Question

If all the solution of the inequality $x^2-6ax+5a^2\le 0$ are also the solutions of inequality $x^2-14x+40\le 0$ then find the number of possible integral value of $a$.

Answer 1

Answer: 0

Let $f(x)=x^2-6ax+5a^2=(x-a)(x-5a)$
and $g(x)=x^2-14x+40=(x-4)(x-10$.
Clearly, $g(a)\le 0$ and $g(5a)\le 0$ must satisfy simultaneously.

Now, $g(x)\le 0\Rightarrow a^2-14a+40\le 0$
$\Rightarrow 4\le a\le 10$....(1)
Also, $g(5a)\le 0\Rightarrow 25a^2-70a-40\le 0$
$\Rightarrow (5a-4)(a-2)\le 0$
$\Rightarrow \frac{4}{5}\le a\le 2$...(2)
$\therefore (1)\cap (2)\text{ gives }a\in \varphi$
Hence no integral value of a can satisfy it.