Question

If $S$ and $P$ be the arithmetic mean and geometric mean of two non-zero 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 $\frac{a}{b}=y$
$y^2+(2-4k)y+1\ge 0\forall y\in R$
Discriminant of the above equation is, $D\le 0$
$\therefore (2-4k{)}^2-4\le 0$
$\Rightarrow k^2-k\le 0$
$\Rightarrow k(k-1)\le 0$
$\Rightarrow k\in [0,1]$
But, the given equation is a quadratic equation, so $k\ne 0$
Hence, $k\in (0,1]$