  1. A
    −2+2- 2 + \sqrt 2−2+2​
  2. B
    4 −-− 3 2\sqrt 22​
  3. C
    2 −3-\sqrt 3−3​
  4. D
    4 −-− 2 3\sqrt 33​
View written solutionFree

Correct answer: B

  1. Given quadratic

    The equation is 3m2x2+m(m−4)x+2=0.3m^2x^2 + m(m-4)x + 2 = 0.3m2x2+m(m−4)x+2=0.

    Let its roots be α,β\alpha, \betaα,β.

    We are told that λ\lambdaλ is the ratio of the roots, so λ=αβorβα.\lambda = \frac{\alpha}{\beta} \quad \text{or} \quad \frac{\beta}{\alpha}.λ=βα​orαβ​. In either case, λ+1λ=αβ+βα=α2+β2αβ.\lambda + \frac{1}{\lambda} = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta}.λ+λ1​=βα​+αβ​=αβα2+β2​.

  2. Use sum and product of roots

    For ax2+bx+c=0,ax^2+bx+c=0,ax2+bx+c=0, we have α+β=−ba,αβ=ca.\alpha+\beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}.α+β=−ab​,αβ=ac​.

    Here, a=3m2,b=m(m−4),c=2.a=3m^2, \quad b=m(m-4), \quad c=2.a=3m2,b=m(m−4),c=2.

    So, α+β=−m(m−4)3m2=−m−43m=4−m3m,\alpha+\beta = -\frac{m(m-4)}{3m^2} = -\frac{m-4}{3m} = \frac{4-m}{3m},α+β=−3m2m(m−4)​=−3mm−4​=3m4−m​, and αβ=23m2.\alpha\beta = \frac{2}{3m^2}.αβ=3m22​.

  3. Condition on λ\lambdaλ

    Given λ+1λ=1.\lambda + \frac{1}{\lambda} = 1.λ+λ1​=1.

    But λ+1λ=α2+β2αβ.\lambda + \frac{1}{\lambda} = \frac{\alpha^2+\beta^2}{\alpha\beta}.λ+λ1​=αβα2+β2​.

    Also, α2+β2=(α+β)2−2αβ.\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta.α2+β2=(α+β)2−2αβ.

    Hence, (α+β)2−2αβαβ=1.\frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} = 1.αβ(α+β)2−2αβ​=1.

    Rearranging, (α+β)2αβ−2=1,\frac{(\alpha+\beta)^2}{\alpha\beta} - 2 = 1,αβ(α+β)2​−2=1, (α+β)2αβ=3,\frac{(\alpha+\beta)^2}{\alpha\beta} = 3,αβ(α+β)2​=3, (α+β)2=3αβ. (\alpha+\beta)^2 = 3\alpha\beta.(α+β)2=3αβ.

  4. Substitute sum and product

    (4−m3m)2=3⋅23m2.\left(\frac{4-m}{3m}\right)^2 = 3\cdot \frac{2}{3m^2}.(3m4−m​)2=3⋅3m22​.

    Simplify RHS: 3⋅23m2=2m2.3\cdot \frac{2}{3m^2} = \frac{2}{m^2}.3⋅3m22​=m22​.

    Therefore, (4−m)29m2=2m2.\frac{(4-m)^2}{9m^2} = \frac{2}{m^2}.9m2(4−m)2​=m22​.

    Since m≠0m\neq 0m=0 (otherwise equation is not quadratic), multiply by 9m29m^29m2: (4−m)2=18. (4-m)^2 = 18.(4−m)2=18.

    So, 4−m=±32.4-m = \pm 3\sqrt{2}.4−m=±32​.

    Hence, m=4∓32.m = 4 \mp 3\sqrt{2}.m=4∓32​.

    Thus the two values are m=4−32,m=4+32.m = 4-3\sqrt{2}, \qquad m = 4+3\sqrt{2}.m=4−32​,m=4+32​.

  5. Least value of mmm

    Clearly, 4−32<4+32.4-3\sqrt{2} < 4+3\sqrt{2}.4−32​<4+32​.

    So the least value is 4−32.\boxed{4-3\sqrt{2}}.4−32​​.

  6. Check options

    This matches Option B.

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