Q95 / 150 · MHT CET 2026 (17 April Shift 2) · Mathematics
The area of the region bounded by curves $y = \sin x, y = \cos x$ and the lines $x = 0$, $x = \dfrac{\pi}{4}$ is
- A$2\sqrt{2} - 1$
- Correct answerB$\sqrt{2} - 1$
- C$2 - \sqrt{2}$
- D$\dfrac{1 - \sqrt{2}}{2}$
Solution
The required area is given by the integral of the upper curve minus the lower curve between the given limits.
In the interval $x \in [0, \dfrac{\pi}{4}]$, $\cos x \ge \sin x$.
Area $= \int_{0}^{\pi/4} (\cos x - \sin x) dx$
$= [\sin x + \cos x]_{0}^{\pi/4}$
$= (\sin \dfrac{\pi}{4} + \cos \dfrac{\pi}{4}) - (\sin 0 + \cos 0)$
$= (\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}) - (0 + 1)$
$= \sqrt{2} - 1$
Answer: $\sqrt{2} - 1$
In the interval $x \in [0, \dfrac{\pi}{4}]$, $\cos x \ge \sin x$.
Area $= \int_{0}^{\pi/4} (\cos x - \sin x) dx$
$= [\sin x + \cos x]_{0}^{\pi/4}$
$= (\sin \dfrac{\pi}{4} + \cos \dfrac{\pi}{4}) - (\sin 0 + \cos 0)$
$= (\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}) - (0 + 1)$
$= \sqrt{2} - 1$
Answer: $\sqrt{2} - 1$
Hence, the correct answer is option: 'B'.