CBSE Class 12 Sample Paper for 2025 Boards
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CBSE Class 12 Sample Paper for 2025 Boards
Last updated at Oct. 3, 2024 by Teachoo
This question is similar to Chapter 8 Class 12 Application of Integrals - Examples
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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
