Title E-Mathematics-Basic-EM10_001-056.pdf ✅

2 MATHEMATICS (BASIC) [018] SOLUTION OF QUESTION PAPER 2 Section A 1. C. 9 7 2. C. 6 3. C. 25 4. D. (2, 2) 5. B. cos2 q 6. B. 2 7. 2 8. parabola opening upwards 9. 7 3 10. 90° 11. 10.6 12. n 2 +1 13. False 14. False 15. True 16. True 17. 10 18. 5 cm 19. 0.6 20. 35 21. (c) 4p r 2 22. (b) 2p rh 23. (b) 2 1 rl 24. (c) 2p r Section B 25. 4s2 – 4s + 1 = 4s2 – 2s – 2s + 1 = 2s (2s – 1) – 1 (2s – 1) = (2s – 1) (2s – 1) So, the value of 4s2 – 4s + 1 is zero, when 2s – 1 = 0 or 2s – 1 = 0, i.e., when s = 2 1 or s = 2 1 . Hence, the zeroes of polynomial 4s2 – 4s + 1 are 2 1 and 2 1 (both equal). Now, Sum of zeroes = 2 1 + 2 1 = 1 = – (– ) 4 4 = ( ) Coefficient of Coefficient of s – s 2 and Product of zeroes = 2 1 × 2 1 = 4 1 = Coefficient of s Constant term 2 26. Let the quadratic polynomial be ax2 + bx + c and its zeroes be a and b. Then, as per given, a + b = 5 = a – b and ab = 4 3 = a c . If a = 4, then b = – 4 5 and c = 3. So, one quadratic polynomial satisfying the given condition is 4x2 – 4 5 x + 3. In general, a quadratic polynomial satisfying the given condition is k (4x2 – 4 5 x + 3), where k is a non-zero real number.
Online Answer Keys – MATHEMATICS (BASIC) 3 27. 6x2 – x – 2 = 0 \ 6x2 – 4x + 3x – 2 = 0 \ 2x (3x – 2) + 1 (3x – 2) = 0 \ (3x – 2) (2x + 1) = 0 \ 3x – 2 = 0 or 2x + 1 = 0 \ x = 3 2 or x = – 2 1 . Thus, the roots of the given quadratic equations are 3 2 and – 2 1 . 28. We have a2 – a1 = 11 – 5 = 6, a3 – a2 = 17 – 11 = 6, a4 – a3 = 23 – 17 = 6 As ak + 1 – ak is the same for k = 1, 2, 3, etc. the given list of numbers is an AP. Now, a = 5 and d = 6. Let 301 be a term, say, the nth term of this AP. We know that an = a + (n – 1) d So, 301 = 5 + (n – 1) × 6 i.e., 301 = 6n – 1 So, n = 6 302 3 151 = But n should be a positive integer as it is the number of a term. So, 301 is not a term of the given list of numbers. 29. For the given AP – 37, – 33, – 29, ..., a = – 37, d = (– 33) – (– 37) = 4, n = 12 and Sn is to be found. Sn = n 2 [2a + (n – 1) d] \ S12 = 2 12 [– 74 + (12 – 1) 4] = 6 (– 30) = – 180 Thus, the sum of first 12 terms of the given AP is – 180. 30. Let P (a, b) and Q ( – a, – b) be the given points. Then, PQ = ( )( ) aa bb + ++ 2 2 = 4 4 a b 2 2 + = 2 a b 2 2 + Thus, the distance between the given points is 2 a b 2 2 + . 31. Let P (x, y) be the required point. Using the section formula, we get x = () () 3 1 38 14 + + = 7, y = () ( ) 3 1 35 1 3 – + + = 3 Therefore, (7, 3) is the required point. 32. Let us consider right triangle ABC, right angled at B. Then, 15 cot A = 8 \ cot A = 15 8 \ BC AB 15 8 =
4 GALA MODEL QUESTION PAPERS : STANDARD 10 (2025) Hence, if AB = 8k, then BC = 15k, where k is some positive number. Now, in D ABC, ) B = 90° \ By Pythagoras theorem, AC = AB BC 2 2 + = () ( ) 8 15 k k 2 2 + = 64 225 k k 2 2 + = 17k Then, sin A = AC BC k k 17 15 17 15 = = and sec A = 7 AB AC k k 8 17 8 1 = = . 33. sin 30° cos 45° + cos 30° sin 45° = 2 1 × 2 1 + 2 3 × 2 1 = 2 2 1 + 2 2 3 = 2 2 3 1 + = ( ) 2 2 3 1 + × 2 2 = 4 6 2 + 34. Here, AB is the pole and AC is the rope. Then, in D ABC, ) B = 90°, ) C = 30° and AC = 20 m. Then, sin C = AC AB \ sin 30° = AB 20 \ AB 2 1 20 = \ AB = 10 Thus, the length (height) of the pole is 10 m. 35. For the cone as well as the hemisphere, radius r = 3.5 cm = 2 7 cm. Lateral height of the cone l = 12.5 cm. Total surface area of the toy = CSA of the cone + CSA of the hemisphere = p rl + 2p r 2 = p r (l + 2r) = 7 22 × 2 7 12 5 2 . 2 7 b l + # cm2 = 11 × 19.5 cm2 = 214.5 cm2 Thus, the total surface area of the toy is 214.5 cm2. 36. The radius of the cylindrical part as well as conical part = r = diameter 2 = . 2 4 2 m = 2.1 m, height of the cylindrical part = h = 8 m and height of the conical part = H = 6 m. Volume of air in the tent = Volume of the tent = Volume of cylindrical part + Volume of conical part