Karnataka 2nd PUC Question Paper 2025 Answer Key Mathematics Basic

Government of Karnataka
Karnataka Secondary Education Examination Board

Question Paper
ANSWER KEY

1

KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD
II PUC EXAM -1 MARCH 2025
SCHEME OF EVALUATION
Subject: 75- BASIC MATHEMATICS Max Marks :80
QN. MARKS
NO.
I PART - A
1. 6 4
d) [ ] 1
4 8
2. a) 15 1
3. 1
c) 1
4

4. d) ~q → ~p 1
5. b) 3: 7 1
6. a) 2
√3
1

7. d) y = -2 1
8. 1
b) 5𝑒 𝑥 − 𝑥 1
9. log(7𝑥+8)
b) 1
7

10. 𝑥6
d) 6 + 𝑐 1

II
11. a) iv 1 1
b) i 7 1
c) Vi 9 1
d) ii
√3−1
2√2 1

e) iii 2 1
III
12. 3 1
13. 35 1
14. 12 1
15. 16 1
16. 1 1

IIPUC EXAM-1, MARCH 2025 KSEAB 75-BASIC MATHS MODEL ANSWERS

2

IV PART B
17. 2 0 1 −1 1 1
2𝐴 = [ ]−[ ]=[ ] 1
−1 3 3 0 −4 3
1 1
𝐴=[ 2 2] 1
3
−2
2
9
18. Total selection = 𝐶5 1
=126 1
𝑃 (𝐴 ∩ 𝐵 ) = 0 1
19.
19
𝑃 (𝐴 ∪ 𝐵 ) = 𝑃 (𝐴 ) + 𝑃 (𝐵 ) = 1
35
20. a : b : c 1
3 : 4 x2
8 : 15
6 : 8 : 15 = a : b: c 1
21. TD=1200 1
F=30000 1
22. Equation is of form 𝑦 2 = −4𝑎𝑥 𝑎=4 1
Equation of parabola 𝑦 2 = −16𝑥 1
23. Taking log logy = x logx 1
𝑑𝑦
Diff and getting answer 𝑑𝑥 = 𝑥 𝑥 (1 + log 𝑥 ) 1
24. 𝑀𝐶 = 3𝑞 2 − 6𝑞 + 15 1
FC = 27 1
1 1
25.
𝐴 = ∫ 𝑦𝑑𝑥 = ∫ 𝑥 2 𝑑𝑥 1
0 0
1
𝑥3 1
= [ ] = 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠 1
3 3 0

V PART C
26. ∆= −7 ∆1 = −7 ∆2 = 7 2
x =1 y = -1 1
27. 11!
Total words = 3!3!2!2! 1
4!
a) 1
3!

9!
b) 3!2!2! 1

IIPUC EXAM-1, MARCH 2025 KSEAB 75-BASIC MATHS MODEL ANSWERS

3

28. Carpenters Rs Days hours
3 360 6 9 1
8 ? 12 6
8 12 6
= 360 × × × 1
3 6 9

= ₹1280 1
29. TD = ₹1200 1
BD =₹ 1224 1
F = ₹61,200 1
30. FV MV I
100 108 7.5
6000 ? ?
MV =6480 1
Income=450 1
FV MV I
100 ? 9 1
6480 720 market value = ₹81
31. MP= x Tax =0.1x 1
Total amount paid : 1.1x=13530 1
X= 12,300 1
32. 𝑉 = 𝑎3 𝐴 = 6𝑎2
𝑑𝑣 𝑑𝑎 𝑑𝐴 𝑑𝑎 1
= 3𝑎2 𝑑𝑡 = 12𝑎 𝑑𝑡
𝑑𝑡 𝑑𝑡

𝑑𝑣 𝑑𝐴
= 1800𝑐𝑚3 /𝑚𝑖𝑛 = 720𝑐𝑚2 /𝑚𝑖𝑛 2
𝑑𝑡 𝑑𝑡

33. 𝑥 𝐴 𝐵
= +
(𝑥 − 1)(𝑥 − 2) 𝑥 − 1 𝑥 − 2
A= -1 B=2 2
𝑥
∫ 𝑑𝑥 = − log(𝑥 − 1) + 2 log(𝑥 − 2) + 𝑐 1
(𝑥 − 1)(𝑥 − 2)
34. 2 2
𝑥
𝑥2
∫ 𝑥 + 𝑒 + 2 𝑑𝑥 = [ + 𝑒 𝑥 + 2𝑥] 1
1 2 1
1
= [2+𝑒 2 +4]-[2 + 𝑒 + 2] 1
7
= 𝑒2 − 𝑒 + 2 1

IIPUC EXAM-1, MARCH 2025 KSEAB 75-BASIC MATHS MODEL ANSWERS

4

VI PART D
35. |𝐴| = −12 1
−7 5 −3
𝐴𝑑𝑗 𝐴 = [ −1 −1 3 ] 2
−11 13 −3
𝑋 = 𝐴−1 𝐵
−7 5 −3 6 −24 1
1 1
[
=−12 −1 −1 3 ] [ 3 ] = −12 −12]
[
−11 13 −3 −1 −24
x=2 y= 1 z= 2 1
36. 9 𝐴 𝐵 𝐶
2
= + + 1
(𝑥 + 1)(𝑥 + 2) 𝑥 + 1 𝑥 + 2 (𝑥 + 2)2
A = 9 B = -9 C = -9 3
Conclusion 1
37. P q ~q 𝑝 ∧ ~q (𝑝 ∧ ~q)∨q p∨q
T T F F T T
T F T T T T
4
F T F F T T
F F T F F F
I Mark 1 Mark 1 Mark 1 Mark

Conclusion: from column V and VI (𝑝 ∧ ~q) ∨ q ≡ p ∨ q 1

38. No of units Total units Average time Total time
1 1 1000 1000
1 2 800 1600 3
2 4 640 2560
4 8 512 4096
Total Time = 4096 hours 1
Total Cost = ₹163840 1
Y
39. (0,80)

(0,50)
(30,20) 3

(40,0) (50,0) X

IIPUC EXAM-1, MARCH 2025 KSEAB 75-BASIC MATHS MODEL ANSWERS

5

Corner Points Z=10500x+9000y
(0,0) 0
(0,50) 450000
2
(30,20) 495000
(40,0) 420000
Maximum value of Z is 495000 at (30,20)
40. (𝑐𝑜𝑠7𝑥 − 𝐶𝑜𝑠𝑥 ) − (𝐶𝑜𝑠5𝑥 − 𝐶𝑜𝑠3𝑥)
1
(𝑆𝑖𝑛7𝑥 + 𝑠𝑖𝑛𝑥 ) − (𝑆𝑖𝑛5𝑥 + 𝑠𝑖𝑛3𝑥)
−2𝑠𝑖𝑛4𝑥𝑠𝑖𝑛3𝑥+2𝑠𝑖𝑛4𝑥𝑠𝑖𝑛𝑥
= 2𝑠𝑖𝑛4𝑥𝑐𝑜𝑠3𝑥−2𝑠𝑖𝑛4𝑥𝑐𝑜𝑠𝑥 1
−2𝑠𝑖𝑛4𝑥(𝑠𝑖𝑛3𝑥 − 𝑠𝑖𝑛𝑥)
= 1
2𝑠𝑖𝑛4𝑥(𝐶𝑜𝑠3𝑥 − 𝐶𝑜𝑠𝑥)
−2𝑐𝑜𝑠2𝑥𝑠𝑖𝑛𝑥
= 1
−2𝑠𝑖𝑛2𝑥𝑠𝑖𝑛𝑥
𝑐𝑜𝑠2𝑥
= = 𝑐𝑜𝑡2𝑥 1
𝑠𝑖𝑛2𝑥
41. 𝑦 = 𝑥 + √𝑥 2 − 1
1
𝑦1 = 1 + . 2𝑥 1
2√𝑥 2 − 1
√𝑥 2 − 1 + 𝑥
𝑦1 = 1
√𝑥 2 − 1
√𝑥 2 − 1𝑦1 = 𝑦
1
(𝑥 2 − 1)𝑦1 2 = 𝑦 2
(𝑥 2 − 1)2𝑦1 𝑦2 + 𝑦1 2 2𝑥 = 2𝑦𝑦1 1
(𝑥 2 − 1)𝑦2 + 𝑥𝑦1 = 𝑦 1
VI PART E
42. Theorem Case 1 2
Case 2 2
Case 3 2
OR
Writing equations (any 3 points)
(2,-4) g - 2f = -5
1
(3,-1) 6g – 2f = -10
(0,0) C=0 1
g = -1 1
f=2 1
Equation of circle 𝑥 2 + 𝑦 2 − 2𝑥 + 4𝑦 = 0 1
Sub 4th point (3,-3) 9+9-6-12=0 , hence points are concyclic 1
IIPUC EXAM-1, MARCH 2025 KSEAB 75-BASIC MATHS MODEL ANSWERS

6

43.

75 300
y 1
X x

600
y

75−𝑥 1 75−𝑥
𝑡𝑎𝑛300 = , = 1
𝑦 √3 𝑦
75 75
𝑡𝑎𝑛600 = 𝑦 , y = 1
√3

X=50 1
OR
(1.1)4 = (1 + 0.1)4 1
= 14 + 4𝐶1 (1)3 (0.1) + 4𝐶2 (1)2 (0.1)2 +
4 4
𝐶3 (1)1 (0.1)3 + 𝐶4 (0.1)4 1

=1+0.4+0.6+0.004+0.0001 1
=1.4641 1
PART F
39 (Only for visually challenged students)
Maximize Z= 6x+8y 2
4x + 2y ≤ 20 1
2x + 5y ≤ 24 1
X,y≥ 0 1

IIPUC EXAM-1, MARCH 2025 KSEAB 75-BASIC MATHS MODEL ANSWERS

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