Question

The condition that the roots of $\frac{x-a}{ax+1}-\frac{x+b}{bx+1}=0$ are reciprocal to each other is

• A. $a=0$
• B. $a+b=0$
• C. $a-b=0$
• D. $b=0$

Answer 1

Answer: (D)

$b=0$

$Thegivenequationcanbewrittenasb{x}^2+x-abx-a-a{x}^2-x-abx-b=0\Rightarrow (b-a)x^2-2abx-(a+b)=0\because therootsarereciprocaltoeachother\therefore \frac{-(a+b)}{b-a}=1(i.e.productoftheroots=1)\Rightarrow -a-b=b-a\Rightarrow 2b=0\Rightarrow b=0Ans-OptionD$