Question

If $A,G$ and $H$ are respectively arithmetic, geometric and harmonic means between $a$ and $b$ both being unequal and positive, then $A=\frac{a+b}{2}\Rightarrow a+b=2A,G=\sqrt{ab}\Rightarrow ab=G^2$ and $H=\frac{2ab}{a+b}\Rightarrow G^2=AH$
From the above discussion we can say that $a,b$ are the roots of the equation $x^2-2Ax+G^2=0$
Now,quadratic equation, $x^2-Px+Q=0$ and quadratic equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ have a root common and satisfy the relation $b=\frac{2ac}{a+c},$ where $a,b,c$ are real numbers.
On the basis of the above information, answer the following questions:

The value of $\left[p\right]$ is (where $[.]$ is the greatest integer function).

• A. $2$
• B. $3$
• C. $5$
• D. $6$

Answer 1

The correct option is A $2$

Given: $a(b-c)x^2+b(c-a)x+c(a-b)=0$ .......$\left(1\right)$

$\because a(b-c)+b(c-a)+c(a-b)=ab-ac+bc-ba+ca-cb=0$

$\therefore x=1$ is a root of eqn$\left(1\right)$

Let the other root be $\alpha$ then $1\times \alpha =\frac{c(a-\frac{2ac}{a+c})}{a(\frac{2ac}{a+c}-c)}$ ......... $\because b=\frac{2ac}{a+c}$

On evaluating, we get the other root as $=\frac{c{a}^2+c^{2}a-2a{c}^2}{2a^{2}c-a^{2}c-c^{2}a}=\frac{c{a}^2-a{c}^2}{c{a}^2-a{c}^2}=\frac{ac(a-c)}{ac(a-c)}=1$

Hence both the roots of eqn$\left(1\right)$ are $1,1$

Then roots of the equation $x^2-Px+Q=0$ is also $1,1$
Then $P=1+1=2$ and $Q=1\times 1=1$

$\therefore P=2,Q=1$

$\therefore \left[P\right]=2$