Question

The number of real values of $k$ for which the equation $x^2-3x+k=0$ has two distinct roots lying in the interval $(0,1)$ are

• A. Three
• B. Two
• C. Infinitely Many
• D. No values of $k$ satisfies the requirement

Answer 1

Answer: (C)

Infinitely Many

$x^2-3x+k=0$.
$D=3^2-4k=9-4k$. For distinct real roots, $9-4k\ge 0$ or $k\le \frac{9}{4}$ or $k\le 2.25$ ...$\left(i\right)$

Now the roots of the above equation is given by
$x_1=\frac{3+\sqrt{9-4k}}{2}$ and $x_2=\frac{3-\sqrt{9-4k}}{2}$

Now
$0\le x_1\le 1$

$0\le \frac{3+\sqrt{9-4k}}{2}\le 1$

$0\le 3+\sqrt{9-4k}\le 2$

$-3\le \sqrt{9-4k}\le -1$ ....$\left(ii\right)$
$0\le x_2\le 1$

$0\le \frac{3-\sqrt{9-4k}}{2}\le 1$

$0\le 3-\sqrt{9-4k}\le 2$

$-3\le -\sqrt{9-4k}\le -1$

$3\ge \sqrt{9-4k}\ge 1$ ....$\left(iii\right)$
From $\left(ii\right)$ and $\left(iii\right)$,

$9\ge 9-4k\ge 1$
$0\ge -4k\ge -8$
$0\le k\le 2$

Also, from $\left(i\right),k\le 2.25$

Hence

$k\in [0,2]$.

Now there are infinite real numbers between $[0,2]$, hence there are infinite possible value of $k$.