Question

The set of value of ${}^{\prime }{a}^{\prime }$ such that $x^2-2ax+a^2-6a\le 0$ in $(1,2)$ is

Answer 1

We have,

$x^2-2ax+a^2-6a\le 0in(1,2)$

Now, Put $x=1$ and we get,

$1-2a+a^2-6a\le 0$

$\Rightarrow a^2-8a+1\le 0$

Using quadratic formula and we get,

$a=\frac{8\pm \sqrt{64-4}}{2}$

$a=\frac{8\pm \sqrt{60}}{2}$

$a=4\pm \sqrt{15}$

If,

$a\le 0$

$4-\sqrt{15}\le 0$

Now,

Put $x=2$ and we get,

$2^2-2a\times 2+a^2-6a\le 0$

$\Rightarrow 4-4a+a^2-6a\le 0$

$\Rightarrow a^2-10a+4\le 0$

Using quadratic formula and we get,

$a=\frac{10\pm \sqrt{100-16}}{2}$

$a=\frac{10\pm \sqrt{84}}{2}$

$a=\frac{10\pm \sqrt{84}}{2}$

$a=5\pm \sqrt{21}$

For,

$a\le 0$

$5\pm \sqrt{21}\le 0$

$5-\sqrt{21}\le 0$

Hence, $[4-\sqrt{15},5-\sqrt{21}]$

Hence, this is the answer.