Let A1 be the area of the region bounded by the curves y= sin x, y= cos x and y -axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y= sin x y = cos x , x -axis and x =(π/2) in the first quadrant. Then,

Question

Let A1 be the area of the region bounded by the curves y= sin x, y= cos x and y -axis in the first quadrant. Also, let A2 be the area of the region bounded by the curves y= sin x y = cos x , x -axis and x =(π/2) in the first quadrant. Then, (A) A 1: A 2=1: √2 and A 1+ A 2=1 (B) A 1= A 2 and A 1+ A 2=√2 (C) 2 A 1= A 2 and A 1+ A 2=1+√2 (D) A 1: A 2=1: 2 and A 1+ A 2=1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/e16935f5b97a7964496261dd1a14184f-.png" /> A1=∫ limits0π / 4( cos x- sin x) d x A1=( sin x+ cos x)0π / 4=√2-1 A2=∫ limits0π / 4 sin x d x+∫ limitsπ / 4π / 2 cos x d x =(- cos x)0π / 4+( sin x)π / 4π / 2 A2=√2(√2-1) A1: A2=1: √2, A1+A2=1

Q. Let $A_{1}$ be the area of the region bounded by the curves $y = sin x, y = cos x$ and $y$ -axis in the first quadrant. Also, let $A_{2}$ be the area of the region bounded by the curves $y = sin x$ $y = cos x, x$ -axis and $x = \frac{π}{2}$ in the first quadrant. Then,

$A_{1} : A_{2} = 1 : 2$ and $A_{1} + A_{2} = 1$

$A_{1} = A_{2}$ and $A_{1} + A_{2} = 2$

$2 A_{1} = A_{2}$ and $A_{1} + A_{2} = 1 + 2$

$A_{1} : A_{2} = 1 : 2$ and $A_{1} + A_{2} = 1$

$A_{1} = 0 \int π /4 (cos x - sin x) d x$
$A_{1} = (sin x + cos x)_{0}^{π /4} = 2 - 1$
$A_{2} = 0 \int π /4 sin x d x + π /4 \int π /2 cos x d x$
$= (- cos x)_{0}^{π /4} + (sin x)_{π /4}^{π /2}$
$A_{2} = 2 (2 - 1)$
$A_{1} : A_{2} = 1 : 2, A_{1} + A_{2} = 1$

Q. Let A1​ be the area of the region bounded by the curves y=sinx,y=cosx and y -axis in the first quadrant. Also, let A2​ be the area of the region bounded by the curves y=sinx y=cosx,x -axis and x=2π​ in the first quadrant. Then,

A1​:A2​=1:2​ and A1​+A2​=1

A1​=A2​ and A1​+A2​=2​

2A1​=A2​ and A1​+A2​=1+2​

A1​:A2​=1:2 and A1​+A2​=1

A1​=0∫π/4​(cosx−sinx)dx A1​=(sinx+cosx)0π/4​=2​−1 A2​=0∫π/4​sinxdx+π/4∫π/2​cosxdx =(−cosx)0π/4​+(sinx)π/4π/2​ A2​=2​(2​−1) A1​:A2​=1:2​,A1​+A2​=1

Q. Let $A_{1}$ be the area of the region bounded by the curves $y = sin x, y = cos x$ and $y$ -axis in the first quadrant. Also, let $A_{2}$ be the area of the region bounded by the curves $y = sin x$ $y = cos x, x$ -axis and $x = \frac{π}{2}$ in the first quadrant. Then,

$A_{1} : A_{2} = 1 : 2$ and $A_{1} + A_{2} = 1$

$A_{1} = A_{2}$ and $A_{1} + A_{2} = 2$

$2 A_{1} = A_{2}$ and $A_{1} + A_{2} = 1 + 2$

$A_{1} : A_{2} = 1 : 2$ and $A_{1} + A_{2} = 1$

$A_{1} = 0 \int π /4 (cos x - sin x) d x$
$A_{1} = (sin x + cos x)_{0}^{π /4} = 2 - 1$
$A_{2} = 0 \int π /4 sin x d x + π /4 \int π /2 cos x d x$
$= (- cos x)_{0}^{π /4} + (sin x)_{π /4}^{π /2}$
$A_{2} = 2 (2 - 1)$
$A_{1} : A_{2} = 1 : 2, A_{1} + A_{2} = 1$