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Can someone help me with this, - Complex numbers and quadratic equations - JEE Main-9

If m is chosen in the quadratic equation \left ( m^{2}+1 \right )x^{2}-3x+\left ( m^{2}+1 \right )^{2}=0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is  :

  • Option 1)

    10\sqrt{5}

  • Option 2)

    8\sqrt{3}

  • Option 3)

    8\sqrt{5}

  • Option 4)

    4\sqrt{3}

 
Answers (1)
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\\\ sum\:\:of\:\:root = \frac{-b}{a} \\\\\:=-\frac{-3}{m^{2}+1}=\frac{3}{(m^{2}+1)}

for sum of root to be greatest   m^{2}+1  should be minimum \neq0

\\f^{'}(m)=m^{2}+1=0\\\\\:f^{'}(m)=2m=0

\therefore m=0

now equation 

=x^{2}-3x+1=0

roots=\frac{3\pm \sqrt{9-4}}{2}= \frac{3\pm \sqrt{5}}{2}

\\\alpha+\beta=\frac{3}{m^{2}+1}=3\\\\\:\alpha\beta=\frac{(m+1)^{2}}{m^{2}+1}=1

\\\left | \alpha^{3}-\beta^{3} \right |=\left | (\alpha-\beta)(\alpha^{2}+\beta^{2}+\alpha\beta) \right |\\\\\:\left | \sqrt{(\alpha+\beta)^{2}-4\alpha\beta} \right | \left | (\alpha+\beta)^{2}-\alpha\beta \right |

=\sqrt{5}.8=8\sqrt{5}

 

 

  


Option 1)

10\sqrt{5}

Option 2)

8\sqrt{3}

Option 3)

8\sqrt{5}

Option 4)

4\sqrt{3}

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