CBSE Class 12 Sample Paper for 2025 Boards
Question 2
Question 3
Question 4
Question 5 Important
Question 6
Question 7
Question 8 Important
Question 9 Important
Question 10 Important
Question 11
Question 12 Important
Question 13 Important
Question 14
Question 15 Important
Question 16 Important
Question 17
Question 18
Question 19 [Assertion Reasoning] Important
Question 20 [Assertion Reasoning] Important
Question 21 Important
Question 22
Question 23 (A)
Question 23 (B)
Question 24 (A)
Question 24 (B) Important
Question 25 Important
Question 26 Important
Question 27 Important
Question 28 (A)
Question 28 (B)
Question 29 (A) Important
Question 29 (B)
Question 30 Important
Question 31 (A) Important
Question 31 (B)
Question 32 Important
Question 33 Important
Question 34 (A)
Question 34 (B)
Question 35 (A)
Question 35 (B) Important
Question 36 (i) [Case Based]
Question 36 (ii)
Question 36 (iii) (A) Important
Question 36 (iii) (B) Important
Question 37 (i) [Case Based]
Question 37 (ii) Important
Question 37 (iii) (A) Important
Question 37 (iii) (B) You are here
Question 38 (i) [Case Based] Important
Question 38 (ii) Important
CBSE Class 12 Sample Paper for 2025 Boards
Last updated at Dec. 13, 2024 by Teachoo
Question 37 (iii) (B) If the track of the final race (for the biker 𝑏_1) follows the curve 𝑥^2=4𝑦; ( where 0≤𝑥≤20√2. & 0≤𝑦≤200), then state whether the track represents a one-one and onto function or not. (Justify).Given 𝑥^2=4𝑦 𝑦=𝑥^2/4 Let 𝑦=𝒇(𝒙)=𝒙^𝟐/𝟒 In [0, 20√2]→[0, 200] 𝒇(𝒙)=𝒙^𝟐/𝟒 Checking one-one 𝒇(𝒙_𝟏 )=〖𝒙_𝟏〗^𝟐/𝟒 𝒇(𝒙_𝟐 )=〖𝒙_𝟐〗^𝟐/𝟒 Putting 𝒇(𝒙_𝟏 )= 𝒇(𝒙_𝟐 ) 〖𝒙_𝟏〗^𝟐/𝟒= 〖𝒙_𝟐〗^𝟐/𝟒 〖𝒙_𝟏〗^𝟐 = 〖𝒙_𝟐〗^𝟐 x1 = x2 or x1 = –x2 But x1 = –x2 is not possible in our domain [0, 20√2] Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 So, x1 = x2 when f (x1) = f (x2) ∴ f(x) is one-one Check onto 𝒇(𝒙)=𝒙^𝟐/𝟒 Let f(x) = y , such that y ∈ R 𝑥^2/4=𝑦 𝑥^2=4𝑦 𝑥=±2√𝑦 Since 0≤𝑦≤200 and where 0≤𝑥≤20√2 For all values of y in [0, 200], there is a corresponding value of x in [0, 20√2] Thus, for every y in [0, 200], there exists a pre-image in [0, 20√2] ∴ f(x) is onto Thus, for every y in [0, 200], there exists a pre-image in [0, 20√2] ∴ f(x) is onto Let f(x) = y , such that y ∈ R x2 = y x = ±√𝑦 Note that y is a real number, so it can be negative also Putting y = −3 x = ±√((−3)) 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑎𝑠 𝑟𝑜𝑜𝑡 𝑜𝑓 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠 𝑛𝑜𝑡 𝑟𝑒𝑎𝑙 Hence, x is not real So, f is not onto