The area of the region A = { ( x , y ) : | cos x - sin x | y sin x , 0 x 2 } is [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The area of the region $A = {(x, y) : | \cos x - \sin x | \leq y \leq \sin x, 0 \leq x \leq \frac{π}{2}}$ is [2023]

1 $\sqrt{5} - 2 \sqrt{2} + 1$

2 $\sqrt{5} + 2 \sqrt{2} - 4.5$

3 $1 - \frac{3}{\sqrt{2}} + \frac{4}{\sqrt{5}}$

4 $\frac{3}{\sqrt{5}} - \frac{3}{\sqrt{2}} + 1$

Ans.
(1)

The area of the region, $A = {(x, y) : | \cos x - \sin x | \leq y \leq \sin x, 0 \leq x \leq \frac{π}{2}}$

$\Rightarrow$ $| \cos x - \sin x |$ $\leq y \leq \sin x$

For finding the intersecting point we must have

$\cos x - \sin x = \sin x$

$\Rightarrow \tan x = \frac{1}{2}$

Let $ϕ = \tan^{- 1} \frac{1}{2}$

$So, \tan ϕ = \frac{1}{2}, \sin ϕ = \frac{1}{\sqrt{5}}, \cos ϕ = \frac{2}{\sqrt{5}}$

$Area = \int_{ϕ}^{π / 2} (\sin x - | \cos x - \sin x |) d x$

$= \int_{ϕ}^{π / 4} (\sin x - (\cos x - \sin x)) d x + \int_{π / 4}^{π / 2} (\sin x - (\sin x - \cos x)) d x$

$= \int_{ϕ}^{π / 4} (2 \sin x - \cos x) d x + \int_{π / 4}^{π / 2} \cos x d x$

$= {[- 2 \cos x - \sin x]}_{ϕ}^{π / 4} + {[\sin x]}_{π / 4}^{π / 2}$

$= \sqrt{5} - 2 \sqrt{2} + 1$

Want Us to Email you About Special Offers & Updates?