Q92 / 150 · MHT CET 2026 (18 April Shift 1) · Mathematics
If $[\cdot]$ denotes the greatest integer function, then $\int_0^{\pi/3}[\tan x]\,dx = $
- Correct answerA$\dfrac{\pi}{12}$
- B$\dfrac{\pi}{6}$
- C$\dfrac{\pi}{4}$
- D$\dfrac{\pi}{3}$
Solution
The given integral is $I = \int_0^{\pi/3} [\tan x] \, dx$.
In the interval $[0, \pi/3]$, the function $\tan x$ is strictly increasing.
For $0 \le x < \pi/4$, $0 \le \tan x < 1 \Rightarrow [\tan x] = 0$.
For $\pi/4 \le x \le \pi/3$, $1 \le \tan x \le \sqrt{3} < 2 \Rightarrow [\tan x] = 1$.
Splitting the integral at $x = \pi/4$, we get:
$I = \int_0^{\pi/4} 0 \, dx + \int_{\pi/4}^{\pi/3} 1 \, dx$
$I = 0 + [x]_{\pi/4}^{\pi/3}$
$I = \dfrac{\pi}{3} - \dfrac{\pi}{4} = \dfrac{\pi}{12}$.
Answer: $\dfrac{\pi}{12}$
In the interval $[0, \pi/3]$, the function $\tan x$ is strictly increasing.
For $0 \le x < \pi/4$, $0 \le \tan x < 1 \Rightarrow [\tan x] = 0$.
For $\pi/4 \le x \le \pi/3$, $1 \le \tan x \le \sqrt{3} < 2 \Rightarrow [\tan x] = 1$.
Splitting the integral at $x = \pi/4$, we get:
$I = \int_0^{\pi/4} 0 \, dx + \int_{\pi/4}^{\pi/3} 1 \, dx$
$I = 0 + [x]_{\pi/4}^{\pi/3}$
$I = \dfrac{\pi}{3} - \dfrac{\pi}{4} = \dfrac{\pi}{12}$.
Answer: $\dfrac{\pi}{12}$
Hence, the correct answer is option: 'A'.