Question

The number of real roots of the equation $\frac{A^2}{x}+\frac{B^2}{x-1}=1$ where A and B are real numbers not equal to zero simultaneously, is :

• A. $1or2$
• B. $1$
• C. $2$
• D. None

Answer 1

The correct option is A $1or2$

Given that $\frac{A^2}{x}+\frac{B^2}{x-1}=1,A,BϵR$

A and B cannot be zero simultaneously

$(A^2+B^2)x-A^2=x^2-x$

$x^2-(A^2+B^2+1)x+A^2=0$

$△={(A^2+B^2+1)}^2-4A^2$

$=(A^2+B^2+1+2A)(A^2+B^2+1-2A)$

$=({(A+1)}^2+B^2)({(A-1)}^2+B^2)$

$△=0$ If $A=\pm 1$ & $B=0$

$\therefore △$ is always $\ge 0$

$\therefore$Number of real roots $1$ (or) $2$