HSC +1 Mathematics-EM

Directorate of Government Examinations, Chennai – 600 006 Higher Secondary First Year Public Examination – May 2022 Mathematics Marking Scheme – English Medium GENERAL INSTRUCTIONS Maximum Marks : 90 1. The answers given in the marking scheme are Text book bound. 2. If a student has given any answer which is different from one given in the marking scheme, but carries the prescribed content meaning (rigorous) such answers should be given full credit with suitable distribution. 3. Follow the footnotes which are given under certain answer schemes. 4. If a particular stage is wrong and if the candidate writes the appropriate formula then award 1 mark for the formula (for the stage mark 2*). This mark (*) is attached with that stage. This is done with the aim that a student who did the problem correctly without writing the formula should not be penalized. 5. In the case of Part II, Part III and Part IV, if the solution is correct then award full mark directly. The stage mark is essential only if the part of the solution is incorrect. 6. Answers written only in Black or Blue ink should be evaluated. 1|Page

PART – I 1. One mark to write the correct option and the corresponding answer. 2. If one of them (answer or option) is wrong, then award zero mark only. CODE A CODE B Q.NO OPTION ANSWER Q.NO OPTION ANSWER 1. (d) discontinuous 1. (c) 2 3 2. (d) + is symmetric 2. (d) 1−2 3. (b) 3. (a) 2 − 8 + 16 + 4. (d) 5. (c) 18 4. (c) 6. (c) 10 7. (c) 5. (b) 8. (d) does not exist ∞ 6. (d) ∞ 9. (c) 2 7. (d) discontinuous 10. (a) 3− + + 11. (a) 0.56 8. (d) 5 12. (b) (or) (c) ec (− ) 9. (b) −2 ̂ − ̂ + 9 −2 ̂ − ̂ + 9 10. (a) (or) (c) ec (− ) 11. (c) 8 12. (d) 8 13. (a) 2 − 8 + 16 + 13. (c) does not exist 1 14. (d) 5 14. (d) 2 , −2 0.56 15. (c) 2 15. (a) 16. (d) 16. (c) 10 3 1 2 , −2 17. (d) '() − *) 17. (d) + is symmetric 18. (d) 8 18. (d) '() − *) + 19. (c) 8 19. (d) 18 20. (d) 1−2 20. (c) 2 3− + 2|Page



Q. NO PART – III MARKS 31. ;( ) = 4 CONTENT 2 ;<℘( )> = 2 = 16 1 1 32. ( + 3)( − 4) = + 3 + − 4 2 = @? + @? 1 2 A 1 33. This word has 13 letters 2 13! 2* 1 2! × 2! × 3! 2 @ 1 34. ( + 2) @ = 2 @ D1 + 2E 2* = 2 @ 5 5 +⋯H 1 F1 − 3 + 36 − 81 1 35. I = J 5(1) + 12(2) − 3 J 2 1 K(5) + (12) 1 1 = 2 1 2* 111 1 0 0 NO→NO NQ M− NR→NR NQ 4|Page 36. L 6 M L = L 6− L M− 6M 6− = ( − 6)(6 − M)(M − ) Note: One can do by different methods 2 37. cos = √50√101 = cos 5 S √2 T 5√101 38. 96 2 + 26 9 = 0 96 − 9 = 6 39. U( ) = 2 − 5 + = 3 U( ) = 2 − 5 + 3 40. = V2,4,6W, = V6W and ∩ = V6W Y( ⁄ ) = 1 3

Q. NO PART – IV MARKS 41.(a) CONTENT 3 The number of persons who knows only language A 1 = 5000 × [ 1 533 2 3 = 1950 2 Note : This problem can be done by property of cardinality 2 1 OR 1 41.(b) 9( + 4 + 6) 1 \\=] +4 +6 2* 1 = ^_`| + 4 + 6| + 1 1 Note: One can do this problem by substitution method 2 1 42.(a) cot cos cos (−cos )(−tan θ) (cosec ) = cot cos (cos ) Dsi_n θθE D si1n θE = cos cot OR 42.(b) d = , 9e = _ 9 e = f; ] _ 9 = f; − ] f; 9 = f; + _ + 43.(a) Y(1) is true. Assuming Y( ) is true. Y( + 1) is true. Remaining part. OR 5|Page

43.(b) 9 = *(1 − _ g) 2 44.(a) 9g 2 1 96 1 9g = * f;g 2* 2 96 f; g 9 = 1 − _ g 1 * = 12, ℎ = , ' = 2, 2` = 11, 2U = −5, =2 2 55 300 242 2 48 − 2 − 4 − 4 − 2 = 0 2 = −5 (or) = 1 OR 2 44.(b) lim f; 2 2 →3 _ 2 ( f; 5 ) 2 1 = lim f; 2 ×2 1+1+1 ×5 1 →3 2 1 f; 5 5 2 =5 45.(a) * ' *' k' *k × (−1) k * 'k *' '* *' −* −' − = k' *k × k * 'k *' ' * 2' − * 2 *−' ' = k * *k ' 2'* − OR 45.(b) ^_` 75 − ^_` 16 − 2 ^_` 5 + 2 ^_` 9 + ^_` 32 − ^_` 243 = ^_` 3 + ^_` 25 − ^_` 16 − ^_` 25 + ^_` 81 + ^_` 16 + ^_` 2 − ^_` 81 − ^_` 3 = ^_` 2 46.(a) = 7, l = 7 and l = √98 + l =l ∴ The given points form a right angled triangle. OR 6|Page

46.(b) cos 15° = √3 + 1 2 2√2 1 tan 120° + tan 45° 2 1 tan 165° = 1 − tan 120° tan 45° 2* 2* 1 − √3 = 2* 1+1 1 + √3 1 47.(a) 6 2 Y( ⁄ 5) = 10 , Y( ⁄ )=4 i) Y( ) = 11 3 ii) Y( 5⁄ ) = o 55 OR 47.(b) −√3 1 2 +26 = 2 5s p = 2 and r = 150° = 6 5s 5s cos 6 + 6 sin 6 = 2 7|Page