Question 33 - CBSE Class 12 Sample Paper for 2025 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Last updated at Sept. 29, 2024 by Teachoo
CBSE Class 12 Sample Paper for 2025 Boards
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CBSE Class 12 Sample Paper for 2025 Boards
Last updated at Sept. 29, 2024 by Teachoo
Question 33 The equation of the path traversed by the ball headed by the footballer is 𝑦=𝑎𝑥^2+𝑏𝑥+𝑐;( where 0≤𝑥≤14 and 𝑎,𝑏,𝑐∈𝑅 and 𝑎≠0) with respect to a XY-coordinate system in the vertical plane. The ball passes through the points (𝟐,𝟏𝟓),(𝟒,𝟐𝟓) and (𝟏𝟒,𝟏𝟓). Determine the values of a,b and c by solving the system of linear equations in a,b and c , using matrix method. Also find the equation of the path traversed by the ball.Given 𝑦=𝑎𝑥^2+𝑏𝑥+𝑐 Since ball passes through (𝟐,𝟏𝟓),(𝟒,𝟐𝟓) and (𝟏𝟒,𝟏𝟓). These 3 points will satisfy our equation So, putting these points in the equation Ball passes through (𝟐,𝟏𝟓) Putting x = 2, y = 15 in equation 𝑦=𝑎𝑥^2+𝑏𝑥+𝑐 15=𝑎(2)^2+𝑏(2)+𝑐 15=4𝑎+2𝑏+𝑐 𝟒𝒂+𝟐𝒃+𝒄=𝟏𝟓 Ball passes through (𝟒,𝟐𝟓) Putting x = 4, y = 25 in equation 𝑦=𝑎𝑥^2+𝑏𝑥+𝑐 25=𝑎(4)^2+𝑏(4)+𝑐 25=16𝑎+4𝑏+𝑐 𝟏𝟔𝒂+𝟒𝒃+𝒄=𝟐𝟓 Ball passes through (𝟏𝟒,𝟏𝟓) Putting x = 14, y = 15 in equation 𝑦=𝑎𝑥^2+𝑏𝑥+𝑐 15=𝑎(14)^2+𝑏(14)+𝑐 15=196𝑎+14𝑏+𝑐 𝟏𝟗𝟔𝒂+𝟏𝟒𝒃+𝒄=𝟏𝟓 Thus, our 3 equations are 4𝑎+2𝑏+𝑐=15 16𝑎+4𝑏+𝑐=25 196𝑎+14𝑏+𝑐=15 Writing equation as AX = B [■8(4&2&1@16&4&1@196&14&1)] [■8(𝑎@𝑏@𝑐)] = [■8(15@25@15)] Hence A = [■8(4&2&1@16&4&1@196&14&1)] , X = [■8(𝑎@𝑏@𝑐)] & B = [■8(15@25@15)] Calculating |A| |A| = |■8(4&2&1@16&4&1@196&14&1)| = 4 |■8(4&1@14&1)| – 2 |■8(16&1@196&1)| + 1 |■8(16&4@196&14)| = 4 (4 – 14) – 2 (16 – 196) + 1 (224 – 784) = 4 (–10) – 2 (–180) + 1 (–560) = –40 + 360 – 560 = –240 ∴ |A|≠ 0 So, the system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 Now, A-1 = 1/(|A|) adj (A) adj (A) = [■8(A11&A12&A13@A21&A22&A23@A31&A32&A33)]^′ = [■8(A11&A21&A31@A12&A22&A32@A13&A23&A33)] A = [■8(4&2&1@16&4&1@196&14&1)] M11 = |■8(4&1@14&1)| = 4 – 14 = –10 M12 = |■8(16&1@196&1)| = (16 – 196) = –180 M13 = |■8(16&4@196&14)| = 224 – 784 = –560 M21 = |■8(2&1@14&1)| = 2 – 14 = –12 M22 = |■8(4&1@196&1)| = 4 – 196 = –192 M23 = |■8(4&2@196&14)| = 56 – 392 = –336 M31 = |■8(2&1@4&1)| = 2 – 4 = –2 M32 = |■8(4&1@16&1)| = 4 – 16 = –12 M33 = |■8(4&2@16&4)| = 16 – 32 = –16 Now, A11 = 〖"(–1)" 〗^(1+1) M11 = (–1)2 . (–10) = –10 A12 = 〖"(–1)" 〗^"1+2" M12 = 〖"(–1)" 〗^3 . (–180) = 180 A13 = 〖(−1)〗^(1+3) M13 = 〖(−1)〗^4 . (–560) = –560 A21 = 〖(−1)〗^(2+1) M21 = 〖(−1)〗^3 . (–12) = 12 A22 = 〖(−1)〗^(2+2) M22 = (–1)4 . (–192) = –192 A23 = 〖(−1)〗^(2+3). M23 = 〖(−1)〗^5. (–336) = 336 A31 = 〖(−1)〗^(3+1). M31 = 〖(−1)〗^4 . (–2) = –2 A32 = 〖(−1)〗^(3+2) . M32 = 〖(−1)〗^5. (–12) = 12 A33 = 〖(−1)〗^(3+3) . M33 = (–1)6 . (–16) = –16 Thus, adj A = [■8(−10&12&−2@180&−192&12@−560&336&−16)] Now, A-1 = 𝟏/(|𝐀|) adj A A-1 = 1/(−240) [■8(−10&12&−2@180&−192&12@−560&336&−16)] A-1 = (−1)/240 [■8(−10&12&−2@180&−192&12@−560&336&−16)] A-1 = 𝟏/𝟐𝟒𝟎 [■8(𝟏𝟎&−𝟏𝟐&𝟐@−𝟏𝟖𝟎&𝟏𝟗𝟐&−𝟏𝟐@𝟓𝟔𝟎&−𝟑𝟑𝟔&𝟏𝟔)] Also, X = A−1 B Putting Values [■8(𝑎@𝑏@𝑐)] = 1/240 [■8(10&−12&2@−180&192&−12@560&−336&16)] [■8(15@25@15)] [■8(𝑎@𝑏@𝑐)] = 1/240 [■8(10&−12&2@−180&192&−12@560&−336&16)] × 5[■8(3@5@3)] [■8(𝑎@𝑏@𝑐)] = 5/240 [■8(10&−12&2@−180&192&−12@560&−336&16)] [■8(3@5@3)] [■8(𝑎@𝑏@𝑐)] = 1/48 [■8(10&−12&2@−180&192&−12@560&−336&16)] [■8(3@5@3)] [■8(𝑎@𝑏@𝑐)] = 1/48 [■8(10(3)−12(5)+2(3)@−180(3)+192(5)+(−12) (3)@560(3)+(−336) (5)+16(3))] [■8(𝑎@𝑏@𝑐)] = 1/48 [■8(30−60+6@−540+960−36@1680−1680+48)] [■8(𝑎@𝑏@𝑐)] = 1/48 [■8(−24@384@48)] [■8(𝑎@𝑏@𝑐)] = [■8((−24)/48@384/48@48/48)] [■8(𝒂@𝒃@𝒄)] = [■8((−𝟏)/𝟐@𝟖@𝟏)] Thus, a = (−𝟏)/𝟐 , b = 8 & c = 1 We also need to find the equation of the path traversed by the ball Thus, 𝑦=𝑎𝑥^2+𝑏𝑥+𝑐 𝒚=−𝟏/𝟐 𝒙^𝟐+𝟖𝒙+𝟏 [■8(𝒂@𝒃@𝒄)] = [■8((−𝟏)/𝟐@𝟖@𝟏)] Thus, a = (−𝟏)/𝟐 , b = 8 & c = 1 We also need to find the equation of the path traversed by the ball Thus, 𝑦=𝑎𝑥^2+𝑏𝑥+𝑐 𝒚=−𝟏/𝟐 𝒙^𝟐+𝟖𝒙+𝟏 Putting u = 𝟏/𝒙 1/2 = 1/𝑥 x = 2 Putting v = 𝟏/𝒚 1/3 = 1/𝑦 y = 3 Putting w = 𝟏/𝒛 1/5 = 1/𝑧 z = 5