Question

If $S$ and $P$ be the arithmetic mean and geometric mean of two positive real numbers $a$ and $b(b>a)$ respectively, then the set of values of $k$ for which the quadratic equation $k{x}^2+2Sx+P^2=0$ has real roots, is

• A. $(-\infty ,-1]\cup [1,\infty )$
• B. $(0,1]$
• C. $(-\infty ,a]\cup [b,\infty )$
• D. $(-\infty ,-1]\cup [\frac{b}{a},\infty )$

Answer 1

The correct option is B $(0,1]$

$S=\frac{a+b}{2}$ and $P=\sqrt{ab}$
$k{x}^2+2Sx+P^2=0$
$\Rightarrow k{x}^2+(a+b)x+ab=0$

For the equation to have real roots,
$D\ge 0$
$\Rightarrow (a+b{)}^2-4kab\ge 0$
$\Rightarrow a^2+(2-4k)ab+b^2\ge 0$
$\Rightarrow b^2[{\left(\frac{a}{b}\right)}^2+(2-4k)\frac{a}{b}+1]\ge 0$
Since, $b\ne 0$
$\Rightarrow {\left(\frac{a}{b}\right)}^2+(2-4k)\frac{a}{b}+1\ge 0$

Let $\frac{a}{b}=y$
Then $y\in (0,1)$
$y^2+(2-4k)y+1\ge 0\forall y\in (0,1)$
$\Rightarrow y+2-4k+\frac{1}{y}\ge 0$
$\Rightarrow 4k\le y+\frac{1}{y}+2$
$\Rightarrow 4k\le {(\sqrt{y}+\frac{1}{\sqrt{y}})}^2$
Above inequality is true when left side is less than or equal to the least value of right side.
$4k\le 2^2$
$\Rightarrow k\le 1$
But for $k=1,y=1$ which is not possible.
Hence, $k<1$