JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Area Under Curves

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,

Answer: (B)

Solution

The curves meet where

tan x = 1

, that is at

x = \frac{π}{4}

, and

cos x > sin x

[0, \frac{π}{4})

.
The first region lies between the two curves from the

y

-axis to their crossing:

A_{1} = \int_{0}^{π /4} (cos x - sin x) d x = [sin x + cos x]_{0}^{π /4} = 2 - 1

The second region lies under whichever curve is lower, up to

x = \frac{π}{2}

A_{2} = \int_{0}^{π /4} sin x d x + \int_{π /4}^{π /2} cos x d x = (1 - \frac{1}{2}) + (1 - \frac{1}{2}) = 2 - 2

Since

2 - 2 = 2 (2 - 1)

, the ratio is

A_{1} : A_{2} = 1 : 2

, and

A_{1} + A_{2} = (2 - 1) + (2 - 2) = 1

A_{1} : A_{2} = 1 : 2

and

A_{1} + A_{2} = 1

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option