JEE MAIN - Mathematics (2019 - 12th January Morning Slot - No. 12)
If $$\lambda $$ 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 $$\lambda + {1 \over \lambda } = 1,$$ is
$$ - 2 + \sqrt 2 $$
4$$-$$3$$\sqrt 2 $$
2 $$-$$ $$\sqrt 3 $$
4 $$-$$ 2$$\sqrt 3 $$
Explanation
3m2x2 + m(m $$-$$ 4) x + 2 = 0
$$\lambda + {1 \over \lambda } = 1,{\alpha \over \beta } + {\beta \over \alpha } = 1,{\alpha ^2} + {\beta ^2} = \alpha \beta $$
($$\alpha $$ + $$\beta $$)2 = 3$$\alpha $$$$\beta $$
$${\left( { - {{m\left( {m - 4} \right)} \over {3{m^2}}}} \right)^2} = {{3\left( 2 \right)} \over {3{m^2}}},{{{{\left( {m - 4} \right)}^2}} \over {9{m^2}}} = {6 \over {3m}}$$
$${\left( {m - 4} \right)^2} = 18,m = 4 \pm \sqrt {18,} \,\,4 \pm 3\sqrt 2 $$
$$\lambda + {1 \over \lambda } = 1,{\alpha \over \beta } + {\beta \over \alpha } = 1,{\alpha ^2} + {\beta ^2} = \alpha \beta $$
($$\alpha $$ + $$\beta $$)2 = 3$$\alpha $$$$\beta $$
$${\left( { - {{m\left( {m - 4} \right)} \over {3{m^2}}}} \right)^2} = {{3\left( 2 \right)} \over {3{m^2}}},{{{{\left( {m - 4} \right)}^2}} \over {9{m^2}}} = {6 \over {3m}}$$
$${\left( {m - 4} \right)^2} = 18,m = 4 \pm \sqrt {18,} \,\,4 \pm 3\sqrt 2 $$
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