Page 1
CBSE XI | Mathematics
Sample Paper – 7 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 7 Solution
SECTION – A
1.
? ?
n
x1
nn
x1
nn
n1
n1
xa
x1
lim
x1
x1
lim
x1
xa
n 1 lim na
xa
n
?
?
?
?
?
?
?
?
?
?
?
??
?
?
2. If you access the website, then you pay a subscription fee.
3. The complex conjugate of a + bi is a – bi. Hence, complex conjugate of -4i – 8 i.e. -8 – 4i is
-8 + 4i.
OR
z = 4 + 4i comparing with z = a + bi we get a = 4 = b
tan ? = b/a = 1
? ? = 45°
4. Standard deviation = 4
Standard deviation = vVariance
4 = vVariance
Variance = 4
2
= 16
SECTION – B
5. X = {a, b, c, d} and Y = {f, b, d, g}
X – Y = {a, c} and Y – X = {f, g}
6. f(x) is defined when x
2
– 1 > 0 ? x
2
> 1 and 3 + x > 0
x < -1 or x > 1 and x > -3 and x ? -2, since base 3 + x ? 1
Domain is (-3, -2) ? (-2, -1) ? (1, 8)
Page 2
CBSE XI | Mathematics
Sample Paper – 7 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 7 Solution
SECTION – A
1.
? ?
n
x1
nn
x1
nn
n1
n1
xa
x1
lim
x1
x1
lim
x1
xa
n 1 lim na
xa
n
?
?
?
?
?
?
?
?
?
?
?
??
?
?
2. If you access the website, then you pay a subscription fee.
3. The complex conjugate of a + bi is a – bi. Hence, complex conjugate of -4i – 8 i.e. -8 – 4i is
-8 + 4i.
OR
z = 4 + 4i comparing with z = a + bi we get a = 4 = b
tan ? = b/a = 1
? ? = 45°
4. Standard deviation = 4
Standard deviation = vVariance
4 = vVariance
Variance = 4
2
= 16
SECTION – B
5. X = {a, b, c, d} and Y = {f, b, d, g}
X – Y = {a, c} and Y – X = {f, g}
6. f(x) is defined when x
2
– 1 > 0 ? x
2
> 1 and 3 + x > 0
x < -1 or x > 1 and x > -3 and x ? -2, since base 3 + x ? 1
Domain is (-3, -2) ? (-2, -1) ? (1, 8)
CBSE XI | Mathematics
Sample Paper – 7 Solution
OR
f(x) =
2
2x 1 x ?
2
2 2 2
x x x
f sin 2sin 1 sin
2 2 2
x x x
f sin 2sin cos 1 sin x cos x
2 2 2
x x x
f sin 2sin cos
2 2 2
x x x
f sin sin x sin x 2sin cos
2 2 2
??
??
??
??
??
? ? ?
??
??
??
?
??
??
??
??
??
??
7. Let s be the length of the arc subtending an angle ?
c
at the centre of a circle of radius r.
Then, s = r?
s =
5
5 15
180 12
??
? ? ?
OR
The angle traced by the hour hand in 12 hours = 360°
The angle traced by the hour hand in 3 hrs 30 min i. e. 7/2 hrs =
360 7
105
12 2
??
? ? ?
??
??
The angle traced by the minute hand in 60 min = 360°
The angle traced by minute hand in 30 min =
360
30 180
60
??
? ? ?
??
??
Hence, the required angle between two hands = 180° - 105° = 75° =
5
75
180 12
??
??
8. The Cartesian product of the set R of all real numbers with itself i. e. R × R is the set of
all ordered pairs (x, y) where x, y ? R. In other words R × R = {(x, y) : x, y ? R }
R × R is the set of all points in XY-plane. The set of R × R is also denoted by R
2
.
R × R × R = {(x, y, z) : x, y, z ? R}
It represents set of all points in space denoted by R
3
.
9.
2
1 cos4x 2cos 2x
cosx sinx
cot x tanx
sinx cosx
?
?
?
?
=
2
22
2cos 2xsinxcosx
cos x sin x ?
=
2
2cos 2xsinxcosx
cos2x
Page 3
CBSE XI | Mathematics
Sample Paper – 7 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 7 Solution
SECTION – A
1.
? ?
n
x1
nn
x1
nn
n1
n1
xa
x1
lim
x1
x1
lim
x1
xa
n 1 lim na
xa
n
?
?
?
?
?
?
?
?
?
?
?
??
?
?
2. If you access the website, then you pay a subscription fee.
3. The complex conjugate of a + bi is a – bi. Hence, complex conjugate of -4i – 8 i.e. -8 – 4i is
-8 + 4i.
OR
z = 4 + 4i comparing with z = a + bi we get a = 4 = b
tan ? = b/a = 1
? ? = 45°
4. Standard deviation = 4
Standard deviation = vVariance
4 = vVariance
Variance = 4
2
= 16
SECTION – B
5. X = {a, b, c, d} and Y = {f, b, d, g}
X – Y = {a, c} and Y – X = {f, g}
6. f(x) is defined when x
2
– 1 > 0 ? x
2
> 1 and 3 + x > 0
x < -1 or x > 1 and x > -3 and x ? -2, since base 3 + x ? 1
Domain is (-3, -2) ? (-2, -1) ? (1, 8)
CBSE XI | Mathematics
Sample Paper – 7 Solution
OR
f(x) =
2
2x 1 x ?
2
2 2 2
x x x
f sin 2sin 1 sin
2 2 2
x x x
f sin 2sin cos 1 sin x cos x
2 2 2
x x x
f sin 2sin cos
2 2 2
x x x
f sin sin x sin x 2sin cos
2 2 2
??
??
??
??
??
? ? ?
??
??
??
?
??
??
??
??
??
??
7. Let s be the length of the arc subtending an angle ?
c
at the centre of a circle of radius r.
Then, s = r?
s =
5
5 15
180 12
??
? ? ?
OR
The angle traced by the hour hand in 12 hours = 360°
The angle traced by the hour hand in 3 hrs 30 min i. e. 7/2 hrs =
360 7
105
12 2
??
? ? ?
??
??
The angle traced by the minute hand in 60 min = 360°
The angle traced by minute hand in 30 min =
360
30 180
60
??
? ? ?
??
??
Hence, the required angle between two hands = 180° - 105° = 75° =
5
75
180 12
??
??
8. The Cartesian product of the set R of all real numbers with itself i. e. R × R is the set of
all ordered pairs (x, y) where x, y ? R. In other words R × R = {(x, y) : x, y ? R }
R × R is the set of all points in XY-plane. The set of R × R is also denoted by R
2
.
R × R × R = {(x, y, z) : x, y, z ? R}
It represents set of all points in space denoted by R
3
.
9.
2
1 cos4x 2cos 2x
cosx sinx
cot x tanx
sinx cosx
?
?
?
?
=
2
22
2cos 2xsinxcosx
cos x sin x ?
=
2
2cos 2xsinxcosx
cos2x
CBSE XI | Mathematics
Sample Paper – 7 Solution
= cos2x sin2x
=
? ?
1
2cos2xsin2x
2
=
1
sin4x
2
1 cos4x 1
sin4x
cot x tanx 2
?
?
?
OR
33
8cos 6cos 2 4cos 3cos
9 9 9 9
? ? ? ? ??
? ? ?
??
??
2cos 3
9
? ??
??
??
??
reason
2cos 1
3
?
??
10. The component statements are
p : All integers are positive.
q : All integers are negative.
p and q both are false. The connecting word is ‘or’.
11. For any x ?domain f we have,
? ?
? ? x4
4x
f x 1
x 4 x 4
??
?
? ? ? ?
??
Hence, the range of the function = {-1}
12.
22
xy
1
36 20
?? comparing with
22
22
xy
1
ab
??
a
2
= 36, b
2
= 20
b
2
= a
2
(1 – e
2
)
20 = 36(1 – e
2
)
e
2
= 16/36
e = 2/3
Distance between the directrices = 2a/e = 18.
Page 4
CBSE XI | Mathematics
Sample Paper – 7 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 7 Solution
SECTION – A
1.
? ?
n
x1
nn
x1
nn
n1
n1
xa
x1
lim
x1
x1
lim
x1
xa
n 1 lim na
xa
n
?
?
?
?
?
?
?
?
?
?
?
??
?
?
2. If you access the website, then you pay a subscription fee.
3. The complex conjugate of a + bi is a – bi. Hence, complex conjugate of -4i – 8 i.e. -8 – 4i is
-8 + 4i.
OR
z = 4 + 4i comparing with z = a + bi we get a = 4 = b
tan ? = b/a = 1
? ? = 45°
4. Standard deviation = 4
Standard deviation = vVariance
4 = vVariance
Variance = 4
2
= 16
SECTION – B
5. X = {a, b, c, d} and Y = {f, b, d, g}
X – Y = {a, c} and Y – X = {f, g}
6. f(x) is defined when x
2
– 1 > 0 ? x
2
> 1 and 3 + x > 0
x < -1 or x > 1 and x > -3 and x ? -2, since base 3 + x ? 1
Domain is (-3, -2) ? (-2, -1) ? (1, 8)
CBSE XI | Mathematics
Sample Paper – 7 Solution
OR
f(x) =
2
2x 1 x ?
2
2 2 2
x x x
f sin 2sin 1 sin
2 2 2
x x x
f sin 2sin cos 1 sin x cos x
2 2 2
x x x
f sin 2sin cos
2 2 2
x x x
f sin sin x sin x 2sin cos
2 2 2
??
??
??
??
??
? ? ?
??
??
??
?
??
??
??
??
??
??
7. Let s be the length of the arc subtending an angle ?
c
at the centre of a circle of radius r.
Then, s = r?
s =
5
5 15
180 12
??
? ? ?
OR
The angle traced by the hour hand in 12 hours = 360°
The angle traced by the hour hand in 3 hrs 30 min i. e. 7/2 hrs =
360 7
105
12 2
??
? ? ?
??
??
The angle traced by the minute hand in 60 min = 360°
The angle traced by minute hand in 30 min =
360
30 180
60
??
? ? ?
??
??
Hence, the required angle between two hands = 180° - 105° = 75° =
5
75
180 12
??
??
8. The Cartesian product of the set R of all real numbers with itself i. e. R × R is the set of
all ordered pairs (x, y) where x, y ? R. In other words R × R = {(x, y) : x, y ? R }
R × R is the set of all points in XY-plane. The set of R × R is also denoted by R
2
.
R × R × R = {(x, y, z) : x, y, z ? R}
It represents set of all points in space denoted by R
3
.
9.
2
1 cos4x 2cos 2x
cosx sinx
cot x tanx
sinx cosx
?
?
?
?
=
2
22
2cos 2xsinxcosx
cos x sin x ?
=
2
2cos 2xsinxcosx
cos2x
CBSE XI | Mathematics
Sample Paper – 7 Solution
= cos2x sin2x
=
? ?
1
2cos2xsin2x
2
=
1
sin4x
2
1 cos4x 1
sin4x
cot x tanx 2
?
?
?
OR
33
8cos 6cos 2 4cos 3cos
9 9 9 9
? ? ? ? ??
? ? ?
??
??
2cos 3
9
? ??
??
??
??
reason
2cos 1
3
?
??
10. The component statements are
p : All integers are positive.
q : All integers are negative.
p and q both are false. The connecting word is ‘or’.
11. For any x ?domain f we have,
? ?
? ? x4
4x
f x 1
x 4 x 4
??
?
? ? ? ?
??
Hence, the range of the function = {-1}
12.
22
xy
1
36 20
?? comparing with
22
22
xy
1
ab
??
a
2
= 36, b
2
= 20
b
2
= a
2
(1 – e
2
)
20 = 36(1 – e
2
)
e
2
= 16/36
e = 2/3
Distance between the directrices = 2a/e = 18.
CBSE XI | Mathematics
Sample Paper – 7 Solution
SECTION – C
13. sin A =
3
5
cos A =
2
94
1 sin A 1
25 5
? ? ? ?
tan A =
3
sin A 3
5
4
cos A 4
5
??
3 4 24
sin2A 2sinAcosA 2
5 5 25
? ? ? ? ?
2
97
cos2A 1 2sin A 1 2
25 25
? ? ? ? ? ?
24
sin2A 24
25
tan2A
7
cos2A 7
25
? ? ?
24 7 336
sin4A 2sin2Acos2A 2
25 25 625
? ? ? ? ?
14. It is given that the set A consists of two positive integers. So, let A = {n, m}. Since, range
of f = {3}.
f(n) = 3 and f(m) = 3
Highest prime factors of n ad m both are equal to 3.
n = 3 and m = 6 or n = 3 and m = 9 or n = 3 and m = 12 or n = 6 and m = 12 etc
A = {3, 6} or A = {3, 9} or A = {3, 12} or A = {6, 12} etc.
Clearly, A is not uniquely determine.
15. Sn = 0.7 + 0.77 + 0.777 + ….+ to n terms
= 7 (0.1 + 0.11 + 0.111 + … to n terms)
= ? ?
7
0.9 0.99 0.999 ....to n terms
9
? ? ?
= ? ? ? ? ? ?
7
1 0.1 1 0.01 1 0.001 ...to n terms
9
?? ? ? ? ? ? ?
??
=
n
0.1 1 0.1
7
n
9 1 0.1
?? ?? ?
??
?? ?
?
??
??
=
n
11
1
7 10 10
n
1
9
1
10
?? ??
?
?? ??
??
?? ?
??
?
??
??
=
n
7 1 1
n1
9 9 10
?? ??
??
?? ??
?? ??
Page 5
CBSE XI | Mathematics
Sample Paper – 7 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 7 Solution
SECTION – A
1.
? ?
n
x1
nn
x1
nn
n1
n1
xa
x1
lim
x1
x1
lim
x1
xa
n 1 lim na
xa
n
?
?
?
?
?
?
?
?
?
?
?
??
?
?
2. If you access the website, then you pay a subscription fee.
3. The complex conjugate of a + bi is a – bi. Hence, complex conjugate of -4i – 8 i.e. -8 – 4i is
-8 + 4i.
OR
z = 4 + 4i comparing with z = a + bi we get a = 4 = b
tan ? = b/a = 1
? ? = 45°
4. Standard deviation = 4
Standard deviation = vVariance
4 = vVariance
Variance = 4
2
= 16
SECTION – B
5. X = {a, b, c, d} and Y = {f, b, d, g}
X – Y = {a, c} and Y – X = {f, g}
6. f(x) is defined when x
2
– 1 > 0 ? x
2
> 1 and 3 + x > 0
x < -1 or x > 1 and x > -3 and x ? -2, since base 3 + x ? 1
Domain is (-3, -2) ? (-2, -1) ? (1, 8)
CBSE XI | Mathematics
Sample Paper – 7 Solution
OR
f(x) =
2
2x 1 x ?
2
2 2 2
x x x
f sin 2sin 1 sin
2 2 2
x x x
f sin 2sin cos 1 sin x cos x
2 2 2
x x x
f sin 2sin cos
2 2 2
x x x
f sin sin x sin x 2sin cos
2 2 2
??
??
??
??
??
? ? ?
??
??
??
?
??
??
??
??
??
??
7. Let s be the length of the arc subtending an angle ?
c
at the centre of a circle of radius r.
Then, s = r?
s =
5
5 15
180 12
??
? ? ?
OR
The angle traced by the hour hand in 12 hours = 360°
The angle traced by the hour hand in 3 hrs 30 min i. e. 7/2 hrs =
360 7
105
12 2
??
? ? ?
??
??
The angle traced by the minute hand in 60 min = 360°
The angle traced by minute hand in 30 min =
360
30 180
60
??
? ? ?
??
??
Hence, the required angle between two hands = 180° - 105° = 75° =
5
75
180 12
??
??
8. The Cartesian product of the set R of all real numbers with itself i. e. R × R is the set of
all ordered pairs (x, y) where x, y ? R. In other words R × R = {(x, y) : x, y ? R }
R × R is the set of all points in XY-plane. The set of R × R is also denoted by R
2
.
R × R × R = {(x, y, z) : x, y, z ? R}
It represents set of all points in space denoted by R
3
.
9.
2
1 cos4x 2cos 2x
cosx sinx
cot x tanx
sinx cosx
?
?
?
?
=
2
22
2cos 2xsinxcosx
cos x sin x ?
=
2
2cos 2xsinxcosx
cos2x
CBSE XI | Mathematics
Sample Paper – 7 Solution
= cos2x sin2x
=
? ?
1
2cos2xsin2x
2
=
1
sin4x
2
1 cos4x 1
sin4x
cot x tanx 2
?
?
?
OR
33
8cos 6cos 2 4cos 3cos
9 9 9 9
? ? ? ? ??
? ? ?
??
??
2cos 3
9
? ??
??
??
??
reason
2cos 1
3
?
??
10. The component statements are
p : All integers are positive.
q : All integers are negative.
p and q both are false. The connecting word is ‘or’.
11. For any x ?domain f we have,
? ?
? ? x4
4x
f x 1
x 4 x 4
??
?
? ? ? ?
??
Hence, the range of the function = {-1}
12.
22
xy
1
36 20
?? comparing with
22
22
xy
1
ab
??
a
2
= 36, b
2
= 20
b
2
= a
2
(1 – e
2
)
20 = 36(1 – e
2
)
e
2
= 16/36
e = 2/3
Distance between the directrices = 2a/e = 18.
CBSE XI | Mathematics
Sample Paper – 7 Solution
SECTION – C
13. sin A =
3
5
cos A =
2
94
1 sin A 1
25 5
? ? ? ?
tan A =
3
sin A 3
5
4
cos A 4
5
??
3 4 24
sin2A 2sinAcosA 2
5 5 25
? ? ? ? ?
2
97
cos2A 1 2sin A 1 2
25 25
? ? ? ? ? ?
24
sin2A 24
25
tan2A
7
cos2A 7
25
? ? ?
24 7 336
sin4A 2sin2Acos2A 2
25 25 625
? ? ? ? ?
14. It is given that the set A consists of two positive integers. So, let A = {n, m}. Since, range
of f = {3}.
f(n) = 3 and f(m) = 3
Highest prime factors of n ad m both are equal to 3.
n = 3 and m = 6 or n = 3 and m = 9 or n = 3 and m = 12 or n = 6 and m = 12 etc
A = {3, 6} or A = {3, 9} or A = {3, 12} or A = {6, 12} etc.
Clearly, A is not uniquely determine.
15. Sn = 0.7 + 0.77 + 0.777 + ….+ to n terms
= 7 (0.1 + 0.11 + 0.111 + … to n terms)
= ? ?
7
0.9 0.99 0.999 ....to n terms
9
? ? ?
= ? ? ? ? ? ?
7
1 0.1 1 0.01 1 0.001 ...to n terms
9
?? ? ? ? ? ? ?
??
=
n
0.1 1 0.1
7
n
9 1 0.1
?? ?? ?
??
?? ?
?
??
??
=
n
11
1
7 10 10
n
1
9
1
10
?? ??
?
?? ??
??
?? ?
??
?
??
??
=
n
7 1 1
n1
9 9 10
?? ??
??
?? ??
?? ??
CBSE XI | Mathematics
Sample Paper – 7 Solution
16.
? ? ? ?
? ? ? ?
? ?
? ?
? ?
? ?
? ?
22
2
22
22
2
2
22
2
2
2
2
1 ix
a ib
1 ix
1 ix 1 ix
a ib 1
1 ix 1 ix a ib 1
2 1 a ib
2ix 1 a ib
1 a ib
ix
1 a ib
1 a ib 1 a ib
ix
1 a ib 1 a ib
1 a b 2ib
ix
1 a i b
1 a b 2ib
ix
1 a b
2ib
ix a b 1
1 a b
2b
x
1 a b
?
??
?
? ? ?
??
?
? ? ? ? ?
??
?
? ? ? ?
??
?
??
? ? ? ?
??
? ? ? ?
? ? ?
?
??
? ? ?
?
??
? ? ?
??
?
??
Hence, x is real as denominator is always positive.
17. Total number of tickets = 100
? Total number of exhaustive, mutually exclusive and equally likely cases n(S) = 100
1. A : getting an even number. Since there are 50 even numbered tickets, n(A) = 50
P(A) = 50/100 = ½
2. B : getting 5 or multiple of 5
The favourable outcomes are 5, 10, 15, 20,….95, 100
n(B) = 20 ? P(B) = 20/100 = 1/5
3. C : getting a number greater than 75
The favourable outcomes are 76, 77,….100
n(C) = 25 ? P(C) = 25/100 = ¼
4. D : getting a square number
The favourable outcomes are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
n(D) = 10 ? P(D) = 10/100 = 1/10
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