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JEE Mains · Maths · STD 12 - 6. Application of derivatives

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

  1. A \(8\sqrt 3 \)
  2. B \(4\sqrt 3 \)
  3. C \(10\sqrt 5 \)
  4. D \(8\sqrt 5 \)
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Answer & Solution

Correct Answer

(D) \(8\sqrt 5 \)

Step-by-step Solution

Detailed explanation

\(\left(m^{2}+1\right) x^{2}-3 x+(m+1)^{2}=0\) \(\Rightarrow \alpha+\beta=\frac{3}{m^{2}+1}\) \(\alpha \beta=\frac{(m+1)^{2}}{m^{2}+1}\) \(\because \alpha+\beta\) is maximum \(\therefore m^{2}+1\) is minimum \(\Rightarrow \mathrm{m}=0\) \(\therefore \alpha+\beta=3\) and…
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