∑»Ê¸¡∑ …ıX√¯Â ò∑\≈ …ÂàËyÊ\ ÆÂÈ¢√Âú, ÆÂÈ£YˇÂZ¬Â¢, ü¢π›ÂͬÂÈ — 560 003 KARNATAKA SECONDARY EDUCATION EXAMINATION BOARD, MALLESWARAM, BANGALORE – 560 003
Ø‚˜Ô.Ø‚˜Ô.Ø£˜Ô.ö. …ÂàËx\, ÆÂ⁄º˜Ô¸/∞éX£˜Ô,Ô
2009 S.S.L.C. EXAMINATION, MARCH/APRIL, 2009
ÆÂ⁄«Âà ©∆ÂO¬Âπ›ÂÈ MODEL ANSWERS å»Ê¢∑
: 31. 03. 2009
Date :
31. 03. 2009
‚¢xË∆ ‚¢zW : 81-K CODE NO. : 81-K ê·ÂŒÈ : πÂä∆ Subject : MATHEMATICS [ …¬ÂÆÂ⁄ÕÂå˘ •¢∑π›ÂÈ : 100 [ Max. Marks : 100 ( Kannada Version )
…ÂX‡R ©∆ÂO¬Â«Â ‚¢zW ∑XÆÂ⁄∑\¬Â
…ÂX‡ÊR»ÂȂʬ Æ¤ı‹WÆÂ⁄…»Â
•¢∑π›ÂÈ
Ø – êü˘Êπ 1.
B
B –A
1
2.
C
P U (Q I R)
1
3.
D
{ 1, 4, 6 }
1
4.
A
21
1
5.
D
φ
1
6.
C
3
1
7.
A
3
1
8.
B
64
1
9.
B
2PQ P + Q
1
10.
D
3
1
11.
C
2 5
12.
A
2n
3
6
1 1 [ Turn over
81-K
2
…ÂX‡R ©∆ÂO¬Â«Â ‚¢zW ∑XÆÂ⁄∑\¬Â
…ÂX‡ÊR»ÂȂʬ Æ¤ı‹WÆÂ⁄…»Â
•¢∑π›ÂÈ
13.
D
190
1
14.
D
6
1
15.
A
n
P r
16.
B
4
P 4
1
17.
A
0·81
1
18.
B
‚ÂÆÂÈÕÊÇ«
1
19.
D
¬ÊÇ
1
20.
C
1
1
21.
B
2x + 2y + 2z
1
22.
D
0
1
23.
A
a 3 + b 3
1
24.
C
18abc
1
25.
C
n
26.
D
9 2
27.
B
28.
A
3 x +3 y
1
29.
D
2x 2 = 16
1
30.
B
‚¬›Â  ‚ÂêÈË∑¬Â≈
1
31.
C
32.
A
7 2
1
33.
B
– q ± q 2 – 4pr 2p
1
34.
D
( x + 4 ) x – 60 = 0
1
35.
A
b 2 – 4ac
1
36.
C
0
1
37.
B
10
1
ÆÂÈ∆ÂÈO
=
n
C r × r
x
p – q
Fr m
1
1 1 1
1
3
…ÂX‡R ©∆ÂO¬Â«Â ‚¢zW ∑XÆÂ⁄∑\¬Â
…ÂX‡ÊR»ÂȂʬ Æ¤ı‹WÆÂ⁄…»Â
81-K
•¢∑π›ÂÈ
38.
A
x 2 + 5x + 4 = 0
1
39.
C
35
1
40.
B
– 4
1
41.
D
m
1
42.
C
6
1
43.
B
11
1
44.
A
4x
ÆÂ⁄»Â
1
45.
C
‹∫Â§È xÍË»Â
1
46.
D
Ø£ÊY ‚ÂÆ ÂÈüÊ„ÂÈ ãXxÍË»Âπ›ÂÈ ‚ÂÆÂȬÂÍ…Â
1
47.
B
AP PC
1
48.
A
16 : 1
1
49.
C
QP BC
1
50.
D
90°, 60°
1
51.
B
Ω¯Òå‚ÂÈ∆ÂOÕ
1
52.
A
8
‚¢.êÈË.
1
53.
C
(x–y)
54.
D
2 πr ( r + h )
1
55.
B
1
1
56.
A
8 : 27
1
57.
B
9
‚¢.êÈË.
1
58.
D
7
‚¢.êÈË.
1
59.
A
4, 6
1
60.
C
4, 2
1
=
BQ QC
‚¢.êÈË.
1
[ Turn over
81-K
4
…ÂX‡R ‚¢zW
…ÂX‡ÊR»ÂȂʬ Æ¤ı‹WÆÂ⁄…»Â
•¢∑π›ÂÈ
è – êü˘Êπ 61.
a = 1000, S n
=
n 2
=
24 2
n = 12 × 2 = 24,
d = 60
[ 2a + ( n – 1 ) d ] [ 2 × 1000 + ( 24 – 1 ) 60 ]
= 12 [ 2000 + 23 × 60 ]
1 2 1 2 1 2
= 12 × [ 2000 + 1380 ]
¬ÂÍ.
=
1 2
40,560.
2
•«˘˚ÂÕÊ Ø¬Â√Â»Ò ê«˘Ê»Â T n
= 1000 + ( 24 – 1 ) 60
1 2
= 1000 + 23 × 60 = 1000 + 1380 T 24 =
¬ÂÍ.
®πÂ,
S 24
1 2
2,380 = =
n (a +l) 2 24 ( 1000 + 2380 ) 2
1 2
= 12 × 3380 = 62.
2 3 A = 5 1 AA l
¬ÂÍ.
40,560
∴ A l
2 3 = 5 1
2 5 = 3 1
2 5 × 3 1
1 2
2
1 2
1 2
4 + 9 10 + 3 = 10 + 3 25 + 1
1 2
13 13 = 13 26 .
1 2
2
5
…ÂX‡R ‚¢zW 63.
…ÂX‡ÊR»ÂȂʬ Æ¤ı‹WÆÂ⁄…» ‚¬ʂÂà
•¢∑π›ÂÈ
X = 15 d
d 2
10
– 5
25
12
– 3
9
14
– 1
1
16
1
1
18
3
9
20
5
•¢∑π›ÂÈ
25 ∑
ÆÂ⁄»Â∑ êºÂ‹»
64.
81-K
(σ) =
d 2
∑ d 2 N
= 70
1 1 2
70 6
=
=
11·66
= 3·4.
1 2
2
a +b +c = 0 LHS = ( b + c ) ( b – c ) + a ( a + 2b ) = b 2 – c 2 + a 2 + 2ab
1 2
= ( a + b ) 2 – c 2
1 2
= (a +b +c) (a +b–c)
1 2
= (0) (a+b–c)
1 2
2
= 0. 65.
( 5 + 3 ) ( 5 + 3 )
1 2
( 5 – 3 ) ( 5 + 3 ) 2
=
2
( 5 ) + ( 3 ) + 2 15 2
( 5 ) – ( 3 )
=
8 + 2 15 5 – 3
=
2 ( 4 + 15 ) 2
2
1 2
1 2
= 4+
15 .
1 2
2
[ Turn over
81-K
6
…ÂX‡R ‚¢zW 66.
…ÂX‡ÊR»ÂȂʬ Æ¤ı‹WÆÂ⁄…» ãXxÍ˻« Ø∆ÂO¬Â
=x
•¢∑π›ÂÈ
¶Ç¬Âë.
∴
…Ê«Â
1 2
x ( x + 5 ) = 150
=x +5 1 2 1 2
x 2 + 5x – 300 = 0 ( x + 20 ) ( x – 15 ) = 0
∴ 67.
∴
x = – 20
Ø∆ÂO¬Â
= 15
•«˘˚ÂÕÊ
‚¢.êÈË. ÆÂÈ∆ÂÈO …Ê«Â
= 20
H = x –3
B =
L = x 3 – 5x 2 – 2x + 24
B =
=
=
1 2
‚¢.êÈË.
A = x 2 – 7x + 12 ?
1 2
x = 15
H × L A
2
1 2
( x – 3 ) ( x 3 – 5x 2 – 2x + 24 )
1 2
x 2 – 7x + 12 ( x – 3 ) ( x 3 – 5x 2 – 2x + 24 )
1
( x – 3 ) ( x – 4 )
(Ω¯Ò«ÂÕ»ÂÈR •…ÂÕÂã¸ö«ÊπÂ) = x –4
x 3 – 5x 2 – 2x + 24 ( x – 4 )
)
x 3 – 5x 2 – 2x + 24
1 2
( x 2
–x –6 1 2
x 3 – 4x 2 – x 2 – 2x
1 2
– x 2 + 4x – 6x + 24 – 6x + 24 0 ∴
èË{ÛËÄO
B = x 2 – x – 6.
•«˘˚ÂÕÊ
1 2
1
12
4
7
…ÂX‡R ‚¢zW
81-K
…ÂX‡ÊR»ÂȂʬ Æ¤ı‹WÆÂ⁄…» B = B =
•¢∑π›ÂÈ
H × L A
1 2
( x – 3 ) ( x 3 – 5x 2 – 2x + 24 )
1 2
x 2 – 7x + 12
ü˘ÊÇö«ÊπÂ, x 2 – 7x + 12
) x 3
– 5x 2 – 2x + 24
(x
+2
1
x 3 – 7x 2 + 12x 2x 2 – 14x + 24 2x 2 – 14x + 24
1
0 ∴
èË{ÛËÄO
B = (x–3)(x +2) = x 2 – x – 6.
1 2 1 2
4
1 2 1 2 1 2 1 2
2
68.
B
O
110°
P
A
‚¢.êÈË. ãXæW«Â ÕÂÎ∆ÂO Ø›ŒÈ‹È 110° xÍ˻«ÂëY ãXæW Ø›ŒÈ‹È BP ‚ÂS‡Â¸«Â Ø›ŒÈ‹È AP ‚ÂS‡Â¸∑ Ø›ŒÈ‹È 3·5
[ Turn over
81-K
8
…ÂX‡R ‚¢zW
…ÂX‡ÊR»ÂȂʬ Æ¤ı‹WÆÂ⁄…»Â
•¢∑π›ÂÈ
69.
¬ÒzÊÑ∆ÂXx@ Ñ∆ÂX«ë  Y ∴
AB
DC ( ABCD ∆ÊXéæW
) 1 2
ABO ||| ∆ CDO
∆ AB CD
=
BO DO
AB CD
=
2 1
=
1 2
AO . CO
1 2
( ˙ ˙ .
BO · OD = 2 : 1
xÍáJ«
)
1 2
2
70. P
O
.
T
Q ∠ PTQ + ∠ POQ = 180 ∴
®π Â
∠ PTQ = 180 – ∠ POQ ∆ OPQ ∴
(i) ¬ÂëY
»ÂëY
…… (i)
1 2
OP = OQ
∠ POQ = 180° – 2 ∠ OPQ
1 2
…… (ii)
1 2
¶«Òòö«Êπ ∠ PTQ = 180 – ( 180 – 2 ∠ OPQ ) = 2 ∠ OPQ.
1 2
2
9
…ÂX‡R ‚¢zW 71.
81-K
…ÂX‡ÊR»ÂȂʬ Æ¤ı‹WÆÂ⁄…» ‚@Ë£˜Ô : 20 êÈË.
=1
•¢∑π›ÂÈ
‚¢.êÈË.
£yÊ@ºÊ¬Âx@ Ñ∆ÂXx@
1 1
2
72.
‚ÂàŒ⁄«Â …ÂXã •√ÂL ‚Êëπ A
B
C
A
2
1
1
B
1
0
2
C
1
2
0
=
=3×
1 2
1
12
2 1 1 1 0 2 1 2 0
‚¢zÊWŒÈ∆« ‚¢zWπ› ÆÍ∆ÂO = ‚¢…Ê∆ 被ÂÈπ› ∑XÆÂÈπ› ÆÍ∆ÂO
1 2
2
[ Turn over
81-K
10
…ÂX‡R ‚¢zW
…ÂX‡ÊR»ÂȂʬ Æ¤ı‹WÆÂ⁄…»Â
•¢∑π›ÂÈ
ÀÚ«˘˚Êπ۬‚» …ÂXÆÈ ËŒÈ :
73.
A
Ñ∆ÂX, «Â∆ÂO ‚Ê«˘ÂçËŒÈ, ¬Âº»Â B
C
D
1
…ÂX
ÆÈˌȫ „¢∆Âπ›ÂÈ
74.
1»Ò
„¢∆Â
BC . DB
2»Ò
„¢∆Â
BC . DC = AC 2
3»Ò
„¢∆Â
BC 2 = AB 2 + AC 2
= AB 2
Õ¬π ‚Êå˘‚Â‹È Õ¬π ‚Êå˘‚Â‹È Ø¢«ÂÈ ‚Êå˘‚Â‹È
1 1 1
4
C 3 = ( C 1 + C 2 )
C 1 , C 2
ÆÂÈ∆ÂÈO
C 3
AB
ÕÊW‚« ÕÂÎ∆ÂO ¬Âт‹È
RL
‚ÂS‡Â¸∑,
PM
ÕÂÎ∆ÂOπ›»ÂÈR ¬Âт‹È
‚ÂS‡Â¸∑ ¬Âт‹È
1
12 1 2
1
‚ÂS‡Â¸∑Õ»ÂÈR •›Â∆ ÆÂ⁄√Â‹È RL, PM = 6·6
‚¢.êÈË.
1
4
11
…ÂX‡R ‚¢zW 75.
81-K
…ÂX‡ÊR»ÂȂʬ Æ¤ı‹WÆÂ⁄…»Â
•¢∑π›ÂÈ
8 T 13 = T 10
1 2
8 × a . r 12 = a . r 9
1 2
1 8
1 2
r =
1 . 2
1 2
S ∞
=
r 3
=
®πÂ,
=
=
a 1 – r
1 2
3
1 2
1 1 – 2 3 1 2
= 3×
1 2
2 1
1 2
= 6.
4
76. x
0
1
– 1
2
– 2
y
0
2
2
8
8
πÊXÀ˜Ô ¬ÂºÂ» 被ÂÈπ›»ÂÈR πÂȬÂÈã‚‹È
1 1
2
[ Turn over