Q114 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics
$[\bullet]$ is the greatest integer function then $\int_0^{2 \pi}[|\sin x+| \cos \mid] d x=$
- A$\frac{ \pi}{2}$
- B${ \pi}{2}$
- C$\frac{3 \pi}{2}$
- Correct answerD$2{ \pi}$
Solution
$\begin{aligned}
& \text { (d) } \int_0^{2 \pi}[|\sin x|+|\cos x|] d x \\
& \because \quad|\sin x| \geq \sin ^2 x \\
& \quad|\cos x| \geq \cos ^2 x \\
& \therefore \quad|\sin x|+|\cos x| \geq \sin ^2 x+\cos ^2 x \\
& \quad|\sin x|+|\cos x| \geq 1
\end{aligned}$
also max. value of $\sin x+\cos x=\sqrt{2}$
$\begin{array}{ll}
\therefore & 1 \leq|\sin x|+|\cos x| \leq \sqrt{2} \\
\therefore & {[|\sin x|+|\cos x|]=1} \\
\therefore & \int_0^{2 \pi} 1 \cdot d x=[x]_0^{2 \pi}=2 \pi .
\end{array}$
Hence, the correct answer is option: 'D'.