Question

Suppose a, b > 0 and $x_1,x_2,x_3(x_1>x_2>x_3)$ are the roots of $\frac{x-b}{a}+\frac{x-a}{b}=\frac{b}{x-a}+\frac{a}{x-b}and{x}_1-x_2-x_3=c,$ then

• A. a, c, b are in H.P. and $x_1=a+b$
• B. a, c, b are in A.P. and $x_2=a+b$
• C. a, c, b are in A.P. and $x_3=0$
• D. a, c, b are in H.P. and $x_3=0$

Answer 1

Answer: (A), (D)

The correct options are
A a, c, b are in H.P. and $x_1=a+b$
D a, c, b are in H.P. and $x_3=0$

$\frac{x-b}{a}+\frac{x-a}{b}=\frac{a}{x-b}+\frac{b}{x-a}\Rightarrow \frac{x-b}{a}-\frac{b}{x-a}=\frac{a}{x-b}-\frac{x-a}{b}\Rightarrow \frac{(x^2-(a+b\left)x\right)}{a(x-a)}=\frac{\left(x\right(a+b)-x^2)}{b(x-b)}\Rightarrow (x^2-(a+b\left)x\right)\left(\right(b+a)x-(a^2+b^2\left)\right)=0\Rightarrow x=0,a+b,\frac{a^2+b^2}{a+b}{x}_3=0,x_2=\frac{a^2+b^2}{a+b},x_1=a+bNow,(a+b)-(a+b)+\frac{2ab}{a+b}=c\Rightarrow c=\frac{2ab}{a+b}hencea,c,bareinH.P.$