Question

Find the values of $k$ for which the inequality $(x-3k)(x-k-3)<0$ is satisfied for all $x$ such that $1\le x\le 3$. The number of intervals in the solution is/are

Answer 1

Case I: Let $3k<k+3\Rightarrow k<\frac{3}{2}$ & from wavy curve method.
$\therefore x\in (3k,k+3)$
Now the original inequation is true for all $x\in (1,3)$
Hence $[1,3]\subseteq (3k,k+3)\therefore 3k<1\&k+3>3\therefore k<\frac{1}{3}\&k>0\therefore kϵ(0,\frac{1}{3})$
Case II: Let $3k>k+3\Rightarrow k>\frac{3}{2}$ & from wavy curve method.
$\therefore x\in (k+3,3k)$
Now the original inequation is true for all $x\in (1,3)$
$[1,3]\subseteq (k+3,3k)\therefore k+3<1\&3k>3\therefore k<-2\&k>1\therefore k\in \varphi$ combining both the cases we get: $k\in (0,1/3)$