Question

If $\alpha$ and $\beta$ are the roots of $x^2-a(x-1)+b=0$, then the value of $1/[{ɑ}^2-aɑ]+1/[{\beta }^2-a\beta ]+2/[a+b]$ is

• A. $4/(a+b)$
• B. $1/(a+b)$
• C. $0$
• D. $-1$

Answer 1

The correct option is C

$0$

Explanation for correct option:

Find the value of $\mathbf{1}\mathbf{}\mathbf{/}\mathbf{}\mathbf{[}{\mathit{ɑ}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\mathit{a}\mathit{ɑ}\mathbf{]}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{}\mathbf{/}\mathbf{}\mathbf{[}{\mathit{\beta }}^{\mathbf{2}}\mathbf{-}\mathbf{}\mathit{a}\mathit{\beta }\mathbf{]}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\mathbf{/}\mathbf{}\mathbf{[}\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}\mathbf{]}$:

Since $\alpha$ and $\beta$ are the roots of $x^2-a(x-1)+b=0$

$$\begin{gathered} \Rightarrow {\beta }^2–a(\beta –1)+b=0 \\ \Rightarrow {\beta }^2–a\beta =–(a+b) \end{gathered}$$

and ${\alpha }^2-a(\alpha -1)+b=0$

$\Rightarrow {\alpha }^2-a\alpha =-(a+b)$

Thus

$\begin{array}{rcl}\frac{1}{\left({ɑ}^2-aɑ\right)}+\frac{1}{\left({\beta }^2-a\beta \right)}+\frac{2}{\left(a+b\right)}& =& -\frac{1}{\left(a+b\right)}-\frac{1}{\left(a+b\right)}+\frac{2}{\left(a+b\right)}\\ & =& 0\end{array}$

Hence, the correct option is C.