Page 1
CBSE XI | Mathematics
Sample Paper – 5 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 5 Solution
SECTION – A
1. We have,
??
??
??
??
? ? ?
a b c
sin A sinB sinC
23
2
sinB
3
3 3sinB
sinB 1
m B 90
2.
? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
25 25 25
12
25 25 24
2
2
12
1 1 i 1
i 1 i 1 i i 1 i i
i i i
i
1 1 i 1 1 i i
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
OR
(1 + i)(1 + 2i) = 1 + 2i + i + 2i
2
= 1 + 3i – 2 = -1 + 3i
Comparing with a + bi we get a = -1 and b = 3
3. p: I have the money;
q: I will buy an i-phone,
p ? q: If I have the money ( ?) then I will buy an i-phone
4. Relation R from P to Q is R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
SECTION – B
5. According to the question,
n(P) = 40, n(P ? Q) = 60 and n(P n Q) = 10
n(P ? Q) = n(P) + n(Q) – n(P n Q)
60 = 40 + n(Q) – 10
n(Q) = 60 – 40 + 10 = 30
Page 2
CBSE XI | Mathematics
Sample Paper – 5 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 5 Solution
SECTION – A
1. We have,
??
??
??
??
? ? ?
a b c
sin A sinB sinC
23
2
sinB
3
3 3sinB
sinB 1
m B 90
2.
? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
25 25 25
12
25 25 24
2
2
12
1 1 i 1
i 1 i 1 i i 1 i i
i i i
i
1 1 i 1 1 i i
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
OR
(1 + i)(1 + 2i) = 1 + 2i + i + 2i
2
= 1 + 3i – 2 = -1 + 3i
Comparing with a + bi we get a = -1 and b = 3
3. p: I have the money;
q: I will buy an i-phone,
p ? q: If I have the money ( ?) then I will buy an i-phone
4. Relation R from P to Q is R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
SECTION – B
5. According to the question,
n(P) = 40, n(P ? Q) = 60 and n(P n Q) = 10
n(P ? Q) = n(P) + n(Q) – n(P n Q)
60 = 40 + n(Q) – 10
n(Q) = 60 – 40 + 10 = 30
CBSE XI | Mathematics
Sample Paper – 5 Solution
6. We know that the equation of the circle described on the line segment joining (x1, y1) and (x2,
y2) as a diameter is,
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
Here, x1 = -1, x2 = 4, y1 = 2 and y2 = -3
Thus, the equation of the required circle is
(x + 1)(x - 4) + (y - 2)(y + 3) = 0
?x
2
+ y
2
- 3x + y -10 = 0
7. One ace can be selected from 4 aces in
4
C1.
Other 4 cards which are non-aces can be selected out of 48 cards in
48
C4 ways.
The total number of ways =
4
C 1 x
48
C4
= 4 x 2 x 47 x 46 x 45 = 778320
OR
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
1. A : Getting a prime number
A = {2, 3, 5}
n(A) = 3
Required probability = 3/6 = ½
2. B : A number more tan 6
As B is null set.
P(B) = 0
8. Let ,
cos 5x = (cos 5x + cos x) – cos x
??
??
??
? ? ? ?
? ? ?
32
53
5x x 5x x
2cos cos cosx
22
2cos3x cos 2x cosx
2(4 cos x 3 cosx)(2cos x 1) cosx
16 cos x 20 cos x 5 cosx
9. 6 beads have to be arranged in a circular fashion which can be done in (6-1)!
But anticlockwise and clockwise arrangement of beads in a necklace are same so
(6-1)!×(1/2) = 60
OR
Page 3
CBSE XI | Mathematics
Sample Paper – 5 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 5 Solution
SECTION – A
1. We have,
??
??
??
??
? ? ?
a b c
sin A sinB sinC
23
2
sinB
3
3 3sinB
sinB 1
m B 90
2.
? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
25 25 25
12
25 25 24
2
2
12
1 1 i 1
i 1 i 1 i i 1 i i
i i i
i
1 1 i 1 1 i i
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
OR
(1 + i)(1 + 2i) = 1 + 2i + i + 2i
2
= 1 + 3i – 2 = -1 + 3i
Comparing with a + bi we get a = -1 and b = 3
3. p: I have the money;
q: I will buy an i-phone,
p ? q: If I have the money ( ?) then I will buy an i-phone
4. Relation R from P to Q is R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
SECTION – B
5. According to the question,
n(P) = 40, n(P ? Q) = 60 and n(P n Q) = 10
n(P ? Q) = n(P) + n(Q) – n(P n Q)
60 = 40 + n(Q) – 10
n(Q) = 60 – 40 + 10 = 30
CBSE XI | Mathematics
Sample Paper – 5 Solution
6. We know that the equation of the circle described on the line segment joining (x1, y1) and (x2,
y2) as a diameter is,
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
Here, x1 = -1, x2 = 4, y1 = 2 and y2 = -3
Thus, the equation of the required circle is
(x + 1)(x - 4) + (y - 2)(y + 3) = 0
?x
2
+ y
2
- 3x + y -10 = 0
7. One ace can be selected from 4 aces in
4
C1.
Other 4 cards which are non-aces can be selected out of 48 cards in
48
C4 ways.
The total number of ways =
4
C 1 x
48
C4
= 4 x 2 x 47 x 46 x 45 = 778320
OR
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
1. A : Getting a prime number
A = {2, 3, 5}
n(A) = 3
Required probability = 3/6 = ½
2. B : A number more tan 6
As B is null set.
P(B) = 0
8. Let ,
cos 5x = (cos 5x + cos x) – cos x
??
??
??
? ? ? ?
? ? ?
32
53
5x x 5x x
2cos cos cosx
22
2cos3x cos 2x cosx
2(4 cos x 3 cosx)(2cos x 1) cosx
16 cos x 20 cos x 5 cosx
9. 6 beads have to be arranged in a circular fashion which can be done in (6-1)!
But anticlockwise and clockwise arrangement of beads in a necklace are same so
(6-1)!×(1/2) = 60
OR
CBSE XI | Mathematics
Sample Paper – 5 Solution
1 1 x
9! 10! 11!
1 1 x
9! 10 9! 11 10 9!
1 1 x 1
1
9! 10 11 10 9!
11 x
10 11 10
x 121
??
??
? ? ?
? ? ? ?
??
? ? ? ?
?
? ? ? ?
?
?
?
10.
Given that the first term is 1.
Also given that each term is the sum of all the terms which follow it.
Let 1, r, r
2
,… be an infinite G.P., where r is the common ratio.
Sum of terms of an infinite G.P.,
a
S
1r
?
?
Here, a = r
Thus,
r
S
1r
?
?
From the given statement of the problem, we have,
?
?
? ? ?
? ? ?
??
??
r
1
1r
1 r r
rr1
2r 1
1
r
2
Thus the required G.P. is:
1 1 1 1
1, , , , ,...
2 4 8 16
11.
27
,k, are in G.P.
72
??
2
2
27
k1
72
k1
k1
27
Whenk 1;G.P.: ,1,
72
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ?
? ? ?
Page 4
CBSE XI | Mathematics
Sample Paper – 5 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 5 Solution
SECTION – A
1. We have,
??
??
??
??
? ? ?
a b c
sin A sinB sinC
23
2
sinB
3
3 3sinB
sinB 1
m B 90
2.
? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
25 25 25
12
25 25 24
2
2
12
1 1 i 1
i 1 i 1 i i 1 i i
i i i
i
1 1 i 1 1 i i
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
OR
(1 + i)(1 + 2i) = 1 + 2i + i + 2i
2
= 1 + 3i – 2 = -1 + 3i
Comparing with a + bi we get a = -1 and b = 3
3. p: I have the money;
q: I will buy an i-phone,
p ? q: If I have the money ( ?) then I will buy an i-phone
4. Relation R from P to Q is R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
SECTION – B
5. According to the question,
n(P) = 40, n(P ? Q) = 60 and n(P n Q) = 10
n(P ? Q) = n(P) + n(Q) – n(P n Q)
60 = 40 + n(Q) – 10
n(Q) = 60 – 40 + 10 = 30
CBSE XI | Mathematics
Sample Paper – 5 Solution
6. We know that the equation of the circle described on the line segment joining (x1, y1) and (x2,
y2) as a diameter is,
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
Here, x1 = -1, x2 = 4, y1 = 2 and y2 = -3
Thus, the equation of the required circle is
(x + 1)(x - 4) + (y - 2)(y + 3) = 0
?x
2
+ y
2
- 3x + y -10 = 0
7. One ace can be selected from 4 aces in
4
C1.
Other 4 cards which are non-aces can be selected out of 48 cards in
48
C4 ways.
The total number of ways =
4
C 1 x
48
C4
= 4 x 2 x 47 x 46 x 45 = 778320
OR
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
1. A : Getting a prime number
A = {2, 3, 5}
n(A) = 3
Required probability = 3/6 = ½
2. B : A number more tan 6
As B is null set.
P(B) = 0
8. Let ,
cos 5x = (cos 5x + cos x) – cos x
??
??
??
? ? ? ?
? ? ?
32
53
5x x 5x x
2cos cos cosx
22
2cos3x cos 2x cosx
2(4 cos x 3 cosx)(2cos x 1) cosx
16 cos x 20 cos x 5 cosx
9. 6 beads have to be arranged in a circular fashion which can be done in (6-1)!
But anticlockwise and clockwise arrangement of beads in a necklace are same so
(6-1)!×(1/2) = 60
OR
CBSE XI | Mathematics
Sample Paper – 5 Solution
1 1 x
9! 10! 11!
1 1 x
9! 10 9! 11 10 9!
1 1 x 1
1
9! 10 11 10 9!
11 x
10 11 10
x 121
??
??
? ? ?
? ? ? ?
??
? ? ? ?
?
? ? ? ?
?
?
?
10.
Given that the first term is 1.
Also given that each term is the sum of all the terms which follow it.
Let 1, r, r
2
,… be an infinite G.P., where r is the common ratio.
Sum of terms of an infinite G.P.,
a
S
1r
?
?
Here, a = r
Thus,
r
S
1r
?
?
From the given statement of the problem, we have,
?
?
? ? ?
? ? ?
??
??
r
1
1r
1 r r
rr1
2r 1
1
r
2
Thus the required G.P. is:
1 1 1 1
1, , , , ,...
2 4 8 16
11.
27
,k, are in G.P.
72
??
2
2
27
k1
72
k1
k1
27
Whenk 1;G.P.: ,1,
72
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ?
? ? ?
CBSE XI | Mathematics
Sample Paper – 5 Solution
17
r
2
2
7
27
Whenk 1;GP : , 1,
72
17
r
2
2
7
? ? ?
?
? ? ? ? ?
?
??
?
OR
3, 6, 12,…
a = 3, r = 2 and n = 7
S7 =
77
r 1 2 1
a 3 3 127 381
r 1 2 1
? ? ? ? ??
? ? ? ?
? ? ? ?
??
? ? ? ?
12. an =
(1 2 3 ....n) n(n 1)
n 2n
? ? ? ?
?
Sn =
n
n
a ?
n
i1
2
1
(n 1)
2
1 n(n 1) n
2 2 2
(n n) n
42
n(n 3)
4
?
? ??
?
??
?
??
?
?
SECTION – C
13.
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
22
22
22
1 i 1 i
1+i 1 i
1 i 1+i 1 i 1 i
1 i 2i 1 i 2i
1 i 1 i
1 1 2i 1 1 2i 2i 2i
1 i 1 i
? ? ?
?
??
? ? ?
? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
??
Page 5
CBSE XI | Mathematics
Sample Paper – 5 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 5 Solution
SECTION – A
1. We have,
??
??
??
??
? ? ?
a b c
sin A sinB sinC
23
2
sinB
3
3 3sinB
sinB 1
m B 90
2.
? ? ? ? ? ? ? ? ? ? ? ?
? ?
? ? ? ? ? ? ? ?
25 25 25
12
25 25 24
2
2
12
1 1 i 1
i 1 i 1 i i 1 i i
i i i
i
1 1 i 1 1 i i
? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ? ?
OR
(1 + i)(1 + 2i) = 1 + 2i + i + 2i
2
= 1 + 3i – 2 = -1 + 3i
Comparing with a + bi we get a = -1 and b = 3
3. p: I have the money;
q: I will buy an i-phone,
p ? q: If I have the money ( ?) then I will buy an i-phone
4. Relation R from P to Q is R = {(9, 3), (9, –3), (4, 2), (4, –2), (25, 5), (25, –5)}
SECTION – B
5. According to the question,
n(P) = 40, n(P ? Q) = 60 and n(P n Q) = 10
n(P ? Q) = n(P) + n(Q) – n(P n Q)
60 = 40 + n(Q) – 10
n(Q) = 60 – 40 + 10 = 30
CBSE XI | Mathematics
Sample Paper – 5 Solution
6. We know that the equation of the circle described on the line segment joining (x1, y1) and (x2,
y2) as a diameter is,
(x - x1)(x - x2) + (y - y1)(y - y2) = 0
Here, x1 = -1, x2 = 4, y1 = 2 and y2 = -3
Thus, the equation of the required circle is
(x + 1)(x - 4) + (y - 2)(y + 3) = 0
?x
2
+ y
2
- 3x + y -10 = 0
7. One ace can be selected from 4 aces in
4
C1.
Other 4 cards which are non-aces can be selected out of 48 cards in
48
C4 ways.
The total number of ways =
4
C 1 x
48
C4
= 4 x 2 x 47 x 46 x 45 = 778320
OR
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
1. A : Getting a prime number
A = {2, 3, 5}
n(A) = 3
Required probability = 3/6 = ½
2. B : A number more tan 6
As B is null set.
P(B) = 0
8. Let ,
cos 5x = (cos 5x + cos x) – cos x
??
??
??
? ? ? ?
? ? ?
32
53
5x x 5x x
2cos cos cosx
22
2cos3x cos 2x cosx
2(4 cos x 3 cosx)(2cos x 1) cosx
16 cos x 20 cos x 5 cosx
9. 6 beads have to be arranged in a circular fashion which can be done in (6-1)!
But anticlockwise and clockwise arrangement of beads in a necklace are same so
(6-1)!×(1/2) = 60
OR
CBSE XI | Mathematics
Sample Paper – 5 Solution
1 1 x
9! 10! 11!
1 1 x
9! 10 9! 11 10 9!
1 1 x 1
1
9! 10 11 10 9!
11 x
10 11 10
x 121
??
??
? ? ?
? ? ? ?
??
? ? ? ?
?
? ? ? ?
?
?
?
10.
Given that the first term is 1.
Also given that each term is the sum of all the terms which follow it.
Let 1, r, r
2
,… be an infinite G.P., where r is the common ratio.
Sum of terms of an infinite G.P.,
a
S
1r
?
?
Here, a = r
Thus,
r
S
1r
?
?
From the given statement of the problem, we have,
?
?
? ? ?
? ? ?
??
??
r
1
1r
1 r r
rr1
2r 1
1
r
2
Thus the required G.P. is:
1 1 1 1
1, , , , ,...
2 4 8 16
11.
27
,k, are in G.P.
72
??
2
2
27
k1
72
k1
k1
27
Whenk 1;G.P.: ,1,
72
? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ?
??
? ? ?
? ? ?
CBSE XI | Mathematics
Sample Paper – 5 Solution
17
r
2
2
7
27
Whenk 1;GP : , 1,
72
17
r
2
2
7
? ? ?
?
? ? ? ? ?
?
??
?
OR
3, 6, 12,…
a = 3, r = 2 and n = 7
S7 =
77
r 1 2 1
a 3 3 127 381
r 1 2 1
? ? ? ? ??
? ? ? ?
? ? ? ?
??
? ? ? ?
12. an =
(1 2 3 ....n) n(n 1)
n 2n
? ? ? ?
?
Sn =
n
n
a ?
n
i1
2
1
(n 1)
2
1 n(n 1) n
2 2 2
(n n) n
42
n(n 3)
4
?
? ??
?
??
?
??
?
?
SECTION – C
13.
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ?
22
22
22
1 i 1 i
1+i 1 i
1 i 1+i 1 i 1 i
1 i 2i 1 i 2i
1 i 1 i
1 1 2i 1 1 2i 2i 2i
1 i 1 i
? ? ?
?
??
? ? ?
? ? ? ? ?
?
??
? ? ? ? ? ? ?
??
??
CBSE XI | Mathematics
Sample Paper – 5 Solution
? ? ? ?
22
2i 2i 4i
2i
1 ( 1) 2
1+i 1 i
2i 0 2 4 2
1 i 1+i
?
? ? ?
??
?
? ? ? ? ? ?
?
OR
1 2 1 2
Consider the given equation:
(a + ib) (c + id) (e + if) (g + ih) = A + iB
Let us take modulus on both sides ,
(a + ib) (c + id) (e + if) (g + ih) = A + iB
We know, z z z z
(a + ib) (c + id) (e + if) (g
?
?
? ? ? ? ? ? ? ?
? ? ? ? ? ?
2 2 2 2 2 2 2 2 2 2
22
2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
+ ih) = A + iB
(a + ib) (c + id) (e + if) (g + ih) = A + iB
a b . c d . e f . g h A B
a b . c d . e f . g h A B
a b . c d . e f . g h A B
Thus, the value of A B a b . c d . e f .
?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ? ? ? ?
? ? ? ? ?
? ?
22
gh ?
14. Consider the given quadratic equation:
2
2
2
9x 12 x + 20 = 0
20
3x 4x 0
3
b b 4ac
x
2a
?
? ? ? ?
? ? ?
?
? ? ? ? ? ? ?
? ? ?
? ? ?
? ? ? ? ? ? ?
? ? ? ?
? ? ?
2
20
( 4) ( 4) 4 3
4 16 4 20
3
x
2.3 6
4 16 80 4 64 4 8 1 2 4i
6 6 6 3
24
xi
33
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