Question

The set of values of $a$ for which the inequality $(x-3a)(x-a-3)<0$ is satisfied for all $x$ in the interval $1\le x\le 3$ are

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

Answer 1

Answer: (B), (C)

The correct options are
B $(-2,0)$
C $(0,\frac{1}{3})$

The given inequality $(x-3a)(x-a-3)<0$ satisfies for all x in interval $1\le x\le 3$

Hence, $x=1,2,3$ should also satisfy the given inequality. We find solutions for $x=1,2,3$

$x=1$ $\Rightarrow$ $(1-3a)(-a-2)<0$ $\Rightarrow$ $(1-3a)(a+2)>0$

$\Rightarrow$ Solution for above inequality are $a>-2$ and $a<1/3$ .......A

$x=2$ $\Rightarrow$ $(2-3a)(-a-1)<0$$\Rightarrow$ $(2-3a)(a+1)>0$

$\Rightarrow$ Solution for above inequality are $a>-1$ and $a<2/3$ .......B

Again,

$x=3$ $\Rightarrow$ $(3-3a)(-a)<0$ $\Rightarrow$ $(3-3a)\left(a\right)>0$

$\Rightarrow$ Solution for above inequality are $a>0$ and $a<1$ .......C

A,B and C equations give solutions to be $a\in (0,\frac{1}{3})$ and $a\in (-2,0)$