Karnataka 2nd PUC Question Paper 2026 Answer Key Mathematics

2026 II PUC EXAM – 1 KSEAB 35- MATHEMATICS
GOVERNMENT OF KARNATAKA
KARNATAKA STATE EXAMINATION AND ASSESSMENT BOARD
2026 II PUC EXAM – 1
SCHEME OF EVALUATION
SUBJECT: 35 – MATHEMATICS 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 .

Q NO: PART-A MARKS
I 1 c) or symmetric and transitive 1

2 d) or (−∞,∞) 1

3 b) or 2 1

4 a) or 25 1
5 b) or (−1 , 1 ) 1
6 c) or 1 1
7 a) or 0 1
8 d) or Statement I is false and Statement II is true 1

9 c) or −𝒄𝒐𝒔𝒆𝒄 𝟏 𝐱 + 𝐜 1

10 b) or 0 1

11 𝟏 1
d) or a =
|𝝀|

12 a) or 3 ̂ + 𝟐 ̂ + 𝒌 1
13 c) or K , 0 , 0 1
14
a) or
𝟏 1
𝟑𝟔

15 b) or 0.12 1
II 16 -1 1

17 0 1

18 3 1

19 2 1

20 1 1

PART-B
21 𝒙 1
𝟏 𝐜𝐨𝐬 𝒙 𝟐𝒔𝒊𝒏𝟐
𝟐
Writing tan-1 = tan-1 𝒙
𝟏 𝐜𝐨𝐬 𝒙 𝟐𝒄𝒐𝒔𝟐
𝟐

𝒙 𝒙
Getting tan-1 ( tan )= 1
𝟐 𝟐

22 𝟏 𝟐 𝟏 1
𝟏
Writing area of triangle 𝚫𝑨𝑩𝑷 = 𝟑 𝟔 𝟏 = 0
𝟐
𝒙 𝒚 𝟏
Getting 𝟐𝒚 − 𝟒𝒙 = 𝟎 OR y = 2x as the equation of line 1

23 𝒅𝒚 𝒅𝒙 1
Writing = - 2 sin 2t and = cos t
𝒅𝒙 𝒅𝒕
𝒅𝒚 𝟐 𝐬𝐢𝐧 𝟐𝒕 1
Getting =
𝒅𝒙 𝐜𝐨𝐬 𝒕

24 Writing C = 2𝝅 r 1

𝒅𝑪 𝒅𝒓 1
Getting = 𝟐𝝅 = 2𝝅 X0.7 = 1.4 𝝅 cm/s
𝒅𝒕 𝒅𝒕

25 𝟏 𝟏 1
Writing G I = ∫ 𝒅𝒙 OR put 1 + log x = t and dx = dt
𝒙( 𝟏 𝐥𝐨𝐠 𝒙 ) 𝒙

Getting 𝑰 = 𝐥𝐨𝐠 | 𝟏 + 𝐥𝐨𝐠 𝒙 | + 𝒄 1

26 𝒅𝒚 𝒅𝒙 1
Writing ∫ 𝟏 𝒚 𝟐 = ∫ 𝟏 𝒙𝟐

Getting tan -1 y = tan -1 x + c 1

27 Getting 𝐚⃗ ∙ 𝐛⃗ = 𝟏𝟎 1

𝒂⃗ ∙ 𝒃⃗ 𝟏𝟎 1
Required = ⃗
=
|𝒃 | √𝟔

28 𝑮𝒆𝒕𝒕𝒊𝒏𝒈 𝒃𝟏⃗ ∙ 𝒃𝟐⃗ = 𝟏𝟖 OR | 𝒃𝟏⃗| =9 and | 𝒃𝟐⃗| = 𝟑 1

𝒃𝟏⃗∙𝒃𝟐⃗ 𝟐 𝟐 1
Getting cos θ = = and θ = 𝐜𝐨𝐬 𝟏 ( )
|𝒃𝟏⃗||∙𝒃𝟐⃗| 𝟑 𝟑

29 Writing S = { ( b , b ) , ( b , g ) , ( g , b ) , ( g , g )} 1

And E = { ( b , b ) and F = {( b , b ) , ( b , g ) , ( g , b ) }

𝑬 𝑷( 𝑬 ∩𝑭 ) 𝟏 1
Writing E ∩ F = { ( b , b ) } and getting P( ) = =
𝑭 𝑷( 𝑭 ) 𝟑

PART-C
30 Every triangle is congruent to itself . ( T , T ) ∈ 𝑹 ∀ T ∈ T 1
∴ R is reflexive
Let ( T1 , T2 ) ∈ 𝑹 => T1 is congruent to T2 1
=> T2 is congruent to T1
( T2 , T1 ) ∈ 𝑹 ∴ R is symmetric
Let ( T1 , T2 ) ∈ 𝑹 and ( T2 , T3) ∈ 𝑹 1
=> T1 is congruent to T2 and T2 is congruent to T3
=> T1 is congruent to T3
( T1 , T3 ) ∈ 𝑹 ∴ R is transitive => R is an equivalence relation .
31 Put 𝐱 = 𝐜𝐨𝐬 𝛉 => 𝛉 = 𝐜𝐨𝐬 𝟏 𝐱 1

𝟏
Getting R𝐇𝐒 = 𝒄𝒐𝒔 (𝟒 𝐜𝐨𝐬 𝟑 𝛉 − 𝟑𝐜𝐨𝐬 𝛉) = 𝐜𝐨𝐬 𝟏 ( 𝒄𝒐𝒔 𝟑 𝛉 ) 1

RHS = 3𝛉 = 3 𝐜𝐨𝐬 𝟏 𝐱 = LHS 1
32 𝑨 𝑨𝟏 𝟏 𝟐 1
𝑾𝒓𝒊𝒕𝒊𝒏𝒈 𝑷= = is symmetric
𝟐 𝟐 𝟐
𝑨 𝑨𝟏 𝟎 𝟑 1
𝑾𝒓𝒊𝒕𝒊𝒏𝒈 𝑸= = is skew- symmetric
𝟐 −𝟑 𝟎

𝟏 𝟓 1
Getting 𝑷+𝑸 = =A
−𝟏 𝟐
33 Writing 𝒍𝒐𝒈 𝒚 = 𝒔𝒊𝒏 𝒙 𝒍𝒐𝒈 𝒙 1

𝟏 𝒅𝒚 𝒔𝒊𝒏 𝒙 1
Getting = + log x . cos x
𝒚 𝒅𝒙 𝒙

𝒅𝒚 𝒔𝒊𝒏 𝒙 1
Getting = 𝒙𝒔𝒊𝒏 𝒙 ( + log x . cos x )
𝒅𝒙 𝒙

34 Writing 𝒇| (𝒙) = 𝟔𝒙𝟐 − 𝟑𝟎𝒙 + 𝟑𝟔 𝑶𝑹 1
𝒕𝒂𝒌𝒊𝒏𝒈 𝒇| (𝒙) = 𝟎 𝒂𝒏𝒅 𝒈𝒆𝒕𝒕𝒊𝒏𝒈 𝒙 = 𝟐 𝒂𝒏𝒅 𝒙 = 𝟑

𝐖𝐫𝐢𝐭𝐢𝐧𝐠 𝐟(𝟏) = 𝟐𝟒, 𝐟(𝟐) = 𝟐𝟗 𝐚𝐧𝐝 𝐟(𝟑) = 𝟐𝟖 and f(5) = 56 1

𝐰𝐫𝐢𝐭𝐢𝐧𝐠 𝐚𝐛𝐬𝐨𝐥𝐮𝐭𝐞 𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐟 𝐢𝐬 𝟓𝟔 𝐨𝐜𝐜𝐮𝐫𝐬 𝐚𝐭 𝐚𝐭 𝐱 = 𝟓 and 1
𝐚𝐛𝐬𝐨𝐥𝐮𝐭𝐞 𝐌𝐢𝐧𝐢𝐦𝐮𝐦 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐟 𝐢𝐬 𝟐𝟒 𝐨𝐜𝐜𝐮𝐫𝐬 𝐚𝐭 𝐚𝐭 𝐱 = 𝟏

35 𝟏 𝑨 𝑩 1
Writing ( 𝒙 𝟏)(𝒙 𝟐 )
=( + (𝒙
𝒙 𝟏) 𝟐)

𝐠𝐞𝐭𝐭𝐢𝐧𝐠 𝐀 = 𝟏 , 𝐁 = −𝟏 1

𝟏 𝟏 1
𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐆𝐈 = ∫ 𝒅𝒙 + ∫ dx
𝒙 𝟏 𝒙 𝟐
𝒙+𝟏
= 𝐥𝐨𝐠 | 𝐱 + 𝟏 |− 𝐥𝐨𝐠 |𝐱 + 𝟐 | + 𝐜 𝐎𝐑 𝐆𝐈 = 𝐥𝐨𝐠 +𝐂
𝒙+𝟐
36 𝐖𝐫𝐢𝐭𝐢𝐧𝐠 𝑨𝑩⃗ = ̂ + 𝟐 𝒌 and 𝑨𝑪⃗ = ̂ + 𝟐 1

̂ ̂ 𝒌 1
𝐆𝐞𝐭𝐭𝐢𝐧𝐠 ⃗ ⃗
𝑨𝑩 𝑿 𝑨𝑪 = 𝟎 𝟏 𝟐 = −𝟒 ̂ + 𝟐 − 𝒌
𝟏 𝟐 𝟎
𝟏
Getting Area = | 𝑨𝑩⃗ 𝑿 𝑨𝑪⃗ | =
√𝟐𝟏 1
𝟐 𝟐

1

37

Correct figure award 1 mark ( Note : wrong figure deduct 1 mark )

𝒘𝒓𝒊𝒕𝒊𝒏𝒈 𝑨𝑷⃗ || 𝒃⃗ 𝒂𝒏𝒅 𝑨𝑷⃗ = 𝝀 𝒃⃗ 1

Getting 𝒓⃗ = 𝒂⃗ + 𝝀 𝒃⃗ 1

38 Let E 1 and E 2 be the events choosing Bag I and Bag II 1
and A be the event of drawing red ball
𝟏
𝐏(𝐄𝟏 ) = 𝐏(𝐄𝟐 ) =
𝟐

𝟑 𝟓 1
𝐠𝐞𝐭𝐭𝐢𝐧𝐠 𝐏(𝐀|𝐄𝟏 ) = and 𝐏(𝐀|𝐄𝟐 ) =
𝟕 𝟏𝟏

𝐏(𝐀|𝐄𝟐 )∙𝐏(𝐄𝟐 ) 𝟑𝟓 1
Getting 𝐏(𝐄𝟐 |𝐀) = =
𝐏(𝐀|𝐄𝟏 )∙ 𝐏(𝐄𝟏 ) 𝐏(𝐀|𝐄𝟐 )∙𝐏(𝐄𝟐 ) 𝟔𝟖

PART-D
39 Define g : R → R by g(y) =
𝒚 𝟑 1
𝟒

Proving gof (x) = x => gof = IR 1

Proving fog (y) = xy => fog = IR 1

gof = fog = IR and Writing f is invertible 1

Writing f-1 = g OR f -1 (x) =
𝒙 𝟑
OR f -1 (y) =
𝒚 𝟑 1
𝟒 𝟒

OR
Let f(x1) = f(x2) x1,x2 ∈ 𝑨 => 4x1 + 3 = 4 x2 + 3 1

=> 𝒙𝟏 = 𝒙𝟐 => f is one - one 1

f(x) = y = 4x + 3 => 𝒙 =
𝒚 𝟑
∈𝑹 1
𝟒

∀ 𝒚∈𝑹 𝒕𝒉𝒆𝒓𝒆 𝒆𝒙𝒊𝒔𝒕𝒔 𝒙 =
𝒚 𝟑
∈ 𝑹 such that f(x) = y => f is onto 1
𝟒

𝐱−𝟑 𝐲−𝟑 1
∴ 𝐟 𝐢𝐬 𝐢𝐧𝐯𝐞𝐫𝐭𝐢𝐛𝐥𝐞 𝐚𝐧𝐝 𝐟 𝟏 (𝐱) = 𝐎𝐑 𝐟 𝟏 (𝐲) =
𝟒 𝟒

40 1
−𝟏 𝟐 𝟏
𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐀𝐁 = 𝟒 −𝟖 −𝟒
−𝟑 𝟔 𝟑

1
−𝟏 𝟒 −𝟑
𝟏
𝐆𝐞𝐭𝐭𝐢𝐧𝐠 (𝐀𝐁) = 𝟐 −𝟖 𝟔
𝟏 −𝟒 𝟑

−𝟏 1
𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝑩𝟏 = 𝟐 and 𝑨𝟏 = [ 1 -4 3 ]
𝟏

−𝟏 𝟒 −𝟑 1
𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝑩 𝑨 = 𝟐 −𝟖 𝟔
𝟏 𝟏

𝟏 −𝟒 𝟑
1
Writing (AB)1 = B1 A1

41 𝟑 −𝟐 𝟑 𝒙 𝟖 1
𝑨= 𝟐 𝟏 −𝟏 , 𝑿 = 𝒚 ,𝑩 = 𝟏 OR Getting |𝑨| = −𝟏𝟕
𝟒 −𝟑 𝟐 𝒛 𝟒

2
−𝟏 −𝟓 −𝟏
𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝐚𝐝𝐣 𝐀 = −𝟖 −𝟔 𝟗
−𝟏𝟎 𝟏 𝟕

NOTE: If any 4 cofactors are correct award 1 mark
−𝟏 −𝟓 −𝟏 𝟖 1
𝒂𝒅𝒋 𝑨 𝟏
𝐖𝐫𝐢𝐭𝐢𝐧𝐠 𝐗 = 𝐀 𝟏 𝐁 = 𝐁= −𝟖 −𝟔 𝟗 𝟏
|𝑨| 𝟏𝟕
−𝟏𝟎 𝟏 𝟕 𝟒

Getting 𝐱 = 𝟏 , 𝐲 = 𝟐 𝐚𝐧𝐝 𝐳 = 𝟑 1

42 𝒅𝒚 1
𝐆𝐞𝐭𝐭𝐢𝐧𝐠 = 𝑨𝒎𝒆𝒎𝒙 + 𝑩𝒏𝒆𝒏𝒙
𝒅𝒙

𝒅𝟐 𝒚 1
𝐆𝐞𝐭𝐭𝐢𝐧𝐠 𝟐
= 𝑨𝒎𝟐 𝒆𝒎𝒙 + 𝑩𝒏𝟐 𝒆𝒏𝒙
𝒅𝒙

Writing LHS = 𝑨𝒎𝟐 𝒆𝒎𝒙 + 𝑩𝒏𝟐 𝒆𝒏𝒙 – ( m + n ) (𝑨𝒎𝒆𝒎𝒙 + 𝑩𝒏𝒆𝒏𝒙 ) 1
+mn (𝑨𝒆𝒎𝒙 + 𝑩𝒆𝒏𝒙 )

Getting LHS = 𝑨𝒎𝟐 𝒆𝒎𝒙 + 𝑩𝒏𝟐 𝒆𝒏𝒙 – 𝑨𝒎𝟐 𝒆𝒎𝒙 − 𝑩𝒎𝒏𝒆𝒏𝒙 − 𝑨𝒎𝒏𝒆𝒎𝒙 1
− 𝑩𝒏𝟐 𝒆𝒏𝒙 +A 𝐦𝐧𝒆𝒎𝒙 + 𝑩𝒎𝒏𝒆𝒏𝒙

𝒅𝟐 𝒚 𝒅𝒚 1
𝐒𝐡𝐨𝐰𝐢𝐧𝐠 𝐋𝐇𝐒 = 𝟎 = RHS OR − (𝒎 + 𝒏 ) + 𝒎𝒏 𝒚 = 𝟎
𝒅𝒙𝟐 𝒅𝒙
43 Writing I =
𝟏 (𝒂 𝒙) (𝒂 𝒙)
𝒅𝒙 1
𝟐𝒂
∫ (𝒂 𝒙)(𝒂 𝒙)

Getting I =
𝟏 𝟏 𝟏 1
𝟐𝒂
∫(𝒂 𝒙
+
𝒂 𝒙
) 𝒅𝒙

Getting I =
𝟏
[ - log | a – x | + log | a + x |] 1
𝟐𝒂
𝟏 𝒂 𝒙 1
Getting I = log +c
𝟐𝒂 𝒂 𝒙
𝟏 𝟏 𝟓 𝒙 1
Getting ∫ 𝟐𝟓 𝒙𝟐 𝒅𝒙 = 𝟏𝟎 𝒍𝒐𝒈 +c
𝟓 𝒙

44 1

Drawing correct figure ( for wrong figure deduct 1 mark )

𝝅 1
Required Area = 2 ∫𝟎 𝒔𝒊𝒏 𝒙 𝒅𝒙

= 2 ( - cos x )𝝅𝟎 1

= - 2 ( cos 𝝅 – cos 0 ) 1

Required area = - 2 ( -1 -1 ) = 4 sq units 1

OR
For Figure 1
𝝅
1
Required Area = 4 ∫𝟎 𝒔𝒊𝒏 𝒙 𝒅𝒙
𝟐

𝝅
𝟐
1
= 4 ( -COS X )𝟎
𝝅
= - 4 ( COS 𝟐 - COS 0 ) 1
= -4 ( -1) = 4 1
45 Writing
𝒅𝒚 𝟐
+ y=x 1
𝒅𝒙 𝒙

𝑷=
𝟐
and Q = x 1
𝒙

𝟐 1
IF = 𝒆∫ 𝒑 𝒅𝒙 = 𝒆∫𝒙 𝒅𝒙 = x2

GS is y ( IF) = ∫ 𝑸. (𝑰𝑭)𝒅𝒙 + 𝒄 OR y . x2 = ∫ 𝒙𝟑 𝒅𝒙 + 𝒄 1

𝒙𝟒 1
Getting y x2 = + 𝒄
𝟒

PART – E
46 𝒂 1
RHS = ∫𝟎 𝒇( 𝒂 − 𝒙 ) 𝒅𝒙
Put a – x = t => dx = - dt and x = 0 , t = a
x=a,t=0

Getting
𝟎
RHS = ∫𝒂 𝒇( 𝒕 ) ( − 𝒅𝒕) 1

𝒂 1
Getting RHS = ∫𝟎 𝒇( 𝒙 ) 𝒅𝒙 = LHS

𝝅
√𝐬𝐢𝐧 𝒙 1
Let I = ∫𝟎𝟐 𝒅𝒙 ---------(1)
√𝐬𝐢𝐧 𝒙 √𝐜𝐨𝐬 𝒙
𝝅
Changing x to − 𝒙
𝟐
𝝅
√𝐜𝐨𝐬 𝒙
Getting I = ∫𝟎 𝟐 𝒅𝒙 ---------(2)
√𝐜𝐨𝐬 𝒙 √𝐬𝐢𝐧 𝒙

𝝅
1
Adding (1) and (2) => 2I= ∫𝟎 𝟏 𝒅𝒙
𝟐

𝝅 1
Getting I=
𝟒

OR

1+1+1

Drawing line x + y = 60 correctly award 1 mark

Drawing line 25x + 5y = 500 or 5x +y = 100 correctly award 1 mark

For Shading the feasible region award 1 mark
Coner points Z = 250x + 1+1
75y
O( 0 , 0) 0
A( 0 , 60) 4500
E( 10 , 50 ) 6250
D(20 , 0 ) 5000

Maximum value of Z is 6250 occurs at the point ( 10 , 50 ) 1

47 𝟕 𝟏𝟐 1
Getting A2 =
𝟒 𝟕

Showing A2 – 4A + I = O 1

Getting A – 4I + A-1 = O OR A -1 = 4I – A 1

𝟐 −𝟑 1
Getting A-1 =
−𝟏 𝟐

OR
Writing At x=2 LHL =RHL 1
5 = 2a + b -------(1)
At x = 10 LHL =RHL 1
10 a + b = 21 -------(2)

Getting a=2 1

Getting b=1 1