JEE Main 12 Jan 2019 Morning Question 17
Question: If X be the ratio of the roots of the quadratic equation in $ x,3m^{2}x^{2}+m(m-4)x+2=0, $ then the least value of m for which $ \lambda +\frac{1}{\lambda }=1, $ is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ -2+\sqrt{2} $
B) $ 4-3\sqrt{2} $
C) $ 4-2\sqrt{3} $
D) $ 2-\sqrt{3} $
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Answer:
Correct Answer: B
Solution:
Here, $ 3m^{2}x^{2}+m(m-4)x+2=0 $
Let $ \alpha ,\beta $ be the roots of the given equation
$ \therefore $ $ \lambda =\frac{\alpha }{\beta },\alpha +\beta =\frac{m(4-m)}{3m^{2}}=\frac{4-m}{3m},\alpha \beta =\frac{2}{3m^{2}} $ Given,
$ \lambda +\frac{1}{\lambda }=1 $
$ \Rightarrow $ $ \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=1\Rightarrow \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=1\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=\alpha \beta $
Now, $ {{(\alpha +\beta )}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta =3\alpha \beta $
$ \Rightarrow $ $ \frac{{{(4-m)}^{2}}}{9m^{2}}=\frac{2}{m^{2}}\Rightarrow {{(4-m)}^{2}}=18 $
$ \Rightarrow $ $ 4-m=\pm 3\sqrt{2}\Rightarrow m=4\pm 3\sqrt{2} $
$ \therefore $ Least value of $ m=4-3\sqrt{2} $
