Question

If $x\in (0,\frac{\pi }{6})$, then the number of solutions of the equation $\left\{2x\right\}+\left\{\tan x\right\}=0$ is

(Here, $\left\{x\right\}$ denotes the fractional part of $x$.)

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$\left\{2x\right\}+\left\{\tan x\right\}=0$
Both the terms of LHS are non-negative.
So, $\left\{2x\right\}=0\cdots \left(1\right)$
$\Rightarrow x=...,-1,\frac{-1}{2},0,\frac{1}{2},1,\frac{3}{2},...$

and $\left\{\tan x\right\}=0\cdots \left(2\right)$
$\Rightarrow x=...,-\pi ,0,\pi ,...$
There is no common value $x\in (0,\frac{\pi }{6})$ which satisfies both $\left(1\right)$ and $\left(2\right)$.
$\therefore$ For $x\in (0,\frac{\pi }{6})$, there is no solution.