Question

If $\alpha ,\beta$ are the roots of the equation $x^2-2x-1=0,$ then the value of $\frac{\alpha }{{\alpha }^3-2{\alpha }^2+1}+\frac{\beta }{{\beta }^3-2{\beta }^2+1}$ is

• A. 0
• B. 3
• C. 1
• D. 2

Answer 1

The correct option is A 0

$\because {\alpha }^2-2\alpha -1=0,{\beta }^2-2\beta -1=0,\alpha +\beta =2,\alpha \beta =-1$

$\Rightarrow \alpha +\beta =2,\alpha \beta =-1$

$x-2x-1=0\Rightarrow {\alpha }^2-2\alpha =1,{\beta }^2-2\beta =1$

$\Rightarrow \frac{\alpha }{{\alpha }^3-2{\alpha }^2+1}+\frac{\beta }{{\beta }^3-2{\beta }^2+1}$

$=\frac{\alpha }{\alpha +1}+\frac{\beta }{\beta +1}=\frac{2\alpha \beta +(\alpha +\beta )}{(\alpha +1)(\beta +1)}=0$