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Page 1
CBSE XII | Mathematics
Board Paper 2016 – Solution
CBSE Board
Class XII Mathematics
Board Paper – 2016 Solution
SECTION – A
1. Consider the given matrix
T
2
cos sin
A0
2 sin cos
A A 2 I
cos sin cos sin 1 0
2
sin cos sin cos 0 1
2cos 0 2 0
0 2cos
02
2cos 2
21
cos
2
2
4
?? ?? ?
? ? ? ?
??
? ? ?
??
??
? ? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
??
? ??
???
??
?
????
??
??
? ? ?
?
??
2. |3 A | = k |A|
|3 A | = 27|A|
k = 27
3. for unique solution |A| ? 0
2 2 1 3 3 1
1
1 1 1
2 1 -1 0
3 2 k
C C C ; C C C
1 0 0
2 -1 -3 0
3 -1 k-3
expansion along R
(k 3) 3 0
k 3 3 0
k0
?
? ? ? ?
?
? ? ? ?
? ? ? ?
?
Page 2
CBSE XII | Mathematics
Board Paper 2016 – Solution
CBSE Board
Class XII Mathematics
Board Paper – 2016 Solution
SECTION – A
1. Consider the given matrix
T
2
cos sin
A0
2 sin cos
A A 2 I
cos sin cos sin 1 0
2
sin cos sin cos 0 1
2cos 0 2 0
0 2cos
02
2cos 2
21
cos
2
2
4
?? ?? ?
? ? ? ?
??
? ? ?
??
??
? ? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
??
? ??
???
??
?
????
??
??
? ? ?
?
??
2. |3 A | = k |A|
|3 A | = 27|A|
k = 27
3. for unique solution |A| ? 0
2 2 1 3 3 1
1
1 1 1
2 1 -1 0
3 2 k
C C C ; C C C
1 0 0
2 -1 -3 0
3 -1 k-3
expansion along R
(k 3) 3 0
k 3 3 0
k0
?
? ? ? ?
?
? ? ? ?
? ? ? ?
?
CBSE XII | Mathematics
Board Paper 2016 – Solution
w
4. r (2i j k) 5 0
? ? ?
? ? ?
?
??
?
??
??
? ? ?
? ? ? ?
? ? ?
in Cartesian form
2x + y - z - 5=0
2x + y - z = 5
2x y z
1
5 5 5
x y z
1
5/2 5 5
5
Intercept cutt of on the axes ,5, 5
2
x y z
1
a b c
5
a b 5 c 5
2
a b c 5/2
5. (i 3j 9k) (3i j k) 0
n
n
i j k
1 3 9 0
3
i(3 9 ) j( 27) k( 9) 0
3 9 0 ...(1)
27 0 ...(2)
9 0 ...(3)
by eq (2) & (3) 27 and 9
, value satisfy the eq (1)
So 27, 9
6. a 4i j k, b 2i 2j k
a b (4i j k) (2i 2j k)
6i 3j 2k
ab
unit vector parallel to (a b)=
ab
6i 3j 2k
36 9 4
6i 3j 2k
49
6 3 2
i j k
7 7 7
Page 3
CBSE XII | Mathematics
Board Paper 2016 – Solution
CBSE Board
Class XII Mathematics
Board Paper – 2016 Solution
SECTION – A
1. Consider the given matrix
T
2
cos sin
A0
2 sin cos
A A 2 I
cos sin cos sin 1 0
2
sin cos sin cos 0 1
2cos 0 2 0
0 2cos
02
2cos 2
21
cos
2
2
4
?? ?? ?
? ? ? ?
??
? ? ?
??
??
? ? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
??
? ??
???
??
?
????
??
??
? ? ?
?
??
2. |3 A | = k |A|
|3 A | = 27|A|
k = 27
3. for unique solution |A| ? 0
2 2 1 3 3 1
1
1 1 1
2 1 -1 0
3 2 k
C C C ; C C C
1 0 0
2 -1 -3 0
3 -1 k-3
expansion along R
(k 3) 3 0
k 3 3 0
k0
?
? ? ? ?
?
? ? ? ?
? ? ? ?
?
CBSE XII | Mathematics
Board Paper 2016 – Solution
w
4. r (2i j k) 5 0
? ? ?
? ? ?
?
??
?
??
??
? ? ?
? ? ? ?
? ? ?
in Cartesian form
2x + y - z - 5=0
2x + y - z = 5
2x y z
1
5 5 5
x y z
1
5/2 5 5
5
Intercept cutt of on the axes ,5, 5
2
x y z
1
a b c
5
a b 5 c 5
2
a b c 5/2
5. (i 3j 9k) (3i j k) 0
n
n
i j k
1 3 9 0
3
i(3 9 ) j( 27) k( 9) 0
3 9 0 ...(1)
27 0 ...(2)
9 0 ...(3)
by eq (2) & (3) 27 and 9
, value satisfy the eq (1)
So 27, 9
6. a 4i j k, b 2i 2j k
a b (4i j k) (2i 2j k)
6i 3j 2k
ab
unit vector parallel to (a b)=
ab
6i 3j 2k
36 9 4
6i 3j 2k
49
6 3 2
i j k
7 7 7
CBSE XII | Mathematics
Board Paper 2016 – Solution
SECTION – B
7.
? ? ? ?
? ? ? ?
? ? ? ? ?
1 1 1 1
Given that tan x 1 tan x tan x 1 tan 3x
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
??
??
? ? ?
?
?
? ? ? ? ? ?
? ??
??
??
?
??
? ??
??
??
?
??
??
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??
??
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??
??
??
?
??
??
??
?
1 1 1 1
-1 1 1
-1 1 1
1 1 1
1
2
tan x 1 tan x 1 tan 3x tan x...(1)
AB
We know that, tan A tan B tan
1 AB
AB
and tan A tan B tan
1 AB
x 1 x 1
Thus, tan x 1 tan x 1 tan
1 x 1 x 1
2x
tan
1 x 1
tan
??
??
?
??
1
2
2x
....(2)
2x
? ?
1 1 1
1
2
11
22
22
22
22
2
2
3x x
Similarly,tan 3x tan x tan
1 3x x
2x
tan ....(3)
1 3x
From equations (1), (2) and (3), we have,
2x 2x
tan tan
2 x 1 3x
2x 2x
2 x 1 3x
11
2 x 1 3x
2 x 1 3x
4x 1
1
x
? ? ?
?
??
??
?
?? ??
??
?
??
??
?
??
?
??
? ? ? ?
?
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??
? ? ? ?
??
??
??
??
? ? ? ?
??
??
4
1
x
2
? ? ?
Page 4
CBSE XII | Mathematics
Board Paper 2016 – Solution
CBSE Board
Class XII Mathematics
Board Paper – 2016 Solution
SECTION – A
1. Consider the given matrix
T
2
cos sin
A0
2 sin cos
A A 2 I
cos sin cos sin 1 0
2
sin cos sin cos 0 1
2cos 0 2 0
0 2cos
02
2cos 2
21
cos
2
2
4
?? ?? ?
? ? ? ?
??
? ? ?
??
??
? ? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
??
? ??
???
??
?
????
??
??
? ? ?
?
??
2. |3 A | = k |A|
|3 A | = 27|A|
k = 27
3. for unique solution |A| ? 0
2 2 1 3 3 1
1
1 1 1
2 1 -1 0
3 2 k
C C C ; C C C
1 0 0
2 -1 -3 0
3 -1 k-3
expansion along R
(k 3) 3 0
k 3 3 0
k0
?
? ? ? ?
?
? ? ? ?
? ? ? ?
?
CBSE XII | Mathematics
Board Paper 2016 – Solution
w
4. r (2i j k) 5 0
? ? ?
? ? ?
?
??
?
??
??
? ? ?
? ? ? ?
? ? ?
in Cartesian form
2x + y - z - 5=0
2x + y - z = 5
2x y z
1
5 5 5
x y z
1
5/2 5 5
5
Intercept cutt of on the axes ,5, 5
2
x y z
1
a b c
5
a b 5 c 5
2
a b c 5/2
5. (i 3j 9k) (3i j k) 0
n
n
i j k
1 3 9 0
3
i(3 9 ) j( 27) k( 9) 0
3 9 0 ...(1)
27 0 ...(2)
9 0 ...(3)
by eq (2) & (3) 27 and 9
, value satisfy the eq (1)
So 27, 9
6. a 4i j k, b 2i 2j k
a b (4i j k) (2i 2j k)
6i 3j 2k
ab
unit vector parallel to (a b)=
ab
6i 3j 2k
36 9 4
6i 3j 2k
49
6 3 2
i j k
7 7 7
CBSE XII | Mathematics
Board Paper 2016 – Solution
SECTION – B
7.
? ? ? ?
? ? ? ?
? ? ? ? ?
1 1 1 1
Given that tan x 1 tan x tan x 1 tan 3x
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
??
??
? ? ?
?
?
? ? ? ? ? ?
? ??
??
??
?
??
? ??
??
??
?
??
??
? ? ?
? ? ? ?
??
??
? ? ?
??
??
??
?
??
??
??
?
1 1 1 1
-1 1 1
-1 1 1
1 1 1
1
2
tan x 1 tan x 1 tan 3x tan x...(1)
AB
We know that, tan A tan B tan
1 AB
AB
and tan A tan B tan
1 AB
x 1 x 1
Thus, tan x 1 tan x 1 tan
1 x 1 x 1
2x
tan
1 x 1
tan
??
??
?
??
1
2
2x
....(2)
2x
? ?
1 1 1
1
2
11
22
22
22
22
2
2
3x x
Similarly,tan 3x tan x tan
1 3x x
2x
tan ....(3)
1 3x
From equations (1), (2) and (3), we have,
2x 2x
tan tan
2 x 1 3x
2x 2x
2 x 1 3x
11
2 x 1 3x
2 x 1 3x
4x 1
1
x
? ? ?
?
??
??
?
?? ??
??
?
??
??
?
??
?
??
? ? ? ?
?
? ? ? ?
??
? ? ? ?
??
??
??
??
? ? ? ?
??
??
4
1
x
2
? ? ?
CBSE XII | Mathematics
Board Paper 2016 – Solution
OR
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?? ? ??
? ?
????
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??
??
?? ??
3
-1 1
22
-1 1 1
3
22
-1
3
22
Consider the left hand side
6x 8x 4x
L.H.S=tan tan
1 12x 1 4x
We know that,
AB
tan A tan B tan
1 AB
6x 8x 4x
1 12x 1 4x
Thus, L.H.S tan
6x 8x 4x
1
1 12x 1 4x
? ? ? ? ? ?
? ? ? ?
? ?
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??
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?
? ? ?
3 2 2
22
-1
3
22
3 2 2
22
-1
2 2 3
22
3 2 2
-1
22
6x 8x 1 4x 4x 1 12x
1 12x 1 4x
tan
4x 6x 8x
1
1 12x 1 4x
6x 8x 1 4x 4x 1 12x
1 12x 1 4x
tan
1 12x 1 4x 4x 6x 8x
1 12x 1 4x
6x 8x 1 4x 4x 1 12x
tan
1 12x 1 4x
? ?
??
??
??
?
??
?? ? ? ? ? ?
?
??
? ? ? ? ?
??
?? ??
?
??
??
??
3
3 3 5 3
-1
2 2 4 2 4
53
-1
42
4x 6x 8x
6x 24x 8x 32x 4x 48x
tan
1 4x 12x 48x 24x 32x
32x 16x 2x
tan
16x 8x 1
? ?
??
??
??
?
?? ??
??
?
42
-1
42
-1
2x 16x 8x 1
tan
16x 8x 1
tan 2x
Thus, L.H.S=R.H.S
Page 5
CBSE XII | Mathematics
Board Paper 2016 – Solution
CBSE Board
Class XII Mathematics
Board Paper – 2016 Solution
SECTION – A
1. Consider the given matrix
T
2
cos sin
A0
2 sin cos
A A 2 I
cos sin cos sin 1 0
2
sin cos sin cos 0 1
2cos 0 2 0
0 2cos
02
2cos 2
21
cos
2
2
4
?? ?? ?
? ? ? ?
??
? ? ?
??
??
? ? ? ? ? ? ? ? ? ? ?
??
? ? ? ? ? ?
? ? ? ? ?
? ? ? ? ? ?
??
? ??
???
??
?
????
??
??
? ? ?
?
??
2. |3 A | = k |A|
|3 A | = 27|A|
k = 27
3. for unique solution |A| ? 0
2 2 1 3 3 1
1
1 1 1
2 1 -1 0
3 2 k
C C C ; C C C
1 0 0
2 -1 -3 0
3 -1 k-3
expansion along R
(k 3) 3 0
k 3 3 0
k0
?
? ? ? ?
?
? ? ? ?
? ? ? ?
?
CBSE XII | Mathematics
Board Paper 2016 – Solution
w
4. r (2i j k) 5 0
? ? ?
? ? ?
?
??
?
??
??
? ? ?
? ? ? ?
? ? ?
in Cartesian form
2x + y - z - 5=0
2x + y - z = 5
2x y z
1
5 5 5
x y z
1
5/2 5 5
5
Intercept cutt of on the axes ,5, 5
2
x y z
1
a b c
5
a b 5 c 5
2
a b c 5/2
5. (i 3j 9k) (3i j k) 0
n
n
i j k
1 3 9 0
3
i(3 9 ) j( 27) k( 9) 0
3 9 0 ...(1)
27 0 ...(2)
9 0 ...(3)
by eq (2) & (3) 27 and 9
, value satisfy the eq (1)
So 27, 9
6. a 4i j k, b 2i 2j k
a b (4i j k) (2i 2j k)
6i 3j 2k
ab
unit vector parallel to (a b)=
ab
6i 3j 2k
36 9 4
6i 3j 2k
49
6 3 2
i j k
7 7 7
CBSE XII | Mathematics
Board Paper 2016 – Solution
SECTION – B
7.
? ? ? ?
? ? ? ?
? ? ? ? ?
1 1 1 1
Given that tan x 1 tan x tan x 1 tan 3x
? ? ? ?
? ? ? ?
? ? ? ?
? ?
? ? ? ?
??
??
? ? ?
?
?
? ? ? ? ? ?
? ??
??
??
?
??
? ??
??
??
?
??
??
? ? ?
? ? ? ?
??
??
? ? ?
??
??
??
?
??
??
??
?
1 1 1 1
-1 1 1
-1 1 1
1 1 1
1
2
tan x 1 tan x 1 tan 3x tan x...(1)
AB
We know that, tan A tan B tan
1 AB
AB
and tan A tan B tan
1 AB
x 1 x 1
Thus, tan x 1 tan x 1 tan
1 x 1 x 1
2x
tan
1 x 1
tan
??
??
?
??
1
2
2x
....(2)
2x
? ?
1 1 1
1
2
11
22
22
22
22
2
2
3x x
Similarly,tan 3x tan x tan
1 3x x
2x
tan ....(3)
1 3x
From equations (1), (2) and (3), we have,
2x 2x
tan tan
2 x 1 3x
2x 2x
2 x 1 3x
11
2 x 1 3x
2 x 1 3x
4x 1
1
x
? ? ?
?
??
??
?
?? ??
??
?
??
??
?
??
?
??
? ? ? ?
?
? ? ? ?
??
? ? ? ?
??
??
??
??
? ? ? ?
??
??
4
1
x
2
? ? ?
CBSE XII | Mathematics
Board Paper 2016 – Solution
OR
? ? ? ?
?
??
?? ? ??
?
??
??
??
??
??
? ??
??
??
?
??
??
?
??
?
? ??
?
?
?? ? ??
? ?
????
?
??
??
?? ??
3
-1 1
22
-1 1 1
3
22
-1
3
22
Consider the left hand side
6x 8x 4x
L.H.S=tan tan
1 12x 1 4x
We know that,
AB
tan A tan B tan
1 AB
6x 8x 4x
1 12x 1 4x
Thus, L.H.S tan
6x 8x 4x
1
1 12x 1 4x
? ? ? ? ? ?
? ? ? ?
? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ?
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??
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??
? ? ? ?
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?? ??
???
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??
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??
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??
? ? ? ?
?
? ? ?
3 2 2
22
-1
3
22
3 2 2
22
-1
2 2 3
22
3 2 2
-1
22
6x 8x 1 4x 4x 1 12x
1 12x 1 4x
tan
4x 6x 8x
1
1 12x 1 4x
6x 8x 1 4x 4x 1 12x
1 12x 1 4x
tan
1 12x 1 4x 4x 6x 8x
1 12x 1 4x
6x 8x 1 4x 4x 1 12x
tan
1 12x 1 4x
? ?
??
??
??
?
??
?? ? ? ? ? ?
?
??
? ? ? ? ?
??
?? ??
?
??
??
??
3
3 3 5 3
-1
2 2 4 2 4
53
-1
42
4x 6x 8x
6x 24x 8x 32x 4x 48x
tan
1 4x 12x 48x 24x 32x
32x 16x 2x
tan
16x 8x 1
? ?
??
??
??
?
?? ??
??
?
42
-1
42
-1
2x 16x 8x 1
tan
16x 8x 1
tan 2x
Thus, L.H.S=R.H.S
CBSE XII | Mathematics
Board Paper 2016 – Solution
8. Let charges for typing one English page be Rs. x.
Letcharges for typing one Hindi page be Rs.y.
Thus from the given statements, we have,
10x+3y=145
3x+10y=180
Thus the above system can be written as,
10 3 x 145
3 10 y 180
10 3
AX B, where, A=
3 10
? ? ? ? ? ?
?
? ? ? ? ? ?
? ? ? ? ? ?
??
-1
-1 -1
-1
-1
-1
x 145
,X and B=
y 180
Multiply A on both the sides, we have,
A AX A B
IX A B
X A B
Thus, we need to find the inverse of the matrix A.
a b d b 1
We know that, if P= then P
c d ad bc
? ? ? ? ? ?
?
? ? ? ? ? ?
? ? ? ? ? ?
??
??
??
? ??
?
??
?
??
-1
ca
10 3 1
Thus, A
10 10 3 3 3 10
10 3 1
100 9 3 10
10 3 1
91 3 10
10 3 145 1
Therefore,X
91 3 10 180
10 145 3 180 1
91 3 145 10 180
910 1
91 1365
10
15
x 10
y 15
??
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??
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??
x 10 and y=15
??
??
??
??
Amount taken from Shyam = 2 × 5 = Rs.10
Actual rate = 15 × 5 =75
Difference amount = Rs.75 – Rs.10 = Rs.65
Rs. 65 less was charged from the poor boy Shyam.
Humanity and sympathy are reflected in this problem.
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FAQs on Past Year Paper - Solutions, Mathematics (Set - 1), 2016, Class 12, Maths - Mathematics (Maths) Class 12 - JEE
| 1. What are the key topics covered in the Mathematics (Set - 1) exam for Class 12? |  |
Ans. The key topics covered in the Mathematics (Set - 1) exam for Class 12 include calculus, algebra, coordinate geometry, probability, and vectors.
| 2. How can I effectively prepare for the Mathematics (Set - 1) exam? | |
Ans. To effectively prepare for the Mathematics (Set - 1) exam, it is important to understand the concepts thoroughly. Practice solving a variety of problems from different topics and refer to the textbook or study materials for additional practice questions. It is also beneficial to solve past year papers and mock tests to familiarize yourself with the exam pattern and improve time management skills.
| 3. Are there any important formulas or theorems that I should focus on for the Mathematics (Set - 1) exam? | |
Ans. Yes, there are several important formulas and theorems that are frequently used in the Mathematics (Set - 1) exam. Some of them include the quadratic formula, Pythagorean theorem, properties of triangles and circles, laws of exponents, differentiation and integration formulas, and probability rules. It is essential to have a good grasp of these formulas and theorems to solve problems effectively.
| 4. How can I improve my problem-solving skills for the Mathematics (Set - 1) exam? | |
Ans. Improving problem-solving skills for the Mathematics (Set - 1) exam requires regular practice. Start by solving basic problems and gradually move on to more complex ones. Break down the problems into smaller steps and try to understand the underlying concepts. Analyze and learn from your mistakes to improve your approach. Additionally, seek help from teachers or classmates if you are stuck on a particular problem.
| 5. What are some common mistakes to avoid during the Mathematics (Set - 1) exam? | |
Ans. Some common mistakes to avoid during the Mathematics (Set - 1) exam include misinterpreting the question, skipping steps, not showing proper working, and using incorrect formulas or theorems. It is important to read the questions carefully and understand what is being asked. Show all the necessary steps and calculations to ensure clarity and accuracy. Double-check your answers before submitting the exam to avoid any careless errors.
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Nov 30, 2024 Last updated |
Document Description: Past Year Paper - Solutions, Mathematics (Set - 1), 2016, Class 12, Maths for JEE 2024 is part of Mathematics (Maths) Class 12 preparation.
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