Question

The set of real values of ' $x$ ' satisfying the equality $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$ (where [ ] denotes the greatest integer function) belongs to the interval $\left( a , \frac{ b }{ c }\right]$ where $a , b , c \in N$ and $\frac{ b }{ c }$ is in its lowest form. Find the value of $a+b+c+a b c$

Answer 1

Case-I : If $x<0$ then $\left[\frac{3}{x}\right]$ and $\left[\frac{4}{x}\right]$ is - ve hence $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]$ can never be equal to 5 Case-II : If $x >0$
we have $\frac{3}{ x }<\frac{4}{ x } ; \therefore \left[\frac{3}{ x }\right] \leq\left[\frac{4}{ x }\right]$
$\left[13^{\text {th }} 30-7-2006\right]$
Since each of $\left[\frac{3}{x}\right]$ and $\left[\frac{4}{x}\right]$ is an integer $\therefore 3$ possibilities are there

now, If $\left[\frac{3}{x}\right]=0 \Rightarrow 0 \leq \frac{3}{x}< 1 \Rightarrow 0 \leq 3< x \Rightarrow x>3$
and $\left[\frac{4}{x}\right]=5 \Rightarrow 5 \leq \frac{4}{x}< 6 \Rightarrow \frac{1}{6}< \frac{x}{4} \leq \frac{1}{5} \Rightarrow \frac{2}{3}< x \leq \frac{4}{5}$ these two equations are not possible. Hence no solutions in these cases.
now, If $\left[\frac{3}{x}\right]=1 \Rightarrow 1 \leq \frac{3}{x}< 2 \Rightarrow \frac{1}{2}< \frac{x}{3} \leq 1 \Rightarrow \frac{3}{2}< x \leq 3$
and $\left[\frac{4}{x}\right]=4 \Rightarrow 4 \leq \frac{4}{x}< 5 \Rightarrow \frac{1}{5}< \frac{x}{4} \leq \frac{1}{4} \Rightarrow \frac{4}{5}< x \leq 1$
not possible simultaneously $\Rightarrow$ no solution
again If $\left[\frac{3}{x}\right]=2 \Rightarrow 2 \leq \frac{3}{x}< 3 \Rightarrow \frac{1}{3}< \frac{x}{3} \leq \frac{1}{2} \Rightarrow 1< x \leq \frac{3}{2}$
and $\left[\frac{4}{x}\right]=3 \Rightarrow 3 \leq \frac{4}{x}<4 \Rightarrow \frac{1}{4}< \frac{x}{4} \leq \frac{1}{3} \Rightarrow 1< x \leq \frac{4}{3}$
common solution $1< x \leq \frac{4}{3}$
Hence $x \in\left(1, \frac{4}{3}\right]$
$\therefore a =1, b =4, c =3 ; $
$ \therefore a + b + c + abc =1+4+3+12=20$