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Past Year Paper - Solutions (Comptt.), Mathematics (Set - 1, 2 and 3), Delhi, 2015, Class 12, Maths | Additional Study Material for JEE PDF Download

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1
Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
July — 2015 (Comptt.)
Marking Scheme — Mathematics (Delhi) 65/1/1, 65/1/2, 65/1/3
General Instructions:
1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking.
The answers given in the Marking Scheme are suggested answers. The content is thus
indicative. If a student has given any other answer which is different from the one given
in the Marking Scheme, but conveys the meaning, such answers should be given full
weightage.
2. Evaluation is to be done as per instructions provided in the marking scheme. It should
not be done according to one’s own interpretation or any other consideration — Marking
Scheme should be strictly adhered to and religiously followed.
3. Alternative methods are accepted. Proportional marks are to be awarded.
4. In question (s) on differential equations, constant of integration has to be written.
5. If a candidate has attempted an extra question, marks obtained in the question attempted
first should be retained and the other answer should be scored out.
6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full
marks if the answer deserves it.
7. Separate Marking Scheme for all the three sets has been given.
8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted
to obtain photocopy of the Answer book on request on payment of the prescribed fee.
All examiners/Head Examiners are once again reminded that they must ensure that
evaluation is carried out strictly as per value points for each answer as given in the
Marking Scheme.
Page 2


1
Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
July — 2015 (Comptt.)
Marking Scheme — Mathematics (Delhi) 65/1/1, 65/1/2, 65/1/3
General Instructions:
1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking.
The answers given in the Marking Scheme are suggested answers. The content is thus
indicative. If a student has given any other answer which is different from the one given
in the Marking Scheme, but conveys the meaning, such answers should be given full
weightage.
2. Evaluation is to be done as per instructions provided in the marking scheme. It should
not be done according to one’s own interpretation or any other consideration — Marking
Scheme should be strictly adhered to and religiously followed.
3. Alternative methods are accepted. Proportional marks are to be awarded.
4. In question (s) on differential equations, constant of integration has to be written.
5. If a candidate has attempted an extra question, marks obtained in the question attempted
first should be retained and the other answer should be scored out.
6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full
marks if the answer deserves it.
7. Separate Marking Scheme for all the three sets has been given.
8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted
to obtain photocopy of the Answer book on request on payment of the prescribed fee.
All examiners/Head Examiners are once again reminded that they must ensure that
evaluation is carried out strictly as per value points for each answer as given in the
Marking Scheme.
2
QUESTION PAPER CODE 65/1/1
EXPECTED ANSWER/V ALUE POINTS
SECTION A
Marks
1.
1 2 7 14
a i j then 7a i j
5 5 5 5
= - = -
? ? ? ? ? ?
½ + ½
2.
( 2a b) ( 2a b) 1
4
p
- · - = ? ? =
   
½ + ½
3.
2 2 2 2 2 2
cos cos cos 1 sin sin sin 2 a + ß + ? = ? a + ß + ? =
½ + ½
4.
1 5
AB | AB | 28
4 8
- ? ?
= ? = -
? ?
? ?
½ + ½
5.
dy dy
1 y x csin x x y xy tan x 0
dx dx
· + = - ? + + = ½ + ½
6. order = 2, degree = 3, sum = 2 + 3 = 5 ½ + ½
SECTION B
7. System of equation is
3x + y + 2z = 1100, x + 2y + 3z = 1400, x + y + z = 600 1½
(i) Matrix equation is
3 1 2 x 1100
1 2 3 y 1400
1 1 1 z 600
? ? ? ? ? ?
? ? ? ? ? ?
=
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
1
(ii) |A| = –3 ? 0, system of equations can be solved. ½
(iii) Any one value with reason. 1
Page 3


1
Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
July — 2015 (Comptt.)
Marking Scheme — Mathematics (Delhi) 65/1/1, 65/1/2, 65/1/3
General Instructions:
1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking.
The answers given in the Marking Scheme are suggested answers. The content is thus
indicative. If a student has given any other answer which is different from the one given
in the Marking Scheme, but conveys the meaning, such answers should be given full
weightage.
2. Evaluation is to be done as per instructions provided in the marking scheme. It should
not be done according to one’s own interpretation or any other consideration — Marking
Scheme should be strictly adhered to and religiously followed.
3. Alternative methods are accepted. Proportional marks are to be awarded.
4. In question (s) on differential equations, constant of integration has to be written.
5. If a candidate has attempted an extra question, marks obtained in the question attempted
first should be retained and the other answer should be scored out.
6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full
marks if the answer deserves it.
7. Separate Marking Scheme for all the three sets has been given.
8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted
to obtain photocopy of the Answer book on request on payment of the prescribed fee.
All examiners/Head Examiners are once again reminded that they must ensure that
evaluation is carried out strictly as per value points for each answer as given in the
Marking Scheme.
2
QUESTION PAPER CODE 65/1/1
EXPECTED ANSWER/V ALUE POINTS
SECTION A
Marks
1.
1 2 7 14
a i j then 7a i j
5 5 5 5
= - = -
? ? ? ? ? ?
½ + ½
2.
( 2a b) ( 2a b) 1
4
p
- · - = ? ? =
   
½ + ½
3.
2 2 2 2 2 2
cos cos cos 1 sin sin sin 2 a + ß + ? = ? a + ß + ? =
½ + ½
4.
1 5
AB | AB | 28
4 8
- ? ?
= ? = -
? ?
? ?
½ + ½
5.
dy dy
1 y x csin x x y xy tan x 0
dx dx
· + = - ? + + = ½ + ½
6. order = 2, degree = 3, sum = 2 + 3 = 5 ½ + ½
SECTION B
7. System of equation is
3x + y + 2z = 1100, x + 2y + 3z = 1400, x + y + z = 600 1½
(i) Matrix equation is
3 1 2 x 1100
1 2 3 y 1400
1 1 1 z 600
? ? ? ? ? ?
? ? ? ? ? ?
=
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
1
(ii) |A| = –3 ? 0, system of equations can be solved. ½
(iii) Any one value with reason. 1
3
8. [ ]
1 2 x
2x 3 0
3 0 3
? ? ? ?
=
? ? ? ?
-
? ? ? ?
[ ]
x
2x 9 4x
3
? ?
-
? ?
? ?
 = 0 1
2
[2x 9x 12x] - +
 = [0] ? 
2
3
2x 3x 0, x 0 or
2
-
+ = =
1+1+1
9.
(a 1) (a 2) a 2 1
(a 2) (a 3) a 3 1
(a 3) (a 4) a 4 1
+ + +
+ + +
+ + +
= 
2 2 1
3 3 1
(a 1) (a 2) a 2 1
R R R
2(a 2) 1 0
R R R
4a 10 2 0
+ + +
? -
+
? -
+
1+1
= 4a + 8 – 4a – 10 = –2. 1+1
10. I = 
/ 2
0
1
dx
1 tan x
p
+
?
I = 
/ 2 / 2 / 2
0 0 0
1 1 tan x
dx dx dx
1 tan( / 2 x) 1 cot x 1 tan x
p p p
= =
+ p - + +
? ? ? 1½
? 2I = 
/ 2 / 2
0 0
1 tan x
dx 1 dx
2 1 tan x
p p
+ p
= · =
+
? ?
1½
? I = 
4
p
1
11.
2 2
x A B C
x 1 x 2
(x 1) (x 2) (x 1)
= + +
- +
- + -
½
A = 
2 1 2
, B ,C
9 3 9
= = -
1½
?
?
?
Page 4


1
Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
July — 2015 (Comptt.)
Marking Scheme — Mathematics (Delhi) 65/1/1, 65/1/2, 65/1/3
General Instructions:
1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking.
The answers given in the Marking Scheme are suggested answers. The content is thus
indicative. If a student has given any other answer which is different from the one given
in the Marking Scheme, but conveys the meaning, such answers should be given full
weightage.
2. Evaluation is to be done as per instructions provided in the marking scheme. It should
not be done according to one’s own interpretation or any other consideration — Marking
Scheme should be strictly adhered to and religiously followed.
3. Alternative methods are accepted. Proportional marks are to be awarded.
4. In question (s) on differential equations, constant of integration has to be written.
5. If a candidate has attempted an extra question, marks obtained in the question attempted
first should be retained and the other answer should be scored out.
6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full
marks if the answer deserves it.
7. Separate Marking Scheme for all the three sets has been given.
8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted
to obtain photocopy of the Answer book on request on payment of the prescribed fee.
All examiners/Head Examiners are once again reminded that they must ensure that
evaluation is carried out strictly as per value points for each answer as given in the
Marking Scheme.
2
QUESTION PAPER CODE 65/1/1
EXPECTED ANSWER/V ALUE POINTS
SECTION A
Marks
1.
1 2 7 14
a i j then 7a i j
5 5 5 5
= - = -
? ? ? ? ? ?
½ + ½
2.
( 2a b) ( 2a b) 1
4
p
- · - = ? ? =
   
½ + ½
3.
2 2 2 2 2 2
cos cos cos 1 sin sin sin 2 a + ß + ? = ? a + ß + ? =
½ + ½
4.
1 5
AB | AB | 28
4 8
- ? ?
= ? = -
? ?
? ?
½ + ½
5.
dy dy
1 y x csin x x y xy tan x 0
dx dx
· + = - ? + + = ½ + ½
6. order = 2, degree = 3, sum = 2 + 3 = 5 ½ + ½
SECTION B
7. System of equation is
3x + y + 2z = 1100, x + 2y + 3z = 1400, x + y + z = 600 1½
(i) Matrix equation is
3 1 2 x 1100
1 2 3 y 1400
1 1 1 z 600
? ? ? ? ? ?
? ? ? ? ? ?
=
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
1
(ii) |A| = –3 ? 0, system of equations can be solved. ½
(iii) Any one value with reason. 1
3
8. [ ]
1 2 x
2x 3 0
3 0 3
? ? ? ?
=
? ? ? ?
-
? ? ? ?
[ ]
x
2x 9 4x
3
? ?
-
? ?
? ?
 = 0 1
2
[2x 9x 12x] - +
 = [0] ? 
2
3
2x 3x 0, x 0 or
2
-
+ = =
1+1+1
9.
(a 1) (a 2) a 2 1
(a 2) (a 3) a 3 1
(a 3) (a 4) a 4 1
+ + +
+ + +
+ + +
= 
2 2 1
3 3 1
(a 1) (a 2) a 2 1
R R R
2(a 2) 1 0
R R R
4a 10 2 0
+ + +
? -
+
? -
+
1+1
= 4a + 8 – 4a – 10 = –2. 1+1
10. I = 
/ 2
0
1
dx
1 tan x
p
+
?
I = 
/ 2 / 2 / 2
0 0 0
1 1 tan x
dx dx dx
1 tan( / 2 x) 1 cot x 1 tan x
p p p
= =
+ p - + +
? ? ? 1½
? 2I = 
/ 2 / 2
0 0
1 tan x
dx 1 dx
2 1 tan x
p p
+ p
= · =
+
? ?
1½
? I = 
4
p
1
11.
2 2
x A B C
x 1 x 2
(x 1) (x 2) (x 1)
= + +
- +
- + -
½
A = 
2 1 2
, B ,C
9 3 9
= = -
1½
?
?
?
4
2
x
dx
(x 1) (x 2) - +
?
 = 
2
2 1 2
dx dx dx
9(x 1) 9(x 2)
3(x 1)
+ -
- +
-
? ? ?
1½
= 
2 1 2
log | x 1| log | x 2 | C
9 3(x 1) 9
- - - + +
-
1½
12. Let X be the number of defective bulbs. Then
X = 0, 1, 2 1
P(X = 0) = 
2
2
C
C
10
3
15 7
=
, 
1 1
2
C C
C
10 5
10
P(X 1)
15 21
·
= = =
1+1
P(X = 2) = 
2
2
C
C
2
15 21
5
=
1
X 0 1 2
P(X)
3
7
10
21
2
21
OR
E
1
: Problem is solved by A.
E
2
: Problem is solved by B.
P(E
1
) = 
2 1 2
1 1 1 2
, P(E ) , P(E ) , P(E )
2 3 2 3
= = = 1
1 2 1 2
1
P(E E ) P(E ) P(E )
6
n = · =
P(problem is solved) = 
1 2
1 P(E ) P(E ) - ·
 = 
1 2 2
1
2 3 3
- · =
1½
P(one of them is solved) = 
1 2 1 2
P(E )P(E ) P(E )P(E ) +
= 
1 2 1 1 1
2 3 2 3 2
· + · =
1½
Page 5


1
Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
July — 2015 (Comptt.)
Marking Scheme — Mathematics (Delhi) 65/1/1, 65/1/2, 65/1/3
General Instructions:
1. The Marking Scheme provides general guidelines to reduce subjectivity in the marking.
The answers given in the Marking Scheme are suggested answers. The content is thus
indicative. If a student has given any other answer which is different from the one given
in the Marking Scheme, but conveys the meaning, such answers should be given full
weightage.
2. Evaluation is to be done as per instructions provided in the marking scheme. It should
not be done according to one’s own interpretation or any other consideration — Marking
Scheme should be strictly adhered to and religiously followed.
3. Alternative methods are accepted. Proportional marks are to be awarded.
4. In question (s) on differential equations, constant of integration has to be written.
5. If a candidate has attempted an extra question, marks obtained in the question attempted
first should be retained and the other answer should be scored out.
6. A full scale of marks - 0 to 100 has to be used. Please do not hesitate to award full
marks if the answer deserves it.
7. Separate Marking Scheme for all the three sets has been given.
8. As per orders of the Hon’ble Supreme Court. The candidates would now be permitted
to obtain photocopy of the Answer book on request on payment of the prescribed fee.
All examiners/Head Examiners are once again reminded that they must ensure that
evaluation is carried out strictly as per value points for each answer as given in the
Marking Scheme.
2
QUESTION PAPER CODE 65/1/1
EXPECTED ANSWER/V ALUE POINTS
SECTION A
Marks
1.
1 2 7 14
a i j then 7a i j
5 5 5 5
= - = -
? ? ? ? ? ?
½ + ½
2.
( 2a b) ( 2a b) 1
4
p
- · - = ? ? =
   
½ + ½
3.
2 2 2 2 2 2
cos cos cos 1 sin sin sin 2 a + ß + ? = ? a + ß + ? =
½ + ½
4.
1 5
AB | AB | 28
4 8
- ? ?
= ? = -
? ?
? ?
½ + ½
5.
dy dy
1 y x csin x x y xy tan x 0
dx dx
· + = - ? + + = ½ + ½
6. order = 2, degree = 3, sum = 2 + 3 = 5 ½ + ½
SECTION B
7. System of equation is
3x + y + 2z = 1100, x + 2y + 3z = 1400, x + y + z = 600 1½
(i) Matrix equation is
3 1 2 x 1100
1 2 3 y 1400
1 1 1 z 600
? ? ? ? ? ?
? ? ? ? ? ?
=
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
1
(ii) |A| = –3 ? 0, system of equations can be solved. ½
(iii) Any one value with reason. 1
3
8. [ ]
1 2 x
2x 3 0
3 0 3
? ? ? ?
=
? ? ? ?
-
? ? ? ?
[ ]
x
2x 9 4x
3
? ?
-
? ?
? ?
 = 0 1
2
[2x 9x 12x] - +
 = [0] ? 
2
3
2x 3x 0, x 0 or
2
-
+ = =
1+1+1
9.
(a 1) (a 2) a 2 1
(a 2) (a 3) a 3 1
(a 3) (a 4) a 4 1
+ + +
+ + +
+ + +
= 
2 2 1
3 3 1
(a 1) (a 2) a 2 1
R R R
2(a 2) 1 0
R R R
4a 10 2 0
+ + +
? -
+
? -
+
1+1
= 4a + 8 – 4a – 10 = –2. 1+1
10. I = 
/ 2
0
1
dx
1 tan x
p
+
?
I = 
/ 2 / 2 / 2
0 0 0
1 1 tan x
dx dx dx
1 tan( / 2 x) 1 cot x 1 tan x
p p p
= =
+ p - + +
? ? ? 1½
? 2I = 
/ 2 / 2
0 0
1 tan x
dx 1 dx
2 1 tan x
p p
+ p
= · =
+
? ?
1½
? I = 
4
p
1
11.
2 2
x A B C
x 1 x 2
(x 1) (x 2) (x 1)
= + +
- +
- + -
½
A = 
2 1 2
, B ,C
9 3 9
= = -
1½
?
?
?
4
2
x
dx
(x 1) (x 2) - +
?
 = 
2
2 1 2
dx dx dx
9(x 1) 9(x 2)
3(x 1)
+ -
- +
-
? ? ?
1½
= 
2 1 2
log | x 1| log | x 2 | C
9 3(x 1) 9
- - - + +
-
1½
12. Let X be the number of defective bulbs. Then
X = 0, 1, 2 1
P(X = 0) = 
2
2
C
C
10
3
15 7
=
, 
1 1
2
C C
C
10 5
10
P(X 1)
15 21
·
= = =
1+1
P(X = 2) = 
2
2
C
C
2
15 21
5
=
1
X 0 1 2
P(X)
3
7
10
21
2
21
OR
E
1
: Problem is solved by A.
E
2
: Problem is solved by B.
P(E
1
) = 
2 1 2
1 1 1 2
, P(E ) , P(E ) , P(E )
2 3 2 3
= = = 1
1 2 1 2
1
P(E E ) P(E ) P(E )
6
n = · =
P(problem is solved) = 
1 2
1 P(E ) P(E ) - ·
 = 
1 2 2
1
2 3 3
- · =
1½
P(one of them is solved) = 
1 2 1 2
P(E )P(E ) P(E )P(E ) +
= 
1 2 1 1 1
2 3 2 3 2
· + · =
1½
5
13.
AB
uuu r
 = 
$
4i 6j 2k - - -
$ $
AC
uuu r
 = 
$
i ( 5) j 3k - + ? - +
$ $
1½
AD
uuur
 = 
$
8i – j 3k - +
$ $
AB (AC AD) · ×
uuu r uuu r uuur
 = 
4 6 2
1 5 3 0
8 1 3
- - -
- ? - =
- -
1
4(3 12) 6(21) 2(8 39) - ? - + - ? - = 0 ? ? = 9 1½
14.
1
a
r
 = 
1
ˆ ˆ
i 2 j k, b i j k + + = - +
r
$ $ $ $
1 1
2
a
r
 = 
$ $
2
2i – j – k, b 2i j 2k = + +
r
$ $ $ $
2 1
a – a
r r
 = 
$
$
1 2
i j k
i – 3j – 2k, b b 1 1 1
2 1 1
× = -
$ $
r r
$ $
 = 
ˆ ˆ
3i 3k - + ½ + 1
1 2
| b b | ×
r r
 = 3 2 ½
2 1 1 2
(a – a ) (b – b ) ·
r r
r r
 = 3 6 9 - - = -
Shortest distance = 
9 3 2
2 3 2
-
=
1
15.
1 1
tan 2x tan 3x
- -
+
 = 
4
p
1
2
5x
tan
1 6x
-
? ?
? ?
- ? ?
 = 
4
p
1
2
5x
1 6x -
 = 1 1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
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FAQs on Past Year Paper - Solutions (Comptt.), Mathematics (Set - 1, 2 and 3), Delhi, 2015, Class 12, Maths - Additional Study Material for JEE

1. What is the format of the Mathematics (Set - 1, 2 and 3) exam in Delhi, 2015 for Class 12 Maths JEE?
Ans. The Mathematics exam in Delhi, 2015 for Class 12 Maths JEE had three sets - Set 1, Set 2, and Set 3. Each set had a different question paper with varying difficulty levels.
2. What are the topics covered in the Mathematics exam for Class 12 Maths JEE in Delhi, 2015?
Ans. The Mathematics exam for Class 12 Maths JEE in Delhi, 2015 covered various topics such as algebra, calculus, trigonometry, coordinate geometry, and statistics.
3. How can I access the past year papers and solutions for the Mathematics exam of Delhi, 2015 Class 12 Maths JEE?
Ans. You can access the past year papers and solutions for the Mathematics exam of Delhi, 2015 Class 12 Maths JEE by searching for them online. There are various websites and educational platforms that provide these papers and solutions for free or for a nominal fee.
4. Are the past year papers and solutions for the Mathematics exam of Delhi, 2015 Class 12 Maths JEE helpful for exam preparation?
Ans. Yes, the past year papers and solutions for the Mathematics exam of Delhi, 2015 Class 12 Maths JEE are extremely helpful for exam preparation. They give you an idea of the exam pattern, types of questions asked, and the level of difficulty. Practicing these papers can greatly improve your problem-solving skills and help you understand the concepts better.
5. Can I rely solely on the past year papers and solutions for the Mathematics exam of Delhi, 2015 Class 12 Maths JEE for my exam preparation?
Ans. While the past year papers and solutions are a valuable resource for exam preparation, it is not advisable to rely solely on them. It is important to study the textbook, understand the concepts, and practice a variety of questions from different sources. This will give you a comprehensive understanding of the subject and better prepare you for the exam.
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