Page 1
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
CBSE Board
Class XII Mathematics
Board Paper 2014 Solution
Delhi
SECTION – A
1. Given that
2
AA ? .
We need to find the value of
? ?
3
7A I A , where I is the identity matrix. ??
Thus,
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ?
? ?
3
3 2 2 3
3
3 2 2 3 2 2 2
3
2
3
3
3
7A I A 7A I 3I A 3IA A
7A I A 7A I 3A 3A A A I I,I A A,IA A
7A I A 7A I 3A 3A A A A
7A I A 7A I 3A 3A A
7A I A 7A I 7A
7A I A I
? ? ? ? ? ? ?
?? ? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ?
2. Given that
x y z 1 4
2x y w 0 5
?? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
We need to find the value of x + y.
?? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
?
? ? ?
??
ij ij
x y z 1 4
2x y w 0 5
Two matrices A and B are equal to each other, if they have the same dimensions
and the same elements a b , for i = 1,2,...,n and j = 1,2,...,m.
x y 1...(1)
2x y 0...(2)
Equa ?
? ? ?
??
tion (2) (1) is x = 1
Substituting the value of x = 1 in equation (1), we have
1 y 1
y2
Therefore, x + y = 1 + 2 = 3
Page 2
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
CBSE Board
Class XII Mathematics
Board Paper 2014 Solution
Delhi
SECTION – A
1. Given that
2
AA ? .
We need to find the value of
? ?
3
7A I A , where I is the identity matrix. ??
Thus,
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ?
? ?
3
3 2 2 3
3
3 2 2 3 2 2 2
3
2
3
3
3
7A I A 7A I 3I A 3IA A
7A I A 7A I 3A 3A A A I I,I A A,IA A
7A I A 7A I 3A 3A A A A
7A I A 7A I 3A 3A A
7A I A 7A I 7A
7A I A I
? ? ? ? ? ? ?
?? ? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ?
2. Given that
x y z 1 4
2x y w 0 5
?? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
We need to find the value of x + y.
?? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
?
? ? ?
??
ij ij
x y z 1 4
2x y w 0 5
Two matrices A and B are equal to each other, if they have the same dimensions
and the same elements a b , for i = 1,2,...,n and j = 1,2,...,m.
x y 1...(1)
2x y 0...(2)
Equa ?
? ? ?
??
tion (2) (1) is x = 1
Substituting the value of x = 1 in equation (1), we have
1 y 1
y2
Therefore, x + y = 1 + 2 = 3
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
3.
11
Given that tan x tan y and xy<1.
4
??
?
??
? ?
11
1
1
We need to find the value of x+y+xy.
tan x tan y
4
xy
tan xy 1
1 xy 4
xy
tan tan tan
1 xy 4
xy
1
1 xy
x y 1 xy
x y xy 1
??
?
?
?
??
?? ??
? ? ?
??
?
??
?? ???? ??
??
?? ?? ??
?
?? ?? ??
?
??
?
? ? ? ?
? ? ? ?
4. Given that
3x 7 8 7
2 4 6 4
?
?
.
We need to find the value of x
? ?
3x 7 8 7
2 4 6 4
12x 14 32 42
12x 14 10
12x 10 14
12x 24
x2
?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ?
? ? ?
Page 3
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
CBSE Board
Class XII Mathematics
Board Paper 2014 Solution
Delhi
SECTION – A
1. Given that
2
AA ? .
We need to find the value of
? ?
3
7A I A , where I is the identity matrix. ??
Thus,
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ?
? ?
3
3 2 2 3
3
3 2 2 3 2 2 2
3
2
3
3
3
7A I A 7A I 3I A 3IA A
7A I A 7A I 3A 3A A A I I,I A A,IA A
7A I A 7A I 3A 3A A A A
7A I A 7A I 3A 3A A
7A I A 7A I 7A
7A I A I
? ? ? ? ? ? ?
?? ? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ?
2. Given that
x y z 1 4
2x y w 0 5
?? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
We need to find the value of x + y.
?? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
?
? ? ?
??
ij ij
x y z 1 4
2x y w 0 5
Two matrices A and B are equal to each other, if they have the same dimensions
and the same elements a b , for i = 1,2,...,n and j = 1,2,...,m.
x y 1...(1)
2x y 0...(2)
Equa ?
? ? ?
??
tion (2) (1) is x = 1
Substituting the value of x = 1 in equation (1), we have
1 y 1
y2
Therefore, x + y = 1 + 2 = 3
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
3.
11
Given that tan x tan y and xy<1.
4
??
?
??
? ?
11
1
1
We need to find the value of x+y+xy.
tan x tan y
4
xy
tan xy 1
1 xy 4
xy
tan tan tan
1 xy 4
xy
1
1 xy
x y 1 xy
x y xy 1
??
?
?
?
??
?? ??
? ? ?
??
?
??
?? ???? ??
??
?? ?? ??
?
?? ?? ??
?
??
?
? ? ? ?
? ? ? ?
4. Given that
3x 7 8 7
2 4 6 4
?
?
.
We need to find the value of x
? ?
3x 7 8 7
2 4 6 4
12x 14 32 42
12x 14 10
12x 10 14
12x 24
x2
?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ?
? ? ?
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
5. Since differentiation operation is the inverse operation of integration, we have
? ? sin ? ? f x x x
Let ? ?
0
sin ?
?
x
f x t tdt
Let us do this by integration by parts.
Therefore assume u = t; du = dt
sin
cos
?
??
??
tdt dv
tv
? ? ? ? ? ?
? ?
? ? ? ?
0
0
Therefore,
= t cos cos
cos sin
Differentiating the above function with respect to x,
f x sin cos cos sin
?? ? ? ?
??
? ? ? ?
??? ? ? ? ? ? ?
??
?
x
x
f x t t dt
f x x x x C
x x x x x x
6. Since the vectors are parallel, we have
? ?
ab
3i 2j 9k i 2pj 3k
3i 2j 9k i 2 pj 3 k
Comparing the respective coefficients, we have
3;
2 p 2
2 3 p 2
1
p
3
??
? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ?
? ? ?
? ? ? ? ?
?
??
7. ? ? The set of natural numbers, N = 1, 2, 3, 4, 5, 6....
? ? ? ?
? ? ? ? ? ? ? ?
? ?
? ?
The relation is given as
R = x, y : 2 8
Thus, R = 6, 1 , 4, 2 , 2, 3
Domain = 6, 4, 2
Range = 1, 2, 3
?? xy
Page 4
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
CBSE Board
Class XII Mathematics
Board Paper 2014 Solution
Delhi
SECTION – A
1. Given that
2
AA ? .
We need to find the value of
? ?
3
7A I A , where I is the identity matrix. ??
Thus,
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ?
? ?
3
3 2 2 3
3
3 2 2 3 2 2 2
3
2
3
3
3
7A I A 7A I 3I A 3IA A
7A I A 7A I 3A 3A A A I I,I A A,IA A
7A I A 7A I 3A 3A A A A
7A I A 7A I 3A 3A A
7A I A 7A I 7A
7A I A I
? ? ? ? ? ? ?
?? ? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ?
2. Given that
x y z 1 4
2x y w 0 5
?? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
We need to find the value of x + y.
?? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
?
? ? ?
??
ij ij
x y z 1 4
2x y w 0 5
Two matrices A and B are equal to each other, if they have the same dimensions
and the same elements a b , for i = 1,2,...,n and j = 1,2,...,m.
x y 1...(1)
2x y 0...(2)
Equa ?
? ? ?
??
tion (2) (1) is x = 1
Substituting the value of x = 1 in equation (1), we have
1 y 1
y2
Therefore, x + y = 1 + 2 = 3
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
3.
11
Given that tan x tan y and xy<1.
4
??
?
??
? ?
11
1
1
We need to find the value of x+y+xy.
tan x tan y
4
xy
tan xy 1
1 xy 4
xy
tan tan tan
1 xy 4
xy
1
1 xy
x y 1 xy
x y xy 1
??
?
?
?
??
?? ??
? ? ?
??
?
??
?? ???? ??
??
?? ?? ??
?
?? ?? ??
?
??
?
? ? ? ?
? ? ? ?
4. Given that
3x 7 8 7
2 4 6 4
?
?
.
We need to find the value of x
? ?
3x 7 8 7
2 4 6 4
12x 14 32 42
12x 14 10
12x 10 14
12x 24
x2
?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ?
? ? ?
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
5. Since differentiation operation is the inverse operation of integration, we have
? ? sin ? ? f x x x
Let ? ?
0
sin ?
?
x
f x t tdt
Let us do this by integration by parts.
Therefore assume u = t; du = dt
sin
cos
?
??
??
tdt dv
tv
? ? ? ? ? ?
? ?
? ? ? ?
0
0
Therefore,
= t cos cos
cos sin
Differentiating the above function with respect to x,
f x sin cos cos sin
?? ? ? ?
??
? ? ? ?
??? ? ? ? ? ? ?
??
?
x
x
f x t t dt
f x x x x C
x x x x x x
6. Since the vectors are parallel, we have
? ?
ab
3i 2j 9k i 2pj 3k
3i 2j 9k i 2 pj 3 k
Comparing the respective coefficients, we have
3;
2 p 2
2 3 p 2
1
p
3
??
? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ?
? ? ?
? ? ? ? ?
?
??
7. ? ? The set of natural numbers, N = 1, 2, 3, 4, 5, 6....
? ? ? ?
? ? ? ? ? ? ? ?
? ?
? ?
The relation is given as
R = x, y : 2 8
Thus, R = 6, 1 , 4, 2 , 2, 3
Domain = 6, 4, 2
Range = 1, 2, 3
?? xy
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
8. Given that the cartesian equation of the line as
? ? ? ? ? ?
? ?
3 4 2 6
5 7 4
That is,
3 4 2 3
5 7 4
4
33
5 7 2
? ? ?
??
? ? ? ? ?
??
??
??
? ? ? ?
?
?
x y z
x y z
y
xz
Any point on the line is of the form:
5 3,7 4,2 3
Thus, the vector equation is of the form:
r , where is the position vector of any
point on the line and b is the vector parallel to the lin
? ? ? ?
?? a b a
? ? ?
?
? ? ? ? ? ?
? ?
e.
Therefore, the vector equation is
r 5 3 7 4 2 3
r 5 7 2 3 4 3
r 3 4 3 5 7 2
? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
i j k
i j k i j k
i j k i j k
? ? ?
? ? ?
?
9.
a
2
0
dx
Given that
4+x 8
?
?
?
a
2
0
-1
0
We need to find the value of a.
dx
Let I=
4+x 8
1
Thus, I= tan
2 2 8
?
??
?
??
??
?
?
?
a
x
1
1
1
1
tan
2 2 8
tan 2
28
tan
24
1
2
2
?
?
?
??
? ? ?
??
??
??
?
?
?
a
a
a
a
a
Page 5
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
CBSE Board
Class XII Mathematics
Board Paper 2014 Solution
Delhi
SECTION – A
1. Given that
2
AA ? .
We need to find the value of
? ?
3
7A I A , where I is the identity matrix. ??
Thus,
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ?
? ?
3
3 2 2 3
3
3 2 2 3 2 2 2
3
2
3
3
3
7A I A 7A I 3I A 3IA A
7A I A 7A I 3A 3A A A I I,I A A,IA A
7A I A 7A I 3A 3A A A A
7A I A 7A I 3A 3A A
7A I A 7A I 7A
7A I A I
? ? ? ? ? ? ?
?? ? ? ? ? ? ? ? ? ? ? ? ?
??
?? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ?
2. Given that
x y z 1 4
2x y w 0 5
?? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
We need to find the value of x + y.
?? ? ? ? ?
?
? ? ? ?
?
? ? ? ?
?
? ? ?
??
ij ij
x y z 1 4
2x y w 0 5
Two matrices A and B are equal to each other, if they have the same dimensions
and the same elements a b , for i = 1,2,...,n and j = 1,2,...,m.
x y 1...(1)
2x y 0...(2)
Equa ?
? ? ?
??
tion (2) (1) is x = 1
Substituting the value of x = 1 in equation (1), we have
1 y 1
y2
Therefore, x + y = 1 + 2 = 3
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
3.
11
Given that tan x tan y and xy<1.
4
??
?
??
? ?
11
1
1
We need to find the value of x+y+xy.
tan x tan y
4
xy
tan xy 1
1 xy 4
xy
tan tan tan
1 xy 4
xy
1
1 xy
x y 1 xy
x y xy 1
??
?
?
?
??
?? ??
? ? ?
??
?
??
?? ???? ??
??
?? ?? ??
?
?? ?? ??
?
??
?
? ? ? ?
? ? ? ?
4. Given that
3x 7 8 7
2 4 6 4
?
?
.
We need to find the value of x
? ?
3x 7 8 7
2 4 6 4
12x 14 32 42
12x 14 10
12x 10 14
12x 24
x2
?
?
? ? ? ? ?
? ? ? ?
? ? ? ?
? ? ?
? ? ?
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
5. Since differentiation operation is the inverse operation of integration, we have
? ? sin ? ? f x x x
Let ? ?
0
sin ?
?
x
f x t tdt
Let us do this by integration by parts.
Therefore assume u = t; du = dt
sin
cos
?
??
??
tdt dv
tv
? ? ? ? ? ?
? ?
? ? ? ?
0
0
Therefore,
= t cos cos
cos sin
Differentiating the above function with respect to x,
f x sin cos cos sin
?? ? ? ?
??
? ? ? ?
??? ? ? ? ? ? ?
??
?
x
x
f x t t dt
f x x x x C
x x x x x x
6. Since the vectors are parallel, we have
? ?
ab
3i 2j 9k i 2pj 3k
3i 2j 9k i 2 pj 3 k
Comparing the respective coefficients, we have
3;
2 p 2
2 3 p 2
1
p
3
??
? ? ? ? ? ? ?
? ? ? ? ? ? ? ? ?
? ? ?
? ? ?
? ? ? ? ?
?
??
7. ? ? The set of natural numbers, N = 1, 2, 3, 4, 5, 6....
? ? ? ?
? ? ? ? ? ? ? ?
? ?
? ?
The relation is given as
R = x, y : 2 8
Thus, R = 6, 1 , 4, 2 , 2, 3
Domain = 6, 4, 2
Range = 1, 2, 3
?? xy
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
8. Given that the cartesian equation of the line as
? ? ? ? ? ?
? ?
3 4 2 6
5 7 4
That is,
3 4 2 3
5 7 4
4
33
5 7 2
? ? ?
??
? ? ? ? ?
??
??
??
? ? ? ?
?
?
x y z
x y z
y
xz
Any point on the line is of the form:
5 3,7 4,2 3
Thus, the vector equation is of the form:
r , where is the position vector of any
point on the line and b is the vector parallel to the lin
? ? ? ?
?? a b a
? ? ?
?
? ? ? ? ? ?
? ?
e.
Therefore, the vector equation is
r 5 3 7 4 2 3
r 5 7 2 3 4 3
r 3 4 3 5 7 2
? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
? ? ? ? ? ? ? ?
i j k
i j k i j k
i j k i j k
? ? ?
? ? ?
?
9.
a
2
0
dx
Given that
4+x 8
?
?
?
a
2
0
-1
0
We need to find the value of a.
dx
Let I=
4+x 8
1
Thus, I= tan
2 2 8
?
??
?
??
??
?
?
?
a
x
1
1
1
1
tan
2 2 8
tan 2
28
tan
24
1
2
2
?
?
?
??
? ? ?
??
??
??
?
?
?
a
a
a
a
a
CBSE XII | Mathematics
Board Paper 2014 – Delhi Set 3 Solution
10. Given that a and b are two perpendicular vectors.
Thus, a b 0
Also given that, a b 13 and a =5.
We need to find the value of b.
??
??
2
2 2 2
2
22
2
2
2
Consider a b :
a b = a 2 a b b
13 5 2 0 b
169 25 b
b 169 25
b 144
b 12
?
? ? ? ?
? ? ? ?
??
??
?
?
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