Bihar Board 12th Question Paper 2023 Maths (Science, Arts)

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BIHAR • EXAINATION BOARD • CLASS 12 •
QUESTION PAPER • PDF • DOWNLOAD

12TH INTER
CLASS Examination

Previous Year
YEAR
2023 QUESTION PAPER
View For Free. Download PDF Online at

INTERMEDIATE EXAMINATION – 2023 (ANNUAL)
Mathematics (ELECTIVE)
xf.kr ¼,sfPNd½ Subject Code:- 121/327
I.Sc. & I.A.

Total no. of Questions : 100+30+8 = 138 Full Marks – 100

Time: 3 Hours 15 Minutes

Instructions for the candidates :

1- ijh{kkFkhZ OMR mÙkj i=d ij viuk iz’u iqfLrdk Øekad ¼10 vadksa dk½ vo’;
fy[ksaA
Candidate must enter his/her Question Booklet Serial No. (10
digits) in the OMR Answer Sheet.

2- ijh{kkFkhZ ;FkklaHko vius 'kCnksa esa gh mÙkj nsaA
Candidates are required to give their answers in own words as far
as practicable.

3- nkfguh vksj gkf’k, ij fn;s gq, vad iw.kkZad fufnZ"V djrs gSaA
Figures in the right hand margin indicate full marks.

4- iz’uksa dks /;kuiwoZd i<+us ds fy, ijh{kkfFkZ;ksa dks 15 feuV dk vfrfjDr le;
fn;k x;k gSA

15 minutes of extra time has been allotted for the candidates to
read the questions carefully.

1

5- ;g iz’u iqfLrdk nks [k.Mksa esa gS & ,oa A
This question booklet is divided into two sections – Section-A and
Section-B.
6- [k.M&v esa 100 oLrqfu"B iz’u gSa] ftuesa ls fdUgha 50 iz’uksa dk mÙkj nsuk
vfuok;Z gS ¼izR;sd ds fy, 1 vad fu/kkZfjr gS½A 50 ls vf/kd iz’uksa ds mÙkj nsus
ij izFke 50 dk gh ewY;kadu dEI;wVj }kjk fd;k tk,xkA lgh mÙkj dks
miyC/k djk, x;s OMR mÙkj i=d esa fn, x, lgh fodYi dks uhys@dkys
ckWy isu ls izxk<+ djsaA fdlh Hkh izdkj ds âkbVuj @ rjy inkFkZ @ CysM @
uk[kwu vkfn dk OMR mÙkj i=d esa iz;ksx djuk euk gS] vU;Fkk ifj.kke
vekU; gksxkA
In Section-A, there are 100 objective type questions, out of which
any 50 questions are to be answered (each carrying 1 mark). First
50 answers will be evaluated by the computer in case more than
50 questions are answered. For answering these darken the circle
with blue / black ball pen against the correct option on OMR
Answer Sheet provided to you. Do not use Whitener / liquid / blade
/ nail etc. on OMR-sheet, otherwise the result will be treated
invalid.

7- [k.M&c esa 30 y?kq mÙkjh; iz’u gSa] ftuesa ls fdUgha 15 iz’uksa dk mÙkj nsuk
vfuok;Z gS ¼izR;sd ds fy, 2 vad fu/kkZfjr gS½A buds vfrfjDr] bl [k.M esa 8
nh?kZ mÙkjh; iz’u fn;s x;s gSa½] ftuesa ls fdUgha 4 iz’uksa dk mÙkj nsuk gS ¼izR;sd
ds fy, 5 vad fu/kkZfjr gSA
In Section-B, there are 30 short answer type questions, out of

which any 15 questions are to be answered (each carrying 2
marks). Apart from this, there are 8 long answer type questions,

2

out of which any 4 questions are to be answered (each carrying 5
marks).

8- fdlh izdkj ds bysDVªkWfud midj.k dk iz;ksx iw.kZr;k oftZr gSA
Use of any electronic appliances is strictly prohibited.

3

[k.M & v @ Section - A
oLrqfu"B iz’u @ Objective Type Questions

iz’u la[;k 1 ls 100 rd ds izR;sd iz’u ds lkFk pkj fodYi fn, x, gSa ftuesa ls ,d
lgh gSA fdUgha 50 iz’uksa ds mÙkj nsaA vius }kjk pqus x, lgh fodYi dks OMR 'khV ij
fpfUgr djsaA 50x1=50
Question nos 1 to 100 have four options, out of which only one is correct.
Answer any 50 questions. You have to mark your selected option on the
OMR-sheet. 50x1=50

2 0 3 5
1. =
0 2 7 1

6 0 3 10
(A) (B)
0 2 14 1

5 5 6 10
(c) (D)
7 3 14 2

4 9 1
2. =
6 3 7

4 63 4 6
(A) (B)
6 21 63 21
67
(C) (D) [67 27]
27

3. ∫ 𝑐𝑜𝑠𝑥𝑑𝑥 =

(A) 0 (B) 1

(C) -2 (D) 3

4. ∫ 𝑐𝑜𝑠 𝜃 sec𝜃d𝜃 =

(A) 𝜃 + C (B) cos𝜃 + C
4

(C) sin𝜃 + C (D) c – tan𝜃

5. ∫(𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝑐𝑜𝑡 𝑥 − 2𝑐𝑜𝑠𝑒𝑐 𝑥𝑐𝑜𝑡 𝑥)dx =

(A) cosecx + cotx + k (B) cosecx – cotx + K

(C) x + K (D) K – x

6. ∫(2𝑠𝑖𝑛𝑥 − 3𝑐𝑜𝑠𝑥)𝑑𝑥 =

(A) K + 2cos𝑥 + 3sin𝑥 (B) K – 2sin𝑥 - 3cos𝑥

(C) K – 2cos𝑥 – 3sin𝑥 (D) K – 2cos𝑥 + 3sin𝑥

7. ∫ d𝑥 =

(A) 2log|𝑠𝑒𝑐𝑥| + K (B) log|𝑠𝑒𝑐𝑥| + K

(C) log|𝑡𝑎𝑛𝑥| + K (D) 2log|𝑡𝑎𝑛𝑥| + K

8. 7∫ dx =

(A) 7log|𝑥 − 49| + K (B) log|𝑥 − 49| + K

(C) log|𝑥 − 49| + K (D) log|𝑥 + 49| + K

9. ∫ =

(A) tan-1( ) + K (B) sin-1( ) + K

(C) tan-1( ) + K (D) cos-1( ) + K

½
10. ∫ ½ log( )dx =

(A) 2log2 (B) 2log3

5

(C) 3log2 (D) 0

11. ∫ 𝑥𝑑𝑥 =

(A) (B)

(C) (D)

12. ∫ 𝑠𝑖𝑛𝑥 𝑐𝑜𝑠 𝑥d𝑥 =

(A) 0 (B) 1

(C) 2 (D) 3

13. ∫ 𝑥 𝑐𝑜𝑠 𝑥𝑑𝑥 =

(A) (B)

(C) O (D) 1

14. ∫ √𝑥 𝑑𝑥 =

(A) 𝑥 +K (B) 𝑥 +K

(C) 𝑥 +K (D) 𝑥 +K

15. ∫ d𝑥 =

(A) log|𝑥 − 9| + K (B) log|𝑥 − 3| + K

(C) log +K (D) log|𝑥 + 3| + K

16. ∫√ =

6

(A) cos-14x + K (B) cos-14x + K

(C) sin-14x + K (D) 4sin-14x + K

17. ∫ 𝑠𝑒𝑐4𝑥. 𝑡𝑎𝑛4𝑥 𝑑𝑥 =

(A) sec4x + K (B) tan4x + K

(C) 4sec4x + K (D) sec4x + K

18. ∫ 𝑠𝑒𝑐 6𝑥 𝑑𝑥 =

(A) 6tan6x + K (B) tan6x + K

(C) tan6x + K (D) 3tan6x + K

19. 𝚥⃗.( 𝚤⃗+𝑘⃗ ) =

(A) 0 (B) 1

(C) 2 (D) -1

20. 5∫ =

(A) tan-15𝑥 + K (B) 5tan-15𝑥 + K

(C) tan-15𝑥 + K (D sin-15𝑥 + K

21. ∫ 𝑠𝑖𝑛 d𝑥 =

(A) K - cos (B) K + cos

(C) K - cos (D) K+ cos

22. ∫ 𝑐𝑜𝑠 d𝑥 =

7

(A) sin +K (B) sin +K

(C) sin +K (D) sin +K

23. ∫ 𝑠𝑒𝑐 dx =

(A) tan +K (B) tan +K

(C) tan +K (D) K - tan

24. ∫ 11 dx =

(A) +K (B) +K

(C) 11 logx + K (D) 11 log11 + K

25. ∫ 𝑥(𝑥 + 5)dx =

(A) x3+5x+K (B) x4+5x2+K

(C) + +K (D) + +K

26. ∫ 𝑒 (𝑐𝑜𝑠 𝑥 − 𝑠𝑖𝑛2𝑥)𝑑𝑥 =

(A) 𝑒 sin2𝑥+K (B) −𝑒 sin2𝑥+K

(C) -𝑒 cos2𝑥+K (D) 𝑒 cos2𝑥+K

27. ∫𝑒 𝑥 +𝑥 𝑑𝑥 =

(A) x7ex + K (B) x6ex + K

(C) x7ex + K (D) x6ex + K

28. ∫ 𝑒 (𝑐𝑜𝑡2𝑥 − 2𝑐𝑜𝑠𝑒𝑐 2𝑥)𝑑𝑥 =

8

(A) K + excot2x (B) k – excot2x

(C) K + excosec2x (D) K – excosec2x

29. 2𝚤⃗.(5𝚥⃗+7𝑘⃗ ) + 7𝚥⃗.(3𝚤⃗-5𝑘⃗ ) =

(A) 10 (B) 5

(C) 7 (D) 0

30. (2sin ) =

(A) 2cos (B) cos

(C) cos (D) cos

31. (cos ) =

(A) sin (B) sin

(C) sin (D) sin

32. (𝑒 )=

(A) 𝑒 (B) 𝑒

(C) 𝑒 (D) 𝑒

33. (2x) =

(A) x2 (B)

(C) 2 log2 (D) 2 logx

9

34. ( )=

(A) log|𝑥 + 1| (B)
( )

(C) (D)
( ) ( )

35. ;fn x = acos𝜃, y =bsin𝜃 rks dk eku gS

(A) cot𝜃 (B) cot𝜃

(C) tan𝜃 (D) tan𝜃

If x = acos𝜃, y =bsin𝜃 then the value of is

(A) cot𝜃 (B) cot𝜃

(C) tan𝜃 (D) tan𝜃

36. vodyu lehdj.k 2xdx – 6y2dy = 0 dk gy gS

(A) 2x – 6y2 = K (B) x2 – 6y2 = K

(C) x2 – 2y3 = K (D) x2 + y3 = K

The solution of the differential equation 2xdx – 6y2dy = 0 is

(A) 2x – 6y2 = K (B) x2 – 6y2 = K

(C) x2 – 2y3 = K (D) x2 + y3 = K

37. (7𝚤⃗ + 5𝚥⃗).(7𝚤⃗+𝚥⃗) =

(A) 74 (B) 54

(C) 50 (D) 48
10

38. vody lehdj.k e-ydx + e-xdy = 0 dk gy gS

(A) ex+y = K (B) e-x + e-y = K

(C) ex + ey = K (D) ex-y = K

The solution of the differential equation e-ydx + e-xdy = 0 is

(A) ex+y = K (B) e-x + e-y = K

(C) ex + ey = K (D) ex-y = K

39. vody lehdj.k - = 0 dk gy gS

(A) - =K (B) - =K

(C) xy3 = K (D) x = Ky3

The solution of the differential equation - = 0 is

(A) - =K (B) - =K

(C) xy3 = K (D) x = Ky3

40. vodyu lehdj.k + ytanx = sinx dk I.F. gS

(A) 𝑡𝑎𝑛𝑥 (B) 𝑠𝑖𝑛𝑥

(C) 𝑠𝑒𝑐𝑥 (D) buesa dksbZ ugha

The I.F. of the differential equation + ytanx = sinx is

(A) 𝑡𝑎𝑛𝑥 (B) 𝑠𝑖𝑛𝑥

(C) 𝑠𝑒𝑐𝑥 (D) none of these

11

41. vodyu lehdj.k + = 3𝑥 5 dk lekdyu xq.kd gS

(A) (B) 10 log𝑥

(C) 𝑥 10 (D)buesa dksbZ ugha

The Integrating factor of the differential equation + = 3𝑥 5 is

(A) (B) 10 log𝑥

(C) 𝑥 10 (D) none of these

42. (2𝚤⃗ + 3𝑘⃗ ) x 5𝚤⃗ =

(A) 15𝚤⃗ (B) 15𝚥⃗

(C) 15𝐾⃗ (D) -15𝚥⃗

43. −2𝚤⃗ − 2𝚥⃗ − 𝑘⃗ =

(A) 3 (B) 4

(C) 5 (D) 9

44. ry x+y-z+7=0 ds vfHkyEc ds fnd~ vuqikr gSa

(A) 1,1,1 (B) 1,1,-1

(C) 1,1,7 (D) 1,-1,7

The direction ratios of the normal to the plane x+y-z+7=0 are

(A) 1,1,1 (B) 1,1,-1

(C) 1,1,7 (D) 1,-1,7

12

45. ljy js[kk = = ds fnd~ vuqikr gS

(A) 7,5,0 (B) 11,3,0

(C) 11,3,2 (D) 7,5,11

The direction ratios of the straight line = = are

(A) 7,5,0 (B) 11,3,0

(C) 11,3,2 (D) 7,5,11

46. ljy js[kk = = fuEufyf[kr esa fdl fcanq ls xqtjrh gS \

(A) (5,6,9) (B) (25,27,70)

(C) (27, 25, 70) (D) buesa dksbZ ugha

Through which of the following point does the st. line

= = pass ?

(A) (5,6,9) (B) (25,27,70)

(C) (27, 25, 70) (D) none of these

47. (𝚤⃗ + 5𝚥⃗ − 9𝑘⃗ ) x (2𝚤⃗ + 10𝚥⃗ − 18𝑘⃗) =

(A) 0⃗ (B) 7𝚤⃗ − 8𝚥⃗ − 11𝑘⃗

(C) 3𝚤⃗ + 11𝚥⃗ − 5𝑘⃗ (D) 13𝚤⃗ + 5𝚥⃗ − 19𝑘

48. (cotx + 2𝑒 ) =

(A) –cosec2x + 4e2x (B) cosec2x + 4e2x

13

(C) –cosecx.cotx + 4e2x (D) cosecx.cotx + 4e2x

49. (7x6 + e3x) =

(A) 42x6 + 3e3x (B) 42x5 + 3e3x

(C) 42x5 + e3x (D) 42x5 + e2x

50. (100x) =

(A) 100 (B) 10

(C) 1 (D) 0

51. ;fn nks lekarj js[kkvksa ds fnd~ vuqikr a,4,8 rFkk 15] 12] 24 gSa rks a dk eku gS

(A) 4 (B) 5

(C) 8 (D) 12

If the direction ratios of two parallel lines are a, 4, 8 and 15, 12, 24

then the value of a is

(A) 4 (B) 5

(C) 8 (D) 12

52. ;fn nks lekarj js[kkvksa ds fnd~ vuqikr a1, a2, a3 rFkk b1, b2, b3 gks rks ¾

(A) (B)

(C) (D)

14

If the direction ratios of two parallel lines be a1, a2, a3 and b1, b2, b3

then =

(A) (B)

(C) (D)

53. ;fn nks ijLij yEc js[kkvksa ds fnd~ vuqikr 3] 5] 7 rFkk a, b, 2 gSa rks 3a + 5b

dk eku gS

(A) 16 (B) -16

(C) 14 (D) -14

If the direction ratios of two mutually perpendicular lines be 3, 5, 7

and a, b, 2 then the value of 3a + 5b is

(A) 16 (B) -16

(C) 14 (D) -14

54. 5𝚤⃗ − 𝚥⃗ + 𝑘⃗ =

(A) 3 (B) 3√3

(C) 2 (D) 5

55. [5𝑎 − 8 2𝑏 − 1] =[𝑎 𝑏 ]  (a, b) =

(A) (1, 2) (B) (2, 1)

(C) (3,1) (D) (-1, 2)

15

14 16 17
56. 3 8 9 =
17 24 26

(A) 0 (B) 1

(C) 1325 (D) 1484

2 1 1
57. 5 2 3 =
7 3 4

(A) 28 (B) 24

(C) 12 (D) 0

𝑠𝑒𝑐𝜃 −𝑐𝑜𝑠𝑒𝑐𝜃
58. =
𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

(A) 0 (B) 1

(C) 2 (D) -1

1 0 5 7
59. =
0 1 3 4

5 −7 5 7
(A) (B)
3 4 3 4

2 7 5 0
(C) (D)
9 11 0 4

3
60. [3 7 ] =
−1

(A) [2] (B) [16]

9

(C) [9 −7] (D)
−7

61. [-4] [5 −6] =

16

(A) [-20 -24] (B) [-20 24]

−20
(C) (D) [4]
24

1 −1
62. 3 =
2 3
3 −3 4 2
(A) (B)
6 9 3 2
3 −3 1 −1
(C) (D)
2 3 6 9

63. O;ojks/kksa x+y≤10, x≥0, y≥0 ds varxZr z = x+2y dk vf/kdre eku gS

(A) 0 (B) 10

(C) 20 (D) 30

The maximum value of z = x+2y subject to constraints x+y≤10, x≥0,

y≥0 is

(A) 0 (B) 10

(C) 20 (D) 30

64. O;ojks/kksa x+y≤25, x≥0, y≥0 ds varxZr z = 5x-y dk vf/kdre eku gS

(A) 100 (B) 125

(C) 150 (D) buesa dksbZ ugha

The maximum value of z = 5x-y subject to constraints x+y≤25, x≥0,

y≥0 is

(A) 100 (B) 125

17

(C) 150 (D) none of these

65. O;ojks/kksa 2x+y≤4, x≥0, y≥0 ds varxZr z = x+y dk U;wure eku gS

(A) -2 (B) -4

(C) 0 (D) buesa dksbZ ugha

The minimum value of z = x+y subject to constraints 2x+y≤4, x≥0,

y≥0 is

(A) -2 (B) -4

(C) 0 (D) none of these

66. (𝚤⃗+𝚥⃗). (𝚥⃗+𝑘⃗ ) =

(A) 1 (B) 2

(C) 3 (D) 0

67. x≥0, cos-1 =

(A) 2sin-1x (B) 2cos-1x

(C) 2tan-1x (D) 2sec-1x

68. x ∈ [-1, 1], sin-1x =

(A) + cos-1x (B) - cos-1x

(C) cos-1x - (D) + cosec-1x

69. xy > -1, tan-1x – tan-1y =

(A) tan-1 (B) tan-1

18

(C) tan-1 (D) tan-1

70. sin-1(sin )

(A) (B)

(C) (D)

√
71. sin[sin-1( )] =
√

√
(A) (B)
√ √

√
(C) (D) 1
√

72. cos-1x + cos-1y =

(A) cos-1{xy-√1 − 𝑥 1−𝑦 } (B) cos-1{xy+√1 − 𝑥 1−𝑦 }

(C) cos-1{x 1 − 𝑦 + 𝑦√1 − 𝑥 } (D) cos-1{x 1 − 𝑦 − 𝑦√1 − 𝑥 }

73. x𝜖[-1, 1], sin(sin-1x+cos-1x) =

(A) 0 (B) 1

(C) (D)

74. x∈R, cos(tan-1x+cot-1x) =

(A) 0 (B) 1

(C) (D)

√

75. |𝑥| ≥1, tan[ (sec-1x+cosec-1x)] =

19

(A) 0 (B)
√

(C) 1 (D) ∞

76. ( 𝑥 - sin5𝑥)=

(A) 𝑥 - cos5𝑥 (B) 𝑥 + cos5𝑥

(C) 𝑥 - cos5𝑥 (D) 6𝑥 - 5cos5𝑥

77. (𝑥 + 𝑒 +sin3𝑥)=

(A) 2𝑥 + 𝑒 + cos3𝑥 (B) 2𝑥 + 3𝑒 + 3cos3𝑥

(C) 2𝑥 + 3𝑒 - 3cos3𝑥 (D) 2𝑥 + 3𝑒 + cos3𝑥

78. (cos23𝑥)=

(A) 2cos3𝑥 (B) 3sin6𝑥

(C) -3sin6𝑥 (D) 2sin6𝑥

79. (loge99𝑥)=

(A) (B)

(C) (D) 99𝑥

80. ry 7x+8y+2z = 12 dh ewy fcanq ls nwjh gS

(A) (B)
√ √

(C) (D)
√ √

20

Distance of the plane 7x+8y+2z = 12 from origin is

(A) (B)
√ √

(C) (D)
√ √

81. ry x-5y+7z = 11 ds lekarj ,d ry dk lehdj.k gS

(A) x-5y+11z = 7 (B) x-5y+7z=12

(C) 5x-y+7z=11 (D) 2x-5y-7z=13

Equation of a plane parallel to the plane x-5y+7z = 11 is

(A) x-5y+11z = 7 (B) x-5y+7z=12

(C) 5x-y+7z=11 (D) 2x-5y-7z=13
82. (7𝚤⃗ -8𝑘⃗ )2=
(A) 90 (B) 113
(C) 1 (D) 12
83. lfn’k 𝚤⃗ - 𝑘⃗ dh fn’kk esa bdkbZ lfn’k gS
⃗ ⃗
(A) (B) 2(𝚤⃗ - 𝑘⃗ )
√

(C) √2(𝚤⃗ - 𝑘⃗ ) (D) 𝚤⃗ + 𝑘⃗
The unit vector in the direction of vector 𝚤⃗ - 𝑘⃗ is
⃗ ⃗
(A) (B) 2(𝚤⃗ - 𝑘⃗ )
√

(C) √2(𝚤⃗ - 𝑘⃗ ) (D) 𝚤⃗ + 𝑘⃗
84. (𝚤⃗ - 2𝚥⃗ + 3𝑘⃗ ).(7𝚤⃗ + 6𝑘⃗) =

(A) 25 (B) 21
(C) 16 (D) 11

21

85. lery x+y-z=7 }kjk x&v{k ij dkVk x;k var% [kaM gS

(A) (B) 7

(C) - (D)

The intercept cut off by the plane x+y-z=7 on the x-axis is

(A) (B) 7

(C) - (D)

86. ;fn x+2y+3z+4=o ry ds lekarj js[kk = = gks rks

(A) 3a+b+2c=0 (B) a+2b+3c=0

(C) 3a+2b+c=0 (D) buesa dksbZ ugha

If the plane x+2y+3z+4=o is parallel to the line = = then

(A) 3a+b+2c=0 (B) a+2b+3c=0

(C) 3a+2b+c=0 (D) none of these

87. ;fn nks ry a1x+b1y+c1z+d1 = 0 rFkk a2x+b2y+c2z+d2=0 ijLij yEc gksa rks

(A) = = (B) a1a2+b1b2+c1c2 = 0

(C) a1c2+a2b1+b2c1=0 (D) buesa dksbZ ugha

If two planes a1x+b1y+c1z+d1 = 0 and a2x+b2y+c2z+d2=0 are mutually

perpendicular then

(A) = = (B) a1a2+b1b2+c1c2 = 0

22

(C) a1c2+a2b1+b2c1=0 (D) none of these

88. (𝚤⃗ + 𝚥⃗ + 𝑘⃗ ).(2𝚤⃗ + 5𝚥⃗ + 10𝑘⃗ ) =

(A) 10 (B) 15

(C) 17 (D) 3

89. P(A) = , P(B) = ,P(A∩B) =  P(A/B) =

(A) (B)

(C) (D)

90. P(A) = , P(B) = ,P(A∪B) =  P(A∩B) =

(A) (B)

(C) (D)

91. Lora= ?kVukvksa A vkSj B ds fy, P(A)=0.7, P(B) = 0.6 rks P(A∩B)=

(A) 0.17 (B) 0.16

(C) 0.23 (D) 0.42

For independent events A and B, P(A)=0.7, P(B) = 0.6 then P(A∩B)=

(A) 0.17 (B) 0.16

(C) 0.23 (D) 0.42

92. vkO;wg 2 25 dk lg[kaMu vkO;wg ¾

1 4

2 1 2 25
(A) (B)
25 4 1 4
23

2 −1
(C) (D) buesa dksbZ ugha
−25 4

2 25
Adjoint matrix of matrix =
1 4
2 1 2 25
(A) (B)
25 4 1 4
2 −1
(C) (D) none of these
−25 4

93. ;fn ,d js[kk dh fnd~ dksT;k,¡ ] ] gSa rks 𝑥 dk ,d eku gS
√ √ √

(A) 2 (B) 4

(C) 6 (D) 8

If the direction cosines of a line be ] ] then a value of x is
√ √ √

(A) 2 (B) 4

(C) 6 (D) 8

1 0 0
5
94. ;fn A= 0 1 0 rks A dk eku gS
0 0 1

(A) 5A (B) 3A

(C) 2A (D) A

1 0 0
If A= 0 1 0 then the value of A5 is
0 0 1

(A) 5A (B) 3A

(C) 2A (D) A

24

95. ;fn lafØ;k ^0* a0b=a+7b ls ifjHkkf"kr gks rks 10¼203½¾

(A) 161 (B) 162

(C) 163 (D) buesa dksbZ ugha

If the operation ‘0’ is defined by a0b=a+7b then 10(203) =

(A) 161 (B) 162

(C) 163 (D) none of these

96. {0, 1, 5} ls {2, 3, 4, 6, 7} esa fHkUu laca/kksa dh dqy la[;k gS

(A) 28 (B) 215

(C) 125 (D) buesa dksbZ ugha

The no. of distinct relations from {0, 1, 5} to {2, 3, 4, 6, 7} is

(A) 28 (B) 215

(C) 125 (D) none of these

97. {0, 1, 2} ls {3, 4, 5, 6, 7, 8} esa Qyuksa dh dqy la[;k gS

(A) 218 (B) 29

(C) 216 (D) buesa dksbZ ugha

Total number of functions from {0, 1, 2} to {3, 4, 5, 6, 7, 8} is

(A) 218 (B) 29

(C) 216 (D) none of these

98. vody lehdj.k 2dx+dy=0 dk gy gS

(A) 2x+y=K (B) x+2y=K

25

(C) 2xy=K (D) x2+y=K

The solution of the differential equation 2dx+dy=0 is

(A) 2x+y=K (B) x+2y=K

(C) 2xy=K (D) x2+y=K

99. 𝑘⃗ . 𝑘⃗=

(A) 0 (B) 1

(C) 2 (D) 3

100. 𝚥⃗ x 𝑘⃗=

(A) 𝚤⃗ (B) −𝚤⃗

(C) 𝑜⃗ (D) 2𝚤⃗

[k.M&c @ Section-B

y?kq mÙkjh; iz’u @ Short Answer Type Questions.

iz'u la[;k 1 ls 30 y?kq mÙkjh; iz’u gSaA buesa ls fdUgha 15 iz’uksa ds mÙkj nsaA izR;sd ds

fy, 2 vad fu/kkZfjr gSA 15x2=30

Question Nos 1 to 30 are short Answer Type. Answer any 15 questions.

Each question carries 2 marks. 15x2=30

1. ;fn y=sin{cos(tan√𝑥)} rks Kkr djsaA

If y=sin{cos(tan√𝑥)} then find .

2. gy djsa % ∫ dx.

26

Solve : ∫ dx

⁄
3. lekdyu djsa % ∫ dx

⁄
Integrate : ∫ dx

4. lekdyu djsa % ∫ 𝑠𝑖𝑛𝑎𝑥. 𝑐𝑜𝑠𝑏𝑥 dx

Integrate : ∫ 𝑠𝑖𝑛𝑎𝑥. 𝑐𝑜𝑠𝑏𝑥 dx

5. ∫ dk lekdyu djsaA

Integrate ∫ .

6. ∫ dx dk lekdyu djsaA

Integrate ∫ dx.

7. ∫ 𝑠𝑖𝑛 𝑥. 𝑐𝑜𝑠 𝑥𝑑𝑥 dk eku Kkr djsaA

Find the value of ∫ 𝑠𝑖𝑛 𝑥. 𝑐𝑜𝑠 𝑥𝑑𝑥.

8. ∫ 𝑑𝑥 dk eku Kkr djsaA

Find the value of ∫ 𝑑𝑥 .

9. ∫ dk eku Kkr djsaA

27

Find the value of ∫

10. gy djsa % cosydx+(1+2e-x)sinydy=0.

Solve : cosydx+(1+2e-x)sinydy=0.

11. ;fn (𝑥 -𝑦) 𝑦n=2√𝑥 rks Kkr djsaA

If (𝑥 -𝑦) 𝑦n=2√𝑥 then find .

12. gy djsa % (1-𝑥 2) + 𝑥𝑦 = a𝑥.

Solve : (1-𝑥 2) + 𝑥𝑦 = a𝑥.

13. ;fn y = 𝑒 rks Kkr djsaA

If y = 𝑒 then find .

14. ;fn x=a(cost+tsint), y=a(sint-tcost) rks Kkr djsaA

If x=a(cost+tsint), y=a(sint-tcost) then find .

15. O;ojks/kksa 3x+11y≤66

X≥0, y≥0

ds varxZr z = 5y+6x dk vf/kdre eku Kkr djsaA

Find the maximum value of z=5y+6x subject to constraints

3x+11y≤66

X≥0, y≥0.

28

4 5 9
16. lkjf.kd 11 9 20 dk eku Kkr djsaA
18 7 25

4 5 9
Find the value of the determinant 11 9 20 .
18 7 25

17. ;fn A = 1 1 vkSj B = 1 2 rks AB vkSj BA Kkr djsaA
1 1 3 4

1 1 1 2
If A = and B = then find AB and BA.
1 1 3 4

18. fl) djsa fd fcanq 𝚤⃗ - 2𝚥⃗+ 3𝑘⃗ , 3𝚤⃗ - 3𝚥⃗+ 4𝑘⃗ vkSj 4𝚤⃗ - 6𝚥⃗ - 𝑘⃗ ,d ledks.k f=Hkqt

cukrs gSaA

Prove that the points 𝚤⃗ - 2𝚥⃗+ 3𝑘⃗ , 3𝚤⃗ - 3𝚥⃗+ 4𝑘⃗ and 4𝚤⃗ - 6𝚥⃗ - 𝑘⃗ form a right

angled triangle.

19. ;fn 𝑎⃗ = 7𝚤⃗ + 8𝚥⃗ − 9𝑘⃗ rFkk 𝑏⃗ = 4𝚤⃗ - 5𝚥⃗ + 𝑘⃗ rks 𝑎⃗x𝑏⃗ Kkr djsaA

If 𝑎⃗ = 7𝚤⃗ + 8𝚥⃗ − 9𝑘⃗ and 𝑏⃗ = 4𝚤⃗ - 5𝚥⃗ + 𝑘⃗ then find 𝑎⃗x𝑏⃗

20. ;fn 𝑎⃗ = 3𝚤⃗ - 2𝑘⃗ , 𝑏⃗=4𝚤⃗ + 3𝚥⃗ - 5𝑘⃗ rFkk 𝑐⃗=7𝚤⃗+5𝚥⃗-11𝑘⃗ rks 2𝑎⃗ + 3𝑏⃗ + 4𝑐⃗ Kkr

djsaA

If 𝑎⃗ = 3𝚤⃗ - 2𝑘⃗ , 𝑏⃗=4𝚤⃗ + 3𝚥⃗ - 5𝑘⃗ and 𝑐⃗=7𝚤⃗+5𝚥⃗-11𝑘⃗

then find 2𝑎⃗ + 3𝑏⃗ + 4𝑐⃗ .

21. fl) djsa fd R ij f(x) = e2x ls iznÙk Qyu fujarj o/kZeku gSA

Prove that the function given by f(x) = e2x is strictly increasing on R.

29

22. oØ y= , 𝑥 ≠2 dk 𝑥 = 10 ij Li’kZ js[kk dh izo.krk Kkr djsaA

Find the slope of the tangent to the curve y= , 𝑥 ≠2 at 𝑥 = 10.

23. fl) djsa fd R esa ;ksx lkgp;Z f}vk/kkjh lafØ;k gSA

Prove the addition is associative binary operation on R.

24. tan-1√3 –cot-1(-√3) dk eq[; eku Kkr djsaA

Find the principal value of tan-1√3 –cot-1(-√3).

25. fl) djsa fd cos-1𝑥 +cos-1𝑦 =cos-1{𝑥𝑦 - (1 − 𝑥 )(1 − 𝑦 )}.

Prove that cos-1𝑥 +cos-1𝑦 =cos-1{𝑥𝑦 - (1 − 𝑥 )(1 − 𝑦 )}.

26. ryksa 2x-2y+z=2 rFkk x-5y-11z+15=0 ds chp dk dks.k Kkr djsaA

Find the angle between the planes 2x-2y+z=2 and x-5y-11z+15=0.

27. ry 𝑟⃗.(3𝚤⃗+4𝚥⃗-12𝑘⃗ )+13=0 ls fcanq (1, 1, 1) dh nwjh Kkr djsaA

Find the distance of the plane 𝑟⃗.(3𝚤⃗+4𝚥⃗-12𝑘⃗ )+13=0 from the point

(1, 1, 1).

28. js[kkvksa = = rFkk = = ds chp dk dks.k Kkr djsaA

Find the angle between the lines = = and = = .

29. ,d FkSys esa 4 lQsn rFkk 8 dkyh xsan gSaA ,d nwljs FkSys esa 3 gjh rFkk 6 yky xsan

gSaA izR;sd FkSys ls ,d xsan fudkyh tkrh gSA ,d dkyh rFkk ,d yky xsan

fudyus dh izkf;drk Kkr djssA

30

A bag contains 4 white and 8 black balls. Another bag contains 3

green and 6 red balls. One ball is taken out from each bag. Find the

probability that one ball is black and the other red.

30. 10 flDdksa dks mNkyk tkrk gSA Bhd 8 'kh"kZ vkus dh izkf;drk Kkr djsaA

10 coins are tossed. Find the probability of the occurrence of exactly

8 heads.

Long Answer Type Questions.

iz'u la[;k 31 ls 38 nh?kZ mÙkjh; iz’u gSaA buesa ls fdUgha 4 iz’uksa ds mÙkj nsaA izR;sd ds

fy, 5 vad fu/kkZfjr gSA 4x5=20

Question Nos. 31 to 38 are Long Answer Type. Answer any 4 questions.

Each question carries 5 marks. 4x5=20

( )
31. gy djsa % =
( )

( )
Solve : =
( )

32. ;fn r2 = x2+y2+z2 rks fl) djsa fd tan-1 + tan-1 + tan-1 = .

If r2 = x2+y2+z2 then prove that tan-1 + tan-1 + tan-1 = .

1 −1 2
-1
33. ;fn vkO;wg A = 3 0 −2 rks A ¼;fn laHko gks½ Kkr djsaA

1 0 3

31

1 −1 2
-1
If the matrix A = 3 0 −2 then find A (if possible).
1 0 3

34. Kkr djsa ;fn y = 𝑥 + 𝑥 ⁄ .

Find when y = 𝑥 + 𝑥 ⁄ .

35. nks iklksa dks Qsadus esa ;fn x NDdksa dh la[;k dks O;Dr djsa rks x dk izlj.k Kkr

djsaA

In throwing two dice if x denotes the number of sixes then find the

variance of x.

36. z = y – 2x dk vf/kdrehdj.k Kkr djsa ;fn x≤2, x+y≤3, -2x+y≤1, x,y≥0

Maximize z = y – 2x subject to x≤2, x+y≤3, -2x+y≤1, x,y≥0

37. ∫ dk eku Kkr djsaA

Find the value of ∫ .

38. [(2𝚤⃗-3𝚥⃗+4𝑘⃗ ) x (𝚤⃗+2𝚥⃗-𝑘⃗ )].(3𝚤⃗-𝚥⃗+2𝑘⃗ ) dk eku Kkr djsaA

Find the value of

[(2𝚤⃗-3𝚥⃗+4𝑘⃗ ) x (𝚤⃗+2𝚥⃗-𝑘⃗ )].(3𝚤⃗-𝚥⃗+2𝑘⃗ ).

32

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