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CBSE XI | Mathematics
Sample Paper – 8 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 8 Solution
SECTION – A
1.
? ? ? ?
2
1
x
2
1
x
2
1
x
2
4x 1
lim
2x 1
2x 1 2x 1
lim
2x 1
lim 2x 1 2x 1 0
2
?
?
?
?
?
??
?
?
? ? ? ?
?
2. The number 2 is not greater than 7.
OR
The number 2 is less than or equal to 7.
OR
It is false that the number 2 is greater than 7.
3.
2
1 1 i i i
i
i i i 1 i
? ? ? ?
? ? ? ? ?
?
OR
z = 1 – 3i
Comparing with a + bi we get a = 1 and b = -3
? ?
2
2
z 1 3 10 ? ? ? ?
4. Variance = 4
Standard deviation = vVariance
Standard deviation = v4
Standard deviation = 2
Page 2
CBSE XI | Mathematics
Sample Paper – 8 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 8 Solution
SECTION – A
1.
? ? ? ?
2
1
x
2
1
x
2
1
x
2
4x 1
lim
2x 1
2x 1 2x 1
lim
2x 1
lim 2x 1 2x 1 0
2
?
?
?
?
?
??
?
?
? ? ? ?
?
2. The number 2 is not greater than 7.
OR
The number 2 is less than or equal to 7.
OR
It is false that the number 2 is greater than 7.
3.
2
1 1 i i i
i
i i i 1 i
? ? ? ?
? ? ? ? ?
?
OR
z = 1 – 3i
Comparing with a + bi we get a = 1 and b = -3
? ?
2
2
z 1 3 10 ? ? ? ?
4. Variance = 4
Standard deviation = vVariance
Standard deviation = v4
Standard deviation = 2
CBSE XI | Mathematics
Sample Paper – 8 Solution
SECTION – B
5. According to the question,
A n B = {b, d}
n(A n B) = 2
n(A’ ? B’) = n(A n B)’ = n(A ? B) – n(A n B) = 8 – 2 = 6
6. f
-1
(-5) = x. Then,
f(x) = -5
x
2
+ 1 = -5
x
2
= -6
This equation is not solvable in R. Therefore, there is no pre-image of -5.
Hence, f
-1
(-5) = ?
OR
f(x) =
x1
x1
?
?
1
1
1
x
f
1
x
1
x
?
??
?
??
??
?
1 1 x
f
x 1 x
? ??
?
??
?
??
? ? x1
1
f
x x 1
??
??
?
??
?
??
? ?
1
f f x
x
??
??
??
??
7. 30’’ =
''
30 1
60 2
? ? ? ?
?
? ? ? ?
? ? ? ?
37’ 30’’ =
''
1 75 75 1 5
37
2 2 2 60 8
??
? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
OR
Let s be the length of the arc subtending an angle ?
c
at the centre of a circle of radius r.
s = r? here r = 5 cm and ? = 15°
5
s r 5 15
180 12
??
? ? ? ? ? ?
Page 3
CBSE XI | Mathematics
Sample Paper – 8 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 8 Solution
SECTION – A
1.
? ? ? ?
2
1
x
2
1
x
2
1
x
2
4x 1
lim
2x 1
2x 1 2x 1
lim
2x 1
lim 2x 1 2x 1 0
2
?
?
?
?
?
??
?
?
? ? ? ?
?
2. The number 2 is not greater than 7.
OR
The number 2 is less than or equal to 7.
OR
It is false that the number 2 is greater than 7.
3.
2
1 1 i i i
i
i i i 1 i
? ? ? ?
? ? ? ? ?
?
OR
z = 1 – 3i
Comparing with a + bi we get a = 1 and b = -3
? ?
2
2
z 1 3 10 ? ? ? ?
4. Variance = 4
Standard deviation = vVariance
Standard deviation = v4
Standard deviation = 2
CBSE XI | Mathematics
Sample Paper – 8 Solution
SECTION – B
5. According to the question,
A n B = {b, d}
n(A n B) = 2
n(A’ ? B’) = n(A n B)’ = n(A ? B) – n(A n B) = 8 – 2 = 6
6. f
-1
(-5) = x. Then,
f(x) = -5
x
2
+ 1 = -5
x
2
= -6
This equation is not solvable in R. Therefore, there is no pre-image of -5.
Hence, f
-1
(-5) = ?
OR
f(x) =
x1
x1
?
?
1
1
1
x
f
1
x
1
x
?
??
?
??
??
?
1 1 x
f
x 1 x
? ??
?
??
?
??
? ? x1
1
f
x x 1
??
??
?
??
?
??
? ?
1
f f x
x
??
??
??
??
7. 30’’ =
''
30 1
60 2
? ? ? ?
?
? ? ? ?
? ? ? ?
37’ 30’’ =
''
1 75 75 1 5
37
2 2 2 60 8
??
? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
OR
Let s be the length of the arc subtending an angle ?
c
at the centre of a circle of radius r.
s = r? here r = 5 cm and ? = 15°
5
s r 5 15
180 12
??
? ? ? ? ? ?
CBSE XI | Mathematics
Sample Paper – 8 Solution
8. B – C = {2, 3, 6} – {2, 3, 7} = {6}
A × (B – C) = {1, 4} × {6} = {(1, 6), (4, 6)}…….(i)
A × B = {1, 4} × {2, 3, 6} = {(1, 2), (1, 3), (1, 6), (4, 2), (4, 3), (4, 6)}
A × C = {1, 4} × {2, 3, 7} = {(1, 2), (1, 3), (1, 7), (4, 2), (4, 3), (4, 7)}
A × B – A × C = {(1, 6), (4, 6)}…..(ii)
From (i) and (ii)
A × (B – C) = A × B – A × C
9. sin
2
x + sin x – 2 = 0
(sinx + 2)(sinx – 1) = 0
Sin x + 2 = 0 or sin x – 1 = 0
sin x = -2 or sin x = 1
sin x cannot be -2 hence, sin x = 1.
x = 90°
OR
22
2
2
tan tan
4 2 4 2
1 tan 1 tan
22
1 tan 1 tan
22
1 tan 1 tan
22
1 tan 1 tan
22
2 1 tan
2
1 tan
2
2
cos
2sec
? ? ? ? ? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
??
??
??
??
??
?? ? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
?? ? ? ? ?
??
? ? ? ?
? ? ? ?
? ??
?
??
??
?
?
?
?
?
??
10. The compound statement is “25 is a multiple of 5 or 8”.
Let us assume that the statement q is false i. e. 25 is not a multiple of 8. Clearly p is true.
Thus, if we assume that q is false then p is true. Hence, compound statement is true.
Page 4
CBSE XI | Mathematics
Sample Paper – 8 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 8 Solution
SECTION – A
1.
? ? ? ?
2
1
x
2
1
x
2
1
x
2
4x 1
lim
2x 1
2x 1 2x 1
lim
2x 1
lim 2x 1 2x 1 0
2
?
?
?
?
?
??
?
?
? ? ? ?
?
2. The number 2 is not greater than 7.
OR
The number 2 is less than or equal to 7.
OR
It is false that the number 2 is greater than 7.
3.
2
1 1 i i i
i
i i i 1 i
? ? ? ?
? ? ? ? ?
?
OR
z = 1 – 3i
Comparing with a + bi we get a = 1 and b = -3
? ?
2
2
z 1 3 10 ? ? ? ?
4. Variance = 4
Standard deviation = vVariance
Standard deviation = v4
Standard deviation = 2
CBSE XI | Mathematics
Sample Paper – 8 Solution
SECTION – B
5. According to the question,
A n B = {b, d}
n(A n B) = 2
n(A’ ? B’) = n(A n B)’ = n(A ? B) – n(A n B) = 8 – 2 = 6
6. f
-1
(-5) = x. Then,
f(x) = -5
x
2
+ 1 = -5
x
2
= -6
This equation is not solvable in R. Therefore, there is no pre-image of -5.
Hence, f
-1
(-5) = ?
OR
f(x) =
x1
x1
?
?
1
1
1
x
f
1
x
1
x
?
??
?
??
??
?
1 1 x
f
x 1 x
? ??
?
??
?
??
? ? x1
1
f
x x 1
??
??
?
??
?
??
? ?
1
f f x
x
??
??
??
??
7. 30’’ =
''
30 1
60 2
? ? ? ?
?
? ? ? ?
? ? ? ?
37’ 30’’ =
''
1 75 75 1 5
37
2 2 2 60 8
??
? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
OR
Let s be the length of the arc subtending an angle ?
c
at the centre of a circle of radius r.
s = r? here r = 5 cm and ? = 15°
5
s r 5 15
180 12
??
? ? ? ? ? ?
CBSE XI | Mathematics
Sample Paper – 8 Solution
8. B – C = {2, 3, 6} – {2, 3, 7} = {6}
A × (B – C) = {1, 4} × {6} = {(1, 6), (4, 6)}…….(i)
A × B = {1, 4} × {2, 3, 6} = {(1, 2), (1, 3), (1, 6), (4, 2), (4, 3), (4, 6)}
A × C = {1, 4} × {2, 3, 7} = {(1, 2), (1, 3), (1, 7), (4, 2), (4, 3), (4, 7)}
A × B – A × C = {(1, 6), (4, 6)}…..(ii)
From (i) and (ii)
A × (B – C) = A × B – A × C
9. sin
2
x + sin x – 2 = 0
(sinx + 2)(sinx – 1) = 0
Sin x + 2 = 0 or sin x – 1 = 0
sin x = -2 or sin x = 1
sin x cannot be -2 hence, sin x = 1.
x = 90°
OR
22
2
2
tan tan
4 2 4 2
1 tan 1 tan
22
1 tan 1 tan
22
1 tan 1 tan
22
1 tan 1 tan
22
2 1 tan
2
1 tan
2
2
cos
2sec
? ? ? ? ? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
??
??
??
??
??
?? ? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
?? ? ? ? ?
??
? ? ? ?
? ? ? ?
? ??
?
??
??
?
?
?
?
?
??
10. The compound statement is “25 is a multiple of 5 or 8”.
Let us assume that the statement q is false i. e. 25 is not a multiple of 8. Clearly p is true.
Thus, if we assume that q is false then p is true. Hence, compound statement is true.
CBSE XI | Mathematics
Sample Paper – 8 Solution
11. The values of x for which f(x) and g(x) are equal given by
f(x) = g(x)
2x
2
– 1 = 1 – 3x
2x
2
+ 3x – 2 = 0
(x + 2) (2x – 1) = 0
x = -2, ½
Thus, f(x) and g(x) are equal on the set {-2, ½}.
12. Suppose the line x + y = 4 divides the join of A(-1, 1) and B(5, 7) in the ratio k : 1.
The coordinate of the point of division are
5k 1 7k 1
,
k 1 k 1
?? ??
??
??
??
It lies on x + y = 4.
5k 1 7k 1
4
k 1 k 1
??
??
??
5k – 1 + 7k + 1 = 4k + 4
12k – 4k = 4
8k = 4
k = ½
SECTION – C
13.
1
1
sec8 1
cos8
1
sec4 1
1
cos4
?
??
?
?
??
?
?
? ?
? ?
1 cos8 cos4
cos8 1 cos4
? ? ?
?
? ? ?
2
2
2sin 4 cos4
cos8 2sin 2
??
?
??
2
2sin4 cos4 sin4
cos8 2sin 2
? ? ?
??
? ?
2
2sin4 cos4 2sin2 cos2
cos8 2sin 2
? ? ? ?
??
? ?
sin8 cos2
cos8 sin2
??
??
??
tan8 cot2 ? ? ? ?
tan8
tan2
?
?
?
Page 5
CBSE XI | Mathematics
Sample Paper – 8 Solution
CBSE Board
Class XI Mathematics
Sample Paper – 8 Solution
SECTION – A
1.
? ? ? ?
2
1
x
2
1
x
2
1
x
2
4x 1
lim
2x 1
2x 1 2x 1
lim
2x 1
lim 2x 1 2x 1 0
2
?
?
?
?
?
??
?
?
? ? ? ?
?
2. The number 2 is not greater than 7.
OR
The number 2 is less than or equal to 7.
OR
It is false that the number 2 is greater than 7.
3.
2
1 1 i i i
i
i i i 1 i
? ? ? ?
? ? ? ? ?
?
OR
z = 1 – 3i
Comparing with a + bi we get a = 1 and b = -3
? ?
2
2
z 1 3 10 ? ? ? ?
4. Variance = 4
Standard deviation = vVariance
Standard deviation = v4
Standard deviation = 2
CBSE XI | Mathematics
Sample Paper – 8 Solution
SECTION – B
5. According to the question,
A n B = {b, d}
n(A n B) = 2
n(A’ ? B’) = n(A n B)’ = n(A ? B) – n(A n B) = 8 – 2 = 6
6. f
-1
(-5) = x. Then,
f(x) = -5
x
2
+ 1 = -5
x
2
= -6
This equation is not solvable in R. Therefore, there is no pre-image of -5.
Hence, f
-1
(-5) = ?
OR
f(x) =
x1
x1
?
?
1
1
1
x
f
1
x
1
x
?
??
?
??
??
?
1 1 x
f
x 1 x
? ??
?
??
?
??
? ? x1
1
f
x x 1
??
??
?
??
?
??
? ?
1
f f x
x
??
??
??
??
7. 30’’ =
''
30 1
60 2
? ? ? ?
?
? ? ? ?
? ? ? ?
37’ 30’’ =
''
1 75 75 1 5
37
2 2 2 60 8
??
? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
OR
Let s be the length of the arc subtending an angle ?
c
at the centre of a circle of radius r.
s = r? here r = 5 cm and ? = 15°
5
s r 5 15
180 12
??
? ? ? ? ? ?
CBSE XI | Mathematics
Sample Paper – 8 Solution
8. B – C = {2, 3, 6} – {2, 3, 7} = {6}
A × (B – C) = {1, 4} × {6} = {(1, 6), (4, 6)}…….(i)
A × B = {1, 4} × {2, 3, 6} = {(1, 2), (1, 3), (1, 6), (4, 2), (4, 3), (4, 6)}
A × C = {1, 4} × {2, 3, 7} = {(1, 2), (1, 3), (1, 7), (4, 2), (4, 3), (4, 7)}
A × B – A × C = {(1, 6), (4, 6)}…..(ii)
From (i) and (ii)
A × (B – C) = A × B – A × C
9. sin
2
x + sin x – 2 = 0
(sinx + 2)(sinx – 1) = 0
Sin x + 2 = 0 or sin x – 1 = 0
sin x = -2 or sin x = 1
sin x cannot be -2 hence, sin x = 1.
x = 90°
OR
22
2
2
tan tan
4 2 4 2
1 tan 1 tan
22
1 tan 1 tan
22
1 tan 1 tan
22
1 tan 1 tan
22
2 1 tan
2
1 tan
2
2
cos
2sec
? ? ? ? ? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
??
??
??
??
??
?? ? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
?
?? ? ? ? ?
??
? ? ? ?
? ? ? ?
? ??
?
??
??
?
?
?
?
?
??
10. The compound statement is “25 is a multiple of 5 or 8”.
Let us assume that the statement q is false i. e. 25 is not a multiple of 8. Clearly p is true.
Thus, if we assume that q is false then p is true. Hence, compound statement is true.
CBSE XI | Mathematics
Sample Paper – 8 Solution
11. The values of x for which f(x) and g(x) are equal given by
f(x) = g(x)
2x
2
– 1 = 1 – 3x
2x
2
+ 3x – 2 = 0
(x + 2) (2x – 1) = 0
x = -2, ½
Thus, f(x) and g(x) are equal on the set {-2, ½}.
12. Suppose the line x + y = 4 divides the join of A(-1, 1) and B(5, 7) in the ratio k : 1.
The coordinate of the point of division are
5k 1 7k 1
,
k 1 k 1
?? ??
??
??
??
It lies on x + y = 4.
5k 1 7k 1
4
k 1 k 1
??
??
??
5k – 1 + 7k + 1 = 4k + 4
12k – 4k = 4
8k = 4
k = ½
SECTION – C
13.
1
1
sec8 1
cos8
1
sec4 1
1
cos4
?
??
?
?
??
?
?
? ?
? ?
1 cos8 cos4
cos8 1 cos4
? ? ?
?
? ? ?
2
2
2sin 4 cos4
cos8 2sin 2
??
?
??
2
2sin4 cos4 sin4
cos8 2sin 2
? ? ?
??
? ?
2
2sin4 cos4 2sin2 cos2
cos8 2sin 2
? ? ? ?
??
? ?
sin8 cos2
cos8 sin2
??
??
??
tan8 cot2 ? ? ? ?
tan8
tan2
?
?
?
CBSE XI | Mathematics
Sample Paper – 8 Solution
14.
1. It is a function. Because normally each country has exactly one capital city.
2. It is not a function.
For each replacement of x there are many replacements of y which makes the
sentence true.
3. It is not a function.
Each Maths teacher ordinarily has a number of pupils.
4. It is a function.
Each replacement of x gives one and only one replacement of y which makes the
sentence true.
15. (x, y) ? R ? x + 2y = 10 ? y =
10 x
2
?
, x, y ? A where A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
x = 1 ? y = 9/2 ? A
This shows that 1 is not related to any element in A. Similarly we can observe that 3, 5,
7, 9 and 10 are not related to any element of A under the defined relation.
Further we find that
For x = 2, y = 4 ? A therefore, (2, 4) ? R
For x = 4, y = 3 ? A therefore, (4, 3) ? R
For x = 6, y = 2 ? A therefore, (6, 2) ? R
For x = 8, y = 1 ? A therefore, (8, 1) ? R
Thus, R = {(2, 4), (4, 3), (6, 2), (8, 1)} ? R
-1
= {(4, 2), (3, 4), (2, 6), (1, 8)}
Clearly, Dom(R) = {2, 4, 6, 8} = Range R
-1
and Range (R) = {4, 3, 2, 1} = Dom (R
-1
)
16. log10 2, log10 (2
x
– 1) and log10 (2
x
+ 3) are in A. P.
2log10 (2
x
– 1) = log10 2 + log10 (2
x
+ 3)
log10 (2
x
– 1)
2
= log10 [2(2
x
+ 3)]
(2
x
– 1)
2
= 2(2
x
+ 3)
(y – 1)
2
= 2(y + 3) ? y = 2
x
y
2
– 2y + 1 = 2y + 6
y
2
– 4y – 5 = 0
(y – 5)(y + 1) = 0
Y = 5, -1
When y = 5, then 2
x
= 5 ? x = log2 5
When y = -1 then 2
x
= -1 which is impossible.
Hence, x = log2 5
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FAQs on Sample Solution Paper 8 - Math, Class 11 - Mathematics (Maths) Class 11 - Commerce
| 1. What are the different types of matrices? |  |
Ans. Matrices can be classified into different types based on their characteristics. Some common types of matrices are:
- Square matrix: A matrix with an equal number of rows and columns is called a square matrix.
- Identity matrix: A square matrix with ones on the main diagonal and zeros elsewhere is called an identity matrix.
- Zero matrix: A matrix in which all the elements are zeros is called a zero matrix.
- Diagonal matrix: A square matrix in which all the non-diagonal elements are zeros is called a diagonal matrix.
- Symmetric matrix: A square matrix that is equal to its transpose is called a symmetric matrix.
- Skew-symmetric matrix: A square matrix in which the transpose of the matrix is equal to the negative of the original matrix is called a skew-symmetric matrix.
| 2. How can determinants be used to find the inverse of a matrix? | |
Ans. The determinant of a matrix plays a crucial role in finding its inverse. To find the inverse of a matrix using determinants, follow these steps:
1. Calculate the determinant of the given matrix.
2. If the determinant is non-zero, the matrix is invertible. If the determinant is zero, the matrix does not have an inverse.
3. Find the adjoint of the given matrix by taking the transpose of the cofactor matrix.
4. Multiply the adjoint matrix by the reciprocal of the determinant.
5. The resulting matrix is the inverse of the original matrix.
| 3. What is the significance of eigenvalues and eigenvectors in linear algebra? | |
Ans. Eigenvalues and eigenvectors are essential concepts in linear algebra. They have various applications and significance, such as:
- Eigenvalues represent the scaling factor of the eigenvector when a linear transformation is applied to it. They give information about the behavior of the transformation.
- Eigenvectors are non-zero vectors that only change by a scalar factor when a linear transformation is applied to them. They define the direction of the transformation.
- Eigenvalues and eigenvectors are used in various fields such as physics, computer science, engineering, and data analysis, for tasks like image compression, principal component analysis, solving differential equations, and solving systems of linear equations.
| 4. How can matrix multiplication be performed? | |
Ans. Matrix multiplication is a fundamental operation in linear algebra. To multiply two matrices, follow these rules:
- The number of columns in the first matrix must be equal to the number of rows in the second matrix.
- Multiply each element of the row of the first matrix with the corresponding element of the column of the second matrix.
- Add the products obtained from the previous step to get the corresponding element of the resulting matrix.
- Repeat the process for each element of the resulting matrix.
- The resulting matrix will have dimensions equal to the number of rows of the first matrix and the number of columns of the second matrix.
| 5. What are the properties of determinants? | |
Ans. Determinants have several properties that are useful in various calculations. Some important properties of determinants include:
- If a matrix has a row or column of zeros, then its determinant is zero.
- Interchanging any two rows or columns of a matrix changes the sign of its determinant.
- Multiplying a row or column of a matrix by a scalar multiplies the determinant by the same scalar.
- If two rows or columns of a matrix are proportional, then its determinant is zero.
- If two rows or columns of a matrix are identical, then its determinant is zero.
- Adding a multiple of one row (or column) to another row (or column) does not change the determinant.
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Nov 24, 2024 Last updated |
Document Description: Sample Solution Paper 8 - Math, Class 11 for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation.
The notes and questions for Sample Solution Paper 8 - Math, Class 11 have been prepared according to the Commerce exam syllabus. Information about Sample Solution Paper 8 - Math, Class 11 covers topics
like and Sample Solution Paper 8 - Math, Class 11 Example, for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Sample Solution Paper 8 - Math, Class 11.
Introduction of Sample Solution Paper 8 - Math, Class 11 in English is available as part of our Mathematics (Maths) Class 11
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Full syllabus notes, lecture & questions for Sample Solution Paper 8 - Math, Class 11 | Mathematics (Maths) Class 11 - Commerce - Commerce | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 11 | Best notes, free PDF download
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Additional Information about Sample Solution Paper 8 - Math, Class 11 for Commerce Preparation
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The content of Sample Solution Paper 8 - Math, Class 11 is prepared as per the latest Commerce syllabus.
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