Karnataka 2nd PUC Question Paper 2026 Basic Mathematics

£ÉÆÃAzÀt ¸ÀASÉå :
Registration No. :

V1 – 2026
«µÀAiÀÄ ¸ÀAPÉÃvÀ /
75 (NS)
Subject Code

ªÀÄÆ®UÀtÂvÀ / BASIC MATHEMATICS
(Kannada and English Versions)

[¸ÀªÀÄAiÀÄ: 3 UÀAmÉUÀ¼ÀÄ] [MlÄÖ ¥Àæ±ÉßUÀ¼À ¸ÀASÉå : 42] [UÀjµÀ× CAPÀUÀ¼ÀÄ : 80]
[Time : 3 Hours] [Total No. of questions : 42] [Max. Marks : 80]

(Kannada Version)
¸ÀÆZÀ£ÉUÀ¼ÀÄ : 1. ¥Àæ±Éß ¥ÀwæPÉAiÀÄÄ 5 ¨sÁUÀUÀ¼À£ÀÄß ºÉÆA¢zÉ. J, ©, ¹, r ªÀÄvÀÄÛ E. J¯Áè
¨sÁUÀUÀ½UÉ GvÀÛj¹.
2. J-¨sÁUÀPÉÌ 20 CAPÀUÀ¼ÀÄ, ©-¨sÁUÀPÉÌ 12 CAPÀUÀ¼ÀÄ, ¹-¨sÁUÀPÉÌ
18 CAPÀUÀ¼ÀÄ, r-¨sÁUÀPÉÌ 20 CAPÀUÀ¼ÀÄ ªÀÄvÀÄÛ E-¨sÁUÀPÉÌ 10 CAPÀUÀ¼ÀÄ
ºÉÆA¢zÉ.
3. ¨sÁUÀ - J £À°ègÀĪÀ ¥Àæ±ÉßUÀ½UÉ ¥ÀæxÀªÀĪÁV §gÉzÀ GvÀÛgÀUÀ¼À£ÀÄß ªÀiÁvÀæ
ªÀiË®åªÀiÁ¥À£ÀzÀ°è ¥ÀjUÀt¸À¯ÁUÀĪÀÅzÀÄ.
4. ¨sÁUÀ - r £À°è£À LPP AiÀÄ 39 £Éà ¥Àæ±ÉßUÉ UÁæ¥sï ²Ãmï£ÀÄß §¼À¸À¨ÉÃPÀÄ.

5. ¥Àæ±Éß ¥ÀwæPÉAiÀÄ°è ¸ÀÆa¹gÀĪÀAvÉ ¥Àæ±Éß ¸ÀASÉåAiÀÄ£ÀÄß ¸ÀjAiÀiÁV §gɬÄj.
6. UÁæ¥sï M¼ÀUÉÆAqÀ ¥Àæ±ÉßUÉ ¥ÀAiÀiÁðAiÀĪÁV ¥Àæ±Éß ¥ÀwæPÉAiÀÄ PÉÆ£ÉAiÀÄ°è
¥ÀævÉåÃPÀ ¨sÁUÀ - J¥sï £À°è zÀ馅 «PÀ®ZÉÃvÀ£À «zÁåyðUÀ½UÉ
«ªÀgÀuÁvÀäPÀ ¥Àæ±ÉßAiÀÄ£ÀÄß PÉüÀ¯ÁVzÉ.

P.T.O.

75 (NS) -2-

¨sÁUÀ - J

I. J¯èÁ §ºÀÄ DAiÉÄÌ ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹. (10 × 1 = 10)

2 3
1) A=  DzÀgÉ, A − A' £À ¨É¯ÉAiÀÄÄ
4 5
 0 1 0 − 1
a) − 1 0 b) 1 0 
   
1 0  − 1 0
c) 0 − 1 d)  0 1
   

2) 10 d£ÀgÀ£ÀÄß MAzÀÄ ªÉÄÃf£À ¸ÀÄvÀÛ PÀÆj¸À§ºÀÄzÁzÀ MlÄÖ §UÉUÀ¼ÀÄ

a) 10 ! b) 9!

10 ! 9!
c) d)
2 2

3) P ( A' ) = 0.65 DzÀgÉ, P ( A) =

a) 1 b) 0

c) 0.35 d) 0.65

4) ~ p ∧ ~ q £À £ÀPÁgÀªÀÅ

a) ~ p∧q b) p∨q
c) p ∨ ~q d) ~ p∨~q

5) 9 : 4 £À EªÀÄär C£ÀÄ¥ÁvÀªÀÅ

a) 9:4 b) 81 : 4
c) 9 : 16 d) 81 : 16

-3- 75 (NS)

3
6) cos A = DzÀgÉ, cos 2 A £À ¨É¯ÉAiÀÄÄ
2
2
a) 2 3 b)
3
1
c) d) 1
2

7) x 2 = 16 y ¥ÀgÀªÀ®AiÀÄzÀ £Á©üAiÀÄÄ

a) (0, 4 ) b) (4, 0 )
c) (16, 0 ) d) (0, 16 )

dy
8) y = 2 x DzÀgÉ, =
dx

a) 2x b) x 2x −1

c) 2 x log 2 d) 2 x log 2

1
9) ªÀiË°åÃPÀj¹ :  dx
5 e−x

ex
a) e +c
x
b) +c
5

1
c) 5 ex + c d) +c
5 ex

10) ªÀiË°åÃPÀj¹ :  4 cosec 2 x dx
4
a) 8 cosec x + c b) cosec 3 x + c
3

c) − 4 cot x + c d) cot x + c

75 (NS) -4-

II. ºÉÆA¢¹ §gɬÄj. (5 × 1 = 5)

11) A B

 6 x + 2   6 1 11
a) 2 = DzÀgÉ, x = i)
 4  2 4 3

5 1
b) p r = 60 DzÀgÉ, r = ii)
2

c) 6, 14, 15 UÀ¼À £Á®Ì£Éà C£ÀÄ¥ÁvÀªÀÅ iii) –1

d) 3 sin 10 − 4 sin3 10 £À ¨É¯ÉAiÀÄÄ iv) –3

x 2 − 4x
e) lim £À ¨É¯ÉAiÀÄÄ v) 35
x →3 x − 2

vi) 3

III. PɼÀV£À DªÀgÀtzÀ°è PÉÆnÖgÀĪÀ GvÀÛgÀUÀ¼À£ÀÄß Dj¹PÉÆAqÀÄ ©lÖ ¸ÀܼÀUÀ¼À£ÀÄß
vÀÄA©j. (5 × 1 = 5)

1 1
(12, , , 252, 6, 16)
3 5

 1 − 1
12) A =   DzÀgÉ, A £À ¨É¯ÉAiÀÄÄ ____________.
2 4 

13) 10 «zÁåyðUÀ½AzÀ, 5 «zÁåyðUÀ½gÀĪÀ ¸À«ÄwAiÀÄ DAiÉÄÌAiÀÄ §UÉUÀ¼ÀÄ
____________.

-5- 75 (NS)

14) 5 : 20 = 3 : x DzÀgÉ, ‘x’ £À ¨É¯ÉAiÀÄÄ ____________.

15) y 2 = 16 x ¥ÀgÀªÀ®AiÀÄzÀ ®A§ £Á©üAiÀÄ GzÀݪÀÅ ____________.

1
2
16)  x dx = ____________.
0

¨sÁUÀ – ©

IV. F PɼÀV£À AiÀiÁªÀÅzÁzÀgÀÆ DgÀÄ ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹. (6 × 2 = 12)

2 3  1 − 2
17) A =   ªÀÄvÀÄÛ B =   DzÀgÉ, 2 A + 3 B AiÀÄ£ÀÄß PÀAqÀÄ»r¬Äj.
 1 − 2 1 3 

18) 20 ©AzÀÄUÀ¼À°è, 8 ©AzÀÄUÀ¼ÀÄ MAzÉà gÉÃSÉAiÀÄ°èzÁÝUÀ JµÀÄÖ wæ¨sÀÄdUÀ¼À£ÀÄß
gÀa¸À§ºÀÄzÀÄ?

1 1 7
19) P ( A) = , P (B ) = , P ( A ∪ B ) = DzÀgÉ, P (B A ) AiÀÄ£ÀÄß PÀAqÀÄ»r¬Äj.
2 3 12

20) 7 : 4 C£ÀÄ¥ÁvÀzÀ°ègÀĪÀ, ¥ÀævÉåÃPÀªÁV ¸ÀASÉåUÀ½UÉ K£À£ÀÄß PÀ¼ÉzÀgÉ
5 : 2 C£ÀÄ¥ÁvÀªÁUÀÄvÀÛzÉ?

21) ºÀÄArAiÀÄ ¤d ¸ÉÆÃrAiÀÄÄ 100 ªÀÄvÀÄÛ ¨ÁåAPÀgï£À ¯Á¨sª
À ÅÀ 10 DzÁUÀ
ºÀÄArAiÀÄ ªÀÄÄR¨É¯ÉAiÀÄ£ÀÄß PÀAqÀÄ»r¬Äj.

75 (NS) -6-

22) MAzÀÄ ¥ÀgÀªÀ®AiÀÄzÀ ±ÀÈAUÀªÀÅ (0, 0) ªÀÄvÀÄÛ CzÀgÀ ¤¢ðµÀÖ gÉÃSÉAiÀÄÄ
y = − 6 DzÀgÉ D ¥ÀgÀªÀ®AiÀÄzÀ ¸À«ÄÃPÀgÀtªÀ£ÀÄß PÀAqÀÄ»r¬Äj.

dy
23) y = x sin x DzÁUÀ, £ÀÄß PÀAqÀÄ»r¬Äj.
dx

24) MlÄÖ DzÁAiÀÄzÀ GvÀà£ÀߪÀÅ R = 400 x − 2x 2 ªÀÄvÀÄÛ MlÄÖ ªÉZÀѪÀÅ
C = 2 x 2 + 40 x + 4000 PÉÆqÀ¯ÁVzÉ, CzÀgÀ ¹Ã«ÄvÀ DzÁAiÀÄ ªÀÄvÀÄÛ ¹Ã«ÄvÀ
ªÉZÀѪÀ£ÀÄß PÀAqÀÄ»r¬Äj.

25) ¥ÀgÀªÀ®AiÀÄ x = 2 y 2 , y-CPëÀgÉÃSÉ ªÀÄvÀÄÛ y = 2, y = 4 ¸ÀgÀ¼ÀgÉÃSÉUÀ¼À
£ÀqÀÄ«£À PÉëÃvÀæ¥sÀ®ªÀ£ÀÄß PÀAqÀÄ»r¬Äj.

¨sÁUÀ – ¹

V. PɼÀV£À AiÀiÁªÀÅzÁzÀgÀÆ DgÀÄ ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹. (6 × 3 = 18)

26) PÉæêÀÄgï£À ¤AiÀĪÀÄ¢AzÀ ©r¹j.
3 x + 4y = 7
7x − y = 6

27) “COMMITTEE” ¥ÀzÀzÀ CPëÀgÀUÀ¼À£ÀÄß JµÀÄÖ §UÉ PÀæªÀÄ¥À®èl£É
ªÀiÁqÀ§ºÀÄzÀÄ?

a) CªÀÅUÀ¼À°è JµÀÄÖ T ¬ÄAzÀ ¥ÁægÀA¨sÀªÁV T ¬ÄAzÀ PÉÆ£ÉUÉƼÀÄîvÀÛªÉ?
b) CªÀÅUÀ¼À°è JµÀÄÖ, J¯Áè ¸ÀégÁPÀëgÀUÀ¼ÀÄ MnÖVgÀÄvÀÛªÉ?

-7- 75 (NS)

28) 3 §qÀVAiÀÄgÀÄ ¢£ÀPÉÌ 9 UÀAmÉUÀ¼ÀAvÉ 6 ¢£ÀUÀ¼ÀÄ PÉ®¸À ªÀiÁrzÁUÀ
360 UÀ¼À£ÀÄß ¸ÀA¥Á¢¸ÀÄvÁÛgÉ. 8 §qÀVAiÀÄgÀÄ ¢£ÀPÉÌ
6 UÀAmÉUÀ¼ÀAvÉ 12 ¢£ÀUÀ¼ÀÄ PÉ®¸À ªÀiÁrzÁUÀ §gÀĪÀ DzÁAiÀĪɵÀÄÖ?

29) ¨ÁåAPÀgï£À DzÁAiÀĪÀÅ 1 £ÉÃAiÀĵÀÄÖ ¨ÁåAPÀgÀ£À ¸ÉÆÃrAiÀiÁVzÉ ªÀÄvÀÄÛ
5
±ÉÃ. 20 gÀµÀÄÖ ªÁ¶ðPÀ §rØAiÀiÁzÁUÀ, ºÀÄArAiÀÄÄ JµÀÄÖ CªÀ¢üAiÉƼÀVzÉ
JAzÀÄ PÀAqÀÄ»r¬Äj.

30) ±ÉÃ. 7.5 gÀ zÁ¸ÁÛ¤£À°è 125 CxÀªÁ ±ÉÃ. 5 gÀ zÁ¸ÁÛ¤£À°è 80.
EªÀÅUÀ¼À°è AiÀiÁªÀÅzÀÄ GvÀÛªÀÄ ºÀÆrPÉAiÀiÁVzÉ?

31) ¨sÀgÀvï£ÀÄ CAVAiÀÄ£ÀÄß 336 UÀ½UÉ ±ÉÃ.12 ªÀiÁgÁl
vÉjUÉAiÀÄ£ÉÆß¼ÀUÉÆAqÀÄ, ªÀÄvÀÄÛ PÉÆgÀ½UÉ PÀlÄÖªÀ PÀgÀªÀ¸ÀÛç 110 UÀ½UÉ
±ÉÃ. 10 gÀ ªÀiÁgÁl vÉjUÉAiÀÄ£ÉÆß¼ÀUÉÆAqÀÄ, Rjâ ªÀiÁqÀÄvÁÛ£É. CAV
ªÀÄvÀÄÛ PÉÆgÀ½UÉ PÀlÄÖªÀ PÀgÀªÀ¸ÀÛçzÀ ªÀÄÄ¢ævÀ ¨É¯ÉAiÀÄ£ÀÄß PÀAqÀÄ»r¬Äj.

32) MAzÀÄ WÀ£ÁPÀÈwAiÀÄ ¨ÁºÀÄ«£À GzÀݪÀÅ 6 ¸ÉA.«ÄÃ./¤«ÄµÀ zÀgÀzÀ°è
ºÉZÁÑUÀÄwÛzÉ. CzÀgÀ ¨ÁºÀÄ«£À GzÀݪÀÅ 10 cm DVzÁÝUÀ, CzÀgÀ
ºÉZÁÑUÀÄwÛgÀĪÀ WÀ£À¥sÀ®ªÀ£ÀÄß ªÀÄvÀÄÛ PÉëÃvÀæ¥sÀ®zÀ «¹ÛÃtðªÀ£ÀÄß
PÀAqÀÄ»r¬Äj.

1
33) ªÀiË°åÃPÀj¹j :  dx .
( x + 1) ( x + 3 )

2
34) ªÀiË°åÃPÀj¹j :  22 x + 5 dx .
1 x + 5x + 3

75 (NS) -8-

¨sÁUÀ – r

VI. PɼÀV£À AiÀiÁªÀÅzÁzÀgÀÆ £Á®ÄÌ ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹. (4 × 5 = 20)

35) ªÀiÁvÀÈPÉ «zsÁ£À¢AzÀ ©r¹j.
3 x − y + 2z = 13
2x + y − z = 3
x + 3 y − 5z = − 8

2x + 1
36) EzÀ£ÀÄß «¨sÀfvÀ ©ü£ÀßgÁ²UÀ¼ÁV ¥ÀjªÀwð¹.
( x − 1) ( x − 2) ( x − 3)

37) ( p ∨ q ) ∧ (~ p ∧ ~ q )
¸ÀAAiÀÄÄPÉÆÛÃQÛAiÀÄÄ C¸ÀªÀÄAd¸ÀvÉAiÉÆà CxÀªÁ
E®èªÉÇà JAzÀÄ ¥Àj²Ã°¹.

38) ABC PÀA¥É¤AiÀÄÄ 30 AiÀÄAvÀæUÀ¼À£ÀÄß GvÁàzÀ£É ªÀiÁqÀ®Ä ¨ÉÃPÁUÀĪÀ
¸ÀªÀÄAiÀÄ 1000 UÀAmÉUÀ¼ÀÄ. PÀ°AiÀÄÄ«PÉAiÀÄ ¥ÀjuÁªÀĪÀÅ ±ÉÃ. 90 DzÀgÉ,
MAzÀÄ UÀAmÉAiÀÄ PÀÆ° 20 gÀAvÉ, 120 AiÀÄAvÀæUÀ¼À£ÀÄß GvÁࢸÀ®Ä
¨ÉÃPÁUÀĪÀ MlÄÖ PÀÆ° ªÉZÀѪÀ£ÀÄß PÀAqÀÄ»r¬Äj.

39) gÉÃSÁ£ÀPÉëAiÀÄ£ÀÄß §¼À¹ F PɼÀV£À ¸ÀgÀ¼ÀgÉÃSÁvÀäPÀ PÁAiÀÄðPÀæªÀÄzÀ
¸ÀªÀĸÉåAiÀÄ£ÀÄß ©r¹j.
UÀjµÀ×UÉƽ¹ : Z = 5 x + 4y
¤§AzsÀ£ÉUÉƼÀ¥ÀlÖAvÉ :
2 x + y ≤ 40
x + 2y ≤ 50
ªÀÄvÀÄÛ x ≥ 0, y ≥ 0

-9- 75 (NS)

sin 6 A + sin 2 A + 2 sin 4 A sin 4 A
40) = JAzÀÄ ¸Á¢ü¹j.
sin 7 A + sin 3 A + 2 sin 5 A sin 5 A

(
41) y = x + 1+ x 2 ) DzÁUÀ, (1+ x )y + x y − m y = 0 JAzÀÄ ¸Á¢ü¹j.
m
2
2 1
2

¨sÁUÀ – E

VII. F PɼÀV£À ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹j. (1 × 10 = 10)

 x n − an 
42) a) J¯èÁ ¨sÁUÀ®§Þ ‘n’ UÀ½UÉ lim   = n a n − 1 JAzÀÄ ¸Á¢ü¹j. (6)
x →a  x − a 

CxÀªÁ

(0, 0 ), (4, 8 ), (8, 6 ) ªÀÄvÀÄÛ (− 1, 3 ) ©AzÀÄUÀ¼ÀÄ MAzÉà ªÀÈvÀÛzÀ ªÀÄÆ®PÀ
ºÁzÀĺÉÆÃUÀÄvÀÛzÉ JAzÀÄ ¸Á¢ü¹j.

b) 20 «ÄÃlgï zÀÆgÀ¢AzÀ MAzÀÄ PÀlÖqÀzÀ ªÉÄÃ¯É MAzÀÄ
zsÀéd¸ÀÛA¨sÀªÀ£ÀÄß ¤°è¸À¯ÁVzÉ, D zsÀéd¸ÀÛA¨sÀzÀ ªÉÄîÄÛ¢ ªÀÄvÀÄÛ PÀlÖqÀzÀ
ªÉÄîÄÛ¢UÀ¼ÀÄ, PÀæªÀĪÁV PÉÆãÀUÀ¼ÀÄ 60° ªÀÄvÀÄÛ 45° UÀ¼ÁVªÉ.
ºÁUÁzÀgÉ, D zsÀéd¸ÀÛA¨sÀzÀ JvÀÛgÀªÀ£ÀÄß PÀAqÀÄ»r¬Äj. (4)

CxÀªÁ

(1. 02)5 £À ¨É¯ÉAiÀÄ£ÀÄß ¢é¥ÀzÀ ¥ÀæªÉÄÃAiÀĪÀ£ÀÄß §¼À¹ £Á®ÄÌ zÀ±ÁA±ÀPÉÌ
PÀAqÀÄ»r¬Äj.

75 (NS) -10-

¨sÁUÀ – J¥sï

(zÀ馅 «PÀ®ZÉÃvÀ£À «zÁåyðUÀ½UÉ ªÀiÁvÀæ) (1 × 5 = 5)

39) PÀA¥À¤AiÀÄÄ P ªÀÄvÀÄÛ Q JA§ JgÀqÀÄ GvÀà£ÀßUÀ¼À£ÀÄß vÀAiÀiÁj¸ÀÄvÀÛzÉ.
¥Àæw ‘P’ UÉ vÉÃAiÀÄĪÀÅzÀPÉÌ 4 UÀAmÉUÀ¼ÀÄ ªÀÄvÀÄÛ ºÉƼÀ¥ÀÅ PÉÆqÀ®Ä 2 UÀAmÉUÀ¼ÀÄ
ºÁUÀÆ ¥Àæw ‘Q’ UÉ vÉÃAiÀÄĪÀÅzÀPÉÌ 2 UÀAmÉUÀ¼ÀÄ ªÀÄvÀÄÛ ºÉƼÀ¥ÀÅ PÉÆqÀ®Ä
5 UÀAmÉUÀ¼ÀÄ ¨ÉÃPÁUÀÄvÀÛzÉ. vÉÃAiÀÄĪÀÅzÀPÉÌ PÉêÀ® 20 UÀAmÉUÀ¼ÀÄ ºÁUÀÆ
ºÉƼÀ¥ÀÅ PÉÆqÀ®Ä PÉêÀ® 24 UÀAmÉUÀ¼ÀÄ ®¨sÀå«gÀĪÀÅzÀÄ. P ªÀÄvÀÄÛ Q UÀ¼À
MAzÀÄ WÀlPÀzÀ ¯Á¨sÀªÀÅ PÀæªÀĪÁV 6 ªÀÄvÀÄÛ 8 DzÀg,É UÀjµÀ×
¯Á¨sÀªÀ£ÉÆß¼ÀUÉÆAqÀ ¸ÀgÀ¼ÀgÉÃSÁvÀäPÀ PÁAiÀÄðPÀæªÀÄ ¸ÀªÀĸÉåAiÀÄ£ÀÄß
¸ÀÆvÀæ gÀÆ¥ÀzÀ°è ªÀåPÀÛ¥Àr¹j.

———————

-11- 75 (NS)

(English Version)

Instructions : 1. The question paper has 5 parts A, B, C, D and E. Answer
all the Parts.
2. Part – A carries 20 marks, Part – B carries 12 marks,
Part – C carries 18 marks, Part – D carries 20 marks
and Part – E carries 10 marks.

3. For Part – A questions, only the first written answers
will be considered for evaluation.
4. In the Part – D, use graph sheet for the question
number 39 on LPP.
5. Write the question numbers properly as indicated in the
question paper.

6. For question having graph, alternate question is given at
the end of the question paper in a separate section in the
Part – F for Visually Challenged Students.

PART – A

I. Answer all the multiple choice questions. (10 × 1 = 10)

2 3
1) If A =   , then the value of A − A'
4 5

 0 1 0 − 1
a) − 1 0 b) 1 0 
   

1 0  − 1 0
c) 0 − 1 d)  0 1
   

75 (NS) -12-

2) The number of ways in which 10 people can be seated around a table
a) 10 ! b) 9!
10 ! 9!
c) d)
2 2

3) If P ( A' ) = 0.65 , then P (A) =
a) 1 b) 0
c) 0.35 d) 0.65

4) Negation of ~ p ∧ ~ q is
a) ~ p∧q b) p∨q
c) p ∨ ~q d) ~ p∨~q

5) The duplicate ratio of 9 : 4 is
a) 9 : 4 b) 81 : 4
c) 9 : 16 d) 81 : 16

3
6) If cos A = , then the value of cos 2 A is
2
2
a) 2 3 b)
3
1
c) d) 1
2

7) The focus of the parabola x 2 = 16 y is
a) (0, 4 ) b) (4, 0 )
c) (16, 0 ) d) (0, 16 )

dy
8) If y = 2 x , then =
dx
a) 2x b) x 2x −1
c) 2 x log 2 d) 2 x log 2

-13- 75 (NS)

1
9) Evaluate :  dx
5 e−x

ex
a) e +c
x
b) +c
5
1
c) 5 ex + c d) +c
5ex

10) Evaluate :  4 cosec 2 x dx

4
a) 8 cosec x + c b) cosec 3 x + c
3
c) − 4 cot x + c d) cot x + c

II. Match the following. (5 × 1 = 5)

11) A B

 6 x + 2   6 1 11
a) If  = , then x = i)
2 4  2 4 3

1
b) If 5 p r = 60 , then r = ii)
2

c) Fourth proportional of 6, 14, 15 is iii) –1

d) The value of 3 sin 10 − 4 sin3 10 is iv) –3

x 2 − 4x
e) The value of lim is v) 35
x →3 x − 2

vi) 3

75 (NS) -14-

III. Fill in the blanks by choosing appropriate answers from those given in the
bracket. (5 × 1 = 5)

1 1
(12, , , 252, 6, 16)
3 5

 1 − 1
12) If A =   , then the value of A is ____________.
2 4 

13) The number of ways a committee of 5 can be chosen from 10 students is
____________.

14) If 5 : 20 = 3 : x, then the value of x is ____________.

15) The length of the latus rectum of the parabola y 2 = 16 x is ___________.

1
2
16)  x dx = ____________.
0

PART – B

IV. Answer any six of the following questions. (6 × 2 = 12)

2 3  1 − 2
17) If A =   and B =   , then find 2 A + 3 B .
 1 − 2 1 3 

18) Find the number of triangles that can be formed out of 20 points in
which 8 are collinear.

1 1 7
19) If P ( A) = , P (B ) = , P ( A ∪ B ) = , then find P (B A ) .
2 3 12

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20) What must be subtracted from each term in the ratio 7 : 4, so that it
becomes 5 : 2?

21) True discount on a bill was 100 and Banker’s gain is 10.
Find the face value of the bill.

22) Find the equation of the parabola whose vertex is (0, 0 ) and directrix is
y =− 6.

dy
23) If y = x sin x , then find .
dx

24) The total revenue function is given by R = 400 x − 2 x 2 and the total cost
function is given by C = 2 x 2 + 40 x + 4000 . Find the marginal revenue and
marginal cost function.

25) Find the area enclosed by the curve x = 2 y 2 , y-axis and the lines
y = 2 and y = 4 .

PART – C

V. Answer any six of the following questions. (6 × 3 = 18)

26) Solve using Cramer’s rule.
3 x + 4y = 7
7x − y = 6

27) Find the number of permutations of the letters of the word “COMMITTEE”.

a) How many of them begin with T and end with T?

b) In how many all the vowels are together?

28) Three carpenters can earn 360 in 6 days working 9 hours a day. How
much will 8 carpenters earn in 12 days working 6 hours a day?

75 (NS) -16-

1
29) The Banker’s gain on a bill is th of the Banker’s discount and the rate of
5
interest is 20% p.a. Find the unexpired period of the bill.

30) Which is a better investment? 7.5% stock at 125 or 5% stock at 80.

31) Bharath bought a shirt for 336 including 12% sales tax and a neck tie
for 110 including 10% sales tax. Find the printed price of shirt and
neck tie together.

32) The edge of a variable cube is increasing at the rate of 6 cm/min.
How fast is the volume and the surface area increasing when the edge is
10 cm long?

1
33) Evaluate :  dx .
( x + 1) ( x + 3 )

2
2x + 5
34) Evaluate :  2
dx .
1 x + 5 x + 3

PART – D

VI. Answer any four of the following questions. (4 × 5 = 20)

35) Solve by matrix method.
3 x − y + 2z = 13
2x + y − z = 3
x + 3 y − 5z = − 8

36) Resolve into partial fractions :
2x + 1
.
( x − 1) ( x − 2) ( x − 3)

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37) Check whether the proposition ( p ∨ q ) ∧ (~ p ∧ ~ q ) is a contradiction
or not.

38) ABC company required 1000 hours to produce first 30 engines. If the
learning effect is 90%, find the total labour cost at 20 per hour to
produce a total of 120 engines.

39) Solve the following LPP by graphical method.
Maximize : Z = 5 x + 4 y
Subject to the constraints
2 x + y ≤ 40
x + 2y ≤ 50
and x ≥ 0, y ≥ 0

40) Prove that :
sin 6 A + sin 2 A + 2 sin 4 A sin 4 A
= .
sin 7 A + sin 3 A + 2 sin 5 A sin 5 A

(
41) If y = x + 1+ x 2 ) then prove that (1+ x )y + x y − m y = 0 .
m
2
2 1
2

PART – E

VII. Answer the following questions. (1 × 10 = 10)

 x n − an 
42) a) Prove that lim   = n a n − 1 for all rational values of n. (6)
x →a  x − a  

OR

Show that the points (0, 0 ), (4, 8 ), (8, 6 ) and (− 1, 3 ) are Concyclic.

75 (NS) -18-

b) A flag staff stands upon the top of a building. At a distance of
20 metres, the angle of elevation of the top of the flag staff and
building are 60° and 45° respectively. Find the height of the
flag staff. (4)

OR
5
Find the value of (1. 02) using Binomial theorem, upto 4 places of
decimal.

PART – F

(Only for Visually Challenged Students) (1 × 5 = 5)

39) A Company produces two products P and Q. Each P requires 4 hours of
grinding and 2 hours of polishing and each Q requires 2 hours of grinding
and 5 hours of polishing. The total available hours for grinding is 20 hours
and for polishing is 24 hours. Profit per unit of P is 6 and that of Q is
8. Formulate the L.P.P to maximize the profit.
———————

-19- 75 (NS)

75 (NS) -20-

V1 – 2026
Subject Name : BASIC MATHEMATICS Subject Code : 75 (NS)

Registration No. of the candidate : ————————————
Invigilator’s Signature