Karnataka 2nd PUC Question Paper 2025 Mathematics Basic

Government of Karnataka
Karnataka Secondary Education Examination Board

Question Papers

£ÉÆÃAzÀt ¸ÀASÉå :
Registration No. :

X1 – 2025
«µÀAiÀÄ ¸ÀAPÉÃvÀ /
75 (NS)
Subject Code

ªÀÄÆ® UÀtÂvÀ±Á¸ÀÛç / BASIC MATHEMATICS
(Kannada and English Versions)

[¸ÀªÀÄAiÀÄ: 3 UÀAmÉUÀ¼ÀÄ] [MlÄÖ ¥Àæ±ÉßUÀ¼À ¸ÀASÉå : 43] [UÀjµÀ× CAPÀUÀ¼ÀÄ : 80]
[Time : 3 Hours] [Total No. of questions : 43] [Max. Marks : 80]

(Kannada Version)
¸ÀÆZÀ£ÉUÀ¼ÀÄ : 1. F ¥Àæ±Éß ¥ÀwæPÉAiÀÄ°è 5 ¨sÁUÀUÀ½ªÉ. ¨sÁUÀ-J, ©, ¹, r ªÀÄvÀÄÛ E JA§
J¯èÁ ¨sÁUÀUÀ¼À£ÀÄß GvÀÛj¹.
2. ¨sÁUÀ - J - 20 CAPÀUÀ¼ÀÄ
¨sÁUÀ - © - 12 CAPÀUÀ¼ÀÄ
¨sÁUÀ - ¹ - 18 CAPÀUÀ¼ÀÄ
¨sÁUÀ - r - 20 CAPÀUÀ¼ÀÄ
¨sÁUÀ - E - 10 CAPÀUÀ¼ÀÄ EgÀÄvÀÛªÉ.
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 £ÀªÀÄÆ¢¹gÀĪÀ ¥Àæ±Éß ¸ÀASÉåAiÀÄ£Éßà GvÀÛgÀ ¥ÀwæPÉAiÀÄ°èAiÀÄÆ
§gÉAiÀĨÉÃPÀÄ.
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. PɼÀV£À J¯èÁ §ºÀÄ DAiÉÄÌ ¥Àæ±ÉßUÀ½UÉ GvÀÛj¹. (10 × 1 = 10)

3 2
1) A=  DzÀgÉ ªÀiÁvÀÈPÉ 2 A′ £À ¨É¯É
2 4 

6 1  6 2
a)  4 2 b) 4 8
   

6 2  6 4
c) 8 4  d) 4 8 
   

2) n
C10 = nC5 DzÀgÉ ‘n’ £À ¨É¯É

a) 15 b) 25

c) 5 d) 10

3) JgÀqÀÄ £ÁtåUÀ¼À£ÀÄß MªÉÄä¯Éà a«ÄäzÁUÀ JgÀqÀÄ vÀ¯ÉUÀ¼ÀÄ §gÀĪÀ
¸ÀA¨sÀªÀ¤ÃAiÀÄvÉ

1 3
a) b)
2 4

1
c) d) 1
4

-3- 75 (NS)

4) ~ p → ~ q ¸ÀAAiÀÄÄPÉÆÛÃQÛAiÀÄ «¯ÉÆêÀĪÀÅ

a) ~p→q b) q→p

c) p→q d) ~q →~p

5) 3 : 4 ªÀÄvÀÄÛ 4 : 7 gÀ ¸ÀAAiÀÄÄPÀÛ C£ÀÄ¥ÁvÀªÀÅ

a) 3 : 12 b) 3:7
c) 7:3 d) 12 : 7

1
6) sin A = DzÀgÉ sin 2 A £À ¨É¯É
2

3 1
a) b)
2 2

c) 0 d) 1

7) x 2 = 8 y ¥ÀgÀªÀ®AiÀÄzÀ ¤¢ðµÀÖ gÉÃSÉAiÀÄ ¸À«ÄÃPÀgÀt

a) y=2 b) x=2
c) x = –2 d) y = −2

dy
8) y = 5 e x − log x DzÀgÉ £À ¨É¯É
dx

1 1
a) e5 x − b) 5 ex −
x x

1
c) 5 ex − x d) 5x −
x

75 (NS) -4-

1
9)  7 x + 8 dx £À ¨É¯É
log (7 x + 8)
a) log (7 x + 8) + C b) +C
7
1
c) 7 log(7 x + 8) + C d) +C
log (7 x + 8)

 x dx £À ¨É¯É
5
10)

4 x5
a) 5x + C b) +C
5

x4 x6
c) +C d) +C
4 6

II. ºÉÆA¢¹ §gɬÄj. (5 × 1 = 5)

11) J ©

3 x
a) = 2 DzÀgÉ ‘x’ £À ¨É¯É i) 7
4 2

3 −1
b) n
p 3 = 210 DzÀgÉ ‘n’ £À ¨É¯É ii)
2 2

c) 4 ªÀÄvÀÄÛ 6 gÀ 3 £Éà C£ÀÄ¥ÁvÀ iii) 2

d) sin 15º £À ¨É¯É iv) 1

sin 4 x + 2 3 +1
e) Lt v)
x → 0 sin 2 x + 1 2 2

vi) 9

-5- 75 (NS)

III. 12 jAzÀ 16 gÀªÀgÉV£À ¥Àæ±ÉßUÀ½UÉ, PɼÀV£À DªÀgÀtzÀ°è PÉÆnÖgÀĪÀ GvÀÛgÀUÀ¼À°è
¸ÀjAiÀiÁzÀ GvÀÛgÀUÀ¼À£ÀÄß DAiÀÄÄÝ §gɬÄj. (5 × 1 = 5)

(35, 12, 1, 3, 30, 16)

− 9  5  − 14
12)   −   =   DzÀgÉ ‘ x ’ £À ¨É¯ÉAiÀÄÄ ____________.
2 −
    1 x 

13) zÀ±À ¨sÀÄdªÀżÀî §ºÀĨsÀÄeÁPÀÈwAiÀÄ°ègÀĪÀ PÀtðUÀ¼ÀÄ ____________.

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

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

π
2
16)  sin x dx £À ¨É¯É ____________.
0

¨sÁUÀ – ©

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

 2 0  1 − 1
17) 2 A + B =   ªÀÄvÀÄÛ B=  DzÀgÉ ªÀiÁvÀÈPÉ A C£ÀÄß
− 1 3 3 0 
PÀAqÀÄ»r¬Äj.

18) M§â ¤¢ðµÀÖ ªÀåQÛ M¼ÀUÉÆAqÀAvÉ 10 ªÀåQÛUÀ½AzÀ 6 ªÀåQÛUÀ¼À£ÀÄß JµÀÄÖ
«zsÀUÀ¼À°è DAiÉÄÌ ªÀiÁqÀ§ºÀÄzÀÄ?

75 (NS) -6-

2 1
19) A ªÀÄvÀÄÛ B ¥ÀgÀ¸ÀàgÀ ¥ÀævÉåÃPÀ WÀl£ÉUÀ¼ÀÄ DVzÀÝ°è ªÀÄvÀÄÛ P ( A) = , P (B ) =
5 7
DVzÁÝUÀ P ( A ∪ B ) £ÀÄß PÀAqÀÄ»r¬Äj.

20) a : b = 3 : 4, b : c = 8 : 15 DzÀgÉ a : b : c PÀAqÀÄ»r¬Äj.

21) ¨ÁåAPÀgï£À ¸ÉÆÃr ªÀÄvÀÄÛ ¨ÁåAPÀgï£À ¯Á¨sÀªÀÅ PÀæªÀĪÁV 1,250 ªÀÄvÀÄÛ

50 DVzÀÄÝ ªÁ¬ÄzÉ EgÀĪÀ ºÀÄArAiÀÄ ªÀÄÄR¨É¯É K£ÀÄ?

22) ±ÀÈAUÀ (0, 0) ªÀÄvÀÄÛ £Á©ü (– 4, 0) DVgÀĪÀ ¥ÀgÀªÀ®AiÀÄzÀ ¸À«ÄÃPÀgÀtªÀ£ÀÄß
PÀAqÀÄ»r¬Äj.

dy
23) y = x x DzÀgÉ £À ¨É¯ÉAiÀÄ£ÀÄß PÀAqÀÄ»r¬Äj.
dx

24) MlÄÖ ªÉZÀÑzÀ ¥sÀ®£ÀªÀÅ C = q 3 − 3 q 2 + 15 q + 27 DVzÉÉ, q JA§ÄzÀÄ
GvÁàzÀ£ÉAiÀiÁzÁUÀ, ¹Ã«ÄvÀ ªÉZÀÑ ªÀÄvÀÄÛ ¹ÜgÀ ªÉZÀѪÀ£ÀÄß PÀAqÀÄ»r¬Äj.

25) y = x 2 ¥ÀgÀªÀ®AiÀÄ, x-CPëÀgÉÃSÉ ªÀÄvÀÄÛ x = 0, x = 1 gÉÃSÉUÀ¼À £ÀqÀÄ«£À
«¹ÛÃtðªÀ£ÀÄß PÀAqÀÄ»r¬Äj.

-7- 75 (NS)

¨sÁUÀ – ¹

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

26) PÉæêÀÄgï£À ¤AiÀĪÀÄ¢AzÀ ©r¹j.
2x + y = 1
x − 3y = 4

27) ENGINEERING ¥ÀzÀzÀ CPëÀgÀUÀ¼É®èªÀ£ÀÆß JµÀÄÖ §UÉAiÀÄ°è PÀæªÀÄ¥À®èl£É
ªÀiÁqÀ§ºÀÄzÀÄ? CzÀgÀ°è

a) JµÀÄÖ GIN ¤AzÀ ±ÀÄgÀĪÁV GRIN ¤AzÀ ªÀÄÄPÁÛAiÀĪÁUÀÄvÀÛzÉ?

b) JµÀÄÖ ¥ÀzÀUÀ¼À°è 3 E UÀ¼ÀÄ MmÁÖVgÀÄvÀÛzÉ?

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

29) 6 wAUÀ¼À ªÁ¬ÄzÉ EgÀĪÀ ±ÉÃ. 4 ªÁ¶ðPÀ §rØ ¤ÃqÀĪÀ MAzÀÄ ºÀÄArAiÀÄ
¨ÁåAPÀgï£À ¯Á¨sÀªÀÅ 24 DVzÀÝgÉ, F ºÀÄArAiÀÄ ¤d¸ÉÆÃr,
¨ÁåAPÀgï£À ¸ÉÆÃr ªÀÄvÀÄÛ CzÀgÀ ªÀÄÄR¨É¯ÉAiÀÄ£ÀÄß PÀAqÀÄ»r¬Äj.

30) ¥ÀæwÃPÀ£ÀÄ 6,000 ¨É¯ÉAiÀÄļÀî ±ÉÃ. 7.5 ¸ÁÖÖPï£ÀÄß 108 gÀAvÉ ªÀiÁj,
D ºÀtªÀ£ÀÄß ±ÉÃ. 9 ¸ÁÖPï£À°è ºÀÆrzÁUÀ CªÀ£À DzÁAiÀÄ 270
ºÉZÁÑUÀÄvÀÛzÉ. ºÁUÁzÀgÉ ±ÉÃ. 9 gÀ ¸ÁÖPï£ÀÄß CªÀ£ÀÄ JµÀÄÖ ¨É¯ÉUÉ
PÉÆAqÀÄPÉÆAqÀ£ÀÄ?

75 (NS) -8-

31) MAzÀÄ §mÉÖ MUÉAiÀÄĪÀ AiÀÄAvÀæzÀ ¨É¯ÉAiÀÄÄ ªÀiÁgÁl vÉjUÉ ¸ÉÃj
13,530 UÀ¼ÁVgÀÄvÀÛzÉ. ªÀiÁgÁl vÉjUÉAiÀÄÄ ±ÉÃ. 10 DVzÀÝgÉ, CzÀgÀ
ªÀÄÆ® ¨É¯ÉAiÀÄ£ÀÄß PÀAqÀÄ»r¬Äj.

32) MAzÀÄ WÀ£ÁPÀÈwAiÀÄ ¨ÁºÀÄ«£À GzÀݪÀÅ 6 cm/min. zÀgÀzÀ°è ºÉZÁÑUÀÄvÀÛzÉ.
CzÀgÀ ¨ÁºÀĪÀÅ 10 cm DVzÁÝUÀ, CzÀgÀ ºÉZÁÑUÀÄwÛgÀĪÀ WÀ£À¥sÀ®ªÀ£ÀÄß
PÀAqÀÄ»r¬Äj ªÀÄvÀÄÛ ºÉZÁÑUÀÄwÛgÀĪÀ PÉëÃvÀæ¥sÀ®ªÀ£ÀÄß PÀAqÀÄ»r¬Äj.

x
33) ªÀiË°åÃPÀj¹ :  dx .
( x − 1) ( x − 2)

2
34) ªÀiË°åÃPÀj¹ :  ( x + e x + 2) dx .
1

¨sÁUÀ – r

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

35) ªÀiÁvÀÈPÉ «zsÁ£À¢AzÀ ©r¹.
3 x + 2y − z = 6
3 x + y − 2z = 3
2 x − 3 y − z = −1

9
36) EzÀ£ÀÄß «¨sÀfvÀ ©ü£ÀßgÁ²UÀ¼ÁV ¥ÀjªÀwð¹.
( x + 1) ( x + 2)2

37) ( p ∧ ~ q ) ∨ q ªÀÄvÀÄÛ p ∨ q vÁQðPÀªÁV ¸ÀªÀiÁ£ÀªÁVzÉAiÉÄà JAzÀÄ ¥ÀjÃQë¹.

-9- 75 (NS)

38) MAzÀÄ JAf¤AiÀÄjAUï PÀA¥É¤AiÀÄ PÀ°AiÀÄÄ«PÉ ±ÉÃ. 80 DVzÀÄÝ, ªÉÆzÀ®
GvÁàzÀ£ÉAiÀÄ WÀlPÀªÀ£ÀÄß vÀAiÀiÁj¸À®Ä D PÀA¥É¤UÉ 1000 PÀÆ° UÀAmÉUÀ¼ÀÄ
¨ÉÃPÁUÀÄvÀÛzÉ. ¥Àæw UÀAmÉUÉ 40 PÀÆ° ªÉZÀѪÁzÀgÉ, 8 WÀlPÀUÀ¼À£ÀÄß
vÀAiÀiÁj¸À®Ä ¨ÉÃPÁUÀĪÀ MlÄÖ PÀÆ° ªÉZÀѪɵÀÄÖ?

39) gÉÃSÁ£ÀPÉëAiÀÄ£ÀÄß §¼À¹ F PɼÀV£À ¸ÀgÀ¼ÀgÉÃSÁvÀäPÀ PÁAiÀÄðPÀæªÀÄzÀ
¸ÀªÀĸÉåAiÀÄ£ÀÄß ©r¹ :
UÀjµÀ×UÉƽ¹ : Z = 10500 x + 9000 y
¤§AzsÀ£ÉUÉƼÀ¥ÀlÖAvÉ
x + y ≤ 50
2 x + y ≤ 80
x, y ≥ 0 .

cos 7 x + cos 3 x − cos 5 x − cos x
40) = cot 2 x JAzÀÄ ¸Á¢ü¹.
sin 7 x − sin 3 x − sin 5 x + sin x

41) y = x + x 2 − 1 DzÀgÉ ( x 2 − 1) y 2 + xy1 − y = 0 JAzÀÄ ¸Á¢ü¹.

¨sÁUÀ – E

VII. PɼÀV£À ¥Àæ±ÉßUÀ¼À£ÀÄß GvÀÛj¹.

J¯èÁ ¨sÁUÀ®§Þ ‘n’ UÀ½UÉ xLt→ a x − a  = n ⋅ a n −1 JAzÀÄ ¸Á¢ü¹.
n n
42) (6)
 x −a 

CxÀªÁ
(2, – 4) (3, –1) (3, – 3) (0, 0) ©AzÀÄUÀ¼ÀÄ MAzÉà ªÀÈvÀÛzÀ ªÀÄÆ®PÀ
ºÁzÀĺÉÆÃUÀÄvÀÛzÉ JAzÀÄ ¸Á¢ü¹.

75 (NS) -10-

43) M§â ªÀåQÛAiÀÄÄ 75 Cr JvÀÛgÀªÀżÀî MAzÀÄ UÉÆÃ¥ÀÅgÀzÀ ªÉÄðAzÀ PɼÀVgÀĪÀ
MAzÀÄ £ÉÃgÀªÁzÀ PÀA§zÀ vÀÄ¢ ªÀÄvÀÄÛ §ÄqÀUÀ¼À£ÀÄß «ÃQë¹zÁUÀ PɼÀªÀÄÄR
PÉÆãÀUÀ¼ÀÄ PÀæªÀĪÁV 30° ªÀÄvÀÄÛ 60° UÀ¼ÁVzÀÝgÉ, D PÀA§zÀ JvÀÛgÀªÉµÀÄÖ?
(4)

CxÀªÁ

(1.1)4 ¨É¯ÉAiÀÄ£ÀÄß ¢é¥À¢ ¥ÀæªÉÄÃAiÀĪÀ£ÀÄß G¥ÀAiÉÆÃV¹ 4 gÀ zÀ±ÀªÀiÁA±ÀPÉÌ
PÀAqÀÄ»r¬Äj.

¨sÁUÀ – J¥sï

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

39) ²æÃ. UÀÄA¦£À MAzÀÄ PÀA¥É¤ P ªÀÄvÀÄÛ Q JA§ JgÀqÀÄ GvÀà£ÀßUÀ¼À£ÀÄß
vÀAiÀiÁj¸ÀÄvÀÛªÉ. ¥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 DzÀgÉ UÀjµÀ× ¯Á¨sª
À £
À Æ
É ß¼ÀUÆ
É AqÀ ¸Àg¼
À g
À ÃÉ SÁvÀäPÀ PÁAiÀÄðPÀæªÀÄ
¸ÀªÀĸÉåAiÀÄ£ÀÄß ¸ÀÆvÀæ gÀÆ¥ÀzÀ°è ªÀåPÀÛ¥Àr¹.

———————

-11- 75 (NS)

(English Version)

Instructions : 1. The question paper has 5 parts namely, 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
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)

3 2
1) If A =   then 2 A′ is
2 4 

6 1  6 2
a)  4 2 b) 4 8
   

6 2  6 4
c) 8 4  d) 4 8
   

75 (NS) -12-

2) If n C10 = nC5 then value of ‘n’ is

a) 15 b) 25

c) 5 d) 10

3) Two coins are tossed simultaneously. The probability of getting exactly

two heads is

1 3
a) b)
2 4

1
c) d) 1
4

4) The converse of ~ p → ~ q is

a) ~p→q b) q→p

c) p→q d) ~q →~p

5) The Compound ratio of 3 : 4 and 4 : 7 is

a) 3 : 12 b) 3:7

c) 7:3 d) 12 : 7

-13- 75 (NS)

1
6) If sin A = then value of sin 2 A is
2

3 1
a) b)
2 2
c) 0 d) 1

7) The equation of the directrix of the parabola x 2 = 8 y is

a) y=2 b) x=2
c) x = –2 d) y = −2

dy
8) If y = 5 e x − log x then is
dx

1 1
a) e5 x − b) 5ex −
x x
1
c) 5 ex − x d) 5x −
x

1
9)  7 x + 8 dx is
log (7 x + 8)
a) log (7 x + 8) + C b) +C
7
1
c) 7 log(7 x + 8) + C d) +C
log (7 x + 8)

 x dx is
5
10)

4 x5
a) 5x + C b) +C
5

x4 x6
c) +C d) +C
4 6

75 (NS) -14-

II. Match the following : (5 × 1 = 5)

11) A B

3 x
a) If = 2 then value of ‘x’ is i) 7
4 2

3 −1
b) If n p 3 = 210 then value of ‘n’ is ii)
2 2

c) Third proportional of 4, 6 is iii) 2

d) sin 15º is iv) 1

sin 4 x + 2 3 +1
e) Lt v)
x → 0 sin 2 x + 1 2 2

vi) 9

III. For questions 12 to 16 choose the appropriate answers from the given
options : (5 × 1 = 5)

(35, 12, 1, 3, 30, 16)

− 9  5  − 14
12) If   −   =   then value of ‘ x ’ is __________.
 2  − 1  x 

13) The number of diagonals in a decagon is __________.

14) If 5 : 20 = 3 : x then ‘x’ = __________.

15) Length of the latus rectum of parabola y 2 = 16 x is __________.

π
2
16)  sin x dx = __________.
0

-15- 75 (NS)

PART – B

IV. Answer any six questions : (6 × 2 = 12)

 2 0  1 − 1
17) Find matrix A if 2 A + B =   where B = 3 0  .
 − 1 3   

18) In how many ways can 6 people be chosen out of 10 people if one
particular person is always included?

2 1
19) If A and B are mutually exclusive events with P ( A) = , P (B ) = .
5 7
Find P ( A ∪ B ) .

20) If a : b = 3 : 4, b : c = 8 : 15 find a : b : c .

21) Banker’s discount and Banker’s gain on a certain bill due after sometime
are 1,250 and 50 respectively. Find the face value of the bill.

22) Find equation of parabola given vertex is (0, 0) and focus is (– 4, 0).

dy
23) If y = x x find .
dx

24) The total cost function is given by C = q 3 − 3 q 2 + 15 q + 27 where q is
output. Find the marginal cost and fixed cost.

25) Find the area bounded by the curve y = x 2 , x-axis and lines x = 0, x = 1.

75 (NS) -16-

PART – C

V. Answer any six questions : (6 × 3 = 18)

26) Solve using Cramer’s rule
2x + y = 1
x − 3y = 4

27) Find the number of permutations of the letters of the word
ENGINEERING. How many of these

a) begin with GIN and end with GRIN

b) have all 3 E’s together.

28) 3 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?

29) Banker’s gain on a bill due after 6 months at 4% p.a. is 24. Find true
discount, Banker’s discount and face value of the bill.

30) Prathik sells out 6,000 of 7.5% stock at 108 and reinvests the
proceeds in 9% stock. If his income increases by 270, at what price did
he buy 9% stock?

31) The price of a washing machine inclusive of sales tax is 13,530 if the
sales tax is 10%. Find the basic price.

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

x
33) Evaluate :  dx .
( x − 1) ( x − 2)

2
34) Evaluate :  ( x + e x + 2) dx .
1

-17- 75 (NS)

PART – D

VI. Answer any four questions : (4 × 5 = 20)

35) Solve by matrix method

3 x + 2y − z = 6
3 x + y − 2z = 3
2 x − 3 y − z = −1

36) Resolve into partial fractions
9
.
( x + 1) ( x + 2)2

37) Verify whether the propositions ( p ∧ ~ q ) ∨ q and p ∨ q are logically
equivalent.

38) An Engineering company has 80% learning effect and spends 1000 hours
to produce 1 lot of the product. Estimate the labour cost of producing
8 lots of the product if the labour cost is 40 per hour.

39) Solve the LPP graphically.
Maximize Z = 10500 x + 9000 y

Subject to the constraints
x + y ≤ 50
2 x + y ≤ 80
x, y ≥ 0 .

75 (NS) -18-

40) Prove that
cos 7 x + cos 3 x − cos 5 x − cos x
= cot 2 x .
sin 7 x − sin 3 x − sin 5 x + sin x

41) If y = x + x 2 − 1 show that ( x 2 − 1) y 2 + xy 1 − y = 0 .

PART – E

VII. Answer the following questions :

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

OR
Show that the points (2, – 4) (3, – 1) (3, – 3) (0, 0) are concyclic.

43) A person is at top of a tower 75 feet high. From there he observes a
vertical pole and finds the angles of depression of top and bottom of the
pole are 30° and 60° respectively. Find the height of the pole. (4)

OR

Find the value of (1.1)4 using binomial theorem upto 4 decimal places.

PART – F

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

39) A company owned by Shree group produces 2 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 LPP to maximize the profit.

———————

-19- 75 (NS)

75 (NS) -20-

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