XI Mathematics Mid-Term Examination 2023-24 PDF

No. of pages - 18 (M) MARKING SCHEME MID-TERM EXAMINATION (2023-24) CLASS : XI SUBJECT: MATHEMATICS (041) Time Allowed...

No. of pages - 18 (M) MARKING SCHEME MID-TERM EXAMINATION (2023-24) CLASS : XI SUBJECT: MATHEMATICS (041) Time Allowed : 3 hours Maximum Marks : 80 GENERAL INSTRUCTIONS: 1. Evaluation is to be done as per instructions provided in the marking scheme. Marking scheme should be strictly adhered to and religiously followed. However, while evaluating, answers which are based on latest information or knowledge and/or are innovative they may be assessed for their correctness otherwise and marks to be absorbed to them. 2. If a student has attempted on extra question, answer of the question deserving more marks should be retained and other answer scored out. 3. A full scale of marks (0-80) has to be used. Please do not hesitate to award full marks if the answer deserves it. ************ SECTION-A 1. (b) (,0] 1 2. (d) R – {4, –4} 1 3. (d) 0 1 4. (b) (0, 1) 1 5. (a) 3 1 6. (c) a2 + b2 = c2 + d2 1 1 XI-MATH-M 7. (b) [3, 5) 1 8. (d) 5 1 9. (b) {2, 3, 4, 5} 1 10. (b) (, ) 1 11. (b) [2, 4) 1 12. (a) {(1, 3)} 1 13. (a) A 1 14. (c) 3 1 15. (d) {1, 2} 1 16. (d) 16 1 17. (b) 3 1 18- (d) 7 1 19- (c) Assertion (A) is true, but Reason (R) is false 1 20. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A) 1 2 XI-MATH-M  SECTION-B 21. Let Z = – 3 + 4i Z  3  4i, | Z | 9  16  5 1 Z 3 4 Multiplicative inverse of Z  2   i 1 | Z| 25 25 2023 22. yx NOTE : 2023= 1 × 7 × 17 × 17 ½ x As factor of 2023 are 1, 7, 17, 119, 289, 2023 So, when x = 1, y = 2024 when x = 7, y = 296 when x = 17, y = 136 when x = 119, y = 136 when x = 289, y = 296 when x = 2023, y = 2024 Thus, range of R = {136, 296, 2024} 1½ 5  3x 23. As, 5   8  10  5  3x  16 ½ 2  15  3x  11 ½ 3 XI-MATH-M  11  5   x  ½ 3 11 5 x  ½ 3  11   x ,5 ½  3  OR As, | 7x  3 | 4  4  7x  3  4 1  1  7x  7 ½ 1    x 1 7  1   x   ,1 ½ 7  24. 0.5, 0.9, 1.3, ….4.5 is an A.P. with common difference 0.4. Such that an = a + (n – 1) d = 0.5 + (n – 1) 0.4 = 0.4n + 0.1 1 Further, when 0.4n + 0.1 = 4.5  n = 11, so A  {x : x  0.4n  0.1, n  N, n  11} 1 OR 1, 2, 3, …. 2023 is an AP with common difference 1 (or we can say its natural number ½) Numerator = n 4 XI-MATH-M  Denominator = n +1  n  A  x : x  , n  N, n  2023 1½  n 1  25. As, x 3  x  x(x 2  1)  0  x(x  1)(x  1)  0 1  x  0,1, 1 So, A = {0, 1, –1} 1 SECTION-C 26. Signum function is defined as 1, x  0  Sgn(x)  0, x  0 1 1, x  0  Graph of signum function 1 Domain = R = (, ) ½ Range = {–1, 0, 1} ½ 5 XI-MATH-M 27. x + iy = (u + iv)1/3 = u + iv = (x + iy)3 ½ u + iv = x3 + i3 y3 + 3x2 (iy) + 3x (iy)2 u + iv = x3 – iy3 + 3x2yi – 3xy2 u + iv = (x3 – 3xy2) + i(3x2y – y3) 1½ On comparing the real part and imaginary part, we get u = x3 – 3xy2, v = 3x2 y – y3 u v x(x 2  3y 2 ) y(3x 2  y 2 ) Thus,     4(x 2  y 2 ) 1 x y x y 28. z1 = 2 – i, z2 = 1 + i z1 + z2 + 1 = 2 – i + 1 + 1 = 4 + 0i 1 z1 – z2 + i = 2 – i – 1 – i + i = 1 – i 1 z1  z 2  1 4  oi | 4  oi | 16  0 16 Now,     2 2 1 z1  z 2  1 1 i |1  i | 11 2 29. Let the amount of water added be ‘x’ litres, so, according to the problem. ½ Case I : 45 0 25 (1125)  x  (1125  x) 100 100 100 6 XI-MATH-M   45(1125)  25(1125  x)  20(1125)  25x  x  900 1 Case II : 45 0 30 (1125)  (x)  (1125  x) 100 100 100  45(1125)  30(1125  x)  15(1125)  30x  562.5  x 1  562.5  x  900 Therefore, the number of litres of water that has to be added will have to more than 562.5 litres but less than 900 litres. ½ OR As  5 (2x – 7) – 3 (2x + 3)  0  10x – 35 – 6x – 9  0  4x  44 x  11 1 and 7 XI-MATH-M  2x + 19  6x + 47  2x – 6x  47 – 19  x  7 1½ Thus, x  [7,11] is required solution for the given system of inequalities. ½ 30. Subsets of given set are : {} {}, {{}}, {{{}}}, {, {}}, {, {{}}} {{}, {{}}} 2 {, {}, {{}}} ½ OR A = {2, 4, 6, 8}, B = {2, 3, 5, 7}, U = {1, 2,3, 4, 5, 6, 7, 8, 9} (i) A  B  {2,3, 4,5,6,7,8} (A  B)  {1,9}...(1) A  B  {1,3,5,7,9}  {1, 4,6,8,9}  {1,9}...(2) From (1) and (2), we get (A  B)  A  B 1½ 8 XI-MATH-M  (ii) A  B  {2} (A  B)  {1,3, 4,5,6,7,8,9}...(3) A  B  {1,3,5,7,9}  {1,4,6,8,9}  {1,3, 4,5,6,7,8,9}...(4) From equation (3) and (4) we get (A  B)  A  B 1½ 31. (cos x – cos y)2 + (sin x – sin y)2 2 2  xy xy   xy xy    2sin   sin      2cos   sin    2   2   2   2   2       x  y 2 x  y  2  x  y  2  x  y   4sin 2   sin    4  cos   sin    2   2    2   2   x  y 2 x  y  2  x  y  2  x  y   4sin 2   sin    4  cos    sin    2   2    2   2   x  y  2  x  y  2  x  y   4sin 2   sin    cos    2   2   2  xy  4sin 2   1  2  xy  4sin 2   1  2  OR 9 XI-MATH-M  x 1  cos x We know that tan  1 2 1  cos x  Put x  4 1 1  2  2 1 tan  1 8 1 2 1 1 2 ( 2  1) 2 2 1   ( 2  1)( 2  1) 1   tan  2 1 1 8 SECTION-D 1 32. (a) f (x)  y 1  2cos x f(x) is defined for all values of x except when 1 – 2 cos x =0 1   cos  x 2 3    So, domain = R    or R  2n   1 3  3 10 XI-MATH-M  Note: Any one of the answer is acceptable as general solution is deleted As 1  cos x  1  2  2cos x  2  1  1  2cos  3 1  1  3 y 1 1 1   0 and 0   3 y y 1 y  1 y 3 1 y  1 and y 3 1 Range  (, 1]  [ , ) 1½ 3 (b) f (x)  y  9  x 2 f(x) is defined for 9  x 2  0  (x  3)(x  3)  0  Domain  [3,3] As y  9  x 2 , y  0  y2  9  x 2  x  9  y2 11 XI-MATH-M  So, 9  y 2  0  (y  3)(y  3)  0 3  y  3, but y  0 Thus, y  [0,3]  Range  [0,3] 1½ n n! 33. Cr 1  36   36...(1) (r  1)!(n  r  1)! n n! Cr  84   84...(2) r!(n  r)! n n! Cr 1  126   126...(3) 1½ (r  1)!(n  r  1)! n  r 1 7 On (2) ÷ (1), we get   10r  3n  3 1 r 3 nr 3 On (3) ÷ (2), we get   2n  5r  3 1 r 1 2 Thus, 20r – 6 = 15r + 9  r  3 ½ So, r C2 3 C2  3 1 OR 12 XI-MATH-M  (a) Let us consider all girls as a group, so (G1, G2, G3, G3, G5), B1, B2, B3, B4, B5 can be arranged in 6! × 5! = 720 × 120 = 86400 ways. 1 (b) No. of ways in which all girls never sit together = Total no. of ways – No. of ways in all girls sit together ½ = 10! – 6! × 5! = 6! [10 × 9 × 8 × 7 – 120] ½ = 6! × 4920 = 3542400 ways 1 (c) No. of ways in which no girls sit together = 6! × 5! = 8640 ways 1 (Means girls will sit between boys) (d) Boys and girls alternate, so B1, G1, B2, G2, B3, G3, B4, G4, B5, G5 = 5! × 5! OR G1, B2, G2, B2, G2, B3, G4, B4, G5, B5 = 5! × 5! No. of ways = (5! × 5!) × 2 = 2 × (5!)2 = 28800 ways 1 4 16 26 34. As, tan x =  sec2 x  1  tan 2 x  1   3 9 9 5 5 sec x = ± but x lies in II quadrant, so, sec x = 3 3 3 cos x  1 5 13 XI-MATH-M  x 3 x Now, cos x  2cos 2  1  1   2cos 2 2 5 2 x 1 cos 2  2 5 x 1 x cos .  (45,90) 1 2 5 2 x x Note : 90  x  180  45  90  lies in I quadrant 1 2 2 x 3 x x 4 cos x  1  2sin 2  1   2sin 2  sin 2  2 5 2 2 5 x 2 sin  1 2 5 x sin x / 2 Thus, tan  2 1 2 cos x / 2 OR 7    As, cos  cos       cos 8  8 8 5  3  3 cos  cos       cos 1 8  8  8    3  5  3  So, 1  cos 1  cos 1  cos 1  cos   8  8  8  8     3  3    1  cos 1  cos 1  cos 1  cos   8  8  8  8 14 XI-MATH-M     3   1  cos 2 1  cos 2  1  8  8   2   2    sin  sin   8  8     3   1  cos 4  1  cos 4     2  2  2      1  1  1  1   1   1  2  2  2 1 1  2   2  4 8       3  5   1  1  cos 1  cos 1  cos 1  cos    8  8  8  8 8 35. As 5x = 4x + x tan 4x  tan x tan 5x  tan(4x  x)  1½ 1  tan 4x.tan x  tan 5x(1  tan 4x.tan x)  tan 4x  tan x  tan 5x  tan 5x tan 4x tan x  tan 4x  tan x  tan 5x  tan 4x  tan x  tan 5x tan 4x tan x...(A) 1½ Put x  10 in eq.(A), we get tan 50  tan 40  tan10  tan 50 tan 40 tan10 1  cot 40 tan 40 tan10 15 XI-MATH-M  tan 50  tan 40  tan10  tan10  tan 50  tan 40  2 tan10 tan 50  1 1 tan 41  2 tan10 SECTION-E 36. From the figure, PS 1 QR 1 tan 1   and tan 2   SU 2 RV 3 1 1 5 (i) tan 1  tan 2    1 2 3 6 1 1 1 (ii) tan 1  tan 2 .  1 2 3 6 5 5 tan 1  tan 2 (iii) tan(1  2 )   6  6 1 2 1  tan 1.tan 2 1  1 5 6 8 OR  As, tan(1  2 )  1  tan 4   1  2  2 4 16 XI-MATH-M 37. (i) As, 2m  2p  112  112  2p  2m  2m (2pm  1)  24 (23  1)  2m (2pm  1)  m  4 and p  m  3  p  7 2 OR x  3  4  x  1 and y  3  7  y  4 2 (ii) n()  x  3  y  10  18 1 (iii) n(A  B)  x  y  10  15 1 38. (a) For selecting atleast one boy and one girls, there are following cases: Boys Girls Total I 1 4 5 II 2 3 5 III 3 2 5 IV 4 1 5 Thus number of always are = 7 C1 4 C4  7 C2 4 C3  7 C3 4 C2  7 C4 4 C1 1  7  1  21  4  35  6  35  4  7  84  210  140 =441 ways 1 17 XI-MATH-M (b) For atleast 3 girls, there are two possibility (3G and 2 boys) or (4G and 1 boy) 1 Thus number of ways = 4 C3 7 C2  4 C4 7 C1 = 21 × 4 + 1 × 7 = 91 1 18 XI-MATH-M