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CBSE Past Year Paper Session (2016) Solutions, Math Class 12 | Mathematics (Maths) Class 12 - JEE PDF Download
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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
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
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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? ??
??
??
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??
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??
??
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??
??
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??
??
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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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?
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??
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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
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
? ? ?
?
??
??
?
?? ??
??
?
??
??
?
??
?
??
? ? ? ?
?
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??
? ? ? ?
??
??
??
??
? ? ? ?
??
??
4
1
x
2
? ? ?
CBSE XII | Mathematics
Board Paper 2016 – Solution
OR
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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
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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?
??
?? ? ??
?
??
??
??
??
??
? ??
??
??
?
??
??
?
??
?
? ??
?
?
?? ? ??
? ?
????
?
??
??
?? ??
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
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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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 CBSE Past Year Paper Session (2016) Solutions, Math Class 12 - Mathematics (Maths) Class 12 - JEE
| 1. What is the CBSE Past Year Paper Session? |  |
Ans. The CBSE Past Year Paper Session refers to a series of practice sessions where students solve past year question papers of CBSE exams. These sessions help students to get familiar with the exam pattern, understand the type of questions asked, and assess their preparation level.
| 2. Why is solving past year papers important for the Math Class 12 exam? | |
Ans. Solving past year papers for the Math Class 12 exam is crucial as it allows students to gain a clear understanding of the exam pattern, marking scheme, and the level of difficulty. It helps them identify their weak areas and work on them, build confidence, and be well-prepared for the actual exam.
| 3. How can I access the CBSE Past Year Paper Session for Math Class 12? | |
Ans. The CBSE Past Year Paper Session for Math Class 12 can be accessed through various platforms. One can find these papers on the official CBSE website, educational websites, or purchase books specifically designed for practicing past year papers. Additionally, some coaching institutes and online learning platforms also provide access to these papers.
| 4. Are the solutions provided for the CBSE Past Year Paper Session accurate and reliable? | |
Ans. Yes, the solutions provided for the CBSE Past Year Paper Session are generally accurate and reliable. However, it is recommended to cross-verify the solutions with the help of subject teachers or experts. Sometimes, there may be different approaches or methods to solve a particular question, and it is essential to understand the concepts rather than just relying on the answers.
| 5. Can practicing past year papers guarantee good marks in the Math Class 12 exam? | |
Ans. Practicing past year papers alone cannot guarantee good marks in the Math Class 12 exam. While solving these papers is an essential part of the preparation process, it is equally important to understand the concepts, practice regularly, and revise the entire syllabus. Past year papers serve as a tool to evaluate preparation and identify areas that need improvement, but overall preparation and understanding of the subject are crucial for scoring well in the exam.
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Dec 22, 2024 Last updated |
Document Description: CBSE Past Year Paper Session (2016) Solutions, Math Class 12 for JEE 2024 is part of Mathematics (Maths) Class 12 preparation.
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Full syllabus notes, lecture & questions for CBSE Past Year Paper Session (2016) Solutions, Math Class 12 | Mathematics (Maths) Class 12 - JEE - JEE | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 12 | Best notes, free PDF download
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Additional Information about CBSE Past Year Paper Session (2016) Solutions, Math Class 12 for JEE Preparation
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