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CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
Mathematics 
Class XII 
Sample Paper – 1 Solution 
 
SECTION – A  
1.  Let ? ? ? ? and a, b R b, c R ?? 
  
? ?
? ?
and
R
a b b c a c
a, c
? ? ? ? ?
??
 
  Hence, R is transitive.                            
 
2.  Let ? ?
1
x tan 1
?
?? 
  
? ? ?
??
?
??
? ??
?? ? ? ? ? ?
?? ??
??
tanx 1
tanx tan
4
tanx tan tan
4
 
  
3 p
tanx = tan
4
3 p
x=
4
                                                     
 
3.  Matrix of order 3 × 3 has 9 elements. 
  Now the entries have to be either 0 or 1 so that each of the 9 places can be filled 
with 2  choices 0 or 1.  
  So 2
9
 = 512 matrices are possible.  
 
4. 
ˆ ˆˆ
a 5i j 3k; ? ? ? 
 
ˆ ˆˆ
b i 3j 5k ? ? ? 
 
ˆ ˆˆ
a b 6i 2j 8k ? ? ? ? ?
 
ab
ˆ ˆˆ
3i j 4k
2
?
? ? ? 
 
 
 
 
 
 
 
 
Page 2


  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
Mathematics 
Class XII 
Sample Paper – 1 Solution 
 
SECTION – A  
1.  Let ? ? ? ? and a, b R b, c R ?? 
  
? ?
? ?
and
R
a b b c a c
a, c
? ? ? ? ?
??
 
  Hence, R is transitive.                            
 
2.  Let ? ?
1
x tan 1
?
?? 
  
? ? ?
??
?
??
? ??
?? ? ? ? ? ?
?? ??
??
tanx 1
tanx tan
4
tanx tan tan
4
 
  
3 p
tanx = tan
4
3 p
x=
4
                                                     
 
3.  Matrix of order 3 × 3 has 9 elements. 
  Now the entries have to be either 0 or 1 so that each of the 9 places can be filled 
with 2  choices 0 or 1.  
  So 2
9
 = 512 matrices are possible.  
 
4. 
ˆ ˆˆ
a 5i j 3k; ? ? ? 
 
ˆ ˆˆ
b i 3j 5k ? ? ? 
 
ˆ ˆˆ
a b 6i 2j 8k ? ? ? ? ?
 
ab
ˆ ˆˆ
3i j 4k
2
?
? ? ? 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
OR 
Given that the magnitude of each of the two vectors a and b have same magnitude. 
2
2
a b ..........(i)
the angle between a and b is 60
60 ........(ii)
9
and a b .........(iii)
2
Since,
a b a b cos
9
a a cos60 From (i),(ii) and (iii)
2
19
a
22
a9
a3
a b 3 ............... From (i)
??
?
? ? ? ?
??
? ? ?
? ? ?
? ? ?
??
??
? ? ?
  
 
 
 
 
SECTION – B  
 
5. Let    I = 
p
x
dx
x p x
0
?
??
              ...(1) 
  According to property,  
  
aa
00
f(x)dx f(a x)dx
??
?? 
  I ?  
p
0
px
dx
p x x
?
?
??
           ...(2) 
  Adding equations (1) and (2), we get  
  2I ?  
p
0
x p x
dx
x p x
?
??
??
 
  
p
0
 1dx
?
? = 
? ?
p
0
x = p – 0 = p 
  Thus, 2I = p I
p
2
??              
 
Page 3


  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
Mathematics 
Class XII 
Sample Paper – 1 Solution 
 
SECTION – A  
1.  Let ? ? ? ? and a, b R b, c R ?? 
  
? ?
? ?
and
R
a b b c a c
a, c
? ? ? ? ?
??
 
  Hence, R is transitive.                            
 
2.  Let ? ?
1
x tan 1
?
?? 
  
? ? ?
??
?
??
? ??
?? ? ? ? ? ?
?? ??
??
tanx 1
tanx tan
4
tanx tan tan
4
 
  
3 p
tanx = tan
4
3 p
x=
4
                                                     
 
3.  Matrix of order 3 × 3 has 9 elements. 
  Now the entries have to be either 0 or 1 so that each of the 9 places can be filled 
with 2  choices 0 or 1.  
  So 2
9
 = 512 matrices are possible.  
 
4. 
ˆ ˆˆ
a 5i j 3k; ? ? ? 
 
ˆ ˆˆ
b i 3j 5k ? ? ? 
 
ˆ ˆˆ
a b 6i 2j 8k ? ? ? ? ?
 
ab
ˆ ˆˆ
3i j 4k
2
?
? ? ? 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
OR 
Given that the magnitude of each of the two vectors a and b have same magnitude. 
2
2
a b ..........(i)
the angle between a and b is 60
60 ........(ii)
9
and a b .........(iii)
2
Since,
a b a b cos
9
a a cos60 From (i),(ii) and (iii)
2
19
a
22
a9
a3
a b 3 ............... From (i)
??
?
? ? ? ?
??
? ? ?
? ? ?
? ? ?
??
??
? ? ?
  
 
 
 
 
SECTION – B  
 
5. Let    I = 
p
x
dx
x p x
0
?
??
              ...(1) 
  According to property,  
  
aa
00
f(x)dx f(a x)dx
??
?? 
  I ?  
p
0
px
dx
p x x
?
?
??
           ...(2) 
  Adding equations (1) and (2), we get  
  2I ?  
p
0
x p x
dx
x p x
?
??
??
 
  
p
0
 1dx
?
? = 
? ?
p
0
x = p – 0 = p 
  Thus, 2I = p I
p
2
??              
 
  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
OR 
Given 
?
?
?
2
2
cos2x 2sin x
I dx
cos x
  
? ?
? ?
?
?
??
? ? ?
?
?
? ? ?
?
??
?
?
?
?
?
2
2
2 2 2
22
2
22
2
22
2
2
cos2x 2sin x
I dx
cos x
cos x sin x 2sin x
I dx ..... cos2x cos x sin x
cos x
cos x sin x
I dx
cos x
1
I dx .... cos x sin x 1
cos x
I sec x dx
I tan x c
  
         
 
6. Area of a parallelogram = Cross product of the vectors representing its adjacent 
sides.  
 So, required area = |(2 i + j + k ) ?  (3 i + j +4 k )| 
 Now (2 i + j + k ) ?  (3 i + j +4 k ) 
 =
i j k
2 1 1
3 1 4
=3 i - 5 j - k 
Area = |3 i -5 j - k | = 9 25 1 35 ? ? ? sq. units.        
 
 
7. tan (cos
–1
4
5
 + tan
–1
2
3
) 
      = tan (tan
–1
3
4
 + tan
–1
 
2
3
) 
 = 
1
32
43
tan tan
1
1
2
?
?? ??
?
?? ??
?? ??
?? ??
?
?? ??
?? ??
 
      =
1
17
tan tan
6
?
?? ??
?? ??
?? ??
 
          =
17
6
                                           
 
Page 4


  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
Mathematics 
Class XII 
Sample Paper – 1 Solution 
 
SECTION – A  
1.  Let ? ? ? ? and a, b R b, c R ?? 
  
? ?
? ?
and
R
a b b c a c
a, c
? ? ? ? ?
??
 
  Hence, R is transitive.                            
 
2.  Let ? ?
1
x tan 1
?
?? 
  
? ? ?
??
?
??
? ??
?? ? ? ? ? ?
?? ??
??
tanx 1
tanx tan
4
tanx tan tan
4
 
  
3 p
tanx = tan
4
3 p
x=
4
                                                     
 
3.  Matrix of order 3 × 3 has 9 elements. 
  Now the entries have to be either 0 or 1 so that each of the 9 places can be filled 
with 2  choices 0 or 1.  
  So 2
9
 = 512 matrices are possible.  
 
4. 
ˆ ˆˆ
a 5i j 3k; ? ? ? 
 
ˆ ˆˆ
b i 3j 5k ? ? ? 
 
ˆ ˆˆ
a b 6i 2j 8k ? ? ? ? ?
 
ab
ˆ ˆˆ
3i j 4k
2
?
? ? ? 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
OR 
Given that the magnitude of each of the two vectors a and b have same magnitude. 
2
2
a b ..........(i)
the angle between a and b is 60
60 ........(ii)
9
and a b .........(iii)
2
Since,
a b a b cos
9
a a cos60 From (i),(ii) and (iii)
2
19
a
22
a9
a3
a b 3 ............... From (i)
??
?
? ? ? ?
??
? ? ?
? ? ?
? ? ?
??
??
? ? ?
  
 
 
 
 
SECTION – B  
 
5. Let    I = 
p
x
dx
x p x
0
?
??
              ...(1) 
  According to property,  
  
aa
00
f(x)dx f(a x)dx
??
?? 
  I ?  
p
0
px
dx
p x x
?
?
??
           ...(2) 
  Adding equations (1) and (2), we get  
  2I ?  
p
0
x p x
dx
x p x
?
??
??
 
  
p
0
 1dx
?
? = 
? ?
p
0
x = p – 0 = p 
  Thus, 2I = p I
p
2
??              
 
  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
OR 
Given 
?
?
?
2
2
cos2x 2sin x
I dx
cos x
  
? ?
? ?
?
?
??
? ? ?
?
?
? ? ?
?
??
?
?
?
?
?
2
2
2 2 2
22
2
22
2
22
2
2
cos2x 2sin x
I dx
cos x
cos x sin x 2sin x
I dx ..... cos2x cos x sin x
cos x
cos x sin x
I dx
cos x
1
I dx .... cos x sin x 1
cos x
I sec x dx
I tan x c
  
         
 
6. Area of a parallelogram = Cross product of the vectors representing its adjacent 
sides.  
 So, required area = |(2 i + j + k ) ?  (3 i + j +4 k )| 
 Now (2 i + j + k ) ?  (3 i + j +4 k ) 
 =
i j k
2 1 1
3 1 4
=3 i - 5 j - k 
Area = |3 i -5 j - k | = 9 25 1 35 ? ? ? sq. units.        
 
 
7. tan (cos
–1
4
5
 + tan
–1
2
3
) 
      = tan (tan
–1
3
4
 + tan
–1
 
2
3
) 
 = 
1
32
43
tan tan
1
1
2
?
?? ??
?
?? ??
?? ??
?? ??
?
?? ??
?? ??
 
      =
1
17
tan tan
6
?
?? ??
?? ??
?? ??
 
          =
17
6
                                           
 
  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
8. 
?? ??
??
? ? ?
??
cos sin
Let A = ,then
sin cos
 
?
?
??
? ? ?
? ? ? ? ? ?
?
? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
22
1
1
cos sin
 A =
sin cos
cos sin 1 0
Since A 0, therefore A exist.
cos sin cos sin AdjA 1
A
sin cos sin cos A1
 
 
 
9. ? ? ?
32
Contentment function : C(x)= x 6x 5x 3   (given) 
?
?
? ? ?
??
? ? ?
??
??
? ? ?
?
x3
2
2
dC(x)
 Marginal contentment =
dx
3x 12x 5
When 3 units of dish are consumed, then 
dC(x)
 3(3) 12(3) 5
dx
27 36 5
68 units.
 
 
 
10. ??
y
On differentiating e (x 1) 1 w.r.t x, we get 
? ? ?
? ? ?
? ? ?
yy
y
y
dy
e (x 1)e 0  
dx
dy
e 0   
dx
dy
e 
dx
 
 
 
 
 
 
 
 
 
 
 
 
Page 5


  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
Mathematics 
Class XII 
Sample Paper – 1 Solution 
 
SECTION – A  
1.  Let ? ? ? ? and a, b R b, c R ?? 
  
? ?
? ?
and
R
a b b c a c
a, c
? ? ? ? ?
??
 
  Hence, R is transitive.                            
 
2.  Let ? ?
1
x tan 1
?
?? 
  
? ? ?
??
?
??
? ??
?? ? ? ? ? ?
?? ??
??
tanx 1
tanx tan
4
tanx tan tan
4
 
  
3 p
tanx = tan
4
3 p
x=
4
                                                     
 
3.  Matrix of order 3 × 3 has 9 elements. 
  Now the entries have to be either 0 or 1 so that each of the 9 places can be filled 
with 2  choices 0 or 1.  
  So 2
9
 = 512 matrices are possible.  
 
4. 
ˆ ˆˆ
a 5i j 3k; ? ? ? 
 
ˆ ˆˆ
b i 3j 5k ? ? ? 
 
ˆ ˆˆ
a b 6i 2j 8k ? ? ? ? ?
 
ab
ˆ ˆˆ
3i j 4k
2
?
? ? ? 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
OR 
Given that the magnitude of each of the two vectors a and b have same magnitude. 
2
2
a b ..........(i)
the angle between a and b is 60
60 ........(ii)
9
and a b .........(iii)
2
Since,
a b a b cos
9
a a cos60 From (i),(ii) and (iii)
2
19
a
22
a9
a3
a b 3 ............... From (i)
??
?
? ? ? ?
??
? ? ?
? ? ?
? ? ?
??
??
? ? ?
  
 
 
 
 
SECTION – B  
 
5. Let    I = 
p
x
dx
x p x
0
?
??
              ...(1) 
  According to property,  
  
aa
00
f(x)dx f(a x)dx
??
?? 
  I ?  
p
0
px
dx
p x x
?
?
??
           ...(2) 
  Adding equations (1) and (2), we get  
  2I ?  
p
0
x p x
dx
x p x
?
??
??
 
  
p
0
 1dx
?
? = 
? ?
p
0
x = p – 0 = p 
  Thus, 2I = p I
p
2
??              
 
  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
OR 
Given 
?
?
?
2
2
cos2x 2sin x
I dx
cos x
  
? ?
? ?
?
?
??
? ? ?
?
?
? ? ?
?
??
?
?
?
?
?
2
2
2 2 2
22
2
22
2
22
2
2
cos2x 2sin x
I dx
cos x
cos x sin x 2sin x
I dx ..... cos2x cos x sin x
cos x
cos x sin x
I dx
cos x
1
I dx .... cos x sin x 1
cos x
I sec x dx
I tan x c
  
         
 
6. Area of a parallelogram = Cross product of the vectors representing its adjacent 
sides.  
 So, required area = |(2 i + j + k ) ?  (3 i + j +4 k )| 
 Now (2 i + j + k ) ?  (3 i + j +4 k ) 
 =
i j k
2 1 1
3 1 4
=3 i - 5 j - k 
Area = |3 i -5 j - k | = 9 25 1 35 ? ? ? sq. units.        
 
 
7. tan (cos
–1
4
5
 + tan
–1
2
3
) 
      = tan (tan
–1
3
4
 + tan
–1
 
2
3
) 
 = 
1
32
43
tan tan
1
1
2
?
?? ??
?
?? ??
?? ??
?? ??
?
?? ??
?? ??
 
      =
1
17
tan tan
6
?
?? ??
?? ??
?? ??
 
          =
17
6
                                           
 
  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
8. 
?? ??
??
? ? ?
??
cos sin
Let A = ,then
sin cos
 
?
?
??
? ? ?
? ? ? ? ? ?
?
? ? ? ? ? ? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
? ? ? ?
22
1
1
cos sin
 A =
sin cos
cos sin 1 0
Since A 0, therefore A exist.
cos sin cos sin AdjA 1
A
sin cos sin cos A1
 
 
 
9. ? ? ?
32
Contentment function : C(x)= x 6x 5x 3   (given) 
?
?
? ? ?
??
? ? ?
??
??
? ? ?
?
x3
2
2
dC(x)
 Marginal contentment =
dx
3x 12x 5
When 3 units of dish are consumed, then 
dC(x)
 3(3) 12(3) 5
dx
27 36 5
68 units.
 
 
 
10. ??
y
On differentiating e (x 1) 1 w.r.t x, we get 
? ? ?
? ? ?
? ? ?
yy
y
y
dy
e (x 1)e 0  
dx
dy
e 0   
dx
dy
e 
dx
 
 
 
 
 
 
 
 
 
 
 
 
  
 
CBSE XII | Mathematics 
Sample Paper – 1 Solution  
 
     
OR 
? ?
? ?
? ?
? ?
? ?
? ?
? ?
? ?
2
x a 2 sin2
y a 1 cos 2
differentiating w.r.t.
dx
a 2 cos2 2
d
dy
a sin2 2
d
sin2 2 sin2
dy
dx 2 cos2 2 1 cos2
dy 2sin cos
dx 2sin
dy cos
cot
dx sin
at,
3
dy 1
cot
dx 3
3
? ? ? ?
? ? ?
?
? ? ? ?
?
? ? ? ?
?
? ? ? ?
??
? ? ? ? ?
? ? ?
?
?
?
? ? ?
?
?
??
?
??
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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FAQs on Sample Solution Paper 1 - Math, Class 12 - Mathematics (Maths) Class 12 - JEE

1. What are the different types of matrices?
Ans. Matrices are classified into various types based on their characteristics. Some common types of matrices include: - Square Matrix: A matrix in which the number of rows is equal to the number of columns. - Diagonal Matrix: A square matrix in which all the elements outside the main diagonal are zero. - Identity Matrix: A diagonal matrix in which all the elements on the main diagonal are equal to 1. - Zero Matrix: A matrix in which all the elements are zero. - Row Matrix: A matrix with only one row. - Column Matrix: A matrix with only one column. - Symmetric Matrix: A square matrix that is equal to its transpose. - Skew-Symmetric Matrix: A square matrix in which the transpose of the matrix is equal to the negative of the matrix.
2. How can determinants be used to solve systems of linear equations?
Ans. Determinants can be used to solve systems of linear equations by using Cramer's Rule. Cramer's Rule states that if a system of linear equations can be represented in matrix form as AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix, then the solution for X can be found using the formula X = (det(Ai))/det(A), where det(A) is the determinant of the coefficient matrix A, and det(Ai) is the determinant of the matrix obtained by replacing the i-th column of A with B.
3. What is the significance of eigenvalues and eigenvectors in linear algebra?
Ans. Eigenvalues and eigenvectors play a significant role in linear algebra. Eigenvalues represent the scalar values that are associated with an eigenvector. They provide information about the behavior of a linear transformation or a matrix. The eigenvalues of a matrix can help determine whether the matrix is invertible or singular, and they also provide insights into the stability of a system described by the matrix. Eigenvectors are the vectors that remain in the same direction, but may change in length, when a linear transformation is applied. They are associated with the eigenvalues of a matrix and provide information about the directions along which the matrix stretches or compresses the vector space. Eigenvectors are widely used in applications such as image processing, data compression, and machine learning.
4. How does matrix multiplication work?
Ans. Matrix multiplication is a binary operation that combines two matrices to produce a third matrix. The number of columns in the first matrix must be equal to the number of rows in the second matrix for the multiplication to be defined. To multiply two matrices A and B, the elements of the resulting matrix C are calculated as follows: C(i,j) = A(i,1) * B(1,j) + A(i,2) * B(2,j) + ... + A(i,n) * B(n,j) Where A(i,j) represents the element in the i-th row and j-th column of matrix A, and B(i,j) represents the element in the i-th row and j-th column of matrix B. The resulting matrix C will have the same number of rows as matrix A and the same number of columns as matrix B.
5. What is the concept of rank in linear algebra?
Ans. The rank of a matrix is a fundamental concept in linear algebra that provides information about the dimensions of the vector space spanned by the columns (or rows) of the matrix. The rank of a matrix is defined as the maximum number of linearly independent rows (or columns) in the matrix. It can be calculated by performing row operations on the matrix and counting the number of non-zero rows (or columns) in the row-echelon form of the matrix. The rank of a matrix is useful in determining whether a system of linear equations has a unique solution, no solution, or infinitely many solutions. If the rank of the coefficient matrix is equal to the rank of the augmented matrix, the system has a unique solution. If the rank of the augmented matrix is greater than the rank of the coefficient matrix, the system has no solution. If the rank of the augmented matrix is less than the rank of the coefficient matrix, the system has infinitely many solutions.
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