Question

The set of values of $a$ for which the inequality, $(x-3a)(x-a-3)>0$ is satisfied for all $x\epsilon [1,3]$

• A. $(\frac{1}{3},3)$
• B. $(0,\frac{1}{3})$
• C. $(-2,0)$
• D. $(-2,3)$

Answer 1

The correct option is B $(0,\frac{1}{3})$

Given: inequality $(x-3a)(x-a-3)>0$, satisfied for all $x\in [1,3]$

To find the set of values of $a$

Sol: $(x-3a)(x-a-3)>0⟹x(x-a-3)-3a(x-a-3)>0⟹x^2-ax-3x-3ax+3a^2+9a>0⟹x^2-(4a+3)x+(3a^2+9a)>0$

This is of the form $a{x}^2+bx+c=0$, where $a=1,b=-(4a+3),c=3a^2+9a$

Hence the value of $x$ is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}⟹x=\frac{-(-(4a+3\left)\right)\pm \sqrt{(-(4a+3){)}^2-4(1\left)\right(3a^2+9a)}}{2\left(1\right)}⟹x=\frac{4a+3\pm \sqrt{16a^2+9+24a-12a^2-36a}}{2}⟹x=\frac{4a+3\pm \sqrt{4a^2-12a+9}}{2}⟹x=\frac{4a+3\pm \sqrt{(2a-3{)}^2}}{2}⟹x=\frac{4a+3\pm (2a-3)}{2}⟹x=\frac{4a+3+2a-3}{2},x=\frac{4a+3-2a+3}{2}⟹x=3a,x=a+3$

To satisfy the condition $x\in [1,3]$, we get $a=(0,\frac{1}{3})$