Bihar Board Class 12th Model Paper 2024 Maths

Bihar Board

SAMPLE
PAPER

2024

INTERMEDIATE EXAMINATION – 2024 (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
Candidates 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

1

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

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 ipkl ls vf/kd iz’uksa ds mÙkj
nsus ij izFke 50 mŸkjksa dk gh ewY;kadu dEI;wVj }kjk fd;k tk,xkA lgh mÙkj
dks miyC/k djk, x, OMR mÙkj i=d esa fn, x, lgh xksys 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 ijh{kk
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
Fifty 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

2

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,
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

1. (𝑥 − 𝑠𝑖𝑛𝑥) =

(A) 1 + 𝑐𝑜𝑠𝑥 (B) 1 − 𝑐𝑜𝑠𝑥

(c) 1 + 𝑠𝑖𝑛𝑥 (D) 𝑥 − 𝑐𝑜𝑠𝑥

2. vody lehdj.k −𝑦𝑐𝑜𝑡𝑥 = 𝑐𝑜𝑠𝑒𝑐 𝑥 dk lekdyu xq.kd gS

(A) 𝑐𝑜𝑠𝑥 (B) 𝑐𝑜𝑠𝑒𝑐𝑥

(C) 𝑠𝑖𝑛𝑥 (D) −𝑡𝑎𝑛𝑥

The integrating factor of the differential equation

−𝑦𝑐𝑜𝑡𝑥 = 𝑐𝑜𝑠𝑒𝑐 𝑥 is

(A) 𝑐𝑜𝑠𝑥 (B) 𝑐𝑜𝑠𝑒𝑐𝑥

(C) 𝑠𝑖𝑛𝑥 (D) −𝑡𝑎𝑛𝑥

3. (2𝚤⃗+3𝚥⃗)x3𝑘⃗ =

(A) 6𝚥⃗+9𝚤⃗ (B) 9𝚤⃗-6𝚥⃗

4

(C) 6𝚥⃗ - 9𝚤⃗ (D) 6𝑘⃗ + 9𝚤⃗

4. 3𝚤⃗ − 5𝑘⃗ + 4𝚥⃗ =

(A) 5 (B) 5√2

(C) 5√3 (D) 7

5. vody lehdj.k 2𝑥𝑑𝑥 + 3𝑦 𝑑𝑦 = 0 dk gy gS

(A) 2𝑥 + 3𝑦 = 𝐾 (B) 𝑥 + 3𝑦 = 𝐾

(C) 𝑥 + 𝑦 = 𝐾 (D) 2𝑥 + 𝑦 = 𝐾

The solution of the differential equation 2𝑥𝑑𝑥 + 3𝑦 𝑑𝑦 = 0 is

(A) 2𝑥 + 3𝑦 = 𝐾 (B) 𝑥 + 3𝑦 = 𝐾

(C) 𝑥 + 𝑦 = 𝐾 (D) 2𝑥 + 𝑦 = 𝐾

6. vody lehdj.k 𝑒 − 𝑒 . = 0 dk gy gS

(A) 𝑒 − 𝑒 = 𝐾 (B) 𝑒 + 𝑒 = 𝐾

(C) 𝑒 + 𝑒 =𝐾 (D) 𝑒 − 𝑒 =𝐾

The solution of the differential equation 𝑒 − 𝑒 . = 0 is

(A) 𝑒 − 𝑒 = 𝐾 (B) 𝑒 + 𝑒 = 𝐾

(C) 𝑒 + 𝑒 =𝐾 (D) 𝑒 − 𝑒 =𝐾

7. (49𝑠𝑖𝑛 ) =

(A) 49𝑐𝑜𝑠 (B) 7𝑐𝑜𝑠𝑥

5

(C) 7𝑐𝑜𝑠 (D) −49𝑐𝑜𝑠

8. (𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛2𝑥) =

(A) 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠2𝑥 (B) −𝑠𝑖𝑛𝑥 + 2𝑐𝑜𝑠2𝑥

(C) – 𝑠𝑖𝑛𝑥 − 2𝑐𝑜𝑠2𝑥 (D) 𝑐𝑜𝑠𝑥 + 𝑠𝑖𝑛2𝑥

9. −𝑐𝑜𝑠𝑥 + 𝑒 =

(A) 𝑠𝑖𝑛𝑥 + 𝑒 (B) 𝑠𝑖𝑛𝑥 + 𝑒

(C) −𝑠𝑖𝑛𝑥 + 𝑒 (D) −𝑐𝑜𝑠𝑥 + 𝑒

10. (𝑒 ) =

(A) 𝑒 (B) 2𝑒

(C) 4𝑒 (D) 8𝑒

11. (3𝑠𝑖𝑛 𝑥 + 3𝑐𝑜𝑠 𝑥) =

(A) 0 (B) 3

(C) 3𝑠𝑖𝑛2𝑥 (D) 3𝑐𝑜𝑠2𝑥

12. (3𝑐𝑜𝑠𝑥. 𝑠𝑒𝑐𝑥) =

(A) 3 (B) 2

(C) 1 (D) 0

13. (4𝑐𝑜𝑠 ) =

6

(A) 4𝑠𝑖𝑛 (B) −4𝑠𝑖𝑛

(C) 𝑠𝑖𝑛 (D) −𝑠𝑖𝑛

14. [𝑙𝑜𝑔 (5𝑥)] =

(A) (B)

(C) (D) 5𝑥

15. [𝑙𝑜𝑔 (3𝑥 )] =

(A) (B)

(C) (D)

16. 𝑡𝑎𝑛5𝑥 =

(A) 𝑠𝑒𝑐 5𝑥 (B) 𝑠𝑒𝑐 5𝑥

(C) 𝑠𝑒𝑐 5𝑥 (D) 5𝑠𝑒𝑐 5𝑥

17. 𝑥 = 𝑎𝑐𝑜𝑠𝜃, 𝑦 = 𝑏𝑠𝑖𝑛𝜃  =

(A) 𝑡𝑎𝑛𝜃 (B) 𝑐𝑜𝑡𝜃

(C) − 𝑡𝑎𝑛𝜃 (D) − 𝑐𝑜𝑡𝜃

18. ∫ 𝑐𝑜𝑠 𝜃. 𝑠𝑒𝑐 𝜃 𝑑𝜃 =

(A) 𝑠𝑖𝑛𝜃 + 𝐾 (B) – 𝑠𝑖𝑛𝜃 + 𝑘

(C) 𝐾 + 𝜃 (D) 𝐾 − 𝜃

7

19. ∫(𝑠𝑖𝑛 𝜃 + 𝑠𝑖𝑛𝜃𝑐𝑜𝑠 𝜃)𝑑𝜃 =

(A) 𝐾 + 𝑐𝑜𝑠𝜃 (B) 𝐾 − 𝑐𝑜𝑠𝜃

(C) 𝐾 + 𝑠𝑖𝑛𝜃 (D) 𝐾 − 𝑠𝑖𝑛𝜃

20. 2∫ =

(A) 𝑙𝑜𝑔|𝑥 + 5| + 𝐾 (B) 2𝑙𝑜𝑔|𝑥 + 5| + 𝐾

(C) 𝑡𝑎𝑛 + 𝐾 (D 𝑡𝑎𝑛 + 𝐾
√ √ √

21. ∫ 𝑑𝑥 =

(A) 𝑙𝑜𝑔|𝑒 − 𝑒 |+𝐾 (B) 𝑙𝑜𝑔|𝑒 + 𝑒 |+𝐾

(C) 𝑒 + 𝑒 +𝐾 (D) 𝑒 − 𝑒 +𝐾

22. |−3𝚤⃗| =

(A) 1 (B) −1

(C) −3 (D) 3

23. 3∫ 𝑠𝑒𝑐2𝑥. 𝑡𝑎𝑛2𝑥 𝑑𝑥 =

(A) 𝑠𝑒𝑐2𝑥 + 𝐾 (B) 6𝑠𝑒𝑐2𝑥 + 𝐾

(C) 3𝑠𝑒𝑐2𝑥 + 𝐾 (D) 𝑡𝑎𝑛2𝑥 + 𝐾

24. ∫ 8 dx =

(A) 8 + 𝐾 (B) 8 +𝐾

(C) +𝐾 (D) +𝐾

8

25. (𝑐𝑜𝑠𝑥. 𝑐𝑜𝑠𝑒𝑐 𝑥 − 𝑐𝑜𝑠𝑥. 𝑐𝑜𝑡 𝑥) =

(A) 𝑠𝑖𝑛𝑥 (B) −𝑠𝑖𝑛𝑥

(C) 𝑐𝑜𝑠𝑥 (D) – 𝑐𝑜𝑠𝑥

26. ;fn 𝑦 = 𝑠𝑖𝑛 𝑥 rks − =
√

(A) 𝑦 (B) 2𝑦

(C) 0 (D) 1

If 𝑦 = 𝑠𝑖𝑛 𝑥 then − =
√

(A) 𝑦 (B) 2𝑦

(C) 0 (D) 1

27. ∫ 𝑒 (𝑡𝑎𝑛𝑥 + 𝑠𝑒𝑐 𝑥)𝑑𝑥 =

(A) 𝑒 𝑠𝑖𝑛𝑥 + 𝐾 (B) 𝑒 𝑐𝑜𝑠𝑥 + 𝐾
(C) 𝑒 𝑡𝑎𝑛𝑥 + 𝐾 (D) 𝑒 𝑠𝑒𝑐 𝑥 + 𝐾

28. ∫𝑒 𝑙𝑜𝑔𝑥 + 𝑑𝑥 =

(A) 𝑥𝑒 + 𝐾 (B) 𝑒 . 𝑙𝑜𝑔𝑥 + 𝐾

(C) 𝑒 + 𝐾 (D) 𝑥𝑒 𝑙𝑜𝑔𝑥 + 𝐾

29. ∫ 𝑒 (𝑐𝑜𝑠 𝑥 − 𝑠𝑖𝑛2𝑥)𝑑𝑥 =

(A) 𝑒 𝑠𝑖𝑛2𝑥 + 𝐾 (B) 𝑒 𝑐𝑜𝑠 𝑥 + 𝐾

(C) −𝑒 𝑐𝑜𝑠 𝑥 + 𝐾 (D)−𝑒 𝑠𝑖𝑛2𝑥 + 𝐾

9

30. ∫𝑒 𝑡𝑎𝑛 𝑥 + 𝑑𝑥 =

(A) 𝑒 . 𝑡𝑎𝑛 𝑥 + 𝐾 (B) + 𝐾

(C) 𝑒 . 𝑠𝑖𝑛 𝑥 + 𝐾 (D) + 𝐾
( )

31. ∫ 𝑥 𝑑𝑥 =

(A) (B)

(C) (D)

32. ∫ 𝑠𝑖𝑛 𝑥𝑑𝑥 =

(A) 0 (B) 1

(C) −1 (D)

33. ∫ 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥𝑑𝑥 =

(A) 0 (B) 1

(C) −1 (D)

34. ∫ 𝑒 𝑑𝑥 =

(A) (B)

( ) ( )
(C) (D)

35. ∫ log(𝑡𝑎𝑛𝑥) 𝑑𝑥 =

10

(A) 0 (B) 1

(C) −1 (D)

√
36. ∫ =

(A) (B)

(C) (D)

37. ∫ =

(A) (B)

(C) (D)

38. ∫ 𝑡𝑎𝑛 𝑑𝑥 =

(A) 1 (B) 0

(C) -1 (D)

39. oØ 𝑦 = 𝑥 , 𝑥 − v{k rFkk dksfV;ksa 𝑥 = −2 rFkk 𝑥 = 1 ls f?kjs {ks= dk

{ks=Qy gS

(A) −9 (B) −

(C) (D)

Area bounded by the curve 𝑦 = 𝑥 , the 𝑥–axis and the ordinates

𝑥 = −2 and 𝑥 = 1 is

11

(A) −9 (B) −

(C) (D)

40. ∫ 𝑥 𝑠𝑒𝑐 𝑥 𝑑𝑥 =

(A) (B)

(C) 0 (D) 1

41. ∫ √ =

(A) 𝑠𝑒𝑐 𝑥+𝐾 (B) 𝑠𝑒𝑐 2𝑥 + 𝐾

(C) 𝑠𝑒𝑐 2𝑥 + 𝐾 (D) 2𝑠𝑒𝑐 2𝑥 + 𝐾

42. √𝑥 − 𝑎 − 𝑙𝑜𝑔 𝑥 + √𝑥 − 𝑎 + 𝐾=

(A) ∫ √𝑥 + 𝑎 𝑑𝑥 (B) ∫ √𝑥 − 𝑎 𝑑𝑥

(C) ∫ √𝑎 − 𝑥 𝑑𝑥 (D) ∫[𝑥 + √𝑥 − 𝑎 ]𝑑𝑥

⁄
43. ∫ 𝑠𝑖𝑛2𝑥 𝑑𝑥

(A) 0 (B) 1

(C) (D)
√

44. ∫ 𝑡𝑎𝑛 𝑥. 𝑠𝑒𝑐 𝑥 𝑑𝑥 =

(A) 0 (B) 1

(C) (D)

12

45. vody lehdj.k 3𝑥 𝑑𝑥 − 𝑐𝑜𝑠𝑦 𝑑𝑦 = 𝑜 dk gy gS

(A) 3𝑥 − 𝑐𝑜𝑠𝑦 = 𝐾 (B) 𝑥 − 𝑠𝑖𝑛𝑦 = 𝐾

(C) 𝑥 + 𝑠𝑖𝑛𝑦 = 𝐾 (D) buesa ls dksbZ ugha

The solution of the differential equation 3𝑥 𝑑𝑥 − 𝑐𝑜𝑠𝑦 𝑑𝑦 = 𝑜 is

(A) 3𝑥 − 𝑐𝑜𝑠𝑦 = 𝐾 (B) 𝑥 − 𝑠𝑖𝑛𝑦 = 𝐾

(C) 𝑥 + 𝑠𝑖𝑛𝑦 = 𝐾 (D) none of these

𝑑𝑦
46. vody lehdj.k (1 − 𝑦2 ) + 𝑦𝑥 = 𝑎𝑦 ; −1 < 𝑦 < 1 dk lekdyu xq.kd
𝑑𝑥

gS

(A) (B)

(C) (D)

The integrating factor of the differential equation

𝑑𝑦
(1 − 𝑦2 ) + 𝑦𝑥 = 𝑎𝑦 ; −1 < 𝑦 < 1
𝑑𝑥

(A) (B)

(C) (D)

−3 13 3 0
47. =
6 9 0 3

0 0 −9 13
(A) (B)
0 0 6 27
−9 39 −9 0
(C) (D)
18 27 0 27
13

2 3
48. [3 −5] =
3 0
9 −9 6 9
(A) (B)
9 0 −15 18
−9 9
(C) (D) xq.ku laHko ugha gS
−15 0

2 3
[3 −5] =
3 0
9 −9 6 9
(A) (B)
9 0 −15 18
−9 9
(C) (D) Multiplication is not possible
−15 0
0
49. [4 −6] =
−1

4 −6 0 −6
(A) (B)
−4 6 0 6

(C) [0 6] (D) [6]

50. [−2][−7 13] =

14
(A) [14 −26] (B)
−26
14 −26 14 0
(C) (D)
0 0 0 −26

3 −5
51. −2 =
5 9

−6 −5 −6 10
(A) (B)
5 9 5 9

−6 10 −6 10
(C) (D)
−10 9 −10 −18

14

52. vkO;wg 7 6 dk lg[kaMt vkO;wg gS
5 4

7 −6 4 5
(A) (B)
−5 4 6 7

4 −6 −4 −6
(C) (D)
−5 7 5 −7

7 6
The adjoint matrix of the matrix is
5 4

7 −6 4 5
(A) (B)
−5 4 6 7

4 −6 −4 −6
(C) (D)
−5 7 5 −7

3 5 9
53. lkjf.kd 6 8 0 dk eku gS
9 13 9

(A) 2430 (B) 2109

(C) 2845 (D) 0

3 5 9
The value of the determinant 6 8 0 is
9 13 9

(A) 2430 (B) 2109

(C) 2845 (D) 0

54. vkO;wg 1 0 dk O;qRØe gS
0 1

−1 0 −1 0
(A) (B)

0 1 0 −1

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

15

1 0
The inverse of the matrix is
0 1
−1 0 −1 0
(A) (B)
0 1 0 −1
0 0 1 0
(C) (D)
0 0 0 1

55. leqPp; 𝐴 = {1, 2, 3, 4} ls Lo;a rd lHkh ,dSdh Qyu dh la[;k gS

(A) 6 (B) 12

(C) 24 (D) buesa dksbZ ugha

The number of all one-one functions from set 𝐴 = {1, 2, 3, 4} to itself

is

(A) 6 (B) 12

(C) 24 (D) none of these

56. ry 3𝑥 − 5𝑦 − 7𝑧 = 6 ds vfHkyEc ds fnd~ vuqikr gSa

(A) 3, 5, 7 (B) 3, -5, 7

(C) 3, 5, -7 (D) 3, -5, -7

The direction ratios of the normal to the plane 3𝑥 − 5𝑦 − 7𝑧 = 6

are

(A) 3, 5, 7 (B) 3, -5, 7

(C) 3, 5, -7 (D) 3, -5, -7

57. js[kk 𝑥 − 1 = 𝑦 + 2 = 𝑧 + 3 ds fnd~ vuqikr gSa

16

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

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

The direction ratios of the line 𝑥 − 1 = 𝑦 + 2 = 𝑧 + 3 are

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

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

58. 3𝚤⃗ + 4𝚥⃗ − 7𝑘⃗ . 11𝚤⃗ − 6𝚥⃗ + 𝑘⃗ =

(A) 0 (B) 1

(C) 2 (D) 3

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

(A) (3, 4, 5) (B) (2, 3, 4)

(C) (2, 5, 6) (D) (4, 5, 6)

Through which of the following points does the straight line

= = pass ?

(A) (3, 4, 5) (B) (2, 3, 4)

(C) (2, 5, 6) (D) (4, 5, 6)

60. ;fn nks lekarj js[kkvksa ds fnd~ vuqikr 40] 9] 8 rFkk 120] 27] 𝑥 gksa rks 𝑥 dk

eku gS

(A) 8 (B) 16

(C) 24 (D) 32

17

If the direction ratios of two parallel lines are 40, 9, 8 and 120, 27, 𝑥

then the value of 𝑥 is

(A) 8 (B) 16

(C) 24 (D) 32

61. ;fn 𝑓: 𝐴 → 𝐵 rFkk 𝑔: 𝐵 → 𝐶 ,dSdh vkPNknd gSa rks 𝑔𝑜𝑓: 𝐴 → 𝐶 gS

(A) ,dSdh vkPNknd (B) cgq,d vkPNknd

(C) ,dSdh ysfdu vkPNknd ugha (D) cgq,d ysfdu vkPNknd ugha

If 𝑓: 𝐴 → 𝐵 and 𝑔: 𝐵 → 𝐶 are one-one onto then 𝑔𝑜𝑓: 𝐴 → 𝐶 is

(A) one-one onto (B) many-one onto

(C) one-one but not onto (D) many-one but not onto

62. 𝚤⃗ + 4𝚥⃗ + 2𝑘⃗ 𝑥 3𝚤⃗ − 2𝚥⃗ + 7𝑘⃗ =

(A) 0⃗ (B) 16𝚤⃗ − 2𝚥⃗ − 32𝑘⃗

(C) 32𝚤⃗ − 𝚥⃗ − 14𝑘⃗ (D) 𝚤⃗ − 𝚥⃗ + 6𝑘⃗

63. ;fn 𝑆 = {1, 2, 3}, 𝑓: 𝑆 → 𝑆 gS rFkk 𝑓 = {(1, 1), (2, 2), (3, 3)} gS rks

(A) 𝑓 cgq,d vkPNknd gSA

(B) 𝑓 ,dSd vkPNknd gSA

(C) 𝑓 cgq,d ysfdu vkPNknd ugha gSA

(D) 𝑓 ,dSd ysfdu vkPNknd ugha gSA

18

If 𝑆 = {1, 2, 3}, 𝑓: 𝑆 → 𝑆 and 𝑓 = {(1, 1), (2, 2), (3, 3)} then

(A) 𝑓 is many-one onto . (B) 𝑓 is one-one onto.

(C) 𝑓 is many-one but not onto. (D) 𝑓 is one-one but not onto.

64. ;fn 𝑋 = {𝑎, 𝑏, 𝑐} rks 𝑋 ls 𝑋 esa lHkh ,dSd Qyuksa dh la[;k gS

(A) 2 (B) 4

(C) 6 (D) 8

If 𝑋 = {𝑎, 𝑏, 𝑐} then the number of all one-one functions from 𝑋 to 𝑋 is

(A) 2 (B) 4

(C) 6 (D) 8

65. ;fn 𝑋 = {1,2,3,4} rks 𝑋 ls Lo;a 𝑋 esa lHkh vkPNknd Qyuksa dh la[;k gS

(A) 4 (B) 16

(C) 24 (D) buesa dksbZ ugha

If 𝑋 = {1, 2, 3, 4} then the number of all onto functions from 𝑋 to 𝑋

itself is

(A) 4 (B) 16

(C) 24 (D) none of these

66. (3𝚤⃗ − 5𝚥⃗ + 7𝑘⃗ ). 2𝚤⃗ + 5𝚥⃗ + 4𝑘⃗ =

(A) 3 (B) 6

(C) 9 (D) 0

19

67. ,sls ikls] ftlds rhu Qydksa ij 1] vU; rhu ij 2 rFkk ,d Qyd ij 5 fy[kk

x;k gS] dks mNkyus ij izkIr la[;kvksa dk ek/; gS

(A) 1 (B) 2

(C) 5 (D) buesa dksbZ ugha

The mean of the numbers obtained on throwing a die having written 1

on three faces, 2 on two faces and 5 on one face is

(A) 1 (B) 2

(C) 5 (D) none of these

68. ,d rk’k dh ,d xM~Mh ls ;kn`PN;k nks iÙks fudkys tkrs gSAa ;fn izkIr bDdksa dh

la[;k 𝑋 gS rks 𝐸(𝑋) dk eku gS

(A) (B)

(C) (D)

Two cards are drawn at random from a deck of cards. If the number

of aces obtained is 𝑋 then the value of 𝐸(𝑋) is

(A) (B)

(C) (D)

69. ,d ;kn`fPNd pj 𝑋 dk izkf;drk caVu uhps fn;k x;k gS %

20

X 0 1 2 3 4

P(x) 0.1 K 2K 2K K

rks K dk eku gS

(A) 1 (B) 0.15

(C) 0.25 (D) 0.35

The probability distribution of a random variable x is given below :

X 0 1 2 3 4

P(x) 0.1 K 2K 2K K

Then the value of K is

(A) 1 (B) 0.15

(C) 0.25 (D) 0.35

70. ,d dy’k esa 5 yky rFkk 2 dkyh xsna gSaA nks xsna ;kn`PN;k fudkyh xbZA ;fn 𝑋

dkyh xsna ksa dh la[;k gks rks fuEufyf[kr esa dkSu 𝑋 dk laHkkfor eku ugha gS \

(A) 0 (B) 1

(C) 2 (D) 3

An urn contains 5 red and 2 black balls. Two balls are randomly

drawn. If 𝑋 is the number of black balls then which of the following is

not a possible value of 𝑋 ?

(A) 0 (B) 1

(C) 2 (D) 3
21

71. ,d U;k¸; flDds dks 10 ckj mNkyus ij Bhd 10 fpÙk vkus dh izkf;drk gS

(A) 10 ( ) (B) 10 ( )

(C) 10 ( ) (D) buesa dksbZ ugha

The probability of getting exactly 10 heads in the toss of a fair coin

ten times is

(A) 10 ( ) (B) 10 ( )

(C) 10 ( ) (D none of these

72. ;fn 𝑃(𝐴) = , 𝑃(𝐵) = rFkk 𝑃(𝐴 ∪ 𝐵) = rks 𝑃(𝐴 ∩ 𝐵) =

(A) (B)

(C) (D)

If 𝑃(𝐴) = , 𝑃(𝐵) = and 𝑃(𝐴 ∪ 𝐵) = then 𝑃(𝐴 ∩ 𝐵) =

(A) (B)

(C) (D)

73. ;fn 𝑃(𝐴 ∪ 𝐵) = , 𝑃(𝐴 ∩ 𝐵) = rFkk 𝑃(𝐴) = gks rks 𝑃(𝐵) =

(A) (B)

(C) (D)

If 𝑃(𝐴 ∪ 𝐵) = , 𝑃(𝐴 ∩ 𝐵) = and 𝑃(𝐴) = then 𝑃(𝐵) =

22

(A) (B)

(C) (D)

74. ;fn 𝑃(𝐴 ∩ 𝐵) = rFkk 𝑃(𝐵) = rks 𝑃 =

(A) (B)

(C) (D)

If 𝑃(𝐴 ∩ 𝐵) = and 𝑃(𝐵) = then 𝑃 =

(A) (B)

(C) (D)

75. nks Lora= ?kVukvksa 𝐸 vkSj 𝐹 ds fy, tc 𝑃(𝐹) ≠ 0 gks rks 𝑃 =

(A) 𝑃(𝐸) (B) 2𝑃(𝐸)

(C) 𝑃(𝐹) (D) 2𝑃(𝐹)

For two independent events 𝐸 and 𝐹 when 𝑃(𝐹) ≠ 0 then 𝑃 =

(A) 𝑃(𝐸) (B) 2𝑃(𝐸)

(C) 𝑃(𝐹) (D) 2𝑃(𝐹)

76. nks Lora= ?kVukvksa 𝐸 vkSj 𝐹 ds fy, 𝑃(𝐸 ∩ 𝐹) =

(A) 𝑃(𝐸) + 𝑃(𝐹) (B) 𝑃(𝐸). 𝑃(𝐹)

( )
(C) 𝑃(𝐸) − 𝑃(𝐹) (D)
( )

23

For two independent events 𝐸 and 𝐹, 𝑃(𝐸 ∩ 𝐹) =

(A) 𝑃(𝐸) + 𝑃(𝐹) (B) 𝑃(𝐸). 𝑃(𝐹)

( )
(C) 𝑃(𝐸) − 𝑃(𝐹) (D)
( )

77. 3𝚤⃗ + 𝚥⃗ + 4𝑘⃗ 𝑋(𝚤⃗ − 𝚥⃗ + 𝑘⃗ ) =

(A) √42 (B) √47

(C) 7 (D) 1

78. 𝚤⃗. 𝚥⃗ 𝑋 𝑘⃗ + 𝚥⃗. 𝚤⃗ 𝑋 𝑘⃗ + 𝑘⃗ . (𝚤⃗ 𝑋 𝚥⃗) =

(A) 3 (B) 0

(C) 1 (D) -1

79. ;fn 𝑥(⃗𝑖 + ⃗𝑗 + 𝑘⃗) ,d bdkbZ lfn’k gks rks 𝑥 dk eku gS

(A) ± (B) ±
√ √

(C) ± (D) ±1
√

If 𝑥(⃗𝑖 + ⃗𝑗 + 𝑘⃗) be the unit vector then the value of 𝑥 is

(A) ± (B) ±
√ √

(C) ± (D) ±1
√

80. ;fn 𝑎⃗ rFkk 𝑏⃗ lekarj gks rks

(A) 𝑎⃗ 𝑋 3𝑏⃗ = 0⃗ (B) 𝑎⃗ 𝑋 2𝑏⃗ = 0⃗

24

(C) 𝑎⃗ 𝑋 𝑏⃗ = 0⃗ (D) buesa ls lHkh

If 𝑎⃗ and 𝑏⃗ are parallel then

(A) 𝑎⃗ 𝑋 3𝑏⃗ = 0⃗ (B) 𝑎⃗ 𝑋 2𝑏⃗ = 0⃗

(C) 𝑎⃗ 𝑋 𝑏⃗ = 0⃗ (D) All of these

81. fuEufyf[kr esa dkSu _.ksÙkj O;ojks/k gS \

(A) 𝑥 ≥ 0, 𝑦 ≥ 0 (B) 𝑧 = 𝑥 + 5𝑦

(C) 𝑥 ≤ 0, 𝑦 ≤ 0 (D) buesa dksbZ ugha

Which of the following are non-negative constraints ?

(A) 𝑥 ≥ 0, 𝑦 ≥ 0 (B) 𝑧 = 𝑥 + 5𝑦

(C) 𝑥 ≤ 0, 𝑦 ≤ 0 (D) none of these

82. fuEufyf[kr esa dkSu mís’; Qyu gS \
(A) 𝑧 = 3𝑥 + 11𝑦 (B) 𝑥 ≥ 0
(C) 𝑦 ≥ 0 (D) 𝑥 + 𝑦 ≤ 7
Which of the following is an objective function ?
(A) 𝑧 = 3𝑥 + 11𝑦 (B) 𝑥 ≥ 0
(C) 𝑦 ≥ 0 (D) 𝑥 + 𝑦 ≤ 7
9 11
83. =
7 9
(A) 1 (B) 2
(C) 3 (D) 4
84. O;ojks/k 𝑥 + 𝑦 ≤ 4, 𝑥 ≥ 0, 𝑦 ≥ 0 ds varxZr 𝑧 = 5𝑥 + 7𝑦 dk vf/kdre eku gS

(A) 20 (B) 28

25

(C) 48 (D) 140

The maximum value of 𝑧 = 5𝑥 + 7𝑦 subject to the constraints

𝑥 + 𝑦 ≤ 4, 𝑥 ≥ 0, 𝑦 ≥ 0 is

(A) 20 (B) 28

(C) 48 (D) 140

85. O;ojks/k 𝑥 + 𝑦 ≤ 2, 𝑥 ≥ 0, 𝑦 ≥ 0 ds varxZr 𝑧 = 4𝑥 − 3𝑦 dk vf/kdre eku gS

(A) 8 (B) -6

(C) 0 (D) 2

The maximum value of 𝑧 = 4𝑥 − 3𝑦 subject to the constraints

𝑥 + 𝑦 ≤ 2, 𝑥 ≥ 0, 𝑦 ≥ 0 is

(A) 8 (B) -6

(C) 0 (D) 2

86. O;ojks/k 2𝑥 + 3𝑦 ≤ 6, 𝑥 ≥ 0, 𝑦 ≥ 0 ds varxZr 𝑧 = 5𝑥 + 7𝑦 dk U;wure eku

gS

(A) 14 (B) 15

(C) 0 (D) -23

The minimum value of 𝑧 = 5𝑥 + 7𝑦 subject to the constraints

2𝑥 + 3𝑦 ≤ 6, 𝑥 ≥ 0, 𝑦 ≥ 0 is

(A) 14 (B) 15

(C) 0 (D) -23
26

87. O;ojks/kksa 3𝑥 + 4𝑦 ≤ 24, 𝑥 ≥ 0, 𝑦 ≥ 0 ds varxZr 𝑧 = 4𝑥 − 5𝑦 dk U;wure

eku gS

(A) 32 (B) -30

(C) 0 (D) buesa dksbZ ugha

The minimum value of 𝑧 = 4𝑥 − 5𝑦 subject to the constraints

3𝑥 + 4𝑦 ≤ 24, 𝑥 ≥ 0, 𝑦 ≥ 0 is

(A) 32 (B) -30

(C) 0 (D) none of these

88. 𝑧 = −7𝑥 − 8𝑦 dk U;wure eku] O;ojks/kksa 𝑥 + 𝑦 ≤ 11, 𝑥 ≥ 0, 𝑦 ≥ 0 ds varxZr gS

(A) 0 (B) -77

(C) -88 (D) buesa dksbZ ugha

The minimum value of 𝑧 = −7𝑥 − 8𝑦 subject to constraints

𝑥 + 𝑦 ≤ 11, 𝑥 ≥ 0, 𝑦 ≥ 0 is

(A) 0 (B) -77

(C) -88 (D) none of these

89. ewy fcUnq ls fcUnq ¼2] 4] 6½ dh nwjh gS

(A) 56 (B) 2√14

(C)12 (D) √102

The distance of a point (2,4,6) from origin is

27

(A) 56 (B) 2√14

(C)12 (D) √102

90. nks leryksa 2𝑥 + 3𝑦 + 4𝑧 = 4 rFkk 4𝑥 + 6𝑦 + 8𝑧 = 12 ds chp dh nwjh

gS

(A) 2 (B) 4

(C) 8 (D)
√

Distance between the two planes 2𝑥 + 3𝑦 + 4𝑧 = 4 and

4𝑥 + 6𝑦 + 8𝑧 = 12 is

(A) 2 (B) 4

(C) 8 (D)
√

91. 𝑠𝑖𝑛 =
√

(A) − (B)

(C) (D)

92. 𝑐𝑜𝑡 − =
√

(A) (B)

(C) (D)

93. 𝑥 ∈ 𝑅, 𝑐𝑜𝑡 (−𝑥) =

28

(A) −𝑐𝑜𝑡 𝑥 (B) 𝜋 − 𝑐𝑜𝑡 𝑥

(C) 2𝜋 − 𝑐𝑜𝑡 𝑥 (D) 𝑐𝑜𝑡 𝑥−𝜋

4 6 7
94. 3 −2 9 =
−1 −8 2

(A) 0 (B) 1269

(C) -2354 (D) 1

95. |𝑥| ≤ 1, 𝑠𝑖𝑛 =

(A) 2𝑠𝑖𝑛 𝑥 (B) 2𝑐𝑜𝑠 𝑥

(C) 2𝑡𝑎𝑛 𝑥 (D) 2𝑠𝑒𝑐 𝑥

96. nks ryksa 2𝑥 + 3𝑦 + 4𝑧 = 9 rFkk 𝑥 − 2𝑦 + 𝑧 = 5 ds chp dk dks.k gS

(A) (B)

√
(C) (D) 𝑠𝑖𝑛

Angle between two planes 2𝑥 + 3𝑦 + 4𝑧 = 9 and 𝑥 − 2𝑦 + 𝑧 = 5 is

(A) (B)

√
(C) (D) 𝑠𝑖𝑛

97. 𝑥𝑦 > 1; 𝑥, 𝑦 > 0 𝑡𝑎𝑛 𝑥 + 𝑡𝑎𝑛 𝑦 =

(A) 𝑡𝑎𝑛 ( ) (B) 𝜋 + 𝑡𝑎𝑛 ( )

(C) 𝑡𝑎𝑛 ( ) (D) 𝑡𝑎𝑛 ( )

29

98. nks js[kkvksa ftuds fnd~ vuqikr (1, 1, 2) vkSj (√3 − 1, −√3 − 1, 4) gS] ds chp

dk U;wudks.k gS

(A) (B)

(C) (D)

The acute angle between the two lines whose direction ratios are

(1, 1, 2) and (√3 − 1, −√3 − 1, 4) is

(A) (B)

(C) (D)

99. fcanqvksa ¼3] 5] 7½ vkSj ¼2] 4] 9½ ls xqtjus okyh js[kk dk lehdj.k gS

(A) = = (B) = =

(C) 𝑥 − 2 = 𝑦 − 4 = 𝑧 − 9 (D) buesa dksbZ ugha

The equation of a line passing through two points (3, 5, 7) and

(2, 4, 9) is

(A) = = (B) = =

(C) 𝑥 − 2 = 𝑦 − 4 = 𝑧 − 9 (D) none of these

2002 2003 2004
100. 2005 2008 2017 =
3 5 13

(A) 21645 (B) 39780

30

(C) 42375 (D) 0

[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 = {𝑛 : 𝑛 ∈ 𝑁}𝑁 rFkk Qyu 𝑓: 𝑁 → 𝑌 tgk¡ 𝑓 (𝑛) = 𝑛 rks fl)

djsa fd 𝑓 O;qRØe.kh; gSA 𝑓 dk izfrykse Hkh Kkr djsAa 2

If 𝑌 = {𝑛 : 𝑛 ∈ 𝑁}𝑁 and the function 𝑓: 𝑁 → 𝑌 as 𝑓 (𝑛) = 𝑛

Show that 𝑓 is invertible. Also find the inverse of 𝑓 .

2. gy djsa % 𝑡𝑎𝑛 = 𝑡𝑎𝑛 𝑥, 𝑥 > 0. 2

Solve : 𝑡𝑎𝑛 = 𝑡𝑎𝑛 𝑥, 𝑥 > 0.

3. 𝑐𝑜𝑠𝑒𝑐 (−2) dk eq[; eku Kkr djsaA 2

Find the principal value of 𝑐𝑜𝑠𝑒𝑐 (−2).

4. izkjafHkd lafØ;kvksa ds iz;ksx }kjk vkO;wg 1 2
dk O;qRØe Kkr djsAa 2
2 −1

31

1 2
Find the inverse of the matrix by elementary operations.
2 −1

𝑎 𝑏 𝑐
5. ;fn 𝑎, 𝑏, 𝑐 /kukRed vkSj fHkUu gSa rks lkjf.kd 𝑏 𝑐 𝑎 dk eku fudkysA 2
𝑐 𝑎 𝑏

If 𝑎, 𝑏, 𝑐 are positive and different then find the value of the

𝑎 𝑏 𝑐
determinant 𝑏 𝑐 𝑎 .
𝑐 𝑎 𝑏

( )( )
6. ;fn 𝑦 = rks Kkr djsAa 2

( )( )
If 𝑦 = , find .

7. ;fn 𝑥 = ,𝑦 = rks 𝑡 = ij Kkr djsAa 2

If 𝑥 = ,𝑦 = then find at 𝑡 = .

8. ;fn 𝑦 = 𝑡𝑎𝑛 𝑥 rks fl) djsa fd (1 + 𝑥 ) + 2𝑥 = 0. 2

If 𝑦 = 𝑡𝑎𝑛 𝑥 then prove that (1 + 𝑥 ) + 2𝑥 = 0.

9. oØ 𝑥 = 𝑎𝑐𝑜𝑠 𝜃, 𝑦 = 𝑎𝑠𝑖𝑛 𝜃 ds 𝜃 = ij vfHkyEc dk lehdj.k Kkr

djsAa 2

Find the equation of the normal to the curve

32

𝑥 = 𝑎𝑐𝑜𝑠 𝜃, 𝑦 = 𝑎𝑠𝑖𝑛 𝜃 at 𝜃 = .

10. √36.6 dk lfUudV eku izkIr djus ds fy, vody dk iz;ksx djsAa 2

Use differentials to find the approximate value of √36.6.

11. varjky [2, 4] esa Qyu 𝑓(𝑥) = 𝑥 ds fy, ek/;eku izes; dks lR;kfir djsAa 2

Verify Mean value theorem for the function 𝑓(𝑥) = 𝑥 in the interval

[2, 4].

( )
12. Kkr djsa % ∫ 𝑑𝑥. 2

( )
Find : ∫ 𝑑𝑥.

13. Kkr djsa % ∫ ( )
𝑑𝑥. 2
( )

Find : ∫ 𝑑𝑥.
( ) ( )

14. Kkr djsa % ∫( ) (
𝑑𝑥. 2
)

Find : ∫( 𝑑𝑥.
) ( )

15. lekdyu djsa % ∫ 𝑑𝑥. 2

Integrate : ∫ 𝑑𝑥.

16. ∫ |𝑥 + 2|𝑑𝑥 dk eku Kkr djsAa 2

33

Find the value of ∫ |𝑥 + 2|𝑑𝑥

⁄
17. ∫ 𝑑𝑥 dk eku Kkr djsAa 2

⁄
Find the value of ∫ 𝑑𝑥 .

18. o`Ùk 𝑥 + 𝑦 = 𝑎 dk {ks=Qy Kkr djsAa 2

Find the area of the circle 𝑥 + 𝑦 = 𝑎 .

19. vody lehdj.k = dks gy djsAa 2

Solve the differential equation = .

20. vody lehdj.k (𝑒 + 1)𝑦𝑑𝑦 = (𝑦 + 1)𝑒 𝑑𝑥 dks gy djsAa 2

Solve the differential equation (𝑒 + 1)𝑦𝑑𝑦 = (𝑦 + 1)𝑒 𝑑𝑥.

21. lfn’k 5𝚤⃗ − 𝚥⃗ + 2𝑘⃗ ds vuqfn’k ekikad 8 okyk lfn’k Kkr djsAa 2

Find the vector of magnitude 8 in the direction of the vector

5𝚤⃗ − 𝚥⃗ + 2𝑘⃗.

22.  vkSj 𝜇 Kkr djsa ;fn 2𝚤⃗ + 6𝚥⃗ + 27𝑘⃗ 𝑋 𝚤⃗ + 𝚥⃗ + 𝜇𝑘⃗ = 0⃗- 2

Find  and 𝜇 if 2𝚤⃗ + 6𝚥⃗ + 27𝑘⃗ 𝑋 𝚤⃗ + 𝚥⃗ + 𝜇𝑘⃗ = 0⃗.

23. ;fn |𝑎⃗ | = 3, 𝑏⃗ = 4, |𝑐⃗| = 2 rFkk 𝑎⃗ + 𝑏⃗ + 𝑐⃗ = 0⃗ rks

𝑎⃗. 𝑏⃗ + 𝑏⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗ dk eku Kkr djsAa 2
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If |𝑎⃗ | = 3, 𝑏⃗ = 4, |𝑐⃗| = 2 and 𝑎⃗ + 𝑏⃗ + 𝑐⃗ = 0⃗ then find the value

of 𝑎⃗. 𝑏⃗ + 𝑏⃗. 𝑐⃗ + 𝑐⃗. 𝑎⃗ .

24. js[kk;qXe 𝑟⃗ = 2𝚤⃗ − 5𝚥⃗ + 𝑘⃗ + (3𝚤⃗ + 2𝚥⃗ + 6𝑘⃗ ) vkSj

𝑟⃗ = 7𝚤⃗ − 6𝑘⃗ + 𝜇(𝚤⃗ + 2𝚥⃗ + 2𝑘⃗) ds chp dk dks.k Kkr djsAa 2

Find the angle between the pair of lines

𝑟⃗ = 2𝚤⃗ − 5𝚥⃗ + 𝑘⃗ + (3𝚤⃗ + 2𝚥⃗ + 6𝑘⃗) and

𝑟⃗ = 7𝚤⃗ − 6𝑘⃗ + 𝜇(𝚤⃗ + 2𝚥⃗ + 2𝑘⃗).

25. fl) djsa fd js[kk,¡ = = rFkk = =

leryh; gSaA 2

Prove that the lines = = and

= = are coplanar.

26. ml lery dk lehdj.k Kkr djsa ftlesa fcanq (1, −1, 2) gS rFkk leryksa

2𝑥 + 3𝑦 − 2𝑧 = 5 rFkk 𝑥 + 2𝑦 − 3𝑧 = 8 esa ls izR;sd ij yEc gSA 2

Find the equation of the plane that contains the point (1, −1, 2) and

is perpendicular to each of the planes 2𝑥 + 3𝑦 − 2𝑧 = 5 and

𝑥 + 2𝑦 − 3𝑧 = 8.

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27. 𝑧 = −3𝑥 + 4𝑦 dk vojks/kksa 𝑥 + 2𝑦 ≤ 8, 𝑥 ≥ 0, 𝑦 ≥ 0 ds varxZr U;wurehdj.k

djsAa 2

Minimize 𝑧 = −3𝑥 + 4𝑦 subject to constraints

𝑥 + 2𝑦 ≤ 8, 𝑥 ≥ 0, 𝑦 ≥ 0.

28. 𝑧 = 6𝑥 + 7𝑦 dk vojks/kksa 𝑥 + 𝑦 ≥ 4, 𝑥 ≥ 0, 𝑦 ≥ 0 ds varxZr vf/kdrehdj.k

djsAa 2

Maximize 𝑧 = 6𝑥 + 7𝑦 subject to constraints 𝑥 + 𝑦 ≥ 4, 𝑥 ≥ 0, 𝑦 ≥ 0 .

29. 𝑃(𝐴 ∪ 𝐵) Kkr djsa ;fn 2𝑃(𝐴) = 𝑃(𝐵) = rFkk 𝑃 𝐴 𝐵 = 2

Find 𝑃(𝐴 ∪ 𝐵) if 2𝑃(𝐴) = 𝑃(𝐵) = rFkk 𝑃 𝐴 𝐵 = .

30. ,d U;k¸; flDds dks 10 ckj mNkyk x;k gSA U;wure ukS fpr vkus dh izkf;drk

Kkr djsaA 2

A fair coin is tossed ten times. Find the probability of getting at least

nine 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

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31. vody lehdj.k + 𝑦𝑐𝑜𝑡𝑥 = 2𝑥 + 𝑥 𝑐𝑜𝑡𝑥(𝑥 ≠ 0) dks gy djsAa 5

Solve the differential equation + 𝑦𝑐𝑜𝑡𝑥 = 2𝑥 + 𝑥 𝑐𝑜𝑡𝑥(𝑥 ≠ 0).

32. eku Kkr djsa % ∫ 𝑑𝑥. 5

Find the value : ∫ 𝑑𝑥.

33. ,d vufHkur ikls dks Qsd
a us ij izkIr la[;kvksa dk izlj.k Kkr djsAa 5

Find the variance of the number obtained on a throw of an unbiased

die.

34. U;wurehdj.k djsa % 𝑧 = −50𝑥 + 20𝑦. 5

tcfd 2𝑥 − 𝑦 ≥ −5

3𝑥 + 𝑦 ≥ 3

2𝑥 − 3𝑦 ≤ 12

𝑥 ≥ 0, 𝑦 ≥ 0

Minimize : 𝑧 = −50𝑥 + 20𝑦 subject to constraints

2𝑥 − 𝑦 ≥ −5

3𝑥 + 𝑦 ≥ 3

2𝑥 − 3𝑦 ≤ 12

𝑥 ≥ 0, 𝑦 ≥ 0

1+𝑎 −𝑏 2𝑎𝑏 −2𝑏
35. fl) djsa fd 2𝑎𝑏 1−𝑎 +𝑏 2𝑎 = (1 + 𝑎 + 𝑏 ) -
2𝑏 −2𝑎 1−𝑎 −𝑏

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1+𝑎 −𝑏 2𝑎𝑏 −2𝑏
Prove that : 2𝑎𝑏 1−𝑎 +𝑏 2𝑎 = (1 + 𝑎 + 𝑏 ) 5
2𝑏 −2𝑎 1−𝑎 −𝑏

36. ,d js[kk ,d ?ku ds fod.kksZa ds lkFk 𝛼, 𝛽, 𝛾, 𝛿 dks.k cukrh gSA fl) djsa fd
4
𝑐𝑜𝑠2 𝛼 + 𝑐𝑜𝑠2 𝛽 + 𝑐𝑜𝑠2 𝛾 + 𝑐𝑜𝑠2 𝛿 = 5
3

A line makes angles 𝛼, 𝛽, 𝛾, 𝛿 with the diagonals of a cube. Prove that

4
𝑐𝑜𝑠 𝛼 + 𝑐𝑜𝑠 𝛽 + 𝑐𝑜𝑠 𝛾 + 𝑐𝑜𝑠 𝛿 =
3

37. fl) djsa fd 𝑡𝑎𝑛 + 𝑡𝑎𝑛 = 𝑐𝑜𝑠 = 𝑠𝑖𝑛 5

Prove that 𝑡𝑎𝑛 + 𝑡𝑎𝑛 = 𝑐𝑜𝑠 = 𝑠𝑖𝑛 .

38. ;fn 𝑦 = 𝑥 + (𝑐𝑜𝑠𝑥) rks Kkr djsAa 5

If 𝑦 = 𝑥 + (𝑐𝑜𝑠𝑥) then find .

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