If a _ 1 be the roots of x ^ 2 - a ( x -1)+ b = 0 , then the value of 1 a^ 2 -a a^ 2 + 1…
Mathematics · JEE Main · NTA Exams — Complex Numbers and Quadratic Equations
If \(\mathrm{a}_{1} \beta\) be the roots of \(\mathbf{x}^{2}-\mathbf{a}(\mathbf{x}-1)+\mathbf{b}=\mathbf{0}\), then the value of \(\frac{1}{a^{2}-a a^{2}}+\frac{1}{\beta^{2}-a \beta}+\frac{2}{a+b}\) is
- \(\frac{4}{a+b}\)
- \(\frac{1}{a+b}\)
- \(0\)
- \(-1\)
Answer
(A) 4 a+b
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