Question

$a,b\in R$m roots of $x^2+ax+b=0$ are $\alpha ,\beta$ and roots of $x^2+bx+a=0$ are $\alpha +k,\beta +k$ where $k\ne 0$
If $ab=\lambda$ then interval for $\lambda$ is

• A. $(-\propto ,4)$
• B. $(-\propto ,4]$
• C. $(-\propto ,0)$
• D. $(4,\propto )$

Answer 1

Answer: (A)

$(-\propto ,4)$

If $\alpha \beta$ are roots of $x^2+ac+b=0\Rightarrow \alpha +\beta =a,\alpha \beta =b$ ........ $\left(1\right)$

If $\alpha +k,\beta +k$ are roots of $x^2+bx+a=0\Rightarrow \alpha +\beta +2k=-b$

$K=\frac{a-b}{2}$......... $\left(2\right)$

from $\left(1\right)$

Also , $(\alpha +k)(\beta +k)=a$

$\alpha \beta +k(\alpha +\beta )+k^2=a$

$b+\frac{a-b}{2}(-a)+\frac{(a-b{)}^2}{4}=a$

$\Rightarrow a=b$ or $\frac{a-b}{4}+\frac{(-a)}{2}=1$

$a=b$ or $a+b+4=0$

$a\ne b$ as $k\ne 0$ (given) $\Rightarrow a+b=-4a^2+b^2+2ab=16$ ........ $\left(3\right)$

Also, $4k^2=(a-b{)}^2=a^2+b^2-2ab$ ..... $\left(4\right)$

$\left(3\right)-\left(4\right),4ab=16-4k^2$

$ab=4-k^2$ ($=\lambda$ given)

Now, $(4-k^2)\le 4$ but $ab\ne 0$

$\Rightarrow ab<4$

$\therefore$ Option $A$ is correct