Question

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

• A. - 5
• B. - 6
• C. - 7
• D. - 2

Answer 1

The correct option is A

- 5

$x^3-x^2-1=0\alpha +\beta +\gamma =1,\alpha \beta +\beta \gamma +\gamma \alpha =0,\alpha \beta \gamma =1.$
Now, $\frac{1+\alpha }{1-\alpha }=\frac{2-(1-\alpha )}{1-\alpha }=\frac{2}{1-\alpha }-1\therefore \sum \frac{1+\alpha }{1-\alpha }=\sum \frac{2}{1-\alpha }-\sum 1=2\sum \frac{1}{1-\alpha }-3\left(i\right)Now,\frac{1}{x-\alpha }+\frac{1}{x-\beta }+\frac{1}{x-\gamma }=\frac{(x-\beta )(x-\gamma )+(x-\alpha )(x-\beta )+(x-\alpha )(x-\gamma )}{(x-\alpha )(x-\beta )(x-\gamma )}\frac{1}{x-\alpha }+\frac{1}{x-\beta }+\frac{1}{x-\gamma }=\frac{3x^2-2x}{x^3-x^2-1}Letx=1=\frac{1}{1-\alpha }+\frac{1}{1-\beta }+\frac{1}{1-\gamma }=\frac{3-2}{1-1-1}=-1\therefore \left(i\right)=2(-1)-3=-5$