Verified Solution • 2025–26 Session moderate

QJEEM21Ma2cd4: JEE Main 2021 Mathematics - Class 12 JEEM21Ma2cd4 Solution: Let A 1 be the area of the region bounded by the curves y =…

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 = π2{\pi \over 2} in the first quadrant. Then,

Multiple Choice Options

  • A A1:A2=1:2{A_1}:{A_2} = 1:\sqrt 2 and A1+A2=1{A_1} + {A_2} = 1
  • B A1=A2{A_1} = {A_2} and A1+A2=2{A_1} + {A_2} = \sqrt 2
  • C 2A1=A22{A_1} = {A_2} and A1+A2=1+2{A_1} + {A_2} = 1 + \sqrt 2
  • D A1:A2=1:2{A_1}:{A_2} = 1:2 and A1+A2=1{A_1} + {A_2} = 1

Step-by-Step Verified Solution

A1+A2=∫0π/2cos⁡x. dx=sin⁡x∣0π/2=1{A_1} + {A_2} = \int\limits_0^{\pi /2} {\cos x.\,dx = \left. {\sin x} \right|_0^{\pi /2}} = 1

JEE Main 2021 (Online) 26th February Evening Shift Mathematics - Area Under The Curves Question 93 English Explanation

A1=∫0π/4(cos⁡x−sin⁡x)dx=(sin⁡x+cos⁡x)∣0π/4=2−1{A_1} = \left. {\int\limits_0^{\pi /4} {(\cos x - \sin x)dx = (\sin x + \cos x)} } \right|_0^{\pi /4} = \sqrt 2 - 1

∴\therefore A2=1−(2−1)=2−2{A_2} = 1 - \left( {\sqrt 2 - 1} \right) = 2 - \sqrt 2

∴\therefore A1A2=2−12(2−1)=12{{{A_1}} \over {{A_2}}} = {{\sqrt 2 - 1} \over {\sqrt 2 \left( {\sqrt 2 - 1} \right)}} = {1 \over {\sqrt 2 }}

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