Bihar Board Class 12 Model Question Paper 2022 Maths

INTERMEDIATE EXAMINATION – 2022 (ANNUAL)
Model Set
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. (of 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 (each carrying
1 mark), out of which any 50 questions are to be answered. First
50 answers will be evaluated by the computer in case more than
50 questions are answered. 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 invalid.

7- [k.M&c esa 30 y?kq mÙkjh; iz’u gSa ¼izR;sd ds fy, 2 vad fu/kkZfjr gS½] ftuesa
ls fdUgha 15 iz’uksa dk mÙkj nsuk vfuok;Z gSA buds vfrfjDr] bl [k.M esa 8
nh?kZ mÙkjh; iz’u fn;s x;s gSa ¼izR;sd ds fy, 5 vad fu/kkZfjr gS½] ftuesa ls
fdUgha 4 iz’uksa dk mÙkj nsuk gSA
In Section-B, there are 30 short answer type questions (each
carrying 2 marks), out of which any 15 questions are to be
answered. Apart from this, there are 8 long answer type questions

2

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

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

1. (sin𝑥0) =

(A) cos𝑥 0 (B) cos( )

(C) 𝜋cos𝑥 0 (D) cos( )

2. [cos(√𝑥 + 5)] =

(A) –sin (√𝑥 + 5) (B) sin (√𝑥 + 5)

(C) sin (√𝑥 + 5) (D) sin (√𝑥 + 5)
√ √

3. [𝑒 √ ] =

(A) 𝑒 √ (B) 𝑒√
√

(C) 𝑥 .𝑒√ (D) 𝑥 .𝑒√

4. [(𝑡𝑎𝑛√5𝑥] =

4

(A) sec2√5𝑥 (B) √5 sec2√5𝑥

√ √
(C) sec2√5𝑥 (D) sec2√5𝑥
√ √

5. ( ) =

(A) - (B) -

(C) (D)

6. [log(𝑥 . 𝑏 )] =

(A) + logb (B) +

(C) + (D)
.

𝑠𝑒𝑐𝑥−1
7.
𝑠𝑒𝑐𝑥+1
=

A. sec2 (B) sec2

(C) tan (D)
√

8. [tan-1( )] =

(A) +𝑥 (B) 1

(C) - 1 (D) tan

5

9. [ sin2𝑥] =

(A) sin2𝑥 (B) sin2𝑥

(C) cos2𝑥 (D) cos2𝑥

9
10. ∫(𝑥 ) d𝑥 =

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

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

11. ∫( ¾
)

(A) K + (B) K -

(C) K – log|𝑥 − 2| (D) K + (𝑥 – 2)

12. ∫ 𝑐𝑜𝑠𝑥 . 𝑑𝑥 ¾

(A) sin𝑥 0 + K (B) sin𝑥 0 + K

(C) K + sin𝑥 0 (D) K – sin𝑥 0

13. ∫ cos (90 − 3𝜃)𝑑𝜃 =

(A) K + (B) sin(900 – 3𝜃) + K

(C) K + (D) K -

14. ∫ 𝑐𝑜𝑠𝑒𝑐𝑥 𝑑𝑥 =

(A) log |𝑐𝑜𝑠𝑒𝑐𝑥 − 𝑐𝑜𝑡𝑥| + K (B) log |𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥| + K
6

(C) log|𝑐𝑜𝑡𝑥| + K (D) sin𝑥 + K

15. ∫ 𝑠𝑒𝑐𝑥 𝑑𝑥 =

(A) K + log|𝑠𝑒𝑐𝑥 − 𝑡𝑎𝑛𝑥| (B) K + log|𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥|

(C) K + log|𝑡𝑎𝑛𝑥| (D) tan𝑥 + K

16. ∫ 𝑠𝑖𝑛4𝑥𝑑𝑥 =

(A) 0 (B) 1

(C) - 1 (D)

17. ∫ 𝑠𝑖𝑛𝑥𝑑𝑥 =

(A) 0 (B) 1

(C) 2 (D) - 2

18. ∫ (5𝑥 + 4)𝑑𝑥 =

(A) (B)

(C) (D) 50

19. ∫ 𝑥 √𝑥𝑑𝑥 =

(A) 12.8 (B) 12.4

(C) 7 (D) 13.6

20. ∫ 𝑑𝑥 ¾

7

(A) +𝑥+K (B) -𝑥+K

(C) log |𝑥 − 1|+ K (D) log |𝑥 + 1| + K

21. ∫ 𝑠𝑖𝑛𝑥𝑑𝑥 ¾

(A) 0 (B) 1

(C) 2 (D) 4

22. ∫ 𝑡𝑎𝑛𝜃𝑑𝜃 ¾

(A) 0 (B) 1

(C) 2 (D)
√

23. ∫( 𝑑𝑥 =
)

(A) +K (B) log |𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥| + K

(C) log |𝑠𝑖𝑛𝑥 − 𝐶𝑜𝑠𝑥| + K (D) 2log|sin𝑥| + K

24. ∫(𝑎𝑥 + 𝑏𝑥 + 𝑐)𝑑𝑥 =

(A) - + c𝑥 + K (B) + + c𝑥 + K

(C) (a + b + c) +K (D) (a + b) 𝑥2 + c𝑥 + K

25. 12∫ √𝑥d𝑥 =

(A) 12𝑥 +K (B) 8𝑥 +K

(C) 6𝑥 +K (D) 2𝑥 +K

8

26. ∫√ ¾

(A) sin-19𝑥 + K (B) sin-16𝑥 + K

(C) sin-13𝑥 + K (D) sin-1𝑥 + K

27. ∫ ¾

(A) tan-1 + K (B) tan-1𝑥 + K

(C) 2tan-1𝑥 + K (D) tan-12𝑥 + K

28. ∫ d𝑥 =

(A) K + cot𝑥 (B) K + tan𝑥

(C) K - cot𝑥 (D) K - sin𝑥

29. 100∫ 𝑑𝑥 =

(A) K + 100𝑥 (B) K – 100𝑥

(C) K + 100 (D) K - 100

30. ∫ 4 d𝑥 =

(A) 4 + K (B) 𝑥4 +K

(C) 4 log4 + K (D) +K

31. vody lehdj.k 𝑥d𝑥 + 𝑦2d𝑦 = 0 dk gy gS &

(A) + =K (B) + =K

(C) 𝑥 2 + 𝑦3 = K (D) 𝑥𝑦2 = K

9

The solution of the differential equation 𝑥d𝑥 + 𝑦2d𝑦 = 0 is

(A) + =K (B) + =K

(C) 𝑥 2 + 𝑦3 = K (D) 𝑥𝑦2 = K

32. ∫ =

(A) log +K (B) tan-1 (𝑥 2 – 9) + K

(C) log +K (D) Sin-1 + K

33. ∫ =

(A) tan-1 + K (B) tan-1𝑥 + K

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

34. 7𝚤⃗ .7𝚥⃗ =

(A) 49𝑘⃗ (B) 49

(C) 0 (D) 1

35. 9𝚥⃗ .3𝑘⃗ =

(A) 27 (B) 27𝚤⃗

(C) 1 (D) 0

36. 4𝚤⃗.(7𝚤⃗ – 8𝚥⃗ + 3𝑘⃗ ) =

(A) 4 (B) 28

(C) -32 (D) 12

10

37. (3𝚤⃗ – 4𝚥⃗) . (2𝚤⃗ – 3𝚥⃗ + 4𝑘⃗ ) =

(A) 22 (B) 16

(C) 18 (D) 25

38. 2𝚥⃗ – 2𝑘⃗ − 𝚤⃗ =

(A) 3 (B) 4

(C) 5 (D) 1

√
39. ∫ =

(A) (B)

C. (D)

40. ;fn 𝑎⃗ = -𝚤⃗ - 2𝚥⃗ - 4𝑘⃗ rFkk 𝑏⃗ = -2𝚤⃗ - 3𝚥⃗ rks 𝑎⃗ + 𝑏⃗ dk eku gS

(A) 2√2 (B) 3√2

(C) 4√2 (D) 5√2

If 𝑎⃗ = -𝚤⃗ - 2𝚥⃗ - 4𝑘⃗ and 𝑏⃗ = -2𝚤⃗ - 3𝚥⃗ then the value of 𝑎⃗ + 𝑏⃗ is

(A) 2√2 (B) 3√2

(C) 4√2 (D) 5√2

41. lfn'k 𝚤⃗ – 3𝚥⃗ + 5𝑘⃗ dh fn’kk esa bdkbZ lfn’k gS &

⃗– ⃗ ⃗ ⃗– ⃗ ⃗
(A) (B)
√ √

⃗– ⃗ ⃗
(C) (D) buesa ls dksbZ ugha
√

11

The unit vector in the direction of the vector 𝚤⃗ – 3𝚥⃗ + 5𝑘⃗ is

⃗– ⃗ ⃗ ⃗– ⃗ ⃗
(A) (B)
√ √

⃗– ⃗ ⃗
(C) (D) None of these
√

42. vody lehdj.k + 4𝑦 = 0 dk gy gS

(A) 𝑦 = 4A𝑒 (B) 𝑦 = A𝑒

(C) 𝑦 + 𝑥 = 4 (D) 𝑦 = 𝑥𝑒 +4

The solution of the differential equation + 4𝑦 = 0 is

(A) 𝑦 = 4A𝑒 (B) 𝑦 = A𝑒

(C) 𝑦 + 𝑥 = 4 (D) 𝑦 = 𝑥𝑒 +4

43. lehdj.k + 4𝑦 = cos𝑥 dk lekdyu xq.kd gS

(A) 𝑒 (B) 𝑒

(C) 𝑒 (D) buesa ls dksbZ ugha

The integrating factor of the equation + 4𝑦 = cos𝑥 is

(A) 𝑒 (B) 𝑒

(C) 𝑒 (D) None of these

44. 11𝚤⃗ x 9𝚥⃗ =

(A) 99 (B) 0

(C) -99𝑘⃗ (D) 99𝑘⃗

12

45. ewy fcanq ls ry 3x + 4y + 5z = 6 dh nwjh gS

(A) (B)
√ √

(C) 11 (D) buesa ls dksbZ ugha

The distance of the plane 3x + 4y + 5z = 6 from origin is

(A) (B)
√ √

(C) 11 (D) None of these

46. ryksa 2x + y + z = 11 rFkk x – 2y + z = 5 ds chp dk dks.k gS

(A) cos-1 (B) cos-1

(C) (D) buesa ls dksbZ ugha

The angle between the planes 2x + y + z = 11 and x – 2y + z = 5 is

(A) cos-1 (B) cos-1

(C) (D) None of these

47. 1] 2] 3 fnd~ vuqikr okys js[kk ds fnd~ dksT;k,¡ gSa

(A) , , (B) , ,
√ √ √ √ √ √

(C) , , (D) buesa ls dksbZ ugha
√ √ √

The direction cosines of the line having direction ratios 1, 2, 3 are

(A) , , (B) , ,
√ √ √ √ √ √

13

(C) , , (D) None of these
√ √ √

48. (𝚤⃗ – 𝚥⃗ - 𝑘⃗ ) . (2𝚤⃗ + 2𝚥⃗ - 𝑘⃗ ) =

(A) 0 (B) -3

(C) 1 (D) -1

49. ∫ 𝑥𝑒 d𝑥 =

(A) (5𝑥 – 1) + K (B) (5𝑥 + 1) + K

(C) 𝑒 (5𝑥 + 1) + K (D) 5𝑥 𝑒5𝑥 + K

50. (7𝚤⃗ – 3𝚥⃗ + 𝑘⃗ ) . (2𝚤⃗ + 3𝚥⃗ - 5𝑘⃗ ) =

(A) 0 (B) 1

(C) 2 (D) 28

51. ;fn fdlh js[kk dh fnd~ dksT;k,¡ √ , √ , x gS rks x dk eku gS

(A) (B)
√ √

(C) (D)
√ √

If the direction cosines of a line are , , x then the value of x
√ √

is -

(A) (B)
√ √

(C) (D)
√ √

14

52. (2𝚤⃗ – 3𝚥⃗ + 4𝑘⃗ ) . (𝚤⃗ + 2𝚥⃗ - 𝑘⃗ ) x (3𝚤⃗ – 𝚥⃗ + 2𝑘⃗ ) =

(A) -6 (B) -7

(C) 8 (D) 0

53. ry 3x – 3y + 3z = 11 ij vfHkyEc ds fnd~ vuqikr gS &

(A) , , (B) 3, -3, 3

(C) 3, 3, 11 (D) 0, 0, 0

The direction ratios of the normal to the plane 3x – 3y + 3z = 11are

(A) , , (B) 3, -3, 3

(C) 3, 3, 11 (D) 0, 0, 0

54. ry z = 3 ds lekarj ,d ry dk lehdj.k gS

(A) x = 3 (B) y = 3

(C) z = 0 (D) y = -3

Equation of a plane parallel to the plane z = 3 is

(A) x = 3 (B) y = 3

(C) z = 0 (D) y = -3

55. ry 2x + y – z = 11 ds lekarj ,d ry dk lehdj.k gS

(A) x + y – z = 11 (B) 2x + y –z = 7

(C) 2x – y + z = 1 (D) buesa ls dksbZ ugha

Equation of a plane parallel to the plane 2x + y – z = 11 is

15

(A) x + y – z = 11 (B) 2x + y –z = 7

(C) 2x – y + z = 1 (D) None of these

56. ry x + 5y + 11z = 7 ij yEc ,d ry dk lehdj.k gS

(A) x + y + z = 3 (B) x + 5y + 11z = 1

(B) x + 2y – z = 3 (D) buesa ls dksbZ ugha

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

(A) x + y + z = 3 (B) x + 5y + 11z = 1

(C) x + 2y – z = 3 (D) None of these

2 5 8
57. −7 13 19 ¾
14 −26 −38

(A) 143 (B) -298

(C) 0 (D) 1

3 5 17
58. 6 7 31 ¾
2 3 11

(A) 1025 (B) -1940

(C) 0 (D) 2160

1 −1
59. &2 ¾
3 5

−2 −1 −2 −2
(A) (B)
3 5 3 5

16

−2 2 −2 2
(C) (D)
3 5 −6 −10

2 −2
60. 4 ¾
1 0

8 −2 8 −8
(A) (B)
1 0 4 0

8 −2 8 −8
(C) (D)
4 10 1 0

1 0 11
61. =
0 1 17

11 0
(A) (B) [11 17]
17 1
11 19
(C) (D)
17 25

2 −3 5
62. 6 0 4 =
3 15 −21

(A) 12 (B) 84

(C) -12 (D) -84

3 5
63. ;fn X = rks X dk adjoint ¾
7 9

−3 5 9 −5
(A) (B)
−7 9 −7 3

−9 5
(C) (D) buesa ls dksbZ ugha
7 −3

17

3 5
If X = then adjoint of X =
7 9

−3 5 9 −5
(A) (B)
−7 9 −7 3

−9 5
(C) (D) None of these
7 −3

64. ;fn 𝑥 = acos2𝜃, y = asin2𝜃 rks cjkcj gS

(A) 1 (B) -1

(C) tan2𝜃 (D) -tan2𝜃

If 𝑥 = acos2𝜃, y = asin2𝜃 then is equal to

(A) 1 (B) -1

(C) tan2𝜃 (D) -tan2𝜃

65. ;fn 𝑥 = at4, 𝑦 = 2at2 rks cjkcj gS

(A) t2 (B)

(C) t (D) buesa ls dksbZ ugha

If 𝑥 = at4, 𝑦 = 2at2 then is equal to

(A) t2 (B)

(C) t (D) None of these

66. (20𝑥 3 + 7𝑥) =

18

(A) 60𝑥 2 (B) 60𝑥 2 + 7

(C) 120 (D) 120𝑥

67. [cot-1 (tan𝑥)] =

(A) 1 (B) -1

(C) -𝑥 (D) -

68. fuEufy[kr esa dkSu mís’kh; Qyu gS \

(A) x > 5 (B) z = 11x + 19y

(C) z ≥ 0 (D) buesa ls dksbZ ugha

Which of the following is an objective function ?

(A) x > 5 (B) z = 11x + 19y

(C) z ≥ 0 (D) None of these

69. O;ojks/kksa 2x + 3y ≤ 6, x ≥ 0, y ≥ 0 ds varxZr 7x + 8y dk U;wure eku gS

(A) 0 (B) 21

(C) 16 (D) 37

The minimum value of 7x + 8y subject to constraints 2x + 3y ≤ 6,

x ≥ 0, y ≥ 0 is

(A) 0 (B) 21

(C) 16 (D) 37

70. O;ojks/kksa x + y ≤ 7, x ≥ 0, y ≥ 0 ds varxZr 5x + 7y dk vf/kdre eku gS

19

(A) 35 (B) 49

(C) 0 (D) buesa ls dksbZ ugha

The maximum value of 5x + 7y subject to constraints x + y ≤ 7,

x ≥ 0, y ≥ 0 is

(A) 35 (B) 49

(C) 0 (D) None of these

0 1
71. X= 𝑋 =
1 0

0 1 0 0
(A) (B)
1 0 0 0

1 1 1 0
(C) (C)
1 1 0 1

72. tan-1 =

(A) 2tan-1𝑥 (B) 2sin-1𝑥

(C) 2cos-1𝑥 (D) 2cot-1𝑥

√
73. cos-1( ) dk eq[; eku gS

(A) (B)

(C) (D) buesa ls dksbZ ugha

√
The principal value of cos-1( ) is

(A) (B)

20

(C) (D) None of these

74. aOb = a + b ls N esa ifjHkkf"kr ,d f}vk/kkjh lafØ;k ^O* ds fy,

fuEufyf[kr esa dkSu lR; gS \

(A) lafØ;k lkgp;Z vkSj Øefofues; nksuksa gSA

(B) lafØ;k lkgp;Z gS fdarq Øefofues; ugha gSA

(C) lafØ;k Øefofues; gS fdarq lkgp;Z ugha gSA

(D) buesa dksbZ ugha

Which one of the following is true for a binary operation ‘O’ defined

on N by aOb = a + b?

(A) Operation is both associative and commutative.

(B) Operation is associative but not commutative.

(C) Operation is commutative but not associative.

(D) None of these

75. leqPp; {a, b} ij f}vk/kkjh lafØ;kvksa dh dqy la[;k gS &

(A) 10 (B) 16

(C) 20 (D) buesa ls dksbZ ugha

Number of binary operations on the set {a, b} is

(A) 10 (B) 16

(C) 20 (D) None of these

21

76. ∫ 𝑥 d𝑥 =

(A) (B)

(C) (D)

77. ;fn ry 𝑥 + 𝑦 + 𝑧 = 2 js[kk = = ds lekarj gks rks

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

(C) 2a + 3b + 4c = 0 (D) buesa ls dksbZ ugha

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

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

(C) 2a + 3b + 4c = 0 (D) None of these

78. ;fn js[kk,¡ = = vkSj = = lekarj gksa rks

(A) = = (B) = =

(C) 𝑎1𝛼 + 𝑏1𝛽 + 𝑐1𝛾 = 0 (D) buesa ls dksbZ ugha

If lines = = and = = are parallel then

(A) = = (B) = =

(C) 𝑎1𝛼 + 𝑏1𝛽 + 𝑐1𝛾 = 0 (D) None of these

79. 𝚤⃗ . (𝚥⃗x𝑘⃗ ) =

(A) 0 (B) 1

22

(C) -1 (D) 2

80. tan-1 + tan-1 =

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

(C) (D)

81. 2tan-1 =

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

(C) tan-1 (D)

82. nks ?kVukvksa A vkSj B ds fy, P(A∩B) =

(A) P(A) + P(B) (B) P(A) . P(𝐵 𝐴)

(C) P(A) . P(𝐴 𝐵) (D) buesa ls dksbZ ugha

For two events A and B, P(A∩B) =

(A) P(A) + P(B) (B) P(A) . P(𝐵 𝐴)

(C) P(A) . P(𝐴 𝐵) (D) None of these

83. nks ?kVuk,¡ A vkSj B Lora= gSa ;fn

(A) A vkSj B ijLij viothZ gSa

(B) P(A^B^) = [1 – P(A)] [1 – P(B)]

23

(C) P(A) = P(B)

(D) P(A) + P(B) = 1

Two events A and B are independent if

(A) A and B are mutually exclusive.

(B) P(A^B^) = [1 – P(A)] [1 – P(B)]

(C) P(A) = P(B)

(D) P(A) + P(B) = 1

−3 9 1 0
84. =
−13 −1 0 1
−3 0 −2 9
(A) (B)
0 −1 −13 −1
−3 9 1 0
(C) (D)
−13 −1 0 1

85. ;fn nks Lora= ?kVukvksa A vkSj B ds fy,] P(A) = x, P(B) = rFkk

P(A∪B) = rks x dk eku gS

(A) (B)

(C) (D) buesa ls dksbZ ugha

For two independent events A and B, if P(A) = x, P(B) = and

P(A∪B) = then the value of x is

(A) (B)

24

(C) (D) None of these

1
86. [2 3] =
0

2
(A) [2 3] (B)
3
2 0 2 3
(C) (D)
0 3 0 0

87. [x-1 y + 2] = [3 5]  (x, y) =

(A) (2, 1) (B) (4, 3)

(C) (3, 4) (D) (1, 2)

88. ;fn oxZ vkO;wg A bl izdkj gks fd A3 + 3A2 - 7A + I = 0 rks A-1 cjkcj gS

(A) A2 + 3A + 7I (B) A2 + 3A – 7I

(C) –A2 – 3A + 7I (D) buesa ls dksbZ ugha

If square matrix A is such that A3 + 3A2 - 7A + I = 0 then A-1 is

equal to

(A) A2 + 3A + 7I (B) A2 + 3A – 7I

(C) –A2 – 3A + 7I (D) None of these

29
89. ∫ 𝑥 (1 - 𝑥) d𝑥 =

(A) (B)

(C) (D)

25

90. ;fn x = rks cot(2tan-1x + cot-1x) dk eku gS

(A) (B) -

(C) - (D)

If x = then the value of cot(2tan-1x + cot-1x) is

(A) (B) -

(C) - (D)

0 −3 1 0 −1 1
91. 2 −1 1 0 1 −1 =
2 −1 1 0 3 −3

0 1 0 1 1 1
(A) 0 0 1 (B) 1 1 1
1 0 0 1 1 1
1 3 7 0 0 0
(C) 1 1 1 (D) 0 0 0
3 −1 7 0 0 0

92. ,d cDls esa 100 cYc gSa ftuesa 10 =qfV;qDr gSaA 5 cYc ds uewus esa ls fdlh Hkh

cYc ds =qfV;qDr u gksus dh izkf;drk gS &

(A) (B) ( )5

(C) ( )5 (D)

A box has 100 bulbs out of which 10 are defective. The probability

that out of a sample of 5 bulbs, none is defective, is

26

(A) (B) ( )5

(C) ( )5 (D)

93. sin-1 + sin-1 =

(A) sin-1 (B) sin-1

(C) sin-1 (D)

94. ∫ 𝑙𝑜𝑔𝑠𝑖𝑛𝑥𝑑𝑥 ¾

(A) 0 (B) 1

(C) –𝜋log2 (D) 𝜋log2

95. ,d flDds dks 8 ckj mNkyk tkrk gSA Bhd 5fpr vkus dh izkf;drk gS

(A) ( )5 (B) ( )8

(C) 8C5 ( )8 (D) 8C5 ( )5

A coin is tossed 8 times. The probability of getting exactly 5 heads

is -

(A) ( )5 (B) ( )8

(C) 8C5 ( )8 (D) 8C5 ( )5

96. x&v{k ds fnd dksT;kvksa ds oxksZa dk ;ksx gS

(A) 1 (B) 4

27

(C) 3 (D) buesa ls dksbZ ugha

The sum of squares of the direction cosines of the x-axis is -

(A) 1 (B) 4

(C) 3 (D) None of these

97. ry 4x – 5y + 6z = 8 }kjk x, y rFkk z v{kksa ij dkVs x, var%[kaM Øe’k% gS&

(A) 2, - , (B) 2, , -

(C) 2, - , (D) 2, ,

The intercepts cutoff by the plane 4x – 5y + 6z = 8 on the x, y and

z - axes respectively are -

(A) 2, - , (B) 2, , -

(C) 2, - , (D) 2, ,

98. O;ojks/kksa 2x + y≤4, x≥o, y≥0 ds varxZr z = 3x – 2y dk U;wure eku gS

(A) 6 (B) -6

(C) -8 (D) -12

The minimum value of z = 3x – 2y subject to constraints

2x + y≤4, x≥o, y≥0 is

(A) 6 (B) -6

(C) -8 (D) -12

99. fcanq ¼1] 2] 5½ ls ry x + y + z + 17 = 0 dh nwjh gS

28

(A) (B) 25√3
√

(C) 25 (D) √3

The distance of the plane x + y + z + 17 = 0 from the point (1, 2, 5)

is

(A) (B) 25√3
√

(C) 25 (D) √3

100. vody lehdj.k tanxdx + tanydy = 0 dk gy gS &

(A) tanx + tany = K (B) secx.secy = K

(C) cosx + cosy = K (D) sinx + siny = K

The solution of the differential equation tanxdx + tanydy = 0 is

(A) tanx + tany = K (B) secx.secy = K

(C) cosx + cosy = K (D) sinx + siny = K

[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 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

29

1- fcanqvksa ¼1] &1] 3½ vkSj ¼2] &4] 5½ dks feykus okyh js[kk dh fnd~ dksT;k,¡ Kkr

djsaA

Find the direction cosines of the line segment joining the points

(1, -1, 3) and (2, -4, 5). 2

2- lekdyu djsa % ∫ 𝑠𝑖𝑛√𝑥d𝑥

Integrate : ∫ 𝑠𝑖𝑛√𝑥d𝑥 2

3- lekdyu djsa % ∫ 𝑠𝑖𝑛 𝑥d𝑥

Integrate : ∫ 𝑠𝑖𝑛 𝑥d𝑥 2

4- fl) djsa fd % sin-1 - cos-1 = sin-1

Prove that : sin-1 - cos-1 = sin-1 2

2
5- ;fn A = 4 rFkk B = [2 3 4] rks fl) djsa fd (AB) = BA.
3

2
If A = 4 and B = [2 3 4] then prove that (AB)  = BA. 2
3

4 9 7
6- lkjf.kd 3 5 7 dk eku Kkr djsaA
5 4 5

4 9 7
Evaluate the determinant 3 5 7. 2
5 4 5

√
7- ;fn 𝑦 = tan-1( ) rks fl) djsa fd =
( )
.

30

√
If 𝑦 = tan-1( ) then prove that = . 2
( )

8- ;fn 𝑦 = 𝑥log( ) rks Kkr djsaA

If 𝑦 = 𝑥log( ) then find . 2

9- ;fn 𝑥 = √1 + 𝑡 , 𝑦 = √1 − 𝑡 rks Kkr djsaA

If 𝑥 = √1 + 𝑡 , 𝑦 = √1 − 𝑡 then find . 2

10- ;fn 𝑦 = (𝑠𝑖𝑛𝑥) rks Kkr djsaA

If 𝑦 = (𝑠𝑖𝑛𝑥) then find . 2

(√ )
11- lekdyu djsa (Integrate) : ∫ 2
√ √

12- lekdyu djsa % ∫

Integrate : ∫ 2

13- lekdyu djsa % ∫ d𝑥

Integrate : ∫ d𝑥 2

14- ∫ 𝑐𝑜𝑠 𝜃 𝑑𝜃 . dk eku Kkr djsaA

Find the value of ∫ 𝑐𝑜𝑠 𝜃 𝑑𝜃. 2

15- fl) djsa fd ∫ 𝑠𝑖𝑛 𝑥𝑑𝑥 = ∫ 𝑐𝑜𝑠 𝑥𝑑𝑥

31

Prove that ∫ 𝑠𝑖𝑛 𝑥𝑑𝑥 = ∫ 𝑐𝑜𝑠 𝑥𝑑𝑥 . 2

16- tk¡psa fd Qyu f(x) = x2 – 4x + 3, x = 1 ij o/kZeku gS ;k âklekuA

Examine whether the function f(x) = x2 – 4x + 3 is increasing or

decreasing at x = 1. 2

17- gy djsa % √𝑎 + 𝑥 +𝑥=0

Solve : √𝑎 + 𝑥 +𝑥=0 2

18- gy djsa % - 𝑦=

Solve : - 𝑦= 2

19- nks ljyjs[kkvksa ds chp dk U;wudks.k Kkr djsa ftuds fnd~ vuqikr ¼1] 1] 0½

vkSj ¼2] 1] 2½ gSaA

Find the acute angle between two lines that have the direction

ratios (1, 1, 0) and (2, 1, 2). 2

20- p ds eku Kkr djsa ftlls ljyjs[kk,¡ = = ,oa = =

ijLij yEc gksaA

Find the values of p so that the lines = = and =

= are perpendicular to each other. 2

32

21- lfn'kksa 𝑎⃗ = 2𝚤⃗ - 3𝚥⃗ - 5𝑘⃗ rFkk 𝑏⃗ = -7𝚤⃗ + 6𝚥⃗ + 8𝑘⃗ dk lfn’k xq.kuQy Kkr

djsaA

Find the vector product of the vectors 𝑎⃗ = 2𝚤⃗ - 3𝚥⃗ - 5𝑘⃗ and

𝑏⃗ = -7𝚤⃗ + 6𝚥⃗ + 8𝑘⃗ . 2

22- ,d flDds dks 3 ckj mNkyk tkrk gSA Bhd nks ckj 'kh"kZ vkus dh izkf;drk

Kkr djsaA

A coin is tossed 3 times. Find the probability that exactly 2 heads

appear. 2

23- fuEu jSf[kd izksxzkeu leL;k dks gy djsa %

fuEu O;ojks/kksa x + y≤8, x≥0, y≥0 ds varxZr

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

Solve the following linear programming problem :

Maximize z = 5x + 6y

Subject to the constraints x + y≤8, x≥0, y≥0. 2

24- r = 6cm f=T;k okys o`Ùk ds {ks=Qy ds ifjorZu dh nj bldh f=T;k r ds

lkis{k Kkr djsaA

Find the rate of change of the area of a circle with radius r = 6cm.

with respect to its radius r. 2

25- eku Kkr djsa % (-2𝚤⃗ + 3𝚥⃗ – 4𝑘⃗ ) . (3𝚤⃗ – 6𝚥⃗ + 6𝑘⃗ ) x (𝚤⃗ - 3𝚥⃗ +2𝑘⃗ )

33

Evaluate : (-2𝚤⃗ + 3𝚥⃗ – 4𝑘⃗ ) . (3𝚤⃗ – 6𝚥⃗ + 6𝑘⃗ ) x (𝚤⃗ - 3𝚥⃗ +2𝑘⃗ ) 2

26- gof vkSj fog Kkr djsa ;fn f(𝑥) = 8𝑥 3 rFkk g(𝑥) = 𝑥 .

Find gof and fog if f(𝑥) = 8𝑥 3 and g(𝑥) = 𝑥 . 2

27- ;fn E vkSj F bl rjg dh ?kVuk,¡ gks fd P(E) = 0.6, P(F) = 0.3 vkSj

P(E∩F) = 0.2 rks P(E/F) Kkr djsaA

If E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E∩F)

= 0.2 then find P(E/F). 2

28- nks iklksa dks ,d lkFk Qsadk tkrk gSA vadksa dk ;ksxQy 9 ;k 11 vkus dh

izkf;drk Kkr djsaA

Two dice are thrown simultaneously. Find the probability of getting

a sum 9 or 11. 2

29- Ekku Kkr djsa % ∫ d𝑥

Evaluate : ∫ d𝑥 2

30- gy djsa %

Solve : (2x + 3y – 5)dx + (3x – 2y – 1)dy = 0 2

34

nh?kZ mÙkjh; iz’u @ Long Answer Type Questions.

iz'u la[;k 31 ls 38 nh?kZ mÙkjh; iz’u gSaA 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

𝑎−𝑏−𝑐 2𝑎 2𝑎
31- xq.ku[kaM fudkysa % 2𝑏 𝑏−𝑐−𝑎 2𝑏
2𝑐 2𝑐 𝑐−𝑎−𝑏

𝑎−𝑏−𝑐 2𝑎 2𝑎
Factorize : 2𝑏 𝑏−𝑐−𝑎 2𝑏 5
2𝑐 2𝑐 𝑐−𝑎−𝑏

32- ;fn ∆ABC esa] A = tan-12 rFkk B=tan-13 rks fl) djsa fd C = .

If in ∆ABC, A = tan-12 and B=tan-13, then prove that C = . 5

( )
33- ;fn sin𝑦 = 𝑥 sin(a+𝑦) rks fl) djsa fd =

( )
If sin𝑦 = 𝑥 sin(a+𝑦) then prove that = 5

34- U;wurehdj.k djsa % z = 3x + 5y

tcfd x + 3y ≥ 3

x+ y≥2

x, y ≥ 0.

Minimize : z = 3x + 5y

35

Subject to x + 3y ≥ 3

x+ y≥2

x, y ≥ 0. 5

35- gy djsa % (1 + 𝑦2) + (𝑥 – 𝑒 ) =0

Solve : (1 + 𝑦2) + (𝑥 – 𝑒 ) =0 5

36- js[kkvksa 𝑟⃗ = 𝚤⃗ + 2𝚥⃗ - 4𝑘⃗ + (2𝚤⃗ + 3𝚥⃗ + 6𝑘⃗ ) rFkk

𝑟⃗ = 3𝚤⃗ + 3𝚥⃗ - 5𝑘⃗ + 𝜇(2𝚤⃗ + 3𝚥⃗ + 6𝑘⃗ ) ds chp dh U;wure nwjh Kkr djsaA

Find the shortest distance between the lines

𝑟⃗ = 𝚤⃗ + 2𝚥⃗ - 4𝑘⃗ + (2𝚤⃗ + 3𝚥⃗ + 6𝑘⃗ ) and

𝑟⃗ = 3𝚤⃗ + 3𝚥⃗ - 5𝑘⃗ + 𝜇(2𝚤⃗ + 3𝚥⃗ + 6𝑘⃗ ). 5

37- ,d flDds dh nks mNkyksa esa 'kh"kksZa dh la[;k dk ek/; Kkr djsaA

Find the mean of the number of heads in two tosses of a coin.5

38- eku fudkysa %

Evaluate : ∫ d𝑥 5

36