Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-p(x+1)-q=0$ then value of $\frac{{\alpha }^2+2\alpha +1}{{\alpha }^2+2\alpha +q}+\frac{{\beta }^2+2\beta +1}{{\beta }^2+2\beta +q}$

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is A $1$

$x^2-p(x+1)-q=0$

$x^2-px+(p+q)$

$\alpha +\beta =p$

$\alpha \beta =-(p+q)$

$⟹\alpha \beta =-(\alpha +\beta )-q$

$⟹q=-(\alpha +\beta )-\alpha \beta$

$⟹q=-\alpha \beta (\frac{1}{\beta }+\frac{1}{\alpha }-1)$

$\therefore \frac{{\alpha }^2+2\alpha +1}{{\alpha }^2+2\alpha +q}+\frac{{\beta }^2+2\beta +1}{{\beta }^2+2\beta +q}$

$=\frac{(\alpha +1{)}^2}{(\alpha +1{)}^2+(q-1)}+\frac{(\beta +1{)}^2}{(\beta +1{)}^2+(q-1)}$

Let, $\alpha +1=m$ and $\beta +1=n$

$\therefore \frac{m^2}{m^2+q-1}+\frac{n^2}{n^2+q-1}$

$=\frac{1}{1+\frac{q-1}{m^2}}+\frac{1}{1+\frac{q-1}{n^2}}$ (equation $01$)

Let $P\left(x\right)=0$

If $a$ and $b$ satisfies this equation then $(a+1)$ and $(b+1)$ will satisfy P(x-1)

$\therefore P(x-1)=(x-1{)}^2-P(x-1+1)-q=0$

$⟹x^2-(p+2)x-(q-1)=0$

So, $mn=-(q-1)$

$\therefore \frac{1}{1-\frac{n}{m}}+\frac{1}{1-\frac{m}{n}}$

$=\frac{m}{m-n}+\frac{n}{n-m}$

$=\frac{m-n}{m-n}$

$=1$