CBSE Class 12 Sample Paper for 2025 Boards
You are here
CBSE Class 12 Sample Paper for 2025 Boards
Last updated at August 14, 2026 by Teachoo
This question is similar to Chapter 8 Class 12 Application of Integrals - Examples
Please check the question here
-lock-
-endlock-
Transcript
Question 32 Draw the rough sketch of the curve 𝑦=20 cos 2𝑥; (where 𝜋/6≤𝑥≤𝜋/3) Using integration, find the area of the region bounded by the curve y=20 cos2x from the ordinates 𝑥=𝜋/6 to 𝑥=𝜋/3 and the 𝑥-axis.Now, 𝑦=20 cos 2𝑥; Since 𝜋/6≤𝑥≤𝜋/3, We find value of y at key points At x = 𝝅/𝟔 𝑦=20 cos 2(𝜋/6) = 20 cos 𝜋/3 = 20 ×1/2 = 𝟏𝟎 At x = 𝝅/𝟑 𝑦=20 cos 2(𝜋/3) = 20 cos 2𝜋/3 = 20 cos(𝜋−𝜋/3) = 20 × − cos 𝜋/3 = 20 ×(−1)/2 = −𝟏𝟎 At x = 𝝅/𝟒 𝑦=20 cos 2(𝜋/4) = 20 cos 𝜋/2 = 20 ×0 = 𝟎 Thus, graph of 𝑦=20 cos 2𝑥 is Now, Area Required = Area ADB + Area BEC + Area DEF Area ADB Area ADB = ∫_(𝜋/6)^(𝜋/( 4))▒〖𝑦 𝑑𝑥〗 𝑦→20 cos2𝑥 = ∫_(𝜋/6)^(𝝅/( 𝟒))▒〖𝟐𝟎 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙〗 = 20[sin2𝑥/2]_(𝜋/6)^(𝜋/4) =10[sin2(𝜋/4)−sin2(𝜋/6) ] =10[sin(𝜋/2)−sin(𝜋/6) ] =10[1−√3/2] =10[(2 − √3)/2] =5(2 − √3) =10−5(2 − √3) =10[(2 − √3)/2] =5(2 − √3) =𝟏𝟎−𝟓√𝟑 Area BEC Area BEC = ∫_(𝜋/4)^(𝜋/( 3))▒〖𝑦 𝑑𝑥〗 𝑦→20 cos2𝑥 = ∫_(𝜋/4)^(𝝅/( 𝟑))▒〖𝟐𝟎 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙〗 = 20[sin2𝑥/2]_(𝜋/4)^(𝜋/3) =10[sin2(𝜋/3)−sin2(𝜋/4) ] =10[sin(2𝜋/3)−sin(𝜋/2) ] =10[sin(𝜋−𝜋/3)−sin(𝜋/2) ] =10[sin(𝜋/3)−sin(𝜋/2) ] =10[√3/2−1] =10 ×√3/2−10 =5√3−10 Since √3 = 1.73, 5√3−10 is negative And, area cannot be negative ∴ Area BEC = 𝟏𝟎−𝟓√𝟑 Therefore Area Required = Area ADB + Area BEC = (10−5√3)+(10−5√3) = 2 ×(10−5√3) = 𝟐𝟎−𝟏𝟎√𝟑 square unit
