Question

The number of integral values of $x$ for which the expression $x(2^x-1)(3^x-9)(x-3)\le 0$ holds true is

Answer 1

Let $E=x(2^x-1)(3^x-9)(x-3)\le 0$
Here $(2^x-1)=0\Rightarrow x=0$ and when $3^x-9=0\Rightarrow x=2$
Therefore, the critical points are $x=0,2,3$
Also, at $x=0$, two factors are $0,x$ and $2^x-1$; hence, the sign of $E$ does not change while crossing $x=0$