Question

The number of solutions of $\left[\cos x\right]+\left|\sin x\right|=1$, $\pi \le x\le 3\pi$ is
(here $[.]$ denotes greatest integer function and $|.|$ denotes modulus function)

• A. 3
• B. 4
• C. 2
• D. 1

Answer 1

Answer: (A)

3

$\left[cosx\right]+\mid sinx\mid =1,\pi \le x\le 3\pi$

If $x\in [\frac{3\pi }{2},\frac{5\pi }{2}]-\left\{2\pi \right\}\Rightarrow \left[cosx\right]=0$

$\mid sinx\mid =1\Rightarrow x=\frac{3\pi }{2},\frac{5\pi }{2}$

If $xϵ(\pi ,\frac{3\pi }{2})\cup (\frac{5\pi }{2},3\pi )\Rightarrow \left[cosx\right]=-1$

$\mid sinx\mid =2\Rightarrow$ there is no solution in the interval

Now check $x=\pi ,2\pi ,3\pi$

clearly $x=2\pi$ is also a solution

Thus, total number of solution is 3.
Hence, option 'A' is correct.