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 $a+b+c+abc$

Answer 1

Answer: 20

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 $\frac{3}{x}<\frac{4}{x};\therefore \left[\frac{3}{x}\right]\le \left[\frac{4}{x}\right]$
$\left[13^{\text{th }}30-7-2006\right]$
Since each of $\left[\frac{3}{x}\right]$ and $\left[\frac{4}{x}\right]$ is an integer $\therefore 3$ possibilities are there

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