Sample Solution Paper 7 - Math, Class 11

# Sample Solution Paper 7 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce PDF Download

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

## Mathematics (Maths) Class 11

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## FAQs on Sample Solution Paper 7 - Math, Class 11 - Mathematics (Maths) Class 11 - Commerce

 1. What is the quadratic formula?
Ans. The quadratic formula is a formula used to find the roots (or solutions) of a quadratic equation. It is given by x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
 2. How is the discriminant related to the solutions of a quadratic equation?
Ans. The discriminant is a term that appears under the square root in the quadratic formula. It is calculated as b^2 - 4ac. The discriminant determines the nature of the solutions of a quadratic equation. If the discriminant is positive, the equation has two distinct real solutions. If it is zero, the equation has one real solution (a double root). And if the discriminant is negative, the equation has two complex conjugate solutions.
 3. Can a quadratic equation have no solutions?
Ans. Yes, a quadratic equation can have no solutions. This occurs when the discriminant is negative, meaning the square root in the quadratic formula would involve imaginary numbers. In other words, if b^2 - 4ac < 0, the equation has no real solutions. However, it is still possible to find complex solutions to the equation.
 4. How can I determine the nature of the solutions without solving the quadratic equation?
Ans. You can determine the nature of the solutions of a quadratic equation by examining the discriminant. If the discriminant is positive, the equation has two distinct real solutions. If it is zero, the equation has one real solution. And if the discriminant is negative, the equation has two complex conjugate solutions.
 5. Can the quadratic formula be used to solve any type of equation?
Ans. No, the quadratic formula can only be used to solve quadratic equations, which are equations of the form ax^2 + bx + c = 0. It cannot be used to solve equations of higher degrees, such as cubic or quartic equations. For those types of equations, different formulas and methods are used.

## Mathematics (Maths) Class 11

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