Question 33 - CBSE Class 12 Sample Paper for 2025 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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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