Roll No.
x……®……ΔEÚ
SS—15—MATHEMATICS
No. of Questions — 30 No. of Printed Pages — 11
=SS… ®……v™… ®…EÚ {…Æ˙“I……,
2014
SENIOR SECONDARY EXAMINATION, 2014
M… h…i… MATHEMATICS ∫…®…™… :
3
{…⁄h……»EÚ
1 P…h]‰ı 4 : 80
{…Æ˙“I……Ãl…™…… E‰Ú ±…B ∫……®……x™… x…nÊ˘∂… : GENERAL INSTRUCTIONS TO THE EXAMINEES : 1. {…Æ˙“I……l…‘ ∫…¥…«|…l…®… +{…x…‰ |…∂x… {…j… {…Æ˙ x……®……ΔEÚ + x…¥……™…«i…:
±…J…Â*
Candidate must write first his / her Roll No. on the question paper compulsorily. 2.
∫…¶…“ |…∂x… EÚÆ˙x…‰ + x…¥……™…« ΩÈ˛* All the questions are compulsory.
3.
|…i™…‰EÚ |…∂x… EÚ… =k…Æ˙ n˘“ M…<« =k…Æ˙ {…÷Œ∫i…EÚ… ®…Â Ω˛“ ±…J…Â* Write the answer to each question in the given answer-book only.
4.
V…x… |…∂x…… ®… +…xi… Æ˙EÚ J…hb˜ ΩÈ˛, =x… ∫…¶…“ E‰Ú =k…Æ˙ BEÚ ∫……l… Ω˛“ ±…J…Â* For questions having more than one part, the answers to those parts are to be written together in continuity.
5.
|…∂x… {…j… E‰Ú Ω˛xn˘“ ¥… +ΔO…‰V…“ ∞¸{……xi…Æ˙ ®… EÚ∫…“ |…EÚ…Æ˙ EÚ“ j…÷ ]ı / +xi…Æ˙ / ¥…Æ˙…‰v……¶……∫… Ω˛…‰x…‰ {…Æ˙ Ω˛xn˘“ ¶……π…… E‰Ú |…∂x… EÚ…‰ Ω˛“ ∫…Ω˛“ ®……x…Â* If there is any error / difference / contradiction in Hindi & English versions of the question paper, the question of Hindi version should be treated valid.
SS—15—Mathematics
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2 6.
J…hb˜
|…∂x… ∫…ΔJ™……
+ΔEÚ |…i™…‰EÚ |…∂x…
+
1 – 10
1
§…
11 – 25
3
∫…
26 – 30
5
Question Nos.
Marks per question
A
1 – 10
1
B
11 – 25
3
C
26 – 30
5
Section
7.
|…∂x… ∫…ΔJ™…… 11, 12, 13, 23, 28 +…ËÆ˙ ∫…‰ +…{…EÚ…‰ BEÚ Ω˛“ ¥…EÚ±{… EÚÆ˙x…… Ω˲*
29
®… +…xi… Æ˙EÚ ¥…EÚ±{… ΩÈ˛*
There are internal choices in Q. Nos. 11, 12, 13, 23, 28 and 29. You have to attempt only one of the alternatives in these questions.
J…hb˜ – + SECTION – A 1.
(
tan−1 3 − cot −1 − 3
) EÚ… ®……x… Y……i… EÚ“ V…B*
(
)
Find the value of tan−1 3 − cot −1 − 3 .
2.
A
Y……i… EÚ“ V…B, ™… n˘
A + B = ⎡5 2⎤ ⎢⎣ 0 9 ⎥⎦
i…l……
A −B = ⎡−3 −6⎤. ⎢⎣ 4 − 1 ⎥⎦
Find A, if A + B = ⎡ 5 2 ⎤ and A − B = ⎡ − 3 − 6 ⎤ . ⎢⎣ 0 9 ⎥⎦ ⎢⎣ 4 − 1 ⎥⎦ SS—15—Mathematics
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3 3.
™… n˘
⎡−2⎤ A=⎢ 4⎥ ⎢ 5⎥ ⎣ ⎦
i…l……
B =[1 4 −6
⎡−2⎤ If A = ⎢ 4 ⎥ and B = [ 1 4 − 6 ⎢ 5⎥ ⎣ ⎦
4.
®……x… Y……i… EÚ“ V…B : ∫ Find the value of
5.
+¥…EÚ±… ∫…®…“EÚÆ˙h…
sec 2 x cosec 2x
] i……‰
AB
Y……i… EÚÆÂ˙*
] then find AB.
dx .
sec 2 x
∫ cosec2x dx .
dy + y = 1 ( y ≠ 1) dx
EÚ… ¥™……{…EÚ Ω˛±… Y……i… EÚ“ V…B*
Find general solution of differential equation
6.
→
∫… n˘∂…
∧
∧
dy + y = 1 ( y ≠ 1) . dx
∧
a = 2 i − 3 j + 4k
E‰Ú +x…÷ n˘∂… ®……j…EÚ (
∧
∧
∧
Find unit vector of a given vector a = 2 i − 3 j + 4 k .
7.
BEÚ Æ‰˙J……
x, y
i…l…… z-+I…… EÚ“ v…x……i®…EÚ n˘∂…… E‰Ú ∫……l… GÚ®…∂…:
90°, 60°
+…ËÆ ˙30° EÚ…
EÚ…‰h… §…x……i…“ Ω˲, i……‰ n˘E¬Ú-EÚ…‰∫……
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4 8.
n˘…‰ ∫…®…i…±……Â
2x + y − 2z = 5
3x − 6y − 2z = 7
+…ËÆ˙
E‰Ú §…“S… EÚ… EÚ…‰h… Y……i… EÚ“ V…B*
Find the angle between planes 2x + y − 2z = 5 and 3x − 6y − 2z = 7 . 9.
x…®x… ¥™…¥…Æ˙…‰v…… E‰Ú +xi…M…«i… ∫…÷∫…ΔM…i… Ω˛±… I…‰j… n˘∂……«
10.
™… n˘ A +…ËÆ˙ B ∫¥…i…Δj… P…]ıx……Bƒ ΩÈ˛ i…l…… E‰Ú ®……x… Y……i… EÚ“ V…B*
P ( A ) = 0 ⋅ 3 +…ËÆ˙ P ( B ) = 0 ⋅ 4
i…§…
P (A ∪ B )
If A and B are independent events with P ( A ) = 0 ⋅ 3 and P ( B ) = 0 ⋅ 4 , then find the value of P ( A ∪ B ) .
J…hb˜ – §… SECTION – B
11.
∫…r˘ EÚ“ V…B EÚ ¥……∫i… ¥…EÚ ∫…ΔJ™……+…Â E‰Ú ∫…®…÷SS…™…
R
®…Â
R = { ( a, b ) : a ≤ b 2 }
u˘…Æ˙…
{… Æ˙¶…… π…i… ∫…Δ§…Δv… R, x… i……‰ ∫¥…i…÷±™… Ω˲, x… ∫…®… ®…i… Ω˲ +…ËÆ˙ x… Ω˛“ ∫…ΔGÚ…®…EÚ Ω˲* +l…¥…… ®……x… ±…“ V…B EÚ
f : N → Y , f ( x ) = 4x + 3
Y = { y ∈ N : y = 4x + 3
EÚ∫…“
x ∈N
E‰Ú ±…B } * ∫…r˘ EÚ“ V…B EÚ
Ω˲* |… i…±……‰®… °Ú±…x… ¶…“ Y……i… EÚ“ V…B*
SS—15—Mathematics
u˘…Æ˙… {… Æ˙¶…… π…i… BEÚ °Ú±…x… Ω˲, V…Ω˛…ƒ
SS–5515
f
¥™…÷iGÚ®…h…“™…
5 Show that the relation R in the set R of real numbers, defined as
R = { ( a , b ) : a ≤ b 2 } is neither reflexive nor symmetric nor transitive. OR
f : N →Y
Let
be a function defined as
f ( x ) = 4x + 3 , where,
Y = { y ∈ N : y = 4x + 3 for some x ∈ N } . Show that f is invertible. Find the inverse function. 12.
°Ú±…x…
⎛ cos x − sin x ⎞ tan−1 ⎜ ⎟, x < π ⎝ cos x + sin x ⎠
EÚ…‰ ∫…Æ˙±…i…®… ∞¸{… ®… ±… J…B*
+l…¥…… ∫…r˘ EÚ“ V…B
tan−1
3 5 63 = sin −1 + cos −1 . 5 13 16
⎛ cos x − sin x ⎞ Write the function tan−1 ⎜ ⎟ , x < π in the simplest form. ⎝ cos x + sin x ⎠
OR Prove that tan−1
13.
™… n˘
63 5 3 = sin −1 + cos −1 . 16 13 5
0 1⎤ ⎡2 A = ⎢2 1 3⎥ ⎢ 1 −1 0 ⎥ ⎣ ⎦
Ω˲ i……‰
A 2 − 5 A + 6I
EÚ… ®……x… Y……i… EÚ“ V…B*
+l…¥…… +…¥™…⁄Ω˛
⎡ 3 10 ⎤ ⎢⎣ 2 7 ⎥⎦
EÚ… |……Æ˙Œ®¶…EÚ ∞¸{……xi…Æ˙h… ∫…‰ ¥™…÷iGÚ®… Y……i… EÚ“ V…B*
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6 0 1⎤ ⎡2 If A = ⎢ 2 1 3 ⎥ , then find the value of A 2 − 5 A + 6I . ⎢ 1 −1 0 ⎥ ⎣ ⎦ OR Using elementary transformations, find the inverse of the ⎡ 3 10 ⎤ . ⎢⎣ 2 7 ⎥⎦ 14.
a
matrix
i…l…… b E‰Ú ®……x……Â EÚ…‰ Y……i… EÚ“ V…B i…… EÚ f(x)=
5,
™… n˘ x ≤ 2
ax + b,
™… n˘ 2 < x < 10
21,
™… n˘ x ≥ 10
u˘…Æ˙… {… Æ˙¶…… π…i… °Ú±…x… ∫…i…i… °Ú±…x… Ω˛…‰* Find the values of a and b such that the function defined by if x ≤ 2
5, f(x)=
ax + b, if 2 < x < 10 21,
if x ≥ 10
is a continuous function.
15.
∫…r˘ EÚ“ V…B EÚ °Ú±…x… ⎛ π ⎞ ⎜ ,π ⎟ ⎝ 2 ⎠
π ⎞ ⎛ f ( x ) = log cos x , ⎜ 0, ⎟ 2 ⎠ ⎝
®…Â x…Æ˙xi…Æ˙ ø…∫…®……x… +…ËÆ˙
®… x…Æ˙xi…Æ˙ ¥…v…«®……x… Ω˲*
Prove that the function f given by f ( x ) = log cos x is strictly decreasing π ⎞ ⎞ ⎛ ⎛ π on ⎜ 0, ⎟ and strictly increasing on ⎜ , π ⎟ . 2 2 ⎠ ⎠ ⎝ ⎝
SS—15—Mathematics
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7 16.
¥…GÚ
x 2 y2 + =1 4 25
{…Æ˙ =x… §…xn÷˘+… EÚ…‰ Y……i… EÚ“ V…B V…x… {…Æ˙ ∫{…∂…« Ɖ˙J……Bƒ x-+I… E‰Ú
∫…®……xi…Æ˙ Ω˛…Â* Find the points on the curve
x 2 y2 + = 1 at which the tangents are 4 25
parallel to x-axis.
17.
2x
®……x… Y……i… EÚÆÂ˙ ∫ e 2x − 1 dx . e +1 Evaluate
e 2x − 1
∫ e 2 x + 1 dx . π
18.
®……x… Y……i… EÚÆÂ˙ ∫ x sin 2x dx . 1 + cos x 0
π
Evaluate
x sin x
∫ 1 + cos2 x dx .
0
19.
¥…GÚ
y = x2
B¥…Δ Æ‰˙J…… y = 4 ∫…‰ P…Ɖ˙ I…‰j… EÚ… I…‰j…°Ú±… Y……i… EÚ“ V…B*
Find the area of the region bounded by the curve y = x 2 and the line y = 4. 20.
∫…®……EÚ±…x… EÚ… ={…™……‰M… EÚÆ˙i…‰ Ω÷˛B BEÚ B‰∫…‰ j…¶…÷V……EÚ…Æ˙ I…‰j… EÚ… I…‰j…°Ú±… Y……i… EÚ“ V…B V…∫…EÚ“ ¶…÷V……+… E‰Ú ∫…®…“EÚÆ˙h…
y = 2x + 1 , y = 3x + 1
B¥…Δ x = 4 ΩÈ˛*
Using integration find the area of triangular region whose sides have the equations y = 2x + 1 , y = 3x + 1 and x = 4. SS—15—Mathematics
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8 21.
A, B, C
™… n˘ §…xn÷˘+… ∧
∧
∧
3 i + 2 j − 3k ∧
∧
∧
+…ËÆ˙
∧
∧
∧
E‰Ú Œ∫l… i… ∫… n˘∂… GÚ®…∂…:
∧
AB
ΩÈ˛, i……‰ ∧
∧
∧
∧
∧
i + j +k , 2 i + 5 j ,
→
→
∧
i − 6 j −k
∧
∧
D
+…ËÆ˙
i…l……
CD
∧
∧
∧
E‰Ú §…“S… EÚ… EÚ…‰h… Y……i… EÚ“ V…B* ∧
If i + j + k , 2 i + 5 j , 3 i + 2 j − 3 k and i − 6 j − k are the position vectors →
of points A, B, C and D respectively, then find the angle between AB →
and CD .
22.
→
™… n˘
∧
∧
→
d
→
∧
∧
∧
→
a
+…ËÆ˙
→
∧
→
∧
+…ËÆ˙
b
∧
∧
∧
c . d = 15 .
→
∧
ΩÈ˛, i……‰ BEÚ ∫… n˘∂…
→ →
n˘…‰x…… {…Æ˙ ±…®§… Ω˲ +…ËÆ˙ ∧
∧
c = 2 i − j + 4k
→
Y……i… EÚ“ V…B V……‰ ∧
→
∧
a = i + 4 j + 2k , b = 3 i − 2 j + 7 k
∧
∧
∧
If a = i + 4 j + 2 k , b = 3 i − 2 j + 7 k and c = 2 i − j + 4 k , then find a →
→
→ →
→
vector d which is perpendicular to both a and b and c . d = 15 .
23.
+…±…‰J…“™… ¥… v… ∫…‰ x…®x… ±… J…i… ÆË˙ J…EÚ |……‰O……®…x… ∫…®…∫™…… EÚ…‰ Ω˛±… EÚ“ V…B : ¥™…¥…Æ˙…‰v……Â
x + 2y ≤ 10 , 3x + y ≤ 15 , x , y ≥ 0
Z = 3x + 2y
E‰Ú +xi…M…«i…
EÚ…
+ v…EÚi…®…“EÚÆ˙h… EÚ“ V…B* +l…¥…… BEÚ |…EÚ…Æ˙ E‰Ú E‰ÚEÚ E‰Ú ±…B
200
i…l…… n⁄˘∫…Æ˙“ |…EÚ…Æ˙ E‰Ú E‰ÚEÚ E‰Ú ±…B
O……®… +…]ı… i…l…… 100
O……®… ¥…∫…… EÚ“ +…¥…∂™…EÚi…… Ω˛…‰i…“ Ω˲
O……®… +…]ı… i…l……
Ω˛…‰i…“ Ω˲* E‰ÚEÚ…Â EÚ“ + v…EÚi…®… ∫…ΔJ™…… §…i……<™…‰ V……‰
SS—15—Mathematics
25
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50
O……®… ¥…∫…… EÚ“ +…¥…∂™…EÚi……
EÚ±……‰ +…]‰ı i…l……
1
EÚ±……‰ ¥…∫…… ∫…‰ §…x…
9
∫…EÚi…‰ ΩÈ˛* ™…Ω˛ ®……x… ±…™…… M…™…… Ω˲ EÚ E‰ÚEÚ…Â EÚ…‰ §…x……x…‰ E‰Ú ±…B +x™… {…n˘…l……Á EÚ“ EÚ®…“ x…Ω˛” Æ˙Ω‰˛M…“* Solve the following linear programming problem by graphical method : Maximize Z = 3x + 2y subject to the constraints x + 2y ≤ 10 , 3x + y ≤ 15 , x , y ≥ 0 .
OR One kind of cake requires 200 gm of flour and 25 gm of fat, and another kind of cake requires 100 gm of flour and 50 gm of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of other ingredients used in making the cakes. 24.
BEÚ |…™……‰M… E‰Ú ∫…°Ú±… Ω˛…‰x…‰ EÚ… ∫…Δ™……‰M… =∫…E‰Ú +∫…°Ú±… Ω˛…‰x…‰ ∫…‰ n˘…‰ M…÷x…… Ω˲* |…… ™…EÚi…… Y……i… EÚ“ V…B EÚ +M…±…‰ UÙ: {…Æ˙“I…h…… ®… EÚ®… ∫…‰ EÚ®… 4 ∫…°Ú±… Ω˛…‰* An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will at least 4 successes.
25.
BEÚ +x… ¶…x…i… {……∫…‰ EÚ…‰ °ÂÚEÚx…‰ {…Æ˙ |……{i… ∫…ΔJ™……+…Â EÚ… |…∫…Æ˙h… Y……i… EÚ“ V…B* Find the variance of numbers obtained on throwing an unbiased die.
J…hb˜ – ∫… SECTION – C
26.
n˘∂……«<™…‰ EÚ
1+a 1 1 1 1+b 1 = abc 1 1 1+c
SS—15—Mathematics
1 1 1⎞ ⎛ ⎜1 + + + ⎟ . a b c⎠ ⎝
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10 1+a 1 1 1 1 1⎞ ⎛ 1 1+b 1 = abc ⎜1 + + + ⎟ . a b c⎠ ⎝ 1 1 1+c
Prove that
27.
™… n˘
x y + y x = ba + a b
Find
dy dx
Y……i… EÚ“ V…B*
dy , if x y + y x = b a + a b . dx 2
28.
i……‰
xe x
®……x… Y……i… EÚ“ V…B ∫
( 1 + x )2 1
dx .
+l…¥…… dx
®……x… Y……i… EÚ“ V…B ∫ 2
Evaluate
sin3 x sin ( x + α )
.
xe x
∫ (1 + x )2 dx .
1
OR Evaluate
29.
∫
dx 3
sin x sin ( x + α )
+¥…EÚ±… ∫…®…“EÚÆ˙h…
.
x dy = y d x + x 2 + y 2 d x
+l…¥……
SS—15—Mathematics
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EÚ…‰ Ω˛±… EÚ“ V…B*
11
+¥…EÚ±… ∫…®…“EÚÆ˙h…
dy + y cot x = 2x + x 2 cot x ( x ≠ 0 ) dx
EÚ“ V…B, n˘™…… Ω÷˛+… Ω˲ EÚ y = 0, V…§…
x =
EÚ… ¥… ∂…π]ı Ω˛±… Y……i…
π . 2
Solve the differential equation x dy = y dx + x 2 + y 2 dx . OR Find the particular solution of the differential equation dy π + y cot x = 2x + x 2 cot x ( x ≠ 0 ) , given that y = 0, when x = . dx 2
30.
→
→
∧
∧
∧
r . ( 2 i + 5 j + 3k ) = 9
∧
∧
∧
r . ( 2 i + 2 j − 3k ) = 7 ,
=∫… ∫…®…i…±… EÚ… ∫… n˘∂… ∫…®…“EÚÆ˙h… Y……i… EÚ“ V…B V……‰ ∫…®…i…±……Â
EÚ“ |… i…SU‰Ùn˘x… Ɖ˙J…… +…ËÆ˙ ( 2, 1, 3 ) ∫…‰ Ω˛…‰EÚÆ˙ V……i…… Ω˲*
Find vector equation of a plane passing through the intersection of the →
∧
∧
∧
→
∧
∧
∧
planes r . ( 2 i + 2 j − 3 k ) = 7 and r . ( 2 i + 5 j + 3 k ) = 9 and the point ( 2, 1, 3 ).
SS—15—Mathematics
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