HBSE Class 12 Sample Paper 2026 Answers Maths

BOARD OF SCHOOL EDUCATION HARYANA
MARKING SCHEME
CLASS: 12th (Sr. Secondary)

Practice Paper 2025 – 26 SET – A

गणित
[MATHEMATICS]
[ ENGLISH MEDIUM ]
• मार्किंग स्कीम में दिए गए हल केवल एक ववधि है इसके अतिरिक्ि सब
ववधियाां भी बिाबि मान्य होंगी यदि वे गणििीय रूप से सही हैं |
• The solution methods adopted in the marking scheme are suggestive.
Different methods are also acceptable if these are mathematically correct.
Section -A : (1 Mark each)
Question Answer Hints/ Solution
No.

1. 3𝜋 1 1
cos −1 (− ) = 𝜋 − cos −1 ( )
4 √2 √2
𝜋 3𝜋
= 𝜋− =
4 4
2. C Skew Symmetric
3. D −8
𝐴−1 𝑒𝑥𝑖𝑠𝑡𝑠 𝑖𝑓 |𝐴| ≠ 0 ⟹ 𝜆 ≠
5
4. C 𝑝𝑢𝑡 ∆= 86 ⟹ 𝑎 = −7,3
5. 1 log (log 𝑥)
𝑥 𝑙𝑜𝑔7 log 𝑥 𝑦 = log 7 (log 𝑥) =
log 7
6. 2 𝑑𝑓 𝑓′(𝑥) −2√1 − 𝑥 2
= = =2
𝑑𝑔 𝑔′(𝑥) −1√1 − 𝑥 2

7. D 𝑔(𝑥)
𝑚𝑎𝑦 𝑏𝑒 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥
𝑓(𝑥)
= 0 ; 𝑎𝑙𝑙 𝑜𝑡ℎ𝑒𝑟 𝑎𝑟𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒𝑙𝑦 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠.
8. 10√3 𝑐𝑚2 /𝑠 𝑑𝐴 √3 𝑑𝑥 √3
= . 2𝑥. = . 10.2 = 10√3 𝑐𝑚2 /𝑠
𝑑𝑡 4 𝑑𝑡 2

9. B 1.𝑑𝑥 𝑠𝑖𝑛2 𝑥+𝑐𝑜𝑠 2 𝑥
∫ 𝑠𝑖𝑛2 𝑥.𝑐𝑜𝑠2 𝑥 = ∫ 𝑠𝑖𝑛2 𝑥.𝑐𝑜𝑠2𝑥 𝑑𝑥
∫(𝑠𝑒𝑐 2 𝑥 + 𝑐𝑜𝑠𝑒𝑐 2 𝑥)𝑑𝑥 = tan 𝑥 − 𝑐𝑜𝑡 𝑥 + 𝑐
10. D
𝑑𝑦 𝑥
= ⟹ ∫ 𝑦 𝑑𝑦 = ∫ 𝑥 𝑑𝑥
𝑑𝑥 𝑦
𝑦2 − 𝑥 2 = 𝑐
11. B Put sin 𝑥 = 𝑡 ⟹ cos 𝑥 𝑑𝑥 = 𝑑𝑡
12. 1 𝑑2𝑦
Since the highest power raised to is one.
𝑑𝑥 2

13. B ±(𝑎⃗ × 𝑏⃗⃗) 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑡𝑤𝑜 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 ⊥ 𝑏𝑜𝑡ℎ 𝑎⃗, 𝑏⃗⃗
14. D For reflection in XY-plane negate the z-coordinate.
15. A 𝑟⃗⃗⃗ = 𝑎⃗ + 𝜆𝑏⃗⃗ ; where
𝑎⃗ 𝑑𝑒𝑛𝑜𝑡𝑒𝑠 𝑡ℎ𝑒 𝑝𝑎𝑠𝑠𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑎𝑛𝑑 𝑏⃗⃗ denotes the
direction vector of the line.
16. B 𝑃(𝑡𝑤𝑜 ℎ𝑖𝑡𝑠) = 𝑃(𝐴)𝑃(𝐵)𝑃(𝐶′) +
𝑃(𝐴)𝑃(𝐵′)𝑃(𝐶) + 𝑃(𝐴′)𝑃(𝐵)𝑃(𝐶)
17. B 𝑑𝑦 3𝑡 𝑑 2 𝑦 3 𝑑𝑡 3
= ⟹ 2= . =
𝑑𝑥 2 𝑑𝑥 2 𝑑𝑥 4𝑡
18. 1 𝑓𝑜𝑟 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑙𝑜𝑐𝑎𝑙 𝑚𝑖𝑛𝑖𝑚𝑎 𝑝𝑢𝑡 𝑓 ′ (𝑥) = 0
⟹ 𝑥 = 4 𝑎𝑛𝑑 𝑓 ′′ (𝑥) = +𝑣𝑒
Then f(4)=1=minimum value
19. D A is false as for mutually exclusive events
𝑃(𝐴⋂𝐵) = 0 𝑎𝑙𝑤𝑎𝑦𝑠 ; but R is true here.
20. A Both A and R are true and correct
explanation.

खंड – ब
SECTION – B (2×5=10)
21. Consider the corresponding eqn. system:
𝑥+𝑦 =8 3𝑥 + 5𝑦 = 15
x 0 8
y 8 0 x 5 0 1
y 0 3

1

No feasible region.
22(a). On multiplying first two ½
𝑥
[2𝑥 − 9 4𝑥 ] [ ] = [0]
8

2𝑥 2 + 23𝑥 = 0 ½
23
𝑥 = 0, −
2 1
OR
22(b). 1 0 1
1
Area of triangle= |6 0 1| 1
2
4 3 1
15
= 𝑠𝑞𝑢𝑎𝑟𝑒 𝑢𝑛𝑖𝑡𝑠 1
2

23(a).
𝑓 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑥 = 3 𝑖𝑓

𝑅. 𝐻. 𝐿. = 𝐿. 𝐻. 𝐿. = 𝑓(0) 1
2𝑠𝑖𝑛2 2𝑥
lim =𝑘
𝑥⟶0 8𝑥 2
sin 2𝑥 2
lim ( ) =𝑘
OR 𝑥⟶0 2𝑥 1
⟹𝑘=1
23(b). 𝑑𝑉 𝑑𝑥
𝑉 = 𝑥3 ⟹ = 3𝑥 2 = 9𝑥 2 1
𝑑𝑡 𝑑𝑡
𝑑𝑉
Put 𝑥 = 10 𝑐𝑚 ⟹ = 900 𝑐𝑚 /𝑠 3 1
𝑑𝑡
24. P (Exactly one of A, B is selected) = 0.6 (given)
P (A∩B′) + P (A′∩B) = 0.6 1
P (A) P (B′) + P (A′) P (B) = 0.6
⇒ (0.7) (1 – p) + (0.3) p = 0.6
⇒ p = 0.25 1
Thus the probability that B gets selected is 0.25

25. ∫ 𝑡𝑎𝑛2 𝑥. 𝑡𝑎𝑛2 𝑥. 𝑑𝑥 = ∫(𝑠𝑒𝑐 2 𝑥 − 1). 𝑡𝑎𝑛2 𝑥
1
2 2 2
= ∫ 𝑠𝑒𝑐 𝑥. 𝑡𝑎𝑛 𝑥. 𝑑𝑥 − ∫ 𝑠𝑒𝑐 𝑑𝑥 + ∫ 1𝑑𝑥
1 1
== 𝑡𝑎𝑛3 𝑥 − tan 𝑥 + 𝑥 + 𝑐
3

खंड – स
SECTION – C (3×6=18)

26.

1.5

𝟐𝟎 𝟒𝟓
O (0, 0), A (2, 0), B (0, 3) and C ( , ) are 0.5
𝟏𝟗 𝟏𝟗
the corner points of the feasible region.
Corner Point Value of Z Remarks
O(0,0) 0
A(2,0) 10
1
B(0,3) 9
C(20/19,45/19) 235/19 MAXIMUM

27(a). R is reflexive since (L1,L1) ∈ 𝑅
R is symmetric since if (L1,L2) ∈ 𝑅, then
1
(L2,L1) ∈ 𝑅
R is transitive since if (L1,L2) ∈ 𝑅, and 1
(L2,L3) ∈ 𝑅 then (L1,L3) ∈ 𝑅
⟹ R is transitive.
Set of all lines related to 𝑦 = 2𝑥 + 4 is given by 1
𝑦 = 2𝑥 + 𝑐
OR
27(b). Put 𝑥 = 𝑡𝑎𝑛𝜃 ⟹ 𝜃 = 𝑡𝑎𝑛−1 𝑥
√1+𝑥 2 −1 𝑠𝑒𝑐𝜃−1 1.5
Then 𝑡𝑎𝑛−1 = tan−1 ( )=
𝑥 tan 𝜃
−1 1−𝑐𝑜𝑠𝜃 𝜃 𝜃 1
= tan ( ) = tan−1 (tan ) = = tan−1 𝑥 1.5
𝑠𝑖𝑛𝜃 2 2 2

28. The total amount of money that will be received from
the sale of all these books can be represented in the
form of matrix multiplication as: 1
(Total Amount )= 12[10 8 10][ 80 60 40 ] 1
=12[10×80+8×60+10×40]
=12[800+480+400] 1
=12[1680]=20160 𝑅𝑠.

29(a). 𝑦 = (tan−1 𝑥)2
𝑑 1½
⟹ 𝑦1 = 2 tan−1 𝑥 (tan−1 𝑥)
𝑑𝑥
1
𝑦1 = 2 tan−1 𝑥 .
1 + 𝑥2
(1 + 𝑥 2 ). 𝑦1 =2 tan−1 𝑥 1½
Diff . again we get the result.

OR 1−𝑥2
𝑦=log( ) = log (1-𝑥) + log(1 + 𝑥) − log (1 + 𝑥 2 ) 1
29(b). 1+x2

𝑑𝑦 −1 1 2𝑥
= + − 1
𝑑𝑥 1 − 𝑥 1 + 𝑥 1 + 𝑥 2
−4𝑥
=
1−𝑥 4 1
30. ∫ 𝑥 log 𝑥 𝑑𝑥 =
Integrating by parts: 1.5
2 2
𝑥 1 𝑥
⟹ log 𝑥 . − ∫ . 𝑑𝑥
2 𝑥 2
𝑥 log 𝑥 𝑥 2
2 1.5
⟹ − +𝑐
2 4
𝜋 1 𝜋
31. Here 𝑚 = 𝑐𝑜𝑠 = , 𝑛 = 𝑐𝑜𝑠 =0
4 √2 2
1
𝑙 2 + 𝑚2 + 𝑛2 = 1
1
𝑙=± 1
√2
𝑟⃗ = 𝑙𝑖̂ + 𝑚𝑗̂ + 𝑛𝑘̂
𝑟⃗ = ±3𝑖̂ + 3𝑗̂ 1

खंड – द
SECTION – D (5×4=20)
𝜋 𝜋
32(a). 2 2 𝜋
∫ log sin 𝑥 𝑑𝑥 = ∫ log sin ( − 𝑥) 𝑑𝑥
0 0 2
𝜋
= ∫0 log cos 𝑥 𝑑𝑥
2 1
𝜋 𝜋
sin 2𝑥
2I=∫02 (log sin 𝑥 + log cos 𝑥 )𝑑𝑥 = ∫02 𝑙𝑜𝑔 ( ) 𝑑𝑥 1
2

𝜋
𝜋
2I= ∫02 sin 2𝑥 𝑑𝑥 - log 2
2
Put 2𝑥 = 𝑡 ⟹ 2𝑑𝑥 = 𝑑𝑡, 𝑤ℎ𝑒𝑛 𝑥 = 0, 𝑡 = 1
𝜋
0 ; 𝑤ℎ𝑒𝑛 𝑥 = , 𝑡 = 𝜋
2
1𝜋 𝜋
2I= ∫0 log sin 𝑡 𝑑𝑡 − log 2
2 2
𝜋
2 2 𝜋
= ∫0 log sin 𝑡 𝑑𝑡 − log 2 1
2 2
𝜋
2I=I- log 2
2 1
−𝜋
OR I= log 2
2
32(b). Given D.E. is of the form:
𝑑𝑦
+ 𝑃𝑦 = 𝑄 𝑤ℎ𝑒𝑟𝑒 𝑃 = −1, 𝑄 = cos 𝑥
𝑑𝑥
I.F. =𝑒 ∫ −1.𝑑𝑥 = 𝑒 −𝑥 1
Solution is :
𝑦𝑒 −𝑥 = ∫ 𝑒 −𝑥 cos 𝑥 𝑑𝑥 + 𝐶 1
Let I=∫ 𝑒 −𝑥 cos 𝑥 𝑑𝑥
= − cos 𝑥 𝑒 −𝑥 − ∫ 𝑠𝑖𝑛𝑥 𝑒 −𝑥 𝑑𝑥 1
I =− cos 𝑥 𝑒 −𝑥 + 𝑠𝑖𝑛𝑥 𝑒 −𝑥 − ∫ 𝑒 −𝑥 cos 𝑥 𝑑𝑥
𝑒 −𝑥 (sin 𝑥−cos 𝑥)
I= 1
2

Using this we get:
sin 𝑥 − cos 𝑥 1
𝑦=( ) + 𝐶𝑒 𝑥
2

33(a). (Let the three amounts are) = 𝑥, 𝑦, 𝑧 Rs.
( According to question):
𝑥 + 𝑦 + 𝑧 = 7000
𝑥−𝑦 =0
5 8 17 1
𝑥+ 𝑦+ 𝑧 = 550
100 100 200
⟹ 10𝑥 + 16𝑦 + 17𝑧 = 110000
This system of equations can be written as:
𝐴𝑋 = 𝐵, 𝑤ℎ𝑒𝑟𝑒 जह ाँ
1 1 1 𝑥 7000
𝑦
𝐴 = [ 1 −1 0 ] , 𝑋 = [ ] , 𝐵 = [ 0 ]
10 16 17 𝑧 110000 1
⟹ |𝐴| = −8 ≠ 0

(Now):
𝐴11 = −17, 𝐴12 = −17, 𝐴13 = 26
𝐴21 = −1, 𝐴22 = 7, 𝐴23 = −6
1
𝐴31 = 1, 𝐴32 = 1, 𝐴33 = −2
−17 −1 1
𝑎𝑑𝑗 𝐴 = [−17 7 1]
26 −6 −2
(Thus):
−1
1 1 −17 −1 1
𝐴 = . 𝑎𝑑𝑗 𝐴 = [−17 7 1] 1
|𝐴| −8
26 −6 −2
(Since):
1 −17 −1 1
−1
7000
𝑋=𝐴 𝐵= [−17 7 1 ][ 0 ]
−8
26 −6 −2 110000
𝑥 1125
𝑋 = [𝑦] = [4750] 1
𝑧 4750

अथवा (OR)
1 1 1
OR Let = 𝑝, = 𝑞, = 𝑟
𝑥 𝑦 𝑧
33(b). The given system of equations can be written as:
𝐴𝑋 = 𝐵, 𝑤ℎ𝑒𝑟𝑒

2 3 10 𝑝 4
𝐴 = [4 −6 5 ] , 𝑋 = [ 𝑞 ] , 𝐵 = [ 1] 1
6 9 −20 𝑟 2
|𝐴| = 1200 ≠ 0
⟹ 𝐴−1 𝑒𝑥𝑖𝑠𝑡𝑠
Co-factors of A are :
𝐴11 = 75 , 𝐴12 = 110 , 𝐴13 = 72

𝐴21 = 150 , 𝐴22 = −100 , 𝐴23 = 0 1

𝐴31 = 75 , 𝐴32 = 30 , 𝐴33 = −24

75 150 75
⟹ 𝑎𝑑𝑗𝐴 = [110 −100 30 ]
72 0 −24 1

𝑎𝑑𝑗 𝐴 1 75 150 75
⟹ 𝐴−1 = = [110 −100 30 ]
|𝐴| 1200
72 0 −24

1 75 150 75 4
−1
⟹𝑋=𝐴 𝐵= [110 −100 30 ] [1]
1200 1
72 0 −24 2
1
2
1 600 1
⟹𝑋= [400] = ⁄3
1200
240 1
[ 5 ]
1 1 1
⟹ 𝑝 = , 𝑞 = , 𝑟 = ⟹ 𝑥 = 2, 𝑦 = 3, 𝑧 = 5
2 3 5 1

34(a) Comparing with 𝑟⃗ = 𝑎⃗ + 𝜆𝑏⃗⃗ we get 𝑎⃗1 , 𝑎⃗2 , 𝑏⃗⃗1 , ⃗⃗⃗⃗⃗
𝑏2 1

Now 𝑎⃗2 − 𝑎⃗1 = −10𝑖̂ − 2𝑗̂ − 3𝑘̂ 1

And
𝑖̂ 𝑗̂ 𝑘̂
𝑏⃗⃗1 × ⃗⃗⃗⃗⃗ 1
𝑏2 = |1 −2 2 | = 8𝑖̂ + 8𝑗̂ + 4𝑘̂
3 −2 −2

⃗⃗1 ×𝑏
(𝑏 ⃗⃗⃗⃗⃗
2 ).(𝑎 ⃗⃗2 −𝑎⃗⃗1 ) 1
Shortest Distance = | ⃗⃗1 ×𝑏 ⃗⃗⃗⃗⃗
|
|𝑏 2|

−108
= | | = 9 𝑢𝑛𝑖𝑡𝑠 1
12

OR
Rewrite the eqn of given line
34(b).
𝑥 𝑦−1 𝑧−2
= = = 𝑘 (𝑠𝑎𝑦)
1 2 3
Then arbitrary D(x,y,z) point on the line is

𝑥 = 𝑘 , 𝑦 = 2𝑘 + 1 , 𝑧 = 3𝑘 + 2
Let this point D is the foot of perpendicular on the line.
1
Now position vector from given point P (1,6,3) to Point
D is given by:
⃗⃗⃗⃗⃗⃗ = (𝑘 − 1 )𝑖̂ + (2𝑘 − 5)𝑗̂ + (3𝑘 − 1) 𝑘̂
𝑃𝐷
Now
Direction vector of line 𝑏⃗⃗ = 1𝑖̂ + 2𝑗̂ + 3𝑘̂

Here 𝑏⃗⃗ ⊥ 𝑃𝐷
⃗⃗⃗⃗⃗⃗ ⟹ 𝑏⃗⃗. 𝑃𝐷
⃗⃗⃗⃗⃗⃗ = 0

⟹ 𝑘 − 1 + 4𝑘 − 10 + 9𝑘 − 3 = 0
⟹ 14𝑘 − 14 = 0 1
⟹𝑘=1
So foot of perpendicular is:

D = (1 , 3, 5 )
Let 𝐸(𝑎 , 𝑏 , 𝑐 ) be the image of P(1,6,3) then D(1,3,5)
will be mid point of PE. 1
So by mid point formula :

𝑎+1 𝑏+6 𝑐+3
=1, =3, =5
2 2 2
⟹ 𝑎 = 1, 𝑏 = 0, 𝑐 = 7
So image of P = E (1,0,7) 1

Also the distance PE =
√(1 − 1)2 + (6 − 0)2 + (3 − 7)2 = √0 + 36 + 16 1
= √52 𝑢𝑛𝑖𝑡𝑠

35(a). Let S denotes the success(getting 6) and F
denotes the failure(not getting 6).
Thus P(S) = , P(F) =
1 5
1
6 6
P( A wins the first throw)= P(S) =
1
6
A gets the third throw ,when the first throw
1
by A and second throw by B results in
failures.
So, P(A wins in the third throw)=P(FFS) 1
5 2
=
5 5 1 1
× × = ( ) ×
6 6 6 6 6
Similarly
5 4
P( A wins in the fifth throw)= ( ) ×
1
6 6 1
Hence
5 2 5 4
P(A wins ) = + ( ) × + ( ) × + ⋯ … ..
1 1 1
6 6 6 6 6
1
1
= 6 6 6 5
25 = , 𝑃(𝐵) = 1 − =
1− 11 11 11
36

OR
35(b). Let E= the chosen coin is two headed
F= the chosen coin is biased
G = the chosen coin is unbiased
Then E,F,G are mutually exclusive and
exhaustive events. 1

P(E)=P(F)=P(G)=
1
3
Let A = the tossed coin shows head

Then P(A/E)=1 , P(A/F) = ¾ , P(A/G)=1/2 2
Using Bayes’ Theorem :

𝐴
𝑃(𝐸).𝑃( )
P(E/A) = 𝐸 4
𝐴 𝐴 𝐴 = 2
𝑃(𝐸).𝑃( )+𝑃(𝐹).𝑃( )+𝑃(𝐺).𝑃( ) 9
𝐸 𝐹 𝐺

खंड – ल
SECTION – E (4×3=12)
36. Here:
(a). Semi circular 1

(b). (−2,0) 𝑎𝑛𝑑 (2,0) 1
2 𝑥
(c). Area = ∫−2 √4 − 𝑥 2 𝑑𝑥 = [ √4 − 𝑥 2 +
2 2
4 𝑥 2
. sin−1 ]
2 2 −2
= 2𝜋 sq. units
37.
If AR= 𝑥 𝑚 ⟹ 𝐵𝑅 = (20 − 𝑥) 𝑚
(a) 1
(b) 𝑆(𝑥) = 𝑅𝑃2 +𝑅𝑄2 = 2𝑥 2 − 40𝑥 + 1140
Here co-eff. Of 𝑥 = −40 1

(c) Using second derivative test :

𝑜𝑛 𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑆 ′ (𝑥) = 0 ⟹ 4𝑥 − 40 = 0 1
⟹ 𝑥 = 10
𝑆 ′′ (𝑥) = 4 = +𝑣𝑒 ⟹ 𝑥 = 10 𝑖𝑠 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑚𝑖𝑛𝑖𝑚𝑎.
𝐴𝑅 = 10 𝑚, 𝐵𝑅 = 10 𝑚 1

38. n(S)= 4 , n(J)= 3
(a) n(S).n(J) = 212 1
𝑁𝑜. 𝑜𝑓 𝑅𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠 = 2

(b) Many one and onto since 𝐽2 ℎ𝑎𝑠 𝑡𝑤𝑜 𝑝𝑟𝑒𝑖𝑚𝑎𝑔𝑒𝑠. 1

(c) 𝑚!
Number of one one functions from S to J = (𝑚−𝑛)! ,

Where m= n(S)=4, n = n(J)=3
So
4!
Number of one one functions from S to J= (4−3)! = 24 2