Question 12 (Choice 1) - CBSE Class 12 Sample Paper for 2022 Boards (For Term 2) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Question 12 (Choice 1) Using integration, find the area of the region in the first quadrant enclosed by the line x + y = 2, the parabola y2 = x and the x-axis. Let’s first draw the Figure Here, 𝒚𝟐 =𝒙 is a Parabola And, x + y = 2 is a straight line Let A be point of intersection of line and parabola And, Point B is (2, 0) Finding point A Since x + y = 2 y = 2 − x Putting y = 2 − x in equation of parabola 𝑦^2=𝑥 (𝟐−𝒙)^𝟐=𝒙 4+𝑥^2−4𝑥=𝑥 𝑥^2−4𝑥−𝑥+4=0 𝒙^𝟐−𝟓𝒙+𝟒=𝟎 𝑥^2−4𝑥−𝑥+4=0 𝑥(𝑥−4)−1(𝑥−4)=0 (𝑥−4)(𝑥−1)=0 So, x = 4, x = 1 Since for point A, x-coordinate will be less than 2 ∴ x = 1 Putting x = 1 in equation of line x + y = 2 1 + y = 2 y = 2 − 1 y = 1 So, Coordinates of point A = (1, 1) Finding Area Area Required = Area OAD + Area ADB Area OAD Area OAD = ∫_0^1▒〖𝑦 𝑑𝑥" " 〗 y → Equation of parabola 𝑦^2 = x 𝑦 = ± √𝑥 Since OAD is in 1st quadrant, value of y is positive ∴ 𝒚 = √𝒙 Now, Area OAD = ∫_0^1▒〖𝑦 𝑑𝑥" " 〗 = ∫_𝟎^𝟏▒〖√𝒙 𝒅𝒙" " 〗 = [𝑥^(1/2 + 1)/(1/2 + 1)]_0^1 = [𝑥^(3/2)/(3/2)]_0^1 = 𝟐/𝟑 [𝒙^(𝟑/𝟐) ]_𝟎^𝟏 = 2/3 [1^(3/2)−0^(3/2) ] = 𝟐/𝟑 Area ADB Area ADB = ∫1_1^2▒〖𝑦 𝑑𝑥〗 y → Equation of line x + y = 2 y = 2 − x Therefore, Area ADB = ∫1_𝟏^𝟐▒(𝟐−𝒙)𝒅𝒙 = [2𝑥−𝑥^2/2]_1^2 = [(2(2)−2^2/2)−(2(1)−1^2/2)] = [2𝑥−𝑥^2/2]_1^2 = [(2(2)−2^2/2)−(2(1)−1^2/2)] = [(4−2)−(2−1/2)] = 2−3/2 = 𝟏/𝟐 Thus, Area Required = Area OAD + Area ADB = 2/3+1/2 = 𝟕/𝟔 square units