Bihar Board Class 12th Model Paper 2025 Maths

BIHAR
BOARD
Model
Paper 2025

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fo"k; dksM%
Subject Code: 121/327

INTERMEDIATE EXAMINATION-2025
bUVjehfM,V ijh{kk-2025
( ANNUAL / okf"kZd )
MATHEMATICS ( ELECTIVE )
xf.kr ¼ ,sfPNd ½
I.Sc. & I.A.
dqy ç”u : 100+30+8 = 138
Total Questions : 100+30+8 = 138
(le; : 3 ?kaVs 15 feuV) ( iw.kkZad :100 )
[Time : 3 Hours 15 Minutes] [Full Marks:100]

ijh{kkfFkZ;ksa ds fy, funsZ'k%
Instructions for the candidates:
1. ijh{kkFkhZ 𝑂𝑀𝑅 mÙkj&i=d ij viuk ç'u iqfLrdk Øekad ¼10 vadks 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 answer in their own words as far as
practicable.
3. nkfguh vksj gkf”k;s ij fn;s gq, vad iw.kkaZd fufnZ"V djrs gSaA
Figures in the right hand margin indicate full marks.
4. ç'uksa dks /;ku iwoZd i<+us ds fy, 15 feuV dk vfrfjä le; fn;k x;k gSA
An extra time of 15 minutes has been allotted for the candidates to read the
questions carefully.

5. ;g ç'u iqfLrdk nks [k.Mksa esa gS – [k.M&v ,oa [k.M&c A
This question booklet is divided into two sections- Section-A and Section-B

Page 1 of 37

6. [k.M&v esa 100 oLrqfu"B ç'u gSa] ftuesa ls fdUgha dsoy 50 ç'uksa dk mÙkj nsuk
vfuok;Z gS ¼ çR;sd ds fy, 1 vad fu/kkZfjr gS ½A ipkl ls vf/kd ç”uksa ds mÙkj nsus
ij ÁFke 50 mÙkjksa dk gh ewY;kadu dEI;wVj }kjk fd;k tk,xkA lgh mÙkj dks miyC/k
djk;s x;s OMR mÙkj&i=d esa fn;s x;s lgh xksys dks uhys@dkys c‚y isu ls izxk<+
djsaA fdlh Hkh çdkj ds OgkbVuj@rjy inkFkZ@CysM@uk[kwu vkfn dk
mÙkj&iqfLrdk esa ç;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 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; ç'u gS]a ftuesa ls fdUgha 15 ç'uksa dk mÙkj nsuk vfuok;Z
gS ¼ çR;sd ç”u ds fy, 2 vad fu/kkZfjr gSa ½A buds vfrfjä] bl [k.M esa 8 nh?kZ
mÙkjh; ç'u fn;s x;s gSa] ftuesa ls fdUgha 4 ç'uksa dk mÙkj nsuk gS ¼ çR;sd ç”u ds
fy, 5 vad fu/kkZfjr gS ½A
In Section-B, there are 30 short answer type questions, out of which any 15
questions are to be answered ( each question carrying 2 marks ). Apart from
these, there are 8 long answer type questions, out of which any 4 questions are
to be answered ( each question 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.

Page 2 of 37

[k.M & v / SECTION-A
oLrqfu"B ç'u / Objective Type Questions

ç'u la[;k 1 ls 100 rd ds ç’u ds lkFk pkj fodYi fn, x, gSa] ftuesa ls ,d
lgh gSA fdUgha 50 ç'uksa ds mÙkj nsAa vius }kjk pqus x, lgh fodYi dks 𝑶𝑴𝑹
'khV ij fpfºur djsaA 𝟓𝟎 × 𝟏 = 𝟓𝟎
Questions 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. 50 × 1 = 50

1. ∫ =

(A) tan 𝑥 + cot 𝑥 + 𝐶 (B) tan2 𝑥 − cot 𝑥 + 𝐶

(C) − cot 𝑥 + tan 𝑥 + 𝐶 (D) tan 𝑥 ∙ cot 𝑥 + 𝐶

2. ∫ =
√

(A) log 𝑥 + √𝑥 − 𝑎 + 𝐶 (B) sin +𝐶

(C) log +𝐶 (D) cot +𝐶

3. ∫ 𝑥 + 1 𝑑𝑥 =

(A) 2𝑥 + 𝑐 (B) +𝑥 +𝐶

(C) 𝑥 −𝑥 +𝐶 (D) 𝑥 +𝑥+𝐶

4. ∫ 𝑑𝑥 =

(A) 5 tan 𝑥 − 3 sec 𝑥 + 𝐶 (B) 3 tan 𝑥 − 5 sec 𝑥 + 𝐶

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(C) 5 tan 𝑥 + 3 sec 𝑥 + 𝐶 (D) 3 tan 𝑥 + 5 sec 𝑥 + 𝐶

5. ∫ 𝑐𝑜𝑠𝑒𝑐 𝑥 𝑑𝑥 =

(A) log|𝑐𝑜𝑠𝑒𝑐 𝑥 + cot 𝑥| + 𝑐 (B) log tan +𝐶

(C) log|sin 𝑥| + 𝐶 (D) log cot +𝐶

( )
6. ∫ 𝑑𝑥

(A) (7 + log 𝑥) + 𝐶 (B) (log 𝑥) + 𝐶

(C) (7 + log 𝑥) + 𝐶 (D) (7 + log 𝑥) + 𝐶

7. ∫ 2(sin (𝑐𝑜𝑠 𝑥)) 𝑑𝑥 =

(A) 𝑥− +𝐶 (B) 𝑥 +𝐶

(C) 𝜋𝑥 + 𝑥 + 𝐶 (D) 𝜋𝑥 − 𝑥 + 𝐶

8. ∫ 𝑒 + sin 𝑥 𝑑𝑥 =
√

(A) 𝑒 sin−1 𝑥 + 𝐶 (B) +𝐶

(C) −𝑒 sin−1 𝑥 + 𝐶 (D) +𝐶

9. ∫ 4𝑥 log 𝑥 𝑑𝑥 =

(A) 4𝑥 log 𝑥 − 4𝑥 + 𝐶 (B) (2 log 𝑥 − 1) + 𝐶

(C) 𝑥 (2 log 𝑥 − 1) + 𝐶 (D) (2 log 𝑥 − 1) + 𝐶

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√
10. ∫ 𝑑𝑥 =
√

(A) 2 sin √𝑥 + 𝐶 (B) sin √𝑥 + 𝐶

(C) sin 2√𝑥 + 𝐶 (D) sin 𝑥 + 𝐶

11. vody lehdj.k 𝑥 ∙ + 𝑥𝑦 − 𝑥𝑦 = 0 dk ?kkr gSa

(A) 1 (B) 2

(C) 0 (D) buesa ls dksbZ ugha

Degree of the differential Equation 𝑥 ∙ + 𝑥𝑦 − 𝑥𝑦 =0

is

(A) 1 (B) 2

(C) 0 (D) None of these

12. vody lehdj.k + 𝑦 log 𝑥 = cos 𝑥 dk lekdyu xq.kd gSa

(A) 𝑥 (B) 𝑒

(C) 𝑒 ( ) (D) buesa ls dksbZ ugha

Integrating factor of the differential Equation + 𝑦 log 𝑥 = cos 𝑥 is

(A) 𝑥 (B) 𝑒

(C) 𝑒 ( ) (D) None of these

13. vody lehdj.k 𝑦 log 𝑦 𝑑𝑥 − 𝑥 𝑑𝑦 = 0 dk gy gS

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(A) 𝑥𝑦 = 𝑒 (B) 𝑦𝑒 = 𝐶

(C) 𝑥=𝑒 (D) 𝑦=𝑒

Solution of Differential equation 𝑦 log 𝑦 𝑑𝑥 − 𝑥 𝑑𝑦 = 0 is

(A) 𝑥𝑦 = 𝑒 (B) 𝑦𝑒 = 𝐶

(C) 𝑥=𝑒 (D) 𝑦=𝑒

14. vody lehdj.k = dk gy gS

(A) 𝑦+ = 𝑥+ +𝑘 (B) =𝑘

(C) tan 𝑥 − tan 𝑦=𝑘 (D) (tan 𝑥)(tan 𝑦) = 𝑘

Solution of differential equation = is

(A) 𝑦+ = 𝑥+ +𝑘 (B) =𝑘

(C) tan 𝑥 − tan 𝑦=𝑘 (D) (tan 𝑥)(tan 𝑦) = 𝑘

15. vody lehdj.k + 3𝑥 = 𝑒 dk lekdyu gS

(A) 𝑒 (B) 𝑒

(C) 3𝑦 (D) 3𝑥

Integrating factor of differential equation + 3𝑥 = 𝑒 is

(A) 𝑒 (B) 𝑒

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(C) 3𝑦 (D) 3𝑥

16. 𝚤̂ ∙ 𝚤̂ − 𝚥̂ ∙ 𝚥̂ =

(A) 1 (B) –1

(C) 0 (D) 2

17. | − 𝚤̂ − 𝚥̂ − 𝑘 =

(A) 3 (B) √3

(C) −√3 (D) 0

18. 2𝚤̂ + 5𝚥̂ − 𝑘 ∙ −𝚤̂ + 7𝚥̂ + 2𝑘 =

(A) 28 (B) 29

(C) 30 (D) 31

19. 𝚤̂ × 𝑘 =

(A) 0⃗ (B) 𝚥̂

(C) −𝚥̂ (D) 𝑘 × 𝚤̂

20. ( 𝚥̂ + 2𝑘 ) × ( 𝚤̂ + 2𝚥̂ ) =

(A) 4𝚤̂ − 2𝚥̂ + 𝑘 (B) −4𝚤̂ + 2𝚥̂ − 𝑘

(C) 4𝚤̂ + 2𝚥̂ + 𝑘 (D) −4𝚤̂ − 2𝚥̂ − 𝑘

21. ,d o`r dh f=T;k 𝑟 = 10 𝑐𝑚. ij 𝑟 ds lkis{k {ks=Qy esa ifjorZu dh nj gS

(A) 20 𝜋 𝑐𝑚 /𝑐𝑚. (B) 22 𝜋 𝑐𝑚 /𝑐𝑚.

(C) 18 𝜋 𝑐𝑚 /𝑐𝑚. (D) 10 𝜋 𝑐𝑚 /𝑐𝑚.

Page 7 of 37

The rate of change of the area of a Circle with respect to its radius 𝑟

at 𝑟 = 10 𝑐𝑚. is

(A) 20 𝜋 𝑐𝑚 /𝑐𝑚. (B) 22 𝜋 𝑐𝑚 /𝑐𝑚.

(C) 18 𝜋 𝑐𝑚 /𝑐𝑚. (D) 10 𝜋 𝑐𝑚 /𝑐𝑚.

22. oØ 𝑦 = 3𝑥 + 7 sin 𝑥 ds 𝑥 = 0 ij yEc js[kk dh ço.krk gSa

(A) –7 (B)

(C) (D) 7

The slope of the normal to the Curve 𝑦 = 3𝑥 + 7 sin 𝑥 at 𝑥 = 0 is

(A) –7 (B)

(C) (D) 7

23. eku ysa fd 𝐸 fdlh çfrn”kZ lef’V 𝑆 dh ?kVuk gSa rks 𝑃 =

(A) (B)

(C) 0 (D) 1

Let 𝐸 be an event of a sample space 𝑆 of an experiement, then

𝑃 =

(A) (B)

(C) 0 (D) 1

24. ;fn 𝑃(𝐴) = ] 𝑃(𝐵) = rFkk 𝑃(𝐴 ∩ 𝐵) = rks 𝑃 =
Page 8 of 37

(A) 1 (B) 0

(C) (D)

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

(A) 1 (B) 0

(C) (D)

25. ,d ifjokj esa nks cPps gSaA ;fn ;g Kkr gks fd cPpksa esa ls de ls de ,d

cPpk yM+dk gS rks nksuksa cPpks ds yM+dk gksus dh çkf;drk gSa

(A) (B)

(C) (D)

A family has two children the probability that both the Children are

boys given that at least one of them is a boy is

(A) (B)

(C) (D)

26. ;fn 𝑃(𝐴) = , 𝑃(𝐵) = vkSj 𝑃(𝐴 ∪ 𝐵) = rks 𝑃 =

(A) (B)

(C) (D)

Page 9 of 37

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

(A) (B)

(C) (D)

27. ;fn 𝐸 vkSj 𝐹 Lora= ?kVuk,¡ gSA rks 𝑃(𝐸 ∩ 𝐹 ) =

(A) 𝑃(𝐸) ∙ 𝑃(𝐹′) (B) 𝑃(𝐸′) ∙ 𝑃(𝐹′)

(C) 𝑃(𝐸) ∙ 𝑃(𝐹) (D) buesa ls lHkh

If 𝐸 and 𝐹 are independent events then 𝑃(𝐸 ∩ 𝐹 ) =

(A) 𝑃(𝐸) ∙ 𝑃(𝐹′) (B) 𝑃(𝐸′) ∙ 𝑃(𝐹′)

(C) 𝑃(𝐸) ∙ 𝑃(𝐹) (D) All of these

28. 𝑍𝑋-ry dk lehdj.k gS

(A) 𝑥=0 (B) 𝑦=0

(C) 𝑧=0 (D) buesa ls dksbZ ugha

The equation of 𝑍𝑋-plane is

(A) 𝑥=0 (B) 𝑦=0

(C) 𝑧=0 (D) None of these

29. 𝑦- v{k dh fnd~-dksT;k,¡ gSa

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

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

Page 10 of 37

The direction Cosines of 𝑦-axis are

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

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

30. fcUnqvksa ¼&2] 6] 7½ vkSj ¼1] 2] 7½ ds chp dh nwjh gS%

(A) 5 (B) 7

(C) 9 (D) 4

The distance between the points (–2, 6, 7) and (1, 2, 7) is

(A) 5 (B) 7

(C) 9 (D) 4

31. ∫ 𝑑𝑥 =

(A) 3 tan 𝑥 +𝐶 (B) log(𝑥 + 1) + 𝐶

(C) tan 𝑥 +𝐶 (D) sin 𝑥 +𝐶

32. ∫ 𝑑𝑥 =
√

(A) √2𝑥 + 𝑥 − 3 + 𝐶 (B) 2√2𝑥 + 𝑥 − 3 + 𝐶

(C) log|2𝑥 + 𝑥 − 3| + 𝐶 (D) (2𝑥 + 𝑥 − 3) + 𝐶

33. ∫ 𝑒 [2 cos 2𝑥 + sin 2𝑥] 𝑑𝑥 =

(A) 𝑒 cos 2𝑥 + 𝐶 (B) 2𝑒 cos 2𝑥 + 𝐶

(C) 2𝑒 sin 2𝑥 + 𝐶 (D) 𝑒 sin 2𝑥 + 𝐶

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34. ∫ =
√

(A) log 𝑥 + √𝑥 + 𝑎 +𝐶 (B) tan +𝐶

(C) log +𝐶 (D) log 𝑥 + √𝑥 − 𝑎 +𝐶

35. ∫ sin 𝑥 cos 𝑥 𝑑𝑥 =

(A) –1 (B) 0

(C) 1 (D) 2

√
36. ∫ 𝑑𝑥 =
√ √

(A) 𝜋 (B)

(C) (D)

37. ∫ 𝑒 𝑑𝑥 =

(A) 𝑒 −1 (B)

(C) 1−𝑒 (D) 𝑒−1

38. ∫ 𝑡𝑎𝑛 𝑥 𝑑𝑥 =

(A) log2 (B) log4

(C) 2log2 (D) log 2

39. ∫ =

(A) (B)

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(C) 0 (D) 𝜋

40. ∫ 𝑥𝑒 𝑑𝑥 =

(A) 1 (B) 2

(C) 3 (D) 4

41. cos =
√

(A) (B)

(C) (D)

42. tan − √3 =

(A) (B)

(C) (D)

43. tan tan =

(A) (B)

(C) (D)

44. sin − 𝑠𝑖𝑛 =

(A) (B) 1
√

(C) (D)
√

45. sin (tan 𝑥) = , |𝑥| < 1

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(A) (B)

(C) (D)

46. tan √3 − cot −√3 =

(A) 𝜋 (B)

(C) 0 (D) 2 √3

47. sin sin =

(A) (B)

(C) (D)

48. 𝑐𝑜𝑠𝑒𝑐 𝑥 dk çkar gS

(A) 𝑅 (B) (−1, 1)

(C) 𝑅 − (−1, 1) (D) [−1, 1]

Domain of 𝑐𝑜𝑠𝑒𝑐 𝑥 is

(A) 𝑅 (B) (−1, 1)

(C) 𝑅 − (−1, 1) (D) [−1, 1]

√
49. sin + cos =

(A) 0 (B)

(C) (D)

Page 14 of 37

50. Qyu 𝑓: 𝑁 → 𝑁, 𝑓(𝑥) = 2𝑥 gS%

(A) ,dSdh rFkk vkPNknd (B) ,dSdh ysfdu vkPNknd ugha

(C) vkPNknd ysfdu ,dSdh ugha (D) buesa ls dksbZ ugha

Function 𝑓: 𝑁 → 𝑁, 𝑓(𝑥) = 2𝑥 is

(A) one-one and onto (B) one-one but not onto

(C) onto but not one-one (D) None of these

51. ;fn 𝑎⃗ = 2𝚤̂ − 𝚥̂ + 2𝑘 rks 𝑎⃗ dh fn”kk esa laxr bdkbZ lfn”k 𝑎 gS%

(A) (−2𝚤̂ + 𝚥̂ − 2𝑘 ) (B) (2𝚤̂ − 𝚥̂ + 2𝑘 )

(C) (−2𝚤̂ + 𝚥̂ − 2𝑘 ) (D) (2𝚤̂ − 𝚥̂ + 2𝑘 )

If 𝑎⃗ = 2𝚤̂ − 𝚥̂ + 2𝑘 then unit vector 𝑎⃗ in the direction of vector 𝑎 is

(A) (−2𝚤̂ + 𝚥̂ − 2𝑘 ) (B) (2𝚤̂ − 𝚥̂ + 2𝑘 )

(C) (−2𝚤̂ + 𝚥̂ − 2𝑘 ) (D) (2𝚤̂ − 𝚥̂ + 2𝑘 )

52. ;fn 𝑎⃗ = 2𝚤̂ + 3𝚥̂ + 4𝑘 , 𝑏⃗ = −𝚥̂ − 2𝑘 rks 𝑎⃗ × 𝑏⃗ =

(A) 2 √6 (B) 6 √2

(C) √12 (D) √36

If 𝑎⃗ = 2𝚤̂ + 3𝚥̂ + 4𝑘, 𝑏⃗ = −𝚥̂ − 2𝑘 then 𝑎⃗ × 𝑏⃗ =

(A) 2 √6 (B) 6 √2

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(C) √12 (D) √36

53. ;fn 3𝚤̂ − 2𝚥̂ + 𝑘 rFkk 2𝚤̂ − 3𝚥̂ − 𝜆𝑘 ijLij yEc gksa rks 𝜆 =

(A) 0 (B) –12

(C) 12 (D) 6

If 3𝚤̂ − 2𝚥̂ + 𝑘 and 2𝚤̂ − 3𝚥̂ − 𝜆𝑘 are perpendicular to each

other then 𝜆 =

(A) 0 (B) –12

(C) 12 (D) 6

54. 𝚥̂ ∙ 𝚤̂ × 𝑘 =

(A) −𝚥̂ (B) 0

(C) –1 (D) 1

55. lfn”k 𝚤̂ − 𝚥̂ dk lfn”k 𝚤̂ + 𝚥̂ ij ç{ksi gSa

(A) 2 (B) 0

(C) 1 (D)

The projection of the vector 𝚤̂ − 𝚥̂ on the vector 𝚤̂ + 𝚥̂ is

(A) 2 (B) 0

(C) 1 (D)

Page 16 of 37

√
56. ;fn |𝑎⃗| = 3] 𝑏⃗ = rFkk 𝑎⃗ × 𝑏⃗ ,d bdkbZ lfn”k gSA rks 𝑎⃗ vkSj 𝑏⃗ ds

chp dk dks.k gS

(A) (B)

(C) (D)

√
If |𝑎⃗| = 3, 𝑏⃗ = and 𝑎⃗ × 𝑏⃗ is a unit vector then angle between

𝑎⃗ and 𝑏⃗ is

(A) (B)

(C) (D)

57. 𝚤̂ ∙ 𝚥̂ × 𝑘 + 𝚥̂ ∙ 𝚤̂ × 𝑘 + 𝑘 ∙ (𝚤̂ × 𝚥̂) =

(A) 0 (B) –1

(C) 1 (D) 3

58. 2𝚤̂ + 3𝚥̂ − 5𝑘 ∙ 7𝚤̂ − 𝚥̂ + 3𝑘 =

(A) 32 (B) 2

(C) 4 (D) –4

59. 𝑧 = 3𝑥 + 4𝑦 dk vf/kdre eku gS] tgk¡ fd O;ojks/k 𝑥 + 𝑦 ≤ 4, 𝑥 ≥

0, 𝑦 ≥ 0

(A) 16 (B) 0

(C) 12 (D) 28

Page 17 of 37

The maximum value of 𝑧 = 3𝑥 + 4𝑦 is, where constraints

𝑥 + 𝑦 ≤ 4, 𝑥 ≥ 0, 𝑦 ≥ 0

(A) 16 (B) 0

(C) 12 (D) 28

60. 𝑧 = 4𝑥 + 𝑦 dk U;wure eku gS] tgk¡ fd O;ojks/k 𝑥 + 𝑦 ≤ 50, 𝑥 + 𝑦 ≤ 90,

𝑥 ≥ 0, 𝑦 ≥ 0

(A) 0 (B) 50

(C) 110 (D) 120

The minimum value of 𝑧 = 4𝑥 + 𝑦 is, where constraints

𝑥 + 𝑦 ≤ 50, 𝑥 + 𝑦 ≤ 90, 𝑥 ≥ 0, 𝑦 ≥ 0

(A) 0 (B) 50

(C) 110 (D) 120

2 𝑥 10 7
61. ;fn = ] rks 𝑥 =
𝑥 8 30 21

(A) ±16 (B) ±4

(C) 0 (D) ±3

2 𝑥 10 7
If = , then 𝑥 =
𝑥 8 30 21

(A) ±16 (B) ±4

(C) 0 (D) ±3

Page 18 of 37

11 7 9
62. 2 −1 5 =
22 14 18

(A) 72 (B) –61

(C) 0 (D) 65

−1 3 0
63. 1 2 4 =
4 1 0

(A) –52 (B) 0

(C) 52 (D) 62

64. 2 × 2 dksfV ds ,sls vkO;wgks dh dqy fdruh la[;k gksxh ftudh çR;sd çfof’V

3 ;k 7 gS

(A) 42 (B) 32

(C) 64 (D) 16

The number of all possible matrices of order 2 × 2 with each entry 3

or 7 is

(A) 42 (B) 32

(C) 64 (D) 16

−2 1
65. 6 =
3 5

12 6 −12 6
(A) (B)
−18 30 18 30

12 6 6 −12
(C) (D)
18 30 18 30

Page 19 of 37

0 1
66. 𝐴 = ⟹𝐴 =
1 0

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

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

3
67. [1 − 2 0] −2 =
11

(A) [7] 3 −6 0
(B) −2 4 0
11 −22 0

(C) [18] 3 6 0
(D) 2 −4 0
11 −22 0

68. ;fn 𝐴 rFkk 𝐵 leku dksfV ds O;qRØe.kh; vkO;wg gksa rks (𝐴𝐵) =

(A) 𝐴−𝐵 (B) 𝐴 𝐵

(C) 𝐵−𝐴 (D) 𝐵 𝐴

If 𝐴 and 𝐵 are invertible matrices of same order then (𝐴𝐵) =

(A) 𝐴−𝐵 (B) 𝐴 𝐵

(C) 𝐵−𝐴 (D) 𝐵 𝐴

69. ,d Qyu 𝑓: 𝑋 → 𝑌 ,dSdh Qyu gS] ;fn 𝑥 , 𝑥 ∈ 𝑋, 𝑓(𝑥 ) = 𝑓(𝑥 ) ⇒

(A) 𝑥 𝑥 =1 (B) 𝑥 𝑥 = −1

(C) 𝑥 =𝑥 (D) buesa ls dksbZ ugha

A function 𝑓: 𝑋 → 𝑌 is one-one if 𝑥 , 𝑥 ∈ 𝑋, 𝑓(𝑥 ) = 𝑓(𝑥 ) ⇒

Page 20 of 37

(A) 𝑥 𝑥 =1 (B) 𝑥 𝑥 = −1

(C) 𝑥 =𝑥 (D) None of these

70. ,d Qyu 𝑓: 𝑋 → 𝑌 O;qRØe.kh; gS] ;fn vkSj dsoy ;fn 𝑓 gks

(A) ,dSdh (B) vkPNknd

(C) ,dSdh rFkk vkPNknd (D) buesa ls dksbZ ugha

A function 𝑓: 𝑋 → 𝑌 invertible if and only if 𝑓 is

(A) one-one (B) onto

(C) one-one and onto (D) None of these

( )
71. =

(A) −𝑐𝑜𝑠𝑥 (B)

(C) (cos 𝑥) (D) sin(𝜋 − 𝑥)

72. log sin(sec 𝑥 + 𝑐𝑜𝑠𝑒𝑐 𝑥) =

(A) 0 (B) 1

(C) cos(sec 𝑥 + 𝑐𝑜𝑠𝑒𝑐 𝑥) (D)cot(sec 𝑥 + 𝑐𝑜𝑠𝑒𝑐 𝑥)

73. ;fn 𝑦 = 𝑡 rks =

(A) 210 𝑡 (B) 420 𝑡

(C) 420 𝑡 (D) 210 𝑡

If 𝑦 = 𝑡 then =

Page 21 of 37

(A) 210 𝑡 (B) 420 𝑡

(C) 420 𝑡 (D) 210 𝑡

74. ;fn 𝑥 = 4𝑡, 𝑦 = rks =

(A) (B)

(C) (D)

If 𝑥 = 4𝑡, 𝑦 = then =

(A) (B)

(C) (D)

( )
75. =

(A) –1 (B)

(C) 1 (D) 0

(√ )
76. =

√ √
(A) (B)
√ √

√ √
(C) (D)
√

( )
77. =

(A) 1 − sin 2𝑥 (B) 2 sin 2𝑥

Page 22 of 37

(C) −2 sin 2𝑥 (D) 2 sin 𝑥

78. ∫ 𝑑𝑥 =

(A) log6 (B) log

(C) log (D) 0

79. ∫ 𝑑𝑥

(A) (sin 𝑥) + 𝐶 (B) (sin 𝑥) + 𝐶

(C) 2(sin 𝑥) + 𝐶 (D) −2(sin 𝑥) + 𝐶

80. ∫ (𝑥 + 𝑥𝑐𝑜𝑠𝑥 + sin 𝑥 + 1)𝑑𝑥 =

(A) 𝜋 (B) 0

(C) 2 (D) 1

81. ;fn ,d js[kk 𝑥, 𝑦 rFkk 𝑧-v{kksa dh /kukRed fn”kk ds lkFk Øe”k% 30 , 60

rFkk 90 dk dks.k cukrh gSA rks fnd~-dkslkbu gSa

(A) 0, 1, 0 (B) , ,
√ √ √

(C) , ,0 (D)
√
, ,0

If a line makes angle 30 , 60 and 90 with the positive direction of

𝑥, 𝑦 and 𝑧-axis respectively, then its direction cosines are

(A) 0, 1, 0 (B) , ,
√ √ √

Page 23 of 37

(C) , ,0 (D)
√
, ,0

82. nks js[kkvksa dk fnd~-vuqikr Øe”k% 𝑎 , 𝑏 , 𝑐 rFkk 𝑎 , 𝑏 , 𝑐 gSA nksuksa js[kk,¡

lekarj gksaxh] ;fn

(A) 𝑎 𝑎 +𝑏 𝑏 +𝑐 𝑐 =0 (B) + + =0

(C) 𝑎 𝑎 +𝑏 𝑏 +𝑐 𝑐 =1 (D) ≠ =

Direction ratios of two lines are 𝑎 , 𝑏 , 𝑐 and 𝑎2 , 𝑏2 , 𝑐2 respectively.

Both lines are parallel, if

(A) 𝑎 𝑎 +𝑏 𝑏 +𝑐 𝑐 =0 (B) + + =0

(C) 𝑎 𝑎 +𝑏 𝑏 +𝑐 𝑐 =1 (D) ≠ =

83. lery 𝑟⃗ ∙ 2𝚤̂ + 3𝚥̂ − 6𝑘 + 1 = 0 ij ewy fcUnq ls Mkys x, yEc bdkbZ

lfn”k dh fnd~-dkslkbu gSa

(A) , , (B) , ,
√ √ √ √ √ √

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

The direction cosines of the unit vector perpendicular to the plane

𝑟⃗ ∙ 2𝚤̂ + 3𝚥̂ − 6𝑘 + 1 = 0 and passing through the origin are

(A) , , (B) , ,
√ √ √ √ √ √

Page 24 of 37

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

84. fdlh ljy js[kk ds fnd~-vuqikr 2, −3, 5 gSA rks blds fnd~-dksT;k,¡ gS

(A) , , (B) , ,
√ √ √

(C) , , (D) , ,
√ √ √ √ √ √

If direction ratios of a straight line are 2, −3, 5 then its direction

cosines are

(A) , , (B) , ,
√ √ √

(C) , , (D) ` , ,
√ √ √ √ √ √

85. ry 2𝑥 − 3𝑦 + 7𝑧 = 9 ds lekarj ry dk lehdj.k gS

(A) 2𝑥 − 3𝑦 + 7𝑧 = 23 (B) 2𝑥 + 3𝑦 + 7𝑧 = 9

(C) 3𝑥 + 2𝑦 − 7𝑧 = 10 (D) buesa ls dksbZ ugha

The plane parallel to the Plane 2𝑥 − 3𝑦 + 7𝑧 = 9 is

(A) 2𝑥 − 3𝑦 + 7𝑧 = 23 (B) 2𝑥 + 3𝑦 + 7𝑧 = 9

(C) 3𝑥 + 2𝑦 − 7𝑧 = 10 (D) None of these

86. ;fn nks ry 𝑥 − 3𝑦 + 2𝑧 = 10 rFkk 7𝑥 + 𝑦 − 𝜆𝑧 = 5 ijLij yEc gksa

rks 𝜆 =

(A) –3 (B) –2

Page 25 of 37

(C) 2 (D) buesa ls dksbZ ugha

If two planes 𝑥 − 3𝑦 + 2𝑧 = 10 and 7𝑥 + 𝑦 − 𝜆𝑧 = 5 are

perpendicular to each other, then 𝜆 =

(A) –3 (B) –2

(C) 2 (D) None of these

87. ;fn js[kk,¡ = = rFkk = = lekarj gks rks

(A) 3𝑎 + 𝑏 + 2𝑐 = 0 (B) = =

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

If the lines = = and = = are parallel

then

(A) 3𝑎 + 𝑏 + 2𝑐 = 0 (B) = =

(C) 3𝑎 = 𝑏 = 2𝑐 (D) None of these

88. ryksa 4𝑥 + 8𝑦 + 𝑧 − 3 = 0 rFkk 𝑦 + 𝑧 − 7 = 0 ds chp dk dks.k gS

(A) (B) 0

(C) (D)

The angle between the planes 4𝑥 + 8𝑦 + 𝑧 − 3 = 0 and 𝑦 + 𝑧 −

7 = 0 is

Page 26 of 37

(A) (B) 0

(C) (D)

89. fcUnq (0, 0, 1) ls ry 2𝑥 − 𝑦 + 𝑧 = 7 dh nwjh gS

(A) 6 √6 (B) √6

(C) (D) 3
√

The distance of the plane 2𝑥 − 𝑦 + 𝑧 = 7 from the point (0, 0, 1) is

(A) 6 √6 (B) √6

(C) (D) 3
√

3 7 1 3
90. 2 +5 =
1 −2 0 4

11 29 29 11
(A) (B)
2 16 2 16

11 29 29 11
(C) (D)
2 −16 −2 −16

2 −1 0
91. ;fn rks 𝐴 =
7 3 5
2 3 2 0
(A) −1 7 (B) −1 7
0 5 3 5

2 7 2 7
(C) −1 3 (D) −1 3
0 5 0 4

If 𝐴 =
2 −1 0 then 𝐴 =
7 3 5

Page 27 of 37

2 3 2 0
(A) −1 7 (B) −1 7
0 5 3 5

2 7 2 7
(C) −1 3 (D) −1 3
0 5 0 4

𝑥+1 3
92. ;fn = 0 rks 𝑥 =
5 𝑥−1

(A) ±16 (B) ±4

(C) 16 (D) 0

𝑥+1 3
If = 0 then 𝑥 =
5 𝑥−1

(A) ±16 (B) ±4

(C) 16 (D) 0

( )
93. =

(A) (B)

(C) (D)

( )
94. =

(A) − tan 𝑥 (B) cot 𝑥

(C) − cot 𝑥 (D) tan 𝑥

1 0
95. ;fn 𝐴 = rks 𝐴 =
0 1

Page 28 of 37

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

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

96. ∫ 𝑑𝑥 =

(A) −𝑥+𝐶 (B) 𝑥− +𝐶

(C) 1−𝑥+𝐶 (D) +𝑥+𝐶

97. ∫ 𝑒 𝑑𝑥 =

(A) −7𝑒 +𝐶 (B) 𝑒 +𝐶

(C) 𝑒 +𝐶 (D) 𝑒 +𝐶

( )
98. =

(A) 2𝑥 + sin 𝑥 (B) 6𝑥 − sin 𝑥

(C) 6𝑥 + sin 𝑥 (D) sin 𝑥 − 6𝑥

99. ∫(5𝑥 + 2𝑥 + sin 𝑥)𝑑𝑥 =

(A) 𝑥 + 𝑥 + cos 𝑥 + 𝐶 (B) 𝑥 + 2 − cos 𝑥 + 𝐶

(C) 𝑥 + 𝑥 − cos 𝑥 + 𝐶 (D) 𝑥 + 𝑥 − cos 𝑥 + 𝐶

100. =

(A) (B) 2𝑥

Page 29 of 37

(C) (D)

[k.M&c @ SECTION-B
y?kq mÙkjh; ç'u @Short Answer Type Questions

ç'u la[;k 1 ls 30 rd y?kq mÙkjh; gSa A buesa ls fdUgha 15 ç'uksa ds mÙkj nsaA
çR;sd ç’u ds fy, 2 vad fu/kkZfjr gSA 𝟏𝟓 × 𝟐 = 𝟑𝟎
Question Nos. 1 to 30 are Short Answer Type. Answer any 15 questions.
Each question carries 2 marks. 15 × 2 = 30
1. 𝑋-v{k ls fcUnq (𝑎, 𝑏, 𝑐) dh nwjh Kkr dhft,A 2

Find the distance of the point (𝑎, 𝑏, 𝑐) from 𝑋-axis.

2. fcUnq (3, 0, 0), (0, −1, 0) rFkk (0, 0, 2) ls xqtjus okys ry dk lehdj.k

Kkr dhft,A 2

Find the equation of a plane passing through the points

(3, 0, 0), (0, −1, 0) and (0, 0, 2).

3. ewyfcUnq ls ,d ry ij Mkys x, yEc dh ikn dk funsZ”kkad (2, −1, 3) gSA

ml ry dk lehdj.k Kkr dhft,A 2

The foot of perpendicular drawn from the origin to a plane is

(2, −1, 3). Find the equation of the plane.

4. ;fn 𝐴 rFkk 𝐵 nks ?kVuk,¡ bl çdkj gS fd 𝑃(𝐴) = , 𝑃(𝐵) = rFkk

𝑃(𝐴 ∩ 𝐵) = rks fuEufyf[kr dks Kkr dhft,A 2

Page 30 of 37

(i) 𝑃 (ii) 𝑃

If 𝐴 and 𝐵 are two events such that 𝑃(𝐴) = , 𝑃(𝐵) = and

𝑃(𝐴 ∩ 𝐵) = then find the following

(i) 𝑃 (ii) 𝑃

5. ,d ;kn`fPNd pj 𝑥 dk çkf;drk caVu fuEufyf[kr gSA 2

𝑋 0.5 1 1.5 2

𝑃(𝑋) 𝐾 𝐾 2𝐾 𝐾

(i) 𝐾 dk eku Kkr djsAa

(ii) caVu dk ek/; Kkr djsaA

The probability distribution of a random variable 𝑥 is following.

𝑋 0.5 1 1.5 2

𝑃(𝑋) 𝐾 𝐾 2𝐾 𝐾

(i) Find the value of 𝐾.

(ii) Determine the mean of the distribution.

6. ;fn cos 𝛼 + cos 𝛽 + cos 𝛾 = 3𝜋 rks 𝛼(𝛽 + 𝛾) +

𝛽(𝛾 + 𝛼) + 𝛾(𝛼 + 𝛽) dk eku Kkr djsaA 2

If cos 𝛼 + cos 𝛽 + cos 𝛾 = 3𝜋, then find the value of

𝛼(𝛽 + 𝛾) + 𝛽(𝛾 + 𝛼) + 𝛾(𝛼 + 𝛽).

Page 31 of 37

7. tan + cot + tan sin dk eku Kkr dhft,A 2
√ √

Find the value of tan + cot + tan sin .
√ √

8. lehdj.k cos(tan 𝑥) = sin cot dks gy dhft,A 2

Solve the Equation cos(tan 𝑥) = sin cot .

9. lekdyu djs%a ∫ 𝑑𝑥 2

Integrate: ∫ 𝑑𝑥

10. lekdyu djsa% ∫ 2

Integrate: ∫

11. ;fn ∫ 𝑑𝑥 = rks 𝑎 dk eku Kkr djsAa 2

If ∫ 𝑑𝑥 = then find the value of 𝑎.

12. lekdyu djsa% ∫ 𝑑𝑥 2

Integrate: ∫ 𝑑𝑥

13. lekdyu djsa% ∫ 𝑑𝑥 2

Integrate: ∫ 𝑑𝑥

14. lekdyu djsa% ∫ |𝑥 − 5|𝑑𝑥 2

Page 32 of 37

Integrate: ∫ |𝑥 − 5|𝑑𝑥

15. lekdyu djsa% ∫ sin 𝑥 𝑑𝑥 2

Integrate: ∫ sin 𝑥 𝑑𝑥

16. ;fn 𝑎⃗ = 𝚤̂ + 𝚥̂ + 2𝑘 rFkk 𝑏⃗ = 2𝚤̂ + 𝚥̂ + 2𝑘 rks lfn”k 2𝑎⃗ − 𝑏⃗ dh

fn”kk esa bdkbZ lfn”k Kkr dhft,A 2

If 𝑎⃗ = 𝚤̂ + 𝚥̂ + 2𝑘 and 𝑏⃗ = 2𝚤̂ + 𝚥̂ + 2𝑘 then find the unit vector in

the direction of vector (2𝑎⃗ − 𝑏⃗).

17. Lfn”k 𝑣⃗ Kkr dhft,A tks fd (2𝚤̂ − 𝚥̂ + 2𝑘) rFkk (4𝚤̂ − 𝚥̂ + 3𝑘 ) ij yEc

gSA rFkk |𝑣⃗| = 6 2 Find

vector 𝑣⃗ which is perpendicular to (2𝚤̂ − 𝚥̂ + 2𝑘) and (4𝚤̂ − 𝚥̂ + 3𝑘 )

and |𝑣⃗| = 6

18. ;fn 𝑎⃗, 𝑏⃗, 𝑐⃗ bdkbZ lfn”k gSA rFkk 𝑎⃗ + 𝑏⃗ + 𝑐⃗ = 0⃗ rks 𝑎⃗ ∙ 𝑏⃗ + 𝑏⃗ ∙ 𝑐⃗

+𝑐⃗ ∙ 𝑎⃗ dk eku Kkr dhft,A 2

If 𝑎⃗, 𝑏⃗, 𝑐⃗ are unit vectors and 𝑎⃗ + 𝑏⃗ + 𝑐⃗ = 0⃗ then find the value

of 𝑎⃗ ∙ 𝑏⃗ + 𝑏⃗ ∙ 𝑐⃗ +𝑐⃗ ∙ 𝑎⃗.

19. ;fn 𝑥 = 3 cos 𝜃 − 2 cos 𝜃, 𝑦 = 3 sin 𝜃 − 2 sin 𝜃 rks Kkr

dhft,A 2

Page 33 of 37

If 𝑥 = 3 cos 𝜃 − 2 cos 𝜃, 𝑦 = 3 sin 𝜃 − 2 sin 𝜃 then find .

20. ;fn tan (𝑥 + 𝑦 ) = 𝐾 rks Kkr djsaA 2

If tan (𝑥 + 𝑦 ) = 𝐾 then find .

21. ;fn 𝑦 = log rks Kkr dhft,A 2

If 𝑦 = log , then find .

22. ;fn √𝑥 + 𝑦 = 1 rks fcUnq , ij Kkr djsAa 2

If √𝑥 + 𝑦 = 1 then find the value of at point , .

23. ;fn oØ 𝑎𝑦 + 𝑥 = 7 rFkk 𝑥 = 𝑦 ,d nwljs dk fcUnq (1, 1) ij yEcor

dkVrs gSa rks 𝑎 dk eku Kkr djsaA 2

If the curves 𝑎𝑦 + 𝑥 = 7 and 𝑥 = 𝑦 cut each other orthogonally

at (1, 1).

24. varjky Kkr djsa ftlesa Qyu 𝑓(𝑥) = 2𝑥 + 9𝑥 + 12𝑥 − 1 gzkleku

gSA

Find the interval in which the function

𝑓(𝑥) = 2𝑥 + 9𝑥 + 12𝑥 − 1 is decreasing.

25. vody lehdj.k (1 − 𝑥 ) − 𝑥𝑦 = 1 dk lekdyu xq.kd Kkr djsaA 2

Find the integrating factor of the differential equation

Page 34 of 37

(1 − 𝑥 ) − 𝑥𝑦 = 1 .

26. gy djsa% +𝑦 =𝑒 2

Solve: +𝑦 =𝑒

7 −1
2 −1 0
27. ;fn 𝐴 = rFkk 𝐵 = 2 3 rks (𝐴𝐵)′ Kkr dhft,A 2
3 4 5
5 0

7 −1
2 −1 0
If 𝐴 = and 𝐵 = 2 3 , then find (𝐴𝐵)′.
3 4 5
5 0

3 2
28. vkO;wg 𝐴 = dk O;qRØe vkO;wg Kkr djsAa 2
5 −1

3 2
Find the inverse matrix of the matrix 𝐴 = .
5 −1

29. gy djsa% 𝑥 − 𝑥 + 1 𝑥 − 1 = 2 2
𝑥+1 𝑥+1

Solve: 𝑥 − 𝑥 + 1 𝑥−1 = 2
𝑥+1 𝑥+1

30. ;fn 𝑓(𝑥) = 4 − (𝑥 − 7) rks 𝑓 (𝑥) Kkr dhft,A 2

If 𝑓(𝑥) = 4 − (𝑥 − 7) then find 𝑓 (𝑥).

nh?kZ mÙkjh; ç’u /Long Answer Type Questions
ç’u la[;k 31 ls 38 nh?kZ mÙkjh; ç’u gSa A buesa ls fdUgha 4 ç'uksa ds mÙkj nsa A
çR;sd ç’u ds fy, 4 vad fu/kkZfjr gSA 𝟒 × 𝟓 = 𝟐𝟎

Question Nos. 31 to 38 are Long Answer Type. Answer any 4 questions. Each
question carries 5 marks. 4 × 5 = 20

Page 35 of 37

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

Prove that sin + cos = tan .

32. gy djsa% ∫ 𝑥 log(1 + 2𝑥)𝑑𝑥 5

Integrate: ∫ 𝑥 log(1 + 2𝑥)𝑑𝑥

33. gy djsa% (1 + tan 𝑦)(𝑑𝑥 − 𝑑𝑦) + 2𝑥𝑑𝑦 = 0 5
Solve: (1 + tan 𝑦)(𝑑𝑥 − 𝑑𝑦) + 2𝑥𝑑𝑦 = 0
34. oØ 3𝑥 − 𝑦 = 8 ij vfHkyEc js[kkvksa dk lehdj.k Kkr djsa tks fd js[kk
𝑥 + 3𝑦 = 4 ds lekarj gSA 5
Find the equation of the normal lines to the curve 3𝑥 − 𝑦 = 8
which are parallel to the line 𝑥 + 3𝑦 = 4.
35. ;fn 𝑎⃗ = 𝚤̂ − 𝚥̂ + 𝑘 rFkk 𝑏⃗ = 𝚥̂ − 𝑘 rks 𝑐⃗ Kkr djsa rkfd 𝑎⃗ × 𝑐⃗ = 𝑏⃗
rFkk 𝑎⃗ ∙ 𝑐⃗ = 3 5

If 𝑎⃗ = 𝚤̂ − 𝚥̂ + 𝑘 and 𝑏⃗ = 𝚥̂ − 𝑘 then find 𝑐⃗ such that 𝑎⃗ × 𝑐⃗ = 𝑏⃗
and 𝑎⃗ ∙ 𝑐⃗ = 3.
36. 𝑧 = 13𝑥 − 15𝑦 dk U;wurehdj.k djsa tcfd 𝑥 + 𝑦 ≤ 7, 2𝑥 − 3𝑦 +
6 ≥ 0 , 𝑥 ≥ 0 rFkk 𝑦 ≥ 0. 5
Minimize 𝑧 = 13𝑥 − 15𝑦 subject to 𝑥 + 𝑦 ≤ 7, 2𝑥 − 3𝑦 + 6 ≥
0 , 𝑥 ≥ 0 and 𝑦 ≥ 0.
37. iklksa ds ,d tksM+s dks rhu ckj mNkyus ij f}dksa dh la[;k dk çkf;drk caVu
Kkr dhft,A 5
Find the probability distribution of number of doublets in three
throws of a pair of dice.
38. vkO;wg fof/k ls fuEukafdr jSf[kd lehdj.kksa dk gy djsAa 5
2𝑥 + 3𝑦 + 3𝑧 = 5 , 𝑥 − 2𝑦 + 𝑧 = −4

Page 36 of 37

3𝑥 − 𝑦 − 2𝑧 = 3.
Solve the following linear equations by matrix method.
2𝑥 + 3𝑦 + 3𝑧 = 5 , 𝑥 − 2𝑦 + 𝑧 = −4
3𝑥 − 𝑦 − 2𝑧 = 3.

Page 37 of 37

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