Karnataka 2nd PUC Question Paper 2025 Answer Key Mathematics

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
Subject: 35-Mathematics SCHEME OF VALUATION MAX. MARKS: 80
Instructions:
a) Any answer by alternate method should be valued and suitably awarded.
b) All answers (including extra, struck off and repeated) should be valued. Answers with maximum marks must
be considered.
Qn PART A Marks
No : I
1 a) or writing (𝑎, 𝑎) ∈ 𝑅 for all 𝑎 ∈ 𝐴 1
𝜋
2 c) or writing 1
4

3 d) or writing A-iii, B-i, C-ii 1
4 b) or writing [
1 0
] 1
3 2
5 1
d) or writing |𝐴|2

6 1
b) or writing − 2
7 1
a) or writing Statement 1 is true and statement 2 is false
8 d) or writing 8 1
9 a) or writing − 𝑒 𝑥 cos 𝑥 1
10 d) or writing 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 1
11 b) or writing
1
,−
1
,
2
1
√6 √6 √6
𝜋
12 c) or writing 4
1
13 b) or writing 𝑥 = 0, 𝑧 = 0 1
1
14 a) or writing 1
3
15 c) or both [A] and [R] are true 1
II
16 1
1
17 −1 1
18 1
6
19 0
1

20 5
1
9

IIPUC EXAM-1, MARCH 2025 KSEAB 35- MATHEMATICS MODEL ANSWERS

2

PART B
21 1
𝑥 𝑦 1 𝑥 𝑦 1 1
Writing 2
|3 1 1| = 0 OR |3 1 1| = 0
9 3 1 9 3 1
1
Getting −4𝑥 + 2𝑦 = 0 OR 4𝑥 − 2𝑦 = 0 OR 2𝑥 − 𝑦 = 0
22 Writing
1 1 dy
+ 2 y dx = 0 1
2√x √
1
𝑑𝑦 𝑦 dy y
Getting = − √ ⟹ dx + √x = 0
𝑑𝑥 √𝑥

23 Writing volume of the sphere, 𝑉 =
4 𝑑𝑉
𝜋 𝑟 3 and 𝑑𝑟 = 4 𝜋𝑟 2 1
3
Getting
𝑑𝑉
= 4 𝜋102 = 400 𝜋 𝑐𝑚3 /𝑐𝑚 1
𝑑𝑟
(Note: Units are not compulsory)
24 Writing 𝑓 ′ (𝑥 ) = 12𝑥 2 − 12𝑥 − 72. OR 1
′(
𝑓 𝑥) = 12(𝑥 − 3)(𝑥 + 2)
1
Getting 𝑥 ∈ (−2, 3)
25 1
Put log(sin 𝑥 ) = 𝑡  𝑑𝑡 = cot 𝑥 𝑑𝑥

𝑡2 (log(sin 𝑥))2 1
Getting ∫ 𝑡 𝑑𝑡 = 2 + 𝐶 = 2
+𝐶

26 𝑑𝑦 𝑑2 𝑦
1
Writing, 𝑑𝑥 = −𝑎𝑠𝑖𝑛𝑥 + 𝑏 𝑐𝑜𝑠𝑥 OR 𝑑𝑥 2 = −𝑎𝑐𝑜𝑠𝑥 − 𝑏 𝑠𝑖𝑛𝑥

𝑑2 𝑦 𝑑2𝑦
1
Getting
𝑑𝑥 2 = −𝑦  𝑑𝑥 2 +𝑦 = 0

𝑑2𝑦
Therefore 𝑦 = 𝑎𝑐𝑜𝑠𝑥 + 𝑏𝑠𝑖𝑛𝑥 is solution of 𝑑𝑥 2
+ 𝑦 = 0.
27 1
Getting 𝑑⃗ = 2𝑎⃗ − 𝑏⃗⃗ + 3𝑐⃗ = 3𝑖̂ − 3𝑗̂ + 2𝑘̂

̂
3𝑖̂−3𝑗̂ +2𝑘 1
Getting |𝑑⃗| = √22 and unit vector = 𝑑̂ = .
√22
28 1
Writing a1a2 + b1b2 + c1c2 = 0 OR (−3)(3𝑘) + (2𝑘)(1) + (2)(−5) = 0

10 1
Getting 𝑘 = − 7

29 10 1
Writin𝑔 P(black ball in first draw) = P(E) = 15 OR
9
P(black ball in second draw) = P(F|E) = 14

10 9 3 1
Getting 𝑃(𝐸 ∩ 𝐹 ) = 𝑃(𝐸 ). 𝑃(𝐹|𝐸) = 15 . 14 = 7

IIPUC EXAM-1, MARCH 2025 KSEAB 35- MATHEMATICS MODEL ANSWERS

3

PART-C

30
Proving not reflexive (by giving suitable counter example) 1
3
[ Say a = 0.1 : 𝑎 ≤ 𝑎 𝑖𝑠 𝑛𝑜𝑡 𝑡𝑟𝑢𝑒 ( ℎ𝑒𝑟𝑒 0 < 𝑎 < 1) ]
Proving not symmetric: By giving any suitable counter example 1

[ say (1,2) ∈ 𝑅 but (2,1) ∉ 𝑅 ⇒ R is not symmetric.]
Proving not transitive: By giving any suitable counter example
1
[ Say a = 9, b = 3 and c=2 , 9 ≤ 33 , 3 ≤ 23 but (9,2) ∉ 𝑅 ]
31 5 3 5 4
1
Writing LHS = sin−1 13 + cos −1 5 = tan−1 12 + tan−1 3

5 4
+
−1 12 3
Writing : = tan ( )
5 4
1−(12)(3) 1

15+48 63 1
Getting: L.H.S = tan−1 ( ) = tan−1 ( )
36−20 16
OR 5 3 5 3
1
Writing A = sin−1 13 , 𝐵 = cos −1 5  sin 𝐴 = 13 , cos 𝐵 = 5

5 4 𝑡𝑎𝑛𝐴+𝑡𝑎𝑛𝐵
Writing : tan 𝐴 = 12 , tan 𝐵 = 3 , tan(𝐴 + 𝐵) = 1−𝑡𝑎𝑛𝐴𝑡𝑎𝑛𝐵 1
63 63
Getting: tan(𝐴 + 𝐵) = 16  A+B = tan−1 (16) 1
OR any other Alternate method allot appropriate marks
32 2 4 1 1 2 1
Getting: (𝐴 + 𝐴′ ) = [ ] OR (𝐴 + 𝐴′ ) = [ ]
4 4 2 2 2
0 6 1 0 3
Getting: (𝐴 − 𝐴′ ) = [ ] OR (𝐴 − 𝐴′ ) = [ ] 1
−6 0 2 −3 0

(𝐴 + 𝐴′ ) + (𝐴 − 𝐴′ ) = [ 1 5 1
1 1
Getting: ]=𝐴
2 2 −1 2
33 𝑑𝑥 1 𝑡 1 cos2 𝑡 1
Getting: 𝑑𝜃 = 𝑎(− sin 𝑡 + 𝑡 . sec2 (2) . 2) = 𝑎
tan(2) sin 𝑡

Getting:
𝑑𝑦
= 𝑎 cos 𝑡 1
𝑑𝜃
𝑑𝑦 𝑎 cos 𝑡 1
Getting: 𝑑𝑥 = cos2 𝑡
OR 𝑑𝑦
𝑑𝑥
= 𝑡𝑎𝑛𝑡
𝑎( sin 𝑡 )

34 Writing: 𝑃 = 𝑥𝑦 3 and 𝑃 = (60 − 𝑦)𝑦 3 OR 𝑃 = 𝑥(60 − 𝑥)3 1

𝑑𝑃
Getting: 𝑑𝑦 = 60 × 3𝑦 2 − 4𝑦 3 OR
1
𝑑𝑃 2 3
= −3𝑥(60 − 𝑥) + (60 − 𝑥)
𝑑𝑥
𝑑2 𝑝 1
Getting: 𝑥 = 15 and 𝑦 = 45 and showing 𝑑𝑦 2 < 0

IIPUC EXAM-1, MARCH 2025 KSEAB 35- MATHEMATICS MODEL ANSWERS

4

35 2𝑥 2𝑥 𝐴 𝐵
= = +
𝑥 2 + 3𝑥 + 2 (𝑥 + 1)(𝑥 + 2) 𝑥 + 1 𝑥 + 2 1
Getting: 𝐴 = −2, 𝐵 = 4 1
−2 4
Getting∫ 𝑥+1 𝑑𝑥 + ∫ 𝑥+2 𝑑𝑥 = −2 log|𝑥 + 1| + 4 log|𝑥 + 2| + c
(without c deduct 1 mark) 1
OR Writing I = ∫ 2𝑥 2𝑥+3−3
𝑑𝑥 = ∫ 𝑥 2 +3𝑥+2 𝑑𝑥 1
𝑥 2+3𝑥+2
𝑑(𝑥 2+3𝑥+2)
Getting: I = ∫ 𝑥2 +3𝑥+2 𝑑𝑥 − ∫
3
𝑑𝑥 1
3 1
(𝑥+ )2−
2 4
3 1
𝑥+ − 1
2
Getting:𝐼 = log(𝑥 + 3𝑥 + 2) − 3 log ( 2 2
3 1 )+c
𝑥+ +
2 2
𝑥+1
=log(𝑥 2 + 3𝑥 + 2) − 3 log (𝑥+2)+c OR −2 log|𝑥 + 1| + 4 log|𝑥 + 2| + 𝑐
36
Writing ∶ ⃗⃗⃗⃗⃗⃗
𝐴𝐵 = ⃗⃗⃗⃗⃗⃗
𝑂𝐵 − 𝑂𝐴 ⃗⃗⃗⃗⃗⃗ = − 𝑖̂ + 3𝑗̂ − 𝑘̂
⃗⃗⃗⃗⃗⃗ = 𝑂𝐶
and 𝐴𝐶 ⃗⃗⃗⃗⃗⃗ = −𝑖̂ + 2𝑗̂ + 𝑘
⃗⃗⃗⃗⃗⃗ − 𝑂𝐴 ̂
1
̂
𝑖 ̂
𝑗̂ 𝑘
̂ 1
⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗
Getting: 𝐴𝐵 × 𝐴𝐶 = |−1 3 −1| = 5𝑖̂ + 2𝑖̂ + 𝑘
−1 2 1
1
⃗⃗⃗⃗⃗⃗ | = √30 square units
⃗⃗⃗⃗⃗⃗ × 𝐴𝐶 1
Writing ∶ Area of △ 𝐴𝐵𝐶 = 2 |𝐴𝐵 2
(unit not Compulsory)
37 Writing correct figure (Writing x, y and z axes necessary)
1

⃗⃗⃗⃗⃗⃗=𝜆𝑏⃗⃗
Writing 𝐴𝑃 where 𝜆 is a scalar
1

Getting 𝑟⃗ = 𝑎⃗+𝜆𝑏⃗⃗ 1

38
1 1 1
Writing: 𝑃(𝐸1 ) = 2 , 𝑃(𝐸2 ) = 2 and 𝑃(𝐴|𝐸1 ) = 1 , 𝑃(𝐴|𝐸2 ) = 2
1
Writing: 𝑃 (𝐸1 |𝐴) =
𝑃(𝐸1 ) 𝑃(𝐴|𝐸1 ) 1
𝑃(𝐸1 ) 𝑃(𝐴|𝐸1 )+𝑃(𝐸2 ) 𝑃(𝐴|𝐸2 )

Getting : 𝑃(𝐸1 |𝐴) =
2 1
3

IIPUC EXAM-1, MARCH 2025 KSEAB 35- MATHEMATICS MODEL ANSWERS

5

PART D
39 Let𝑥1 , 𝑥2 ∈ A = R − {3} such that
𝑥 −2 𝑥 −2
𝑓 (𝑥1 ) = 𝑓 (𝑥2 ) ⟹ 𝑥1 −3 = 𝑥2 −3 1
1 2

⟹ (𝑥1 − 2)(𝑥2 − 3) = (𝑥2 − 2)(𝑥1 − 3) 1
⟹ 𝑥1 𝑥2 − 3𝑥1 − 2𝑥2 + 6 = 𝑥1 𝑥2 − 3𝑥2 − 2𝑥1 + 6
⟹ 3𝑥2 − 2𝑥2 = 3𝑥1 − 2𝑥1 ⟹ 𝑥1 = 𝑥2. ∴ f is one-one. 1
𝑥− 2
Take y  B and let f(x) = y⟹ 𝑥 − 3 = 𝑦
1
2−3𝑦
Getting 𝑥 = ∈ 𝐴 𝑓 is onto.
1−𝑦 1
OR
𝑥− 2
𝑓 (𝑥 ) = 𝑥 − 3 = 1 + 𝑥−3
1 1

1 1
𝑓(𝑥1 ) = 𝑓(𝑥2 ) ⟹ 1 + 𝑥 −3 = 1 + 𝑥 −3 ⟹ 𝑥 −3 = 𝑥 −3
1 1 1
1 2 1 2

Writing ⟹ 𝑥1 = 𝑥2. ∴ f is one-one. 1

1
Take y  B and let f(x) = y ⟹ 𝑦 = 1 + 𝑥 − 3
1
1 1
Writing 𝑦 − 1 = ⟹ 𝑥 = 3 + 𝑦−1 ∈ 𝐴 ∴ f is one-one. 1
𝑥 − 3

40 1 −1 2 1 1
Getting AB =[−4] [−1 2 1] = [ 4 −8 −4]
3 −3 6 3
−1 4 −3
writing (𝐴𝐵)| =[ 2 −8 6 ] …….. (1) 1
1 −4 3
−1 −1
writing 𝐴| = [1 −4 3] and 𝐵| = [ 2 ] OR 𝐵| 𝐴| = [ 2 ] [1 −4 3] 1
1 1
−1 4 −3
Getting 𝐵 | 𝐴| = [ 2 −8 6 ] ……. (2) 1
1 −4 3
Comparing (1) and (2) (𝐴𝐵)| =𝐵| 𝐴|
1
41 4 3 2 𝑥 60
Writing 𝐴 = [2 4 6] , 𝑋 = [𝑦] 𝑎𝑛𝑑 𝐵 = [90]
6 2 3 𝑧 70
OR Getting |𝐴| = 50 ≠ 0 1
Note: Award a mark, if student writes directly |𝐴| = 50.
0 −5 10
Getting adj(A)=[ 30 0 −20]
−20 10 10 2

Note: If any 4 cofactors are correct award 1 mark..
0 −5 10 60
1 1
Writing 𝑋 = 𝐴−1 𝐵 = |𝐴| (𝑎𝑑𝑗𝐴)𝐵 OR 𝑋 = 50 [ 30 0 −20] [90] 1
−20 10 10 70
1
Getting x =5, y =8, z =8

IIPUC EXAM-1, MARCH 2025 KSEAB 35- MATHEMATICS MODEL ANSWERS

6

42 1 1
y = (tan-1 x)2 Diff. w.r.to x, 𝑦1 = 2 tan−1 𝑥. 1+𝑥 2,
1
multiply by (1 + 𝑥 2 )  (1 + 𝑥 2 )𝑦1 = 2 tan−1 𝑥

2
diff. again w. r. t. x, and getting (1 + 𝑥 2 )𝑦2 + 2𝑥𝑦1 = (1+𝑥 2),
1

1
multiply by (1 + 𝑥 2 ) OR writing (1 + 𝑥 2 )[(1 + 𝑥 2 )𝑦2 + 2𝑥𝑦1 ] = 2

1
Writing (1 + 𝑥 2 )2
𝑦2 + 2𝑥(1 + 𝑥 𝑦1 = 2 2)

𝑥
43 Taking x=atan 𝜃  tan−1 = 𝜃 and dx=a𝑠𝑒𝑐 2 𝜃𝑑𝜃 1
𝑎

𝑑𝑥 𝑎𝑠𝑒𝑐 2 𝜃𝑑𝜃 𝑎𝑠𝑒𝑐 2 𝜃𝑑𝜃 1
Getting ∫ =∫ 2 =∫ 2
𝑥 2 +𝑎 2 𝑎 𝑡𝑎𝑛2 𝜃+𝑎 2 𝑎 (𝑡𝑎𝑛2 𝜃+1)
𝑎𝑠𝑒𝑐 2 𝜃𝑑𝜃 1 1
=∫ = ∫ 1𝑑𝜃 = 𝑎 (θ) +c
𝑎2 𝑠𝑒𝑐 2 𝜃 𝑎
𝑑𝑥 1 𝑥 1
Getting ∫ = tan−1 ( )+c
𝑥 2 +𝑎2 𝑎 𝑎
Writing 𝑥 2 − 6𝑥 + 13=(𝑥 − 3)2 + 22 1
1
𝑑𝑥 1 𝑥−3
Getting ∫ 𝑥2 −6𝑥+13 = 2 tan−1 ( 2 ) + 𝑐

44 Writing correct figure

1

𝑎
Writing y = √𝑎2 − 𝑥 2 OR Writing Area=4 ∫0 𝑦𝑑𝑥
OR Area = 4 times shaded area 1

𝑎
Writing Area =4 ∫0 √𝑎2 − 𝑥 2 𝑑𝑥 1

𝑎
𝑥 𝑎2 𝑥 1
Getting Area = 4 [ √𝑎2 − 𝑥 2 + sin−1 𝑎]
2 2 0

Getting Area = 𝜋𝑎2 square units 1
Note: Units are not compulsory

IIPUC EXAM-1, MARCH 2025 KSEAB 35- MATHEMATICS MODEL ANSWERS

7

45 Writing
𝑑𝑦
+ sec2x . y = tanx.sec2x OR P =sec2x , Q = tanx sec2x 1
𝑑𝑥

Getting
2
I.F = 𝑒 ∫ 𝑃 𝑑𝑥 = 𝑒 ∫ sec x = 𝑒 tanx 1

Writing y(I.F) = ∫ 𝑄 (𝐼. 𝐹 )𝑑𝑥+c OR 1

y etan x = ∫ 𝑡𝑎𝑛𝑥. sec2x etanx dx + c
Put tanx = t  sec2x dx = dt  yetanx = ∫ 𝑡 et dt +c 1

Getting yetanx = etanx (tanx-1) +c OR y = (tanx - 1) + c.e-tanx 1
(without c deduct 1 marrk )
PART E
𝑎
46 Let 𝐼 = ∫0 𝑓 (𝑥)𝑑𝑥
Putting x = a − t , then dx = − dt
𝑥 = 0 ⟹ 𝑡 = 𝑎 𝑎𝑛𝑑 𝑥 = 𝑎 ⟹ 𝑡 = 0 1

0
Getting I = − f ( a − t ) dt

a 1
a
Getting I =  f ( a − x ) dx
0 1
𝜋/4 𝜋
Writing Let 𝐼 = ∫0 log(1 + 𝑡𝑎𝑛𝑥)𝑑𝑥 replace x by 4 − 𝑥
𝜋/4 1+𝑡𝑎𝑛𝑥 1
𝐼 = ∫0 log (1 + 1−𝑡𝑎𝑛𝑥 ) 𝑑𝑥

𝜋/4 2 𝜋/4
𝐼 = ∫0 log (1+𝑡𝑎𝑛𝑥 ) 𝑑𝑥 = ∫0 (log(2) − log(1 + 𝑡𝑎𝑛𝑥))𝑑𝑥
1
𝜋 𝜋
2𝐼 = (𝑙𝑜𝑔2) ⟹ 𝐼 = 𝑙𝑜𝑔2
4 8 1
OR

2

Drawing the graph of (any two lines 1mark) all 3 lines award 2 marks

IIPUC EXAM-1, MARCH 2025 KSEAB 35- MATHEMATICS MODEL ANSWERS

8

Getting corner points (60, 0), (120, 0), ( 40, 20) 𝑎𝑛𝑑 (60, 30)
1

Corner points Z=5x+10y
(60, 0) 300
(120, 0) 600
1
(40, 20) 400
(60, 30) 600

1
Writing the minimum value of Z is 300 at
( 60, 0) .
The maximum value of Z is 600 at all the points on the line segment joining
1
( 60, 0) and (120,0)
47
8 5
Getting 𝐴2 = [ ] 1
−5 3

1
Proving 𝐴 − 5𝐴 + 7𝐼 = 𝑂
2

1
1
Getting 7𝐴−1 = 5𝐼 − 𝐴 𝑜𝑟 𝐴−1 = 7 (5𝐼 − 𝐴))

2
−7
1 1
1 2 −1
Getting 𝐴 −1
= 7[ ] OR 𝐴−1 = [71 3 ]
1 3
7 7
OR 1
𝑙𝑖𝑚 𝜋
Writing condition for continuity 𝑥 → 𝜋(𝑓(𝑥)) = 𝑓 ( 2 )
2

π t π
Put π − 2x = t ⟹x = 2 − 2 When x = 2 , t = 0
1
π t
k cos( 2− 2) k sin2
t
k 1
Getting lim t
= lim t = 2
t →0 t →0 2 . 2

Getting
k
= 3 ⟹k = 6. [ Any other alternate method award marks] 1
2
PART F
7
a) or writing Statement 1 is true and statement 2 is false
1

***************************

IIPUC EXAM-1, MARCH 2025 KSEAB 35- MATHEMATICS MODEL ANSWERS

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