JAC Class 12 Model Question Paper 2026 Mathematics

2026

MODEL
PAPERS
JAC BOARD JHARKHAND
EXAM ACADEMIC
COUNCIL MODEL
PREPARATION QUESTION
PAPER 2026



JHARKHAND ACADEMIC COUNCIL

Jharkhand Council of Educational Research and Training, Ranchi
झारखण्ड शैक्षिक अनुसंधान एवं प्रक्षशिण पररषद् ,रााँची
MODEL QUESTION PAPER
मॉडल प्रश्न पत्र
Session: 2025-26 (सत्र: 2025-26)

Class – 12 Subject – Mathematics F. M. – 80 Time – 3 Hours
(वर्ग -12) (क्षवषय-र्क्षणत) (पण
ू ाां क-80) (समय-3 घंटा)

Instructions / निर्देश :
1. Examinees are required to answer in their own words as far as practicable. The
booklet contains 10 printed pages.
परीक्षार्थी यर्थासंभव अपिे शब्र्दों में ही उत्तर र्दें। पुनतिका में 10 मुद्रिि पृष्ठ है ।
2. This question paper has four sections: A, B, C, and D. The total number of
questions is 52.
इस प्रश्न पत्र में चार खण्ड - A, B, C, एवं D है। कु ल प्रश्नों की संख्या 52 है।
3. There are 30 multiple-choice questions in Section A. Four options are given for
each question, choose one of the correct options. Each question carries 1 marks.
खण्ड A में कु ल 30 बहुनवकल्पीय प्रश्न हैं। प्रत्येक प्रश्न के चार नवकल्प द्रर्दए गए हैं, इिमें से एक
सही नवकल्प का चयि कीनिए। प्रत्येक प्रश्न का माि 1 अंक निर्ााररि है।

4. Section B – Question numbers 31 – 38 are very short answer type. Answer any
six of these questions. Each question carries 2 marks.
खण्ड B में प्रश्न संख्या 31 - 38 अनि लघु उत्तरीय प्रश्न हैं। इिमें से द्रकन्ही छह प्रश्नों के उत्तर
र्दीनिए। प्रत्येक प्रश्न का माि 2 अंक निर्ााररि है।

5. Section C – Question numbers 39 – 46 are short answer type. Answer any six
of these questions. Each question carries 3 marks.
खण्ड C में प्रश्न संख्या 39 - 46 लघु उत्तरीय प्रश्न हैं। इिमें से द्रकन्ही छह प्रश्नों के उत्तर
र्दीनिए। प्रत्येक प्रश्न का माि 3 अंक निर्ााररि है।
6. Section D – Question numbers 47 – 52 are long answer type. Answer any four
of these questions. Each question carries 5 marks.
खण्ड D में प्रश्न संख्या 47 - 52 र्दीघा उत्तरीय प्रश्न हैं। इिमें से द्रकन्ही चार प्रश्नों के उत्तर
र्दीनिए। प्रत्येक प्रश्न का माि 5 अंक निर्ााररि है।

Section-A (खण्ड - क) (𝟏 × 𝟑𝟎 = 𝟑𝟎)

1. A relation R in a set A is said to be an equivalence relation if and only if-
(a) Only Reflexive (b) Only Symmetric
(c) Only Transitive (d) All of them i.e., Reflexive, Symmetric and Transitive
,d lac/a k R fdlh leqPp; A ij rqY;rk laca/k dgykrk gSa] ;fn vkSj dsoy ;fn&
(a) dsoy LorqY; (b) dsoy lefer
(c) dsoy laØked (d) mijksDr lHkh vFkkZr~ LorqY;] lefer ,oa laØked

2. If f : R → R, given by f(x) = 4x + 3 then find the f-1 (f inverse).
x−3
(a) f-1(x) = (b) f-1(x) does not exist
4
x−4
(c) f-1(x) = (d) None of them
3

;fn f : R → R, esa ifjHkkf"kr Qyu f(x) = 4x + 3 gSa] rks f-1 dk eku fudkysa A
x−3
(a) f-1(x) = (b) f-1(x) izkIr ugh gSA
4
x−4
(c) f-1(x) = (d) buesa ls dksbZ ughaA
3

3. If f : R → R, defined by f(x) = x 2 + 2 then find fof(x).
;fn f : R → R, f(x) = x 2 + 2 }kjk ifjHkkf"kr gSa rks fof(x) dk eku fudkysa A
(a) x 4 + 4 x 2 + 6 (b) 4 x 2 + 6
(c) x 4 + 6 (d) x 4 + 6 x 2 + 4

4. cosec -1 x + sec -1 x = ……… ; |𝑥| ≥ 1
π π
(a) (b)
2 4
3π
(c) 𝜋 (d)
2
1
5. If cot -1(− ) = x then find the value of sin x .
5
1
;fn cot -1(− 5 ) = x] rks sin x eku D;k gksxk ?
1 5
(a) (b)
√26 √26
1
(c) (d) Not possible/ laHko ugh gSaA
√24

JCERT, Ranchi Class-12_Model Question Paper Session: 2025 -26 Subject- Maths Page 2

6. A square matric B is said to be symmetric matrix if -
,d oxZ vkO;wg B lefer vkO;wg gksxk ;fn &
(a) B = B ′ (b) B = −B ′
(c) B = −B (d) None of these/ dksbZ ugha

7. If A and B are two matrices, then (AB)-1 is ……..
;fn A vkSj B nks vkO;wg gks rks (AB)-1 ………… gksxkA
(a) B-1 A-1 (b) A-1 B-1
(c) (BA)-1 (d) Undefined/ vifjHkkf"kr

x 2
8. If | | then x is equal to ……..
18 x

;fn | x 2
| rks x dk eku ……..gSA
18 x
(a) 6 (b) ± 6

(c) -6 (d) Zero

2 0 0
9. |0 3 0| = ?
0 0 8
(a 8 (b) 40
(c) 48 (d) 50

2x − 1, x < 0
10. The function f(x) = { is discontinuous at x = ?
2x + 1, x ≥ 0
(a) 1 (b) 2
(c) 0 (d) None of these
2x − 1, x < 0
Qyu f(x) = { , x =……. larr ugha gSA
2x + 1, x ≥ 0
(a) 1 ij (b) 2 ij

(c) 0 ij (d) buesa ls dksbZ ugha
11. Let f(x) = x3/2, Then f’(o) = ?
ekuk f(x) = x3/2 rks f’(o) = ?

JCERT, Ranchi Class-12_Model Question Paper Session: 2025 -26 Subject- Maths Page 3

3 1
(a) (b) 2
2

(c) does not exist (d) None of these
¼izkIr ugha gSa½ ¼buesa ls dksbZ ugha½
12. Find the least value of k for which f(x) = x2 + kx + 1 is increasing on (1, 2).
k ds fdl eku ds fy, f(x) = x2 + kx + 1, vUrjky (1, 2) ij o/kZeku gSa \
(a) −2 (b) −1
(c) 1 (d) 2

13. Find the slope of the curve x2 + y2 = 4 at the point (1, √3).
fcUnw (1, √3) ij oØ x2 + y2 = 4 dk <ky fudkysa A
1 1
(a) − √3 (b) −
2

(c) −√3 (d) 3
dx
14. ∫ =
sin2 x.cos2 x
(a) tan x + cot x + c (b) tan x – cot x + c

(c) tan x . cot x + c (d) tan x - cot 2 x + c

1
15. ∫ d𝑥 is equal to ¼cjkcj gSa½
𝑥 + √𝑥

(a) log |𝑥| + log (1+ √𝑥) + c (b) 2log (1+ √𝑥) + c
(c) log (1+ √𝑥) + c (d) log|𝑥| + c
cos2𝑥+2cos2 𝑥
16. ∫ dx =
cos2 𝑥

(a) tan𝑥 + c (b) cot𝑥 + c
(c) sin𝑥 + c (d) cos𝑥 + c
𝑥/2 4 + 3sin𝑥
17. ∫0 log( )d𝑥 =
4 + 3cos𝑥
3
(a) 2 (b)
4
(c) 0 (d) −2

1 2𝑥 − 1
18. ∫0 tan-1 ( )dx =
1 + 𝑥 − 𝑥2

JCERT, Ranchi Class-12_Model Question Paper Session: 2025 -26 Subject- Maths Page 4

(a) 1 (b) 0
π
(c) −1 (d)
4
1/2 dx
19. ∫0 =
√1−𝑥 2
𝜋 𝜋
(a) (b)
6 3
𝜋
(c) (d) None of these ¼buesa ls dksbZ ugha½
2
d4 y dy
20. The order and degree of the differential equation = y + ( )4 are respectively
dx4 dx
d4 y dy
vody lehdj.k dx4 = y + (dx)4 dh dksfV ,oa ?kkr Øe'k% …….. gSA
(a) 2,2 (b) 4,1
(c) 2,4 (d) 4,2
𝑑𝑦
21. What is the integrating factor of + 𝑦 𝑠𝑒𝑐𝑥 = 𝑡𝑎𝑛𝑥 ?
𝑑𝑥
dy
+ y secx = tanx dk lekdyu xq.kkad D;k gSa\
dx

(a) secx + tanx (b) log(secx + tanx)
(c) esecx (d) secx

𝑎 = 2𝑖̂ + 3𝑗̂ + 2𝑘̂ 𝑜𝑛 ⃗⃗⃗
22. Find the projection of vector ⃗⃗⃗ 𝑏 = 𝑖̂ + 2𝑗̂ + 𝑘̂.
lfn'k 𝑎⃗⃗⃗ = 2𝑖̂ + 3𝑗̂ + 2𝑘̂ dk lfn'k ⃗⃗⃗𝑏 = 𝑖̂ + 2𝑗̂ + 𝑘̂ ij iz{ksi fudkysa A
10 15
(a) (b) √
√6 6

5√6 6√5
(c) (d)
3 3

𝑎 = 5𝑖̂ + 7𝑗̂ − 3𝑘̂ on ⃗⃗⃗
23. If ⃗⃗⃗ 𝑏 = 2𝑖̂ − 3𝑗̂ − 𝑘̂ then find the value of ⃗⃗⃗ 𝑎 − ⃗⃗⃗𝑏 .
;fn ⃗⃗⃗𝑎 = 5𝑖̂ + 7𝑗̂ − 3𝑘̂ vkSj 𝑏⃗⃗⃗ = 2𝑖̂ − 3𝑗̂ − 𝑘̂ rks 𝑎⃗⃗⃗ − ⃗⃗⃗𝑏 dk eku fudkysa A
(a) 7𝑖̂ + 10𝑗̂ − 4𝑘̂ b) 3𝑖̂ + 10𝑗̂ − 2𝑘̂

(c) 7𝑖̂ + 4𝑗̂ − 2𝑘̂ (d) 10𝑖̂ − 21𝑗̂ + 3𝑘̂

24. If |𝑎| = 1, |𝑏⃗| = 2 and 𝑎
⃗⃗⃗ . ⃗⃗⃗
𝑏 = 1 then find the angle between ⃗⃗⃗ ⃗⃗⃗ .
𝑎 and 𝑏

;fn |𝑎| = 1, |𝑏⃗| = 2 vkSj 𝑎⃗⃗⃗ . ⃗⃗⃗𝑏 = 1 gks rks ⃗⃗⃗𝑎 vkSj 𝑏⃗⃗⃗ ds chp dk dks.k fudkysa A
𝜋 𝜋
(a) b) 4
3

JCERT, Ranchi Class-12_Model Question Paper Session: 2025 -26 Subject- Maths Page 5

𝜋 2𝜋
(c) (d)
2 3

25. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑐𝑜𝑠𝑖𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖̂ + 2𝑗̂ + 3𝑘̂.
lfn'k 𝑖̂ + 2𝑗̂ + 3𝑘̂ dk fnd~&dkslkbu fudkysa A
̂
𝑖̂ + 2𝑗̂ + 3𝑘
(a) 1,2,3 b)
√14
1 2 3 −1 −2 −3
(c)
14
, 14
, (d) , ,
√ √ √14 √ 14 √ 14 √14

26. A line makes equal angles with the axes. Then find the direction cosines.
,d js[kk dh fnd~&dkslkbu fudkysa , tks v{kksa ds lkFk leku dks.k cukrh gSaA
1 1 1
(a) ±1,±1, ±1 (b) ±
√2
, ± , ±
√2 √2
1 1 1
(c) 0, 0, 0 (d) ± , ± , ±
√3 √3 √3

27. Find the vector equation of the line passing through the points (-1, 0, 2) and (3, 4, 6).
fcUnw (-1, 0, 2) vkSj (3, 4, 6) ls gksdj tkus oky js[kk dk lfn'k lehdj.k fudkysa A
(a) 𝑖̂ + 2𝑘̂ + 𝜆̂ (4𝑖̂ + 4𝑗̂ + 4𝑘̂) ̂+ 𝜆
(b) 𝑖̂ − 2𝑘 ̂)
̂ (4𝑖̂ + 4𝑗̂ + 4𝑘

(c)− 𝑖̂ + 2𝑘̂ + 𝜆̂ (4𝑖̂ + 4𝑗̂ + 4𝑘̂) (d) − 𝑖̂ + 2𝑘̂ + 𝜆̂ (4𝑖̂ − 4𝑗̂ − 4𝑘̂)
28. The vector equation
𝑟⃗⃗ = (−3𝑖̂ + 5𝑗̂ − 6𝑘̂) + 𝜆̂ (2𝑖̂ + 4𝑗̂ + 2𝑘̂) then the cartesian form of the equation is
lfn'k lehdj.k 𝑟⃗⃗ = (−3𝑖̂ + 5𝑗̂ − 6𝑘̂) + 𝜆̂ (2 𝑖̂ + 4𝑗̂ + 2𝑘̂) dks dkrhZ; lehdj.k #i
gksxkA
x−3 y+5 z−6 x+3 y−5 z+6
(a) = = (b) = =
2 4 2 2 4 2
x+3 y+5 z+6
(c) = = (d) None of these ¼buesa ls dksbZ ugha½
−2 −4 −2

29. A card is drawn from a pack of 52 cards. What is the probability of getting a king of
black suit?
52 rk'k ds iRRkksa dh ,d xìh ls ,d dkyk jktk feyus dh izkf;drk D;k gSa\
1 1
(a) (b)
52 26
3 7
(c) (d)
26 52
30. A die is thrown once, then find the probability of getting a number greater than 3.
,d ikls dks ,d Qsadus ij 3 ls cM+h la[;k vkus dh izkf;drk fudkysaA

JCERT, Ranchi Class-12_Model Question Paper Session: 2025 -26 Subject- Maths Page 6

1 2
(a) (b)
2 3

(c) 3 (d) 0

Section-B (खण्ड – ख) (𝟐 × 𝟔 = 𝟏𝟐)

31. If f(x) = sin x and g(x) = x2 then find the value of fog(x).
;fn f(x) = sin x vkSj g(x) = x2 gSa rks fog(x) dk eku Kkr djsAa
32. If a function f : R → R, defined by f(x) = l x l, x ∈ R then examine whether the
function is one-one or many one.

Qyd f : R → R, dks ,dSd ;k cgq,d ds fy, tk¡ps(a tcfd f(x) = l x l, x ∈ R
2 7 1
33. Prove that, tan-1
11
+ tan-1 24 = tan-1 2
2 7 1
fl) djsa fd tan-1 + tan-1 = tan-1
11 24 2
5 60 7
34. Evluate |3 36 2|
2 24 4
5 60 7
|3 36 2| dk eku Kkr djsaA
2 24 4
dy
35. If y = b sin , x = a cos 𝜃 then find the value of .
dx
dy
;fn y = b sin 𝜃 , x = a cos 𝜃 ,rks dx dk eku Kkr djsaA
1 dx
36. Evluate ∫0
√1−x2

1 dx
∫0 √1−x2 dk eku Kkr djsaA
dy
37. Solve the differential equation = 1- x + y – xy.
dx
dy
= 1- x + y – xy vody lehdj.k dk eku Kkr djsaA
dx
5 𝐴 2
38. Evluate P(A ∪ B), if 2 P(A) = P(B) =
13
and P( ) =
𝐵 5
.
5 𝐴 2
P(A ∪ B) Kkr dhft, ;fn 2 P(A) = P(B) = vkSj P(𝐵) = 5 .
13

JCERT, Ranchi Class-12_Model Question Paper Session: 2025 -26 Subject- Maths Page 7

Section-C (खण्ड – ग) (𝟑 × 𝟔 = 𝟏𝟖)

39. Show that f(x) = lx - 2l is continuous but not differentiable at x = 2.
fl) dhft, fd x = 2 ij f(x) = lx - 2l larr gSa] ijUrq vodyuh; ugha gSaA

−1 −2 −2
40. If A = [ 2 1 −2] , Show that adj A = 3A′ .
2 −2 1
−1 −2 −2
;fn A = [ 2 1 −2] , rks fl) djsa adj A = 3A′ .
2 −2 1

41. Find the intervals in which the function f given by f(x) = 2x2 – 3x.
varjky Kkr dhft, ftuesa f(x) = 2x2 – 3x ls iznRr Qyu f –
(a) strictly increasing@fujarj o/kZeku
(b) strictly decreasing@fujarj ãleku
dy
42. Find of (cos y)x = (cos x)y.
dx
dy
;fn (cos y)x = (cos x)y , rks dx Kkr djsAa
x+3
43. Evaluate ∫ dx.
√5−4x−x2
x+3
∫ √5−4x−x2 dx dk eku Kkr djsaA
x/3 dx
44. Evaluate ∫x/6 .
1+√tanx
x/3 dx
∫x/6 1+√tanx dk eku Kkr djsaA

45. If 𝑎, 𝑏⃗ , 𝑐⃗⃗ be three vectors such that 𝑎 + 𝑏⃗ + 𝑐⃗⃗ = ⃗0 and |𝑎 | = 3, |𝑏⃗ | = 5, |𝑐⃗⃗ | =
7 . Find the angle between 𝑎 and 𝑏⃗.
;fn 𝑎, 𝑏⃗ , 𝑐⃗⃗ rhu lfn'k bl izdkj gSa fd 𝑎 + 𝑏⃗ + 𝑐⃗⃗ = ⃗0 vkSj |𝑎 | = 3, |𝑏⃗ | = 5,
|𝑐⃗⃗ | = 7 gSa rks lfn'k 𝑎 vkSj 𝑏⃗ ds chp dks.k dk eku Kkr djsAa
46. Show that the points A(1,2,7), B(2,6,3) and C(3,10,-1) are collinear.
fl) djsa fd fcanq A(1,2,7), B(2,6,3) vkSj C(3,10,-1) ,djSf[kd gSA

JCERT, Ranchi Class-12_Model Question Paper Session: 2025 -26 Subject- Maths Page 8

Section-D (खण्ड – घ) (𝟓 × 𝟒 = 𝟐𝟎)

cos x − sin x 0
47. If f(x) = | sin x cos x 0| then prove that f(x+y) = f(x), (y).
0 0 1
cos x − sin x 0
;fn f(x) = | sin x cos x 0| fl) djsa fd f(x+y) = f(x), (y).
0 0 1
48. Find the maximum and minimum values of 3x4 – 8x3 + 12x2 – 48x + 25 in [0,3]?
varjky [0,3] ij 3x4 – 8x3 + 12x2 – 48x + 25 ds mPpRre eku vkSj fuUere eku Kkr
dhft,\

49. Find the angle between the line
x−2 y−1 z+3 x+2 y−4 z−5
= = and
−1
= =
2 5 −3 8 4
x−2 y−1 z+3 x+2 y−4 z−5
fn, x;s js[kkvksas = = vkSj −1
= = ds chp dk dks.k Kkr djsAa
2 5 −3 8 4
5 A 2
50. Evaluate P(A U B), if 2P(A) = P(B) = and P ( ) = .
13 B 5
5 A 2
P(A U B) Kkr dhft,] ;fn 2P(A) = P(B) = vkSj P ( ) = .
13 B 5

51. Find maximum and minimum value of
Z = 5x + 10y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0 x,y ≥ 0
fuEu vojks/kksa ds varxZr Z = 5x + 10y dk U;wurehdj.k rFkk vf/kdrehdj.k dhft,%
x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0 x,y ≥ 0

52. Find the area lying above the x-axis and included between the curves x2 + y2 = 8x
and y2 = 4x.
x-d{k ds Åij o`r x2 + y2 = 8x ,oa ijoy; y2 = 4x ds e/;orhZ {ks= dk {ks=Qy Kkr
dhft,A
***************************************************************

JCERT, Ranchi Class-12_Model Question Paper Session: 2025 -26 Subject- Maths Page 9

Answer keys:-

01 D 11 C 21 A
02 A 12 A 22 A
03 A 13 A 23 B
04 A 14 B 24 A
05 B 15 B 25 C
06 A 16 A 26 D
07 A 17 C 27 C
08 B 18 B 28 B
09 C 19 A 29 B
10 C 20 B 30 A

JCERT, Ranchi Class-12_Model Question Paper Session: 2025 -26 Subject- Maths Page 10

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