Sample Solution Paper 5 - Math, Class 11

# Sample Solution Paper 5 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce PDF Download

``` 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

```

## Mathematics (Maths) Class 11

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## FAQs on Sample Solution Paper 5 - Math, Class 11 - Mathematics (Maths) Class 11 - Commerce

 1. What is the Pythagorean theorem?
Ans. The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. It can be expressed as a^2 + b^2 = c^2, where a and b are the lengths of the two legs and c is the length of the hypotenuse.
 2. How is the Pythagorean theorem useful in real life?
Ans. The Pythagorean theorem is widely used in various fields such as construction, architecture, engineering, and navigation. It helps in calculating distances, finding the length of diagonals, determining the height of buildings, and solving problems related to angles and distances in real-life situations.
 3. Can the Pythagorean theorem be applied to non-right-angled triangles?
Ans. No, the Pythagorean theorem is only applicable to right-angled triangles. In non-right-angled triangles, other trigonometric functions such as sine, cosine, and tangent are used to find the relationships between the sides and angles.
 4. How can I use the Pythagorean theorem to find the length of the hypotenuse?
Ans. To find the length of the hypotenuse using the Pythagorean theorem, you need to know the lengths of the other two sides (legs) of the right-angled triangle. Square the lengths of the legs, add them together, and then take the square root of the sum. The result will be the length of the hypotenuse.
 5. Are there any practical applications of the Pythagorean theorem?
Ans. Yes, the Pythagorean theorem has numerous practical applications. It is used in construction to ensure that buildings are structurally sound and to calculate the lengths of diagonals. It is also used in navigation to determine distances and angles. Additionally, it is used in various fields of science and engineering to solve problems involving right-angled triangles.

## Mathematics (Maths) Class 11

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