Past Year Paper - Solutions, Mathematics (Set - 1, 2 and 3), Outside Delhi, 2016, Class 12, Maths | Mathematics (Maths) Class 12 - JEE PDF Download

 Page 1


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/C, 65/2/C, 65/3/C
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


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/C, 65/2/C, 65/3/C
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.
QUESTION PAPER CODE 65/1/C
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. (x + 3)2x – (–2) (–3x) = 8
1
2
x = 2
1
2
2.
2 5
2 4
? ?
? ?
? ?
 = 
3 1 1 2
2 0 1 1
? ?? ?
? ?? ?
- -
? ?? ?
 1
2
 + 
1
2
3. No. of possible matrices = 3
4
1
or 81
4.
2 (2a 3b) 1(3a – 2b)
2 1
+ +
+

 


 

          1
2
= 
7 4
a b
3 3
+




       (or enternal division may also be considered)
1
2
5. 2 1
6.
x y z
3 4 2
+ +
-
     = 1
1
2
?
ˆ ˆ ˆ
r (4i 3j 6k) · - +


     = 12   or  
ˆ ˆ ˆ
i j k
r 1
3 4 2
? ?
· - + =
? ?
? ?
? ?


1
2
SECTION B
7. Equation of line through A(3, 4, 1) and B(5, 1, 6)
x 3
2
-
   = 
y 4 z 1
k(say)
3 5
- -
= =
-
        1
General point on the line:
x = 2k + 3, y = – 3k + 4, z = 5k + 1
1
2
line crosses xz plane i.e. y = 0 if –3k + 4 = 0
? k = 
4
3
1
Co-ordinate of required point 
17 23
, 0,
3 3
? ?
? ?
? ?
  1
2
Angle, which line makes with xz plane:
sin ? = 
2 (0) ( 3) (1) 5(0) 3
4 9 25 1 38
+ - +
=
+ +
              ? ? = 
1
3
sin
38
-
? ?
? ?
? ?
 1
?
?
?
?
?
65/1/C (1)
65/1/C
Page 3


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/C, 65/2/C, 65/3/C
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.
QUESTION PAPER CODE 65/1/C
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. (x + 3)2x – (–2) (–3x) = 8
1
2
x = 2
1
2
2.
2 5
2 4
? ?
? ?
? ?
 = 
3 1 1 2
2 0 1 1
? ?? ?
? ?? ?
- -
? ?? ?
 1
2
 + 
1
2
3. No. of possible matrices = 3
4
1
or 81
4.
2 (2a 3b) 1(3a – 2b)
2 1
+ +
+

 


 

          1
2
= 
7 4
a b
3 3
+




       (or enternal division may also be considered)
1
2
5. 2 1
6.
x y z
3 4 2
+ +
-
     = 1
1
2
?
ˆ ˆ ˆ
r (4i 3j 6k) · - +


     = 12   or  
ˆ ˆ ˆ
i j k
r 1
3 4 2
? ?
· - + =
? ?
? ?
? ?


1
2
SECTION B
7. Equation of line through A(3, 4, 1) and B(5, 1, 6)
x 3
2
-
   = 
y 4 z 1
k(say)
3 5
- -
= =
-
        1
General point on the line:
x = 2k + 3, y = – 3k + 4, z = 5k + 1
1
2
line crosses xz plane i.e. y = 0 if –3k + 4 = 0
? k = 
4
3
1
Co-ordinate of required point 
17 23
, 0,
3 3
? ?
? ?
? ?
  1
2
Angle, which line makes with xz plane:
sin ? = 
2 (0) ( 3) (1) 5(0) 3
4 9 25 1 38
+ - +
=
+ +
              ? ? = 
1
3
sin
38
-
? ?
? ?
? ?
 1
?
?
?
?
?
65/1/C (1)
65/1/C
8. let 
1
d


 & 
2
d


 be the two diagonal vectors:
?
1
d


 = 
ˆ ˆ ˆ
4i 2j 2k - -
    , 
2
d


 = 
ˆ ˆ
6j 8k - -
  1 1
2 2
+
  or
2
ˆ ˆ
d 6j 8k = +


Unit vectors parallel to the diagonals are:
1
ˆ
d = 
2 1 1
ˆ ˆ ˆ
i j k
6 6 6
- -
       1
2
2
ˆ
d
 = 
3 4
ˆ ˆ
j k
5 5
- -
        2
3 4
ˆ ˆ ˆ
or d j k
5 5
? ?
= +
? ?
? ?
1
2
1 2
d d ×

 

   = 
ˆ ˆ ˆ
i j k
ˆ ˆ ˆ
4i 32j 24k
4 2 2
0 6 8
= + -
- -
- -
      1
Area of parallelogram = 
1 2
1
404 or 2 101 sq. units
d d
2
=
×

 

       1
9. let X = Amount he wins then x = ` 5, 4, 3, – 3 1
P = Probability of getting a no. >4 = 
1
3
, q = 
2
1 p
3
- =
    1
2
X: 5 4 3 –3
P(x)
1
3
2 1
3 3
·
 = 
2
9
2
1 2
3 3
? ?
·
? ?
? ?
  = 
4
27
3
2
3
? ?
? ?
? ?
 = 
8
27
2
Expected amount he wins = 
5 8 12 24
XP(X)
3 9 27 27
S = + + -
        = 
19 1
or 2
9 9
    ` `
1
2
OR
E
1
 = Event that all balls are white,
E
2
 = Event that 3 balls are white and 1 ball is non white
E
3
 = Event that 2 balls are white and 2 balls are non-white
1
A = Event that 2 balls drawn without replacement are white
P(E
1
) = P(E
2
) = P(E
3
) = 
1
3
1
2
P(A/E
1
) = 1, P(A/E
2
) = 
3 2 1
4 3 2
· =
  , P(A/E
3
) = 
2 1 1
4 3 6
· =
  1
1
2
P(E
1
/A) = 
1
3
1 1 1 1 1
3 3 2 3 6
1. 3
1 5
=
· + · + ·
        1
?
?
?
?
?
?
?
65/1/C (2)
65/1/C
Page 4


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/C, 65/2/C, 65/3/C
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.
QUESTION PAPER CODE 65/1/C
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. (x + 3)2x – (–2) (–3x) = 8
1
2
x = 2
1
2
2.
2 5
2 4
? ?
? ?
? ?
 = 
3 1 1 2
2 0 1 1
? ?? ?
? ?? ?
- -
? ?? ?
 1
2
 + 
1
2
3. No. of possible matrices = 3
4
1
or 81
4.
2 (2a 3b) 1(3a – 2b)
2 1
+ +
+

 


 

          1
2
= 
7 4
a b
3 3
+




       (or enternal division may also be considered)
1
2
5. 2 1
6.
x y z
3 4 2
+ +
-
     = 1
1
2
?
ˆ ˆ ˆ
r (4i 3j 6k) · - +


     = 12   or  
ˆ ˆ ˆ
i j k
r 1
3 4 2
? ?
· - + =
? ?
? ?
? ?


1
2
SECTION B
7. Equation of line through A(3, 4, 1) and B(5, 1, 6)
x 3
2
-
   = 
y 4 z 1
k(say)
3 5
- -
= =
-
        1
General point on the line:
x = 2k + 3, y = – 3k + 4, z = 5k + 1
1
2
line crosses xz plane i.e. y = 0 if –3k + 4 = 0
? k = 
4
3
1
Co-ordinate of required point 
17 23
, 0,
3 3
? ?
? ?
? ?
  1
2
Angle, which line makes with xz plane:
sin ? = 
2 (0) ( 3) (1) 5(0) 3
4 9 25 1 38
+ - +
=
+ +
              ? ? = 
1
3
sin
38
-
? ?
? ?
? ?
 1
?
?
?
?
?
65/1/C (1)
65/1/C
8. let 
1
d


 & 
2
d


 be the two diagonal vectors:
?
1
d


 = 
ˆ ˆ ˆ
4i 2j 2k - -
    , 
2
d


 = 
ˆ ˆ
6j 8k - -
  1 1
2 2
+
  or
2
ˆ ˆ
d 6j 8k = +


Unit vectors parallel to the diagonals are:
1
ˆ
d = 
2 1 1
ˆ ˆ ˆ
i j k
6 6 6
- -
       1
2
2
ˆ
d
 = 
3 4
ˆ ˆ
j k
5 5
- -
        2
3 4
ˆ ˆ ˆ
or d j k
5 5
? ?
= +
? ?
? ?
1
2
1 2
d d ×

 

   = 
ˆ ˆ ˆ
i j k
ˆ ˆ ˆ
4i 32j 24k
4 2 2
0 6 8
= + -
- -
- -
      1
Area of parallelogram = 
1 2
1
404 or 2 101 sq. units
d d
2
=
×

 

       1
9. let X = Amount he wins then x = ` 5, 4, 3, – 3 1
P = Probability of getting a no. >4 = 
1
3
, q = 
2
1 p
3
- =
    1
2
X: 5 4 3 –3
P(x)
1
3
2 1
3 3
·
 = 
2
9
2
1 2
3 3
? ?
·
? ?
? ?
  = 
4
27
3
2
3
? ?
? ?
? ?
 = 
8
27
2
Expected amount he wins = 
5 8 12 24
XP(X)
3 9 27 27
S = + + -
        = 
19 1
or 2
9 9
    ` `
1
2
OR
E
1
 = Event that all balls are white,
E
2
 = Event that 3 balls are white and 1 ball is non white
E
3
 = Event that 2 balls are white and 2 balls are non-white
1
A = Event that 2 balls drawn without replacement are white
P(E
1
) = P(E
2
) = P(E
3
) = 
1
3
1
2
P(A/E
1
) = 1, P(A/E
2
) = 
3 2 1
4 3 2
· =
  , P(A/E
3
) = 
2 1 1
4 3 6
· =
  1
1
2
P(E
1
/A) = 
1
3
1 1 1 1 1
3 3 2 3 6
1. 3
1 5
=
· + · + ·
        1
?
?
?
?
?
?
?
65/1/C (2)
65/1/C
10. let y = u + v, u = x
sin x 
, v = (sin x)
cos x
log u = sin x.log x ? 
sin x
du sin x
x cos x.log x
dx x
? ?
= · +
? ?
? ?
         1
1
2
+
  log v = cos x.log (sin x) ? { }
cos x
dv
(sin x) cos x cot x sin x log(sin x)
dx
= · · - ·
        1
1
2
+
  dy
dx
 = 
sin x cos x
du dv sin x
x cos x log x (sin x) {cos x cot x sin x log(sin x)}
dx dx x
? ?
+ = · · + + · - ·
? ?
? ?
                    1
2
+
1
2
OR
dy
dx
 = 
2 sin (log x) 3 cos (log x)
x x
-
+
       1
? x
dy
dx
 = –2 sin (log x) + 3 cos (log x), differentiate w.r.t ‘x’
1
2
?
2
2
d y dy
x
dx
dx
+
   = 
2cos (log x) 3 sin (log x)
x x
-
-
       2
?
2
2
2
d y dy
x x
dx
dx
+
    = – y  ?  
2
2
2
d y dy
x x y
dx
dx
+ +
      = 0
1
2
11.
dx
dt
 = 2a cos 2t (1 + cos 2t) – 2a sin 2t·sin 2t
1
1
2
dy
dt
 = –2b sin 2t (1 – cos 2t) + 2b cos 2t·sin 2t 1
t
4
dy
dx
p
=
?
?
?
   = 
t
4
2b cos 2t sin 2t 2b sin 2t (1 cos 2t) b
2a cos 2t (1 cos 2t) 2a sin 2t sin 2t a
p
=
· - - ?
=
?
+ - ·
?
                          1
2
 + 1
12. y
2
 = ax
3
 + b ? 
dy
2y
dx
  = 3ax
2
 ? 
dy
dx
 = 
2
3a x
2 y
 1
Slope of tangent at (2, 3) = 
(2, 3)
3a 4 dy
2a
2 3 dx
?
= · =
?
?
     1
Comparing with slope of tangent  y = 4x – 5, we get, 2a = 4 ? 
a 2 =
  1
Also (2, 3) lies on the curve ? 9 = 8a + b, put a = 2, we get b = –
 
7 1
13. Let x
2
 = t ? 
2
4 2
x
x x 2 + -
     = 
2
2 2
x t A B
(t 1) (t 2) t 1 t 2
(x 1) (x 2)
= = +
- + - +
- +
                    1
Solving for A and B to get,  A = 
1 2
, B
3 3
=
   1
2
4 2
x
dx
x x 2 + -
?
      = 
1
2 2
1 1 2 1 1 2 x x 1
dx dx log tan C
3 3 6 3 x 1 2 x 1 x 2
-
-
+ = + +
+ - +
? ?
                        1 + 1
65/1/C (3)
65/1/C
Page 5


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/1/C, 65/2/C, 65/3/C
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.
QUESTION PAPER CODE 65/1/C
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. (x + 3)2x – (–2) (–3x) = 8
1
2
x = 2
1
2
2.
2 5
2 4
? ?
? ?
? ?
 = 
3 1 1 2
2 0 1 1
? ?? ?
? ?? ?
- -
? ?? ?
 1
2
 + 
1
2
3. No. of possible matrices = 3
4
1
or 81
4.
2 (2a 3b) 1(3a – 2b)
2 1
+ +
+

 


 

          1
2
= 
7 4
a b
3 3
+




       (or enternal division may also be considered)
1
2
5. 2 1
6.
x y z
3 4 2
+ +
-
     = 1
1
2
?
ˆ ˆ ˆ
r (4i 3j 6k) · - +


     = 12   or  
ˆ ˆ ˆ
i j k
r 1
3 4 2
? ?
· - + =
? ?
? ?
? ?


1
2
SECTION B
7. Equation of line through A(3, 4, 1) and B(5, 1, 6)
x 3
2
-
   = 
y 4 z 1
k(say)
3 5
- -
= =
-
        1
General point on the line:
x = 2k + 3, y = – 3k + 4, z = 5k + 1
1
2
line crosses xz plane i.e. y = 0 if –3k + 4 = 0
? k = 
4
3
1
Co-ordinate of required point 
17 23
, 0,
3 3
? ?
? ?
? ?
  1
2
Angle, which line makes with xz plane:
sin ? = 
2 (0) ( 3) (1) 5(0) 3
4 9 25 1 38
+ - +
=
+ +
              ? ? = 
1
3
sin
38
-
? ?
? ?
? ?
 1
?
?
?
?
?
65/1/C (1)
65/1/C
8. let 
1
d


 & 
2
d


 be the two diagonal vectors:
?
1
d


 = 
ˆ ˆ ˆ
4i 2j 2k - -
    , 
2
d


 = 
ˆ ˆ
6j 8k - -
  1 1
2 2
+
  or
2
ˆ ˆ
d 6j 8k = +


Unit vectors parallel to the diagonals are:
1
ˆ
d = 
2 1 1
ˆ ˆ ˆ
i j k
6 6 6
- -
       1
2
2
ˆ
d
 = 
3 4
ˆ ˆ
j k
5 5
- -
        2
3 4
ˆ ˆ ˆ
or d j k
5 5
? ?
= +
? ?
? ?
1
2
1 2
d d ×

 

   = 
ˆ ˆ ˆ
i j k
ˆ ˆ ˆ
4i 32j 24k
4 2 2
0 6 8
= + -
- -
- -
      1
Area of parallelogram = 
1 2
1
404 or 2 101 sq. units
d d
2
=
×

 

       1
9. let X = Amount he wins then x = ` 5, 4, 3, – 3 1
P = Probability of getting a no. >4 = 
1
3
, q = 
2
1 p
3
- =
    1
2
X: 5 4 3 –3
P(x)
1
3
2 1
3 3
·
 = 
2
9
2
1 2
3 3
? ?
·
? ?
? ?
  = 
4
27
3
2
3
? ?
? ?
? ?
 = 
8
27
2
Expected amount he wins = 
5 8 12 24
XP(X)
3 9 27 27
S = + + -
        = 
19 1
or 2
9 9
    ` `
1
2
OR
E
1
 = Event that all balls are white,
E
2
 = Event that 3 balls are white and 1 ball is non white
E
3
 = Event that 2 balls are white and 2 balls are non-white
1
A = Event that 2 balls drawn without replacement are white
P(E
1
) = P(E
2
) = P(E
3
) = 
1
3
1
2
P(A/E
1
) = 1, P(A/E
2
) = 
3 2 1
4 3 2
· =
  , P(A/E
3
) = 
2 1 1
4 3 6
· =
  1
1
2
P(E
1
/A) = 
1
3
1 1 1 1 1
3 3 2 3 6
1. 3
1 5
=
· + · + ·
        1
?
?
?
?
?
?
?
65/1/C (2)
65/1/C
10. let y = u + v, u = x
sin x 
, v = (sin x)
cos x
log u = sin x.log x ? 
sin x
du sin x
x cos x.log x
dx x
? ?
= · +
? ?
? ?
         1
1
2
+
  log v = cos x.log (sin x) ? { }
cos x
dv
(sin x) cos x cot x sin x log(sin x)
dx
= · · - ·
        1
1
2
+
  dy
dx
 = 
sin x cos x
du dv sin x
x cos x log x (sin x) {cos x cot x sin x log(sin x)}
dx dx x
? ?
+ = · · + + · - ·
? ?
? ?
                    1
2
+
1
2
OR
dy
dx
 = 
2 sin (log x) 3 cos (log x)
x x
-
+
       1
? x
dy
dx
 = –2 sin (log x) + 3 cos (log x), differentiate w.r.t ‘x’
1
2
?
2
2
d y dy
x
dx
dx
+
   = 
2cos (log x) 3 sin (log x)
x x
-
-
       2
?
2
2
2
d y dy
x x
dx
dx
+
    = – y  ?  
2
2
2
d y dy
x x y
dx
dx
+ +
      = 0
1
2
11.
dx
dt
 = 2a cos 2t (1 + cos 2t) – 2a sin 2t·sin 2t
1
1
2
dy
dt
 = –2b sin 2t (1 – cos 2t) + 2b cos 2t·sin 2t 1
t
4
dy
dx
p
=
?
?
?
   = 
t
4
2b cos 2t sin 2t 2b sin 2t (1 cos 2t) b
2a cos 2t (1 cos 2t) 2a sin 2t sin 2t a
p
=
· - - ?
=
?
+ - ·
?
                          1
2
 + 1
12. y
2
 = ax
3
 + b ? 
dy
2y
dx
  = 3ax
2
 ? 
dy
dx
 = 
2
3a x
2 y
 1
Slope of tangent at (2, 3) = 
(2, 3)
3a 4 dy
2a
2 3 dx
?
= · =
?
?
     1
Comparing with slope of tangent  y = 4x – 5, we get, 2a = 4 ? 
a 2 =
  1
Also (2, 3) lies on the curve ? 9 = 8a + b, put a = 2, we get b = –
 
7 1
13. Let x
2
 = t ? 
2
4 2
x
x x 2 + -
     = 
2
2 2
x t A B
(t 1) (t 2) t 1 t 2
(x 1) (x 2)
= = +
- + - +
- +
                    1
Solving for A and B to get,  A = 
1 2
, B
3 3
=
   1
2
4 2
x
dx
x x 2 + -
?
      = 
1
2 2
1 1 2 1 1 2 x x 1
dx dx log tan C
3 3 6 3 x 1 2 x 1 x 2
-
-
+ = + +
+ - +
? ?
                        1 + 1
65/1/C (3)
65/1/C
14. Let I = 
2 2 2
2
2 2
0 0 0
sin x
sin x cos x
2
dx, Also I dx dx
sin x cos x cos x sin x
sin cos x x
2 2
p p p
p ? ?
-
? ?
? ?
= =
p p + ? ? ? ? +
+ - -
? ? ? ?
? ? ? ?
? ? ?
                           1
Adding to get, 2I = 
2 2
4
0 0
1 1 1
dx dx
sin x cos x cos (x ) 2
p p
p
=
+ -
? ?
          1
1
2
+
  ? 2I = 
2
2
4
0
1 1
sec (x ) dx log sec tan x x
4 4 2 2
p
p
p
p p ? ? ? ?
- = + - -
? ? ? ?
? ? ? ?
?
0                 1
? 2I = { }
1
log log
2 1 2 1
2
-
+ -
         ? I = 
1 1
2 1
log log or log
2 1 2 1
2 2 2 2
2 1
?
? +
-
+ - ?
- ?
?
            1
2
OR
3/2
0
| x cos x | dx p
?
    = 
1/2 3/2
0 1/2
x cos x dx x cos x dx p - p
? ?
         1
1
2
= 
1/2 3/2
2 2
0 1/2
x sin x cos x x sin x cos x p p p p ? ? ? ?
+ - +
? ? ? ?
p p
p p ? ? ? ?
            1
1
2
= 
2 2
1 1 5 1 3 1
2 2 2 2
? ?
- - = - - -
? ?
p p p p ? ? p p
          1
15.
2
(3x 1) 4 3x 2x dx + - -
?
         = 
2 2
3 5
( 4x 3) 4 3x 2x dx 4 3x 2x dx
4 4
- - - - - - - -
? ?
                 1
= 
3
2
2
2
2
1 5 3
41
(4 3x 2x ) 2 dx x
2 4 4
4
? ?
? ?
- - - - - +
? ?
? ?
? ?
? ?
?
              1 + 1
3
2
3
2
2
2 1
2 2 1
4x 3 1 5 4x 3 41 41 3
(4 3 2x ) 2 sin C x
2 4 8 16 32 4 41
4x 3 1 5 4x 3 41 2
(4 3x 2x ) 4 3x 2x sin C
2 4 8 32 41
-
-
? ? ?
+ + ? ? ? ? ? ?
?
= - - - - - + · + +
? ? ? ? ? ?
? ? ? ? ?
? ? ?
? ?
?
?
? ?
+ + ? ? ? ?
= - - - - - - + · +
? ? ?
? ?
? ? ? ? ? ? ? ?
                                           1
16. The differential equation can be re-written as:
dy
dx
 = 
x y dy dv
, put y vx, v x
x y dx dx
-
= = +
+
              1
2
 + 
1
2
? 
dv
v x
dx
+
    = 
2
1 v 1 v 1
dv dx
1 v x
1 2v v
- +
? =
+
- -
                1
integrating we get
? 
2
1 2V 2
dv
2
V 2V 1
+
+ -
?
         = –
2
1 1
dx log | V 2V 1| log x log C
x 2
= + - = - +
?
               1
1
2
? Solution of the differential equation is:
2
2
1
y 2y
log
1
2
x
x
+ -
      = 
2 2 2
log C log x or, y 2xy x C - + - =
            1
2
65/1/C (4)
65/1/C