Karnataka 2nd PUC Answer Key 2026 Basic Mathematics (Model Answers)

2026 II PUC EXAM – 1 KSEAB 75 BASIC MATHS
KARNATAKA SCHOOL EXAMINATION AND ASSESSMENT BOARD
2026 II PUC EXAM - 1
SUBJECT: 75 - BASIC MATHS SCHEME OF EVALUATION MAX MARKS: 80
QN. NO. MARKS
I PART - A
1. 0 −1
b) [ ] 1
1 0
2. b ) (n-1)! = 9! 1
3. c) 0.35 1
4. b ) pVq 1
5. d ) 81 : 16 1
1
6. c)2 1
7. a ) (0,4) 1
8. 𝑑)2𝑥 log 2 1
9. 𝑒𝑥
b) 5 +𝑐 1
10. 𝒄 ) −4cot 𝑥+c 1
II
11. a) (iii) −1 1
b) (vi) r=3 1
c) (v) 35 1
1
d) (ii) 2 1
e) (iv) −3 1
III
12. 6 1
13. 252 1
14. 12 1
15. 16 1
16. 1
1
3

IV PART B
17. 4 6 3 −6
2𝐴 = [ ] 3B=[ ] 1
2 −4 3 9
7 0
2A+3B=[ ]
5 5 1
18. Total number of triangles = 20𝐶 3 −8𝐶 3
1
=1084 1
1
𝑃(𝐴 ∩ 𝐵) = 1
19. 4
1
𝑃(𝐵 𝐴)=2
1
20. 7−𝑥 5 1
=
4−𝑥 2
X=2 1
SCHEME OF EVALUATION Page 1 of 5

2026 II PUC EXAM – 1 KSEAB 75 BASIC MATHS
21. BD=110 1
𝐵𝐷𝑋𝑇𝐷
F= =1100
𝐵𝐺 1

22. writing 𝑥 2 = 4𝑎𝑦 1
a=6, Equation of parabola 𝑥 2 = 24𝑦 1
23. log 𝑦=sin 𝑥 log 𝑥 1
𝑑𝑦 sin 𝑥
Diff and getting answer 𝑑𝑥 = 𝑥 sin 𝑥 [ 𝑥 + cos 𝑥 log 𝑥]
1

24. 𝑀𝐶 = 4𝑥 + 40 1
MR=400-4x 1
4
25.
𝐴 = ∫ 2𝑦 2 𝑑𝑦
2
1

112
= 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠 1
3
V PART C
26. Finding ∆= −31 1
Finding ∆1 = −31 ∆2 = −31 1
Finding x = 1 y = 1 1
27. 9!
Total number of permutations of words = 1
2! 2! 2!
7!
a ) 1
2!2!
6! 4! 720 24
b) X = X = 2160 1
2!2! 2! 2𝑋2 2
28. Carpenters Days Hours Earnings
3 6 9 360 1
8 12 6 x
3X6X 9X𝑥=8X12X6X360 1
x =₹ 1280 1
𝐵𝐷𝑋𝑇𝐷
29. writing F= 1
5𝐵𝐷 1
𝐹𝑡𝑟
F=5(1+𝑡𝑟)
1

t=1.25 years 1
30. Yield of 7.5 % stock at 125=6% 1
Yield of 5 % stock at 80=6.25% 1
5 % stock at 80 is better investment 1
336
31. The printed price of shirt =x=1.12=300
1
110
The printed price of neck tie =y= 1.1 =100
1

The printed price of shirt and neck tie together=x +y= 400 1

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2026 II PUC EXAM – 1 KSEAB 75 BASIC MATHS
32. 𝑑𝑥
Let the edge of a cube =x then, = 6 𝑐𝑚/𝑚𝑖𝑛 𝑑𝑡 1
𝑑𝑉
= 1800 𝑐𝑚3 /𝑚𝑖𝑛 1
𝑑𝑡
𝑑𝐴
=720 cm2 /min 1
𝑑𝑡

33. 1 𝐴 𝐵
= + 1
(𝑥 + 1)(𝑥 + 3) 𝑥 + 1 𝑥 + 3
1 1
Finding A= 2 B=− 2 1
1 1 1
∫ 𝑑𝑥 = log(𝑥 + 1) − log(𝑥 + 3) + 𝑐 1
(𝑥 + 1)(𝑥 + 3) 2 2
2 2𝑥+5 17 1
34. ∫1 𝑥 2 +5𝑥+3 𝑑𝑥 = ∫9 dt
𝑡
1

log 𝑡] 17
9 1
17
= log 9 1
VI PART D
35. |𝐴| = −5 1
−2 1 −1
𝐴𝑑𝑗 𝐴 = [ 9 −17 7] 2
5 −10 5
𝑋 = 𝐴−1 𝐵
−2 1 −1 13 −15
1 1 1
=−5 [ 9 −17 7 ] [ 3 ] = [ 10 ]
−5
5 −10 5 −8 −5
⇒x=3 y=−2 z= 1 1
36. 2𝑥 + 1 𝐴 𝐵 𝐶
= + + 1
(𝑥 − 1)(𝑥 − 2)(𝑥 − 3) 𝑥 − 1 𝑥 − 2 𝑥 − 3
3 7
A=2 B=−5 C=2 3

( Finding A-1 mark , Finding B-1 mark , Finding C-1 mark)
2𝑥+1 3 5 7 1
Conclusion :(𝑥−1)(𝑥−2)(𝑥−3) = 2(𝑥−1) − (𝑥−2) + 2(𝑥−3)

37. P q (𝑝⋁𝑞) ~𝑝 ~𝑞 (~𝑝 ∧ ~𝑞) (𝑝⋁𝑞) ∧ (~𝑝 ∧ ~𝑞)
(1) (2) (3) (4) (5) (6) (7)
T T T F F F F
T F T F T F F
5
F T T T F F F
F F F T T T F
up to column 3-1mark and 4and5,6,7 and conclusion each carries
1-mark

SCHEME OF EVALUATION Page 3 of 5

2026 II PUC EXAM – 1 KSEAB 75 BASIC MATHS
38. No of units Total units Average time Total time
1 1 1000 1000
1 2 900 1800 3
2 4 810 3240
up to column 2-1mark and 3 and 4 each carries 1-mark
Total Time = 3240 hours 1
Total Cost = ₹ 64800 1

Y
39.
(0,40) (Finding points 1-mark)
(Drawing lines 1-mark)
(Finding feasible region 1-mark)
3
A(0,25) B(10,20)
######
(0,0) #########
C (20,0) (50,0) X
(Finding Z at corner points -1 MARK, Conclusion -1 MARK)
Corner Points Z=5x+4y
O(0,0) 0
A(0,25) 100
2
B(10,20) 130(Max)
C(20,0) 100
Maximum value of Z is 130 at (10,20)
40.
(Sin6A+sin2A)+2Sin4A
LHS=(Sin7A+sin3A)+2Sin5A 1

2Sin4A Cos2A.+2Sin4A
= 1
2Sin5ACos2A+2Sin5A
2Sin4A (Cos2A.+1)
= 2
2Sin5A(Cos2A+1)
Sin4A
= 1
Sin5A

41.
2𝑥
𝑦1 =m(𝑥 + √1 + 𝑥 2 )𝑚−1 X(1+ ) 1
2√1+𝑥 2
(√1 + 𝑥 2 )𝑦1 =m(𝑥 + √1 + 𝑥 2 )𝑚 1
(1 + 𝑥 2 )𝑦1 2=(my)2 1
2 2
(1 + 𝑥 )( 2𝑦1 𝑦2 )+ 𝑦1 (2x)=m2 (2y𝑦 ) 1 1
(1 + 𝑥 2 ) 𝑦2+ x𝑦1 −m2 y=0 1
VI PART E
42. a) Theorem Case 1 2
Case 2 2
Case 3 2
OR

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2026 II PUC EXAM – 1 KSEAB 75 BASIC MATHS
Writing equation
(0,0) C=0 1
(-1,3) g - 3f = 5
1
(4,8) g +2f = -10
g = -4 1
f = -3 1
Equation of circle 𝑥 2 + 𝑦 2 − 8𝑥 − 6𝑦 = 0 1
Sub 4th point (8,6) 64+36-64-36=0 , hence points are concyclic 1
b)

C

y
D 300 E

1
x

A 450 600 B
20mts
𝑥+𝑦 1
𝑡𝑎𝑛600 = 20 , x+y=20√3
𝑥
𝑡𝑎𝑛450 = 20 , x=20m
1

DC= Height of the flag staff from building=y=20√3 -20 1

OR
5 5 1
(1.02) = (1 + 0.02)
=15+5C1(0.02)+ 5C2(0.02)2+ 5C3(0.02)3+ 5C4(0.02)4+ 5C5(0.02)5 1
=1+0.01+0.004+0.00008 1
=1.10408 1
PART F
39 (Only for visually challenged students)
Maximize Z= 6x+8y 2
s.t. 4x +2 y ≤ 20 1
2x + 5y ≤ 24 1
X≥,y≥ 0 1

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