Question

The inequality $(x-3m)(x-m-3)<0$ is satisfied for x in $[1,3]$. Determine the values of m for this to hold good.

Answer 1

Case I. Choose $\alpha -3m$ and $\beta =m+3$ where
$\alpha <\beta$ or $3m<m+3$ or $m<3/2$.......(1)
$(x-\alpha )(x-\beta )<0\Rightarrow \alpha <x<\beta$
But we are given that the inequality holds $\forall xϵ[1,3]$
and hence $[1,3]$ should be a subset of $]\alpha ,\beta [$
$\therefore$ $\alpha <1$ and $\beta >3$
$3m<1$ and $m+3>3$
$\therefore$ $m<\frac{1}{3}$ and $m>0$ i.e. $+ive$
$\therefore$ $mϵ(0,\frac{1}{3})$.
This also satisfies the condition I.
Case II : If $\alpha >\beta$ i.e. $3m>m+3$ or $m>3/2$.......(2)
then $\beta <x<\alpha$ ans as above $\beta <1$ and $\alpha >3$
or $m+3<1$ and $3m>3$ $\therefore$ $m<-2$ and $m>1$
There can't be any value of m satisfying both the above and also (2). Thus from case I, we get $mϵ(0,1/3)$.