Q69 / 180 · COMEDK 2015 · Mathematics

The number of solutions of the equation $|\cot x|=\cot x+\frac{1}{\sin x},(0 \leq x \leq 2 \pi)$ is

  • A
    0
  • Correct answer
    B
    1
  • C
    2
  • D
    3

Solution

With the help of group of cot $x$, we know that When $x \in\left(0, \frac{\pi}{2}\right] \cup\left(\pi, \frac{3 \pi}{2}\right]$, then $\cot x \geq 0$ $\therefore \quad|\cot x|=\cot x+\frac{1}{\sin x}$ $\Rightarrow \cot x=\cot x+\frac{1}{\sin x} \Rightarrow \frac{1}{\sin x}=0$ $\Rightarrow$ No solution exist. $\quad \text{...(i)}$ When $x \in\left(\frac{\pi}{2}, \pi\right) \cup\left(\frac{3 \pi}{2}, 2 \pi\right)$, then $\cot x < 0$ $\begin{aligned} & &|\cot x|=\cot x+\frac{1}{\sin x} \\ \therefore \quad &-\cot x &=\cot x+\frac{1}{\sin x} \\ \Rightarrow \quad &-2 \cot x &=\frac{1}{\sin x} \\ \Rightarrow \quad & & \cos x=\frac{-1}{2} \Rightarrow x=\frac{2 \pi}{3} \quad \text{...(ii)} \end{aligned}$ $\therefore$ From Eqs. (i) and (ii), only one solution exists.

Hence, the correct answer is option: 'B'.

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