Past Year Paper - Solutions, Mathematics (Set - 1, 2 and 3 ), Foreign, 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 65/2/1/F, 65/2/2/F, 65/2/3/F
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 65/2/1/F, 65/2/2/F, 65/2/3/F
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 65/2/1/F
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. 1 × 1 1
2. Expanding we get
x
3
 = – 8 ? x = – 2
1 1
2 2
+
  3. P = 
3 6
1
(A A ) P
6 9 2
? ?
+ ' ? =
? ?
? ?
       1 1
2 2
+
  4. (a b c) (a b c) + + · + +

 


 
 
 

         = 0
1
2
?
2 2 2
| a | | b | | c | 2 (a b b c c a) + + + · + · + ·

 
 


 
 
 
 
 

            = 0
?
a b b c c a · + · + ·

 


 
 
 

     = 
3
2
-
 1
2
5. a
2
 b
2
 sin
2
 ? + a
2
 b
2
 cos
2
 ? = 400
1
2
? | b |


 = 4
1
2
6.
x y z
3 3 3
+ +
     = 5 3 or x y z 15 + + =
          
1
mark for dc's of normal
2
? ?
? ?
? ?
1
SECTION B
7. LHS = 
1
x x x x
cos sin cos sin
2 2 2 2
cot
x x x x
cos sin cos sin
2 2 2 2
-
? ? ? ? ? ?
+ + -
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ?
? ?
+ - -
? ? ? ?
? ?
? ? ? ? ? ?
                     1+1
= 
1
x
cot cot
2
-
? ?
? ?
? ?
  1
= 
x
RHS
2
=
  1
OR
1
x 2 x 2
x 1 x 1
tan
x 2 x 2
1
x 1 x 1
-
- + ? ?
+
? ?
- +
? ?
- +
? ? - ·
? - + ?
                      = 
4
p 1
1
2
?
2
2x 4
3
-
   = 
tan
4
p
 1
1
2
? x = 
7
2
±
 1
65/2/1/F (1)
65/2/1/F
Page 3


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/2/1/F, 65/2/2/F, 65/2/3/F
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 65/2/1/F
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. 1 × 1 1
2. Expanding we get
x
3
 = – 8 ? x = – 2
1 1
2 2
+
  3. P = 
3 6
1
(A A ) P
6 9 2
? ?
+ ' ? =
? ?
? ?
       1 1
2 2
+
  4. (a b c) (a b c) + + · + +

 


 
 
 

         = 0
1
2
?
2 2 2
| a | | b | | c | 2 (a b b c c a) + + + · + · + ·

 
 


 
 
 
 
 

            = 0
?
a b b c c a · + · + ·

 


 
 
 

     = 
3
2
-
 1
2
5. a
2
 b
2
 sin
2
 ? + a
2
 b
2
 cos
2
 ? = 400
1
2
? | b |


 = 4
1
2
6.
x y z
3 3 3
+ +
     = 5 3 or x y z 15 + + =
          
1
mark for dc's of normal
2
? ?
? ?
? ?
1
SECTION B
7. LHS = 
1
x x x x
cos sin cos sin
2 2 2 2
cot
x x x x
cos sin cos sin
2 2 2 2
-
? ? ? ? ? ?
+ + -
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ?
? ?
+ - -
? ? ? ?
? ?
? ? ? ? ? ?
                     1+1
= 
1
x
cot cot
2
-
? ?
? ?
? ?
  1
= 
x
RHS
2
=
  1
OR
1
x 2 x 2
x 1 x 1
tan
x 2 x 2
1
x 1 x 1
-
- + ? ?
+
? ?
- +
? ?
- +
? ? - ·
? - + ?
                      = 
4
p 1
1
2
?
2
2x 4
3
-
   = 
tan
4
p
 1
1
2
? x = 
7
2
±
 1
65/2/1/F (1)
65/2/1/F
8. Let each poor child pay ` x per month and each rich child pay ` y per month.
? 20x + 5y = 9000
1
2
5x + 25y = 26000
1
2
In matrix form,
20 5 x
5 25 y
? ? ? ?
? ? ? ?
? ? ? ?
  = 
9000
26000
? ?
? ?
? ?
1
AX = B  ?  X = A
–1
 B
A
–1
 = 
25 5
1
5 20 475
- ? ?
? ?
-
? ?
 x
y
? ?
? ?
? ?
 = 
25 5 9000 200
1
5 20 26000 1000 475
- ? ? ? ? ? ?
=
? ? ? ? ? ?
-
? ? ? ? ? ?
    ? x = 200, y = 1000 1
V alue: Compassion or any relevant value 1
9. f '
1–
 = 2x + 3 = 5
f '
1+
 = b
f'
1–
 = 
1
f b 5
'
+
? =
    1+1
x 1
lim f (x)
-
?
  = 
x 1
f (1) lim f (x)
+
?
 = ? 4 + a = b + 2 1
? a 3 =
  1
10. Let u = 
2
1
1 x 1
tan
x
-
+ -
     Put x = tan ? ? ? = tan
–1
 x
1
2
? u = 
1
sec 1
tan
tan
-
? - ? ?
? ?
?
? ?
     = 
1
1 cos
tan
sin
-
- ? ? ?
? ?
?
? ?
     = 
1
tan tan
2
-
? ? ?
? ?
? ?
  = 
1
1
tan x
2 2
-
?
=
   1
?
du
dx
= 
2
1
2 (1 x ) +
   1
v = 
1
2
2x
sin
1 x
-
? ?
? ?
? + ?
   = 2 tan
–1
 x
65/2/1/F (2)
65/2/1/F
Page 4


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/2/1/F, 65/2/2/F, 65/2/3/F
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 65/2/1/F
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. 1 × 1 1
2. Expanding we get
x
3
 = – 8 ? x = – 2
1 1
2 2
+
  3. P = 
3 6
1
(A A ) P
6 9 2
? ?
+ ' ? =
? ?
? ?
       1 1
2 2
+
  4. (a b c) (a b c) + + · + +

 


 
 
 

         = 0
1
2
?
2 2 2
| a | | b | | c | 2 (a b b c c a) + + + · + · + ·

 
 


 
 
 
 
 

            = 0
?
a b b c c a · + · + ·

 


 
 
 

     = 
3
2
-
 1
2
5. a
2
 b
2
 sin
2
 ? + a
2
 b
2
 cos
2
 ? = 400
1
2
? | b |


 = 4
1
2
6.
x y z
3 3 3
+ +
     = 5 3 or x y z 15 + + =
          
1
mark for dc's of normal
2
? ?
? ?
? ?
1
SECTION B
7. LHS = 
1
x x x x
cos sin cos sin
2 2 2 2
cot
x x x x
cos sin cos sin
2 2 2 2
-
? ? ? ? ? ?
+ + -
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ?
? ?
+ - -
? ? ? ?
? ?
? ? ? ? ? ?
                     1+1
= 
1
x
cot cot
2
-
? ?
? ?
? ?
  1
= 
x
RHS
2
=
  1
OR
1
x 2 x 2
x 1 x 1
tan
x 2 x 2
1
x 1 x 1
-
- + ? ?
+
? ?
- +
? ?
- +
? ? - ·
? - + ?
                      = 
4
p 1
1
2
?
2
2x 4
3
-
   = 
tan
4
p
 1
1
2
? x = 
7
2
±
 1
65/2/1/F (1)
65/2/1/F
8. Let each poor child pay ` x per month and each rich child pay ` y per month.
? 20x + 5y = 9000
1
2
5x + 25y = 26000
1
2
In matrix form,
20 5 x
5 25 y
? ? ? ?
? ? ? ?
? ? ? ?
  = 
9000
26000
? ?
? ?
? ?
1
AX = B  ?  X = A
–1
 B
A
–1
 = 
25 5
1
5 20 475
- ? ?
? ?
-
? ?
 x
y
? ?
? ?
? ?
 = 
25 5 9000 200
1
5 20 26000 1000 475
- ? ? ? ? ? ?
=
? ? ? ? ? ?
-
? ? ? ? ? ?
    ? x = 200, y = 1000 1
V alue: Compassion or any relevant value 1
9. f '
1–
 = 2x + 3 = 5
f '
1+
 = b
f'
1–
 = 
1
f b 5
'
+
? =
    1+1
x 1
lim f (x)
-
?
  = 
x 1
f (1) lim f (x)
+
?
 = ? 4 + a = b + 2 1
? a 3 =
  1
10. Let u = 
2
1
1 x 1
tan
x
-
+ -
     Put x = tan ? ? ? = tan
–1
 x
1
2
? u = 
1
sec 1
tan
tan
-
? - ? ?
? ?
?
? ?
     = 
1
1 cos
tan
sin
-
- ? ? ?
? ?
?
? ?
     = 
1
tan tan
2
-
? ? ?
? ?
? ?
  = 
1
1
tan x
2 2
-
?
=
   1
?
du
dx
= 
2
1
2 (1 x ) +
   1
v = 
1
2
2x
sin
1 x
-
? ?
? ?
? + ?
   = 2 tan
–1
 x
65/2/1/F (2)
65/2/1/F
?
dv
dx
= 
2
2
1 x +
  1
?
du
dv
= 
du / dx 1
dv / dx 4
=
  1
2
OR
x = 
dx
sin t cos t
dt
? =
      1
2
y = 
dy
sin pt p cos pt
dt
? =
       1
2
dy
dx
= 
p cos pt
cos t
  1
2
2
d y
dx
= 
2
2
cos t ( p sin pt) p cos pt ( sin t) dt
dx
cos t
- - -
·
          = 
2
3
p sin pt cos t p cos pt sin t
cos t
- +
          1
Now
2
2 2
2
d y dy
(1 x ) x p y
dx
dx
- - +
         = 0 
2
2
dy d y
Substituting values of y, &
dx
dx
? ?
? ?
? ?
? ?
     1
11. Eqn of given curves
y
2
 = 4ax and x
2
 = 4by
Their point of intersections are (0, 0) and ( )
1 2 2 1
3 3 3 3
4a b , 4a b
   1
y
2
 = 
1
3
1
3
dy 2a a
4ax , slope
dx y
2 b
? = =
         ...(i) 1
x
2
 = 
1/3
1/3
dy x 2a
4by , slope
dx 2b
b
? = =
        ...(ii) 1
At (0, 0), angle between two curves is 90°
                        or
Acute angle ? between (i) and (ii) is
? = 
1 1
3 3
2 2
3 3
1
3 a b
tan
2
a b
-
? ? ? ?
? ?
? ?
? ?
? ?
+ ? ?
? ? ? ?
     12. I = 
0
( x)
dx
1 sin sin ( x)
p
p -
+ a p -
?
          1
2I = 
0
dx
1 sin sin x
p
p
+ a
?
= 
2
0
dx
2
1 sin sin x
p/
p
+ a
?
? I = 
/2
0
2
dx
x
2 tan
2
1 sin
x
1 tan
2
p
p
+ a
+
?
       1
?
?
?
?
?
?
?
?
?
?
?
1
65/2/1/F (3)
65/2/1/F
Page 5


Strictly Confidential — (For Internal and Restricted Use Only)
Senior School Certificate Examination
March 2016
Marking Scheme — Mathematics 65/2/1/F, 65/2/2/F, 65/2/3/F
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 65/2/1/F
EXPECTED ANSWER/VALUE POINTS
SECTION A
1. 1 × 1 1
2. Expanding we get
x
3
 = – 8 ? x = – 2
1 1
2 2
+
  3. P = 
3 6
1
(A A ) P
6 9 2
? ?
+ ' ? =
? ?
? ?
       1 1
2 2
+
  4. (a b c) (a b c) + + · + +

 


 
 
 

         = 0
1
2
?
2 2 2
| a | | b | | c | 2 (a b b c c a) + + + · + · + ·

 
 


 
 
 
 
 

            = 0
?
a b b c c a · + · + ·

 


 
 
 

     = 
3
2
-
 1
2
5. a
2
 b
2
 sin
2
 ? + a
2
 b
2
 cos
2
 ? = 400
1
2
? | b |


 = 4
1
2
6.
x y z
3 3 3
+ +
     = 5 3 or x y z 15 + + =
          
1
mark for dc's of normal
2
? ?
? ?
? ?
1
SECTION B
7. LHS = 
1
x x x x
cos sin cos sin
2 2 2 2
cot
x x x x
cos sin cos sin
2 2 2 2
-
? ? ? ? ? ?
+ + -
? ? ? ?
? ?
? ? ? ?
? ?
? ? ? ?
? ?
+ - -
? ? ? ?
? ?
? ? ? ? ? ?
                     1+1
= 
1
x
cot cot
2
-
? ?
? ?
? ?
  1
= 
x
RHS
2
=
  1
OR
1
x 2 x 2
x 1 x 1
tan
x 2 x 2
1
x 1 x 1
-
- + ? ?
+
? ?
- +
? ?
- +
? ? - ·
? - + ?
                      = 
4
p 1
1
2
?
2
2x 4
3
-
   = 
tan
4
p
 1
1
2
? x = 
7
2
±
 1
65/2/1/F (1)
65/2/1/F
8. Let each poor child pay ` x per month and each rich child pay ` y per month.
? 20x + 5y = 9000
1
2
5x + 25y = 26000
1
2
In matrix form,
20 5 x
5 25 y
? ? ? ?
? ? ? ?
? ? ? ?
  = 
9000
26000
? ?
? ?
? ?
1
AX = B  ?  X = A
–1
 B
A
–1
 = 
25 5
1
5 20 475
- ? ?
? ?
-
? ?
 x
y
? ?
? ?
? ?
 = 
25 5 9000 200
1
5 20 26000 1000 475
- ? ? ? ? ? ?
=
? ? ? ? ? ?
-
? ? ? ? ? ?
    ? x = 200, y = 1000 1
V alue: Compassion or any relevant value 1
9. f '
1–
 = 2x + 3 = 5
f '
1+
 = b
f'
1–
 = 
1
f b 5
'
+
? =
    1+1
x 1
lim f (x)
-
?
  = 
x 1
f (1) lim f (x)
+
?
 = ? 4 + a = b + 2 1
? a 3 =
  1
10. Let u = 
2
1
1 x 1
tan
x
-
+ -
     Put x = tan ? ? ? = tan
–1
 x
1
2
? u = 
1
sec 1
tan
tan
-
? - ? ?
? ?
?
? ?
     = 
1
1 cos
tan
sin
-
- ? ? ?
? ?
?
? ?
     = 
1
tan tan
2
-
? ? ?
? ?
? ?
  = 
1
1
tan x
2 2
-
?
=
   1
?
du
dx
= 
2
1
2 (1 x ) +
   1
v = 
1
2
2x
sin
1 x
-
? ?
? ?
? + ?
   = 2 tan
–1
 x
65/2/1/F (2)
65/2/1/F
?
dv
dx
= 
2
2
1 x +
  1
?
du
dv
= 
du / dx 1
dv / dx 4
=
  1
2
OR
x = 
dx
sin t cos t
dt
? =
      1
2
y = 
dy
sin pt p cos pt
dt
? =
       1
2
dy
dx
= 
p cos pt
cos t
  1
2
2
d y
dx
= 
2
2
cos t ( p sin pt) p cos pt ( sin t) dt
dx
cos t
- - -
·
          = 
2
3
p sin pt cos t p cos pt sin t
cos t
- +
          1
Now
2
2 2
2
d y dy
(1 x ) x p y
dx
dx
- - +
         = 0 
2
2
dy d y
Substituting values of y, &
dx
dx
? ?
? ?
? ?
? ?
     1
11. Eqn of given curves
y
2
 = 4ax and x
2
 = 4by
Their point of intersections are (0, 0) and ( )
1 2 2 1
3 3 3 3
4a b , 4a b
   1
y
2
 = 
1
3
1
3
dy 2a a
4ax , slope
dx y
2 b
? = =
         ...(i) 1
x
2
 = 
1/3
1/3
dy x 2a
4by , slope
dx 2b
b
? = =
        ...(ii) 1
At (0, 0), angle between two curves is 90°
                        or
Acute angle ? between (i) and (ii) is
? = 
1 1
3 3
2 2
3 3
1
3 a b
tan
2
a b
-
? ? ? ?
? ?
? ?
? ?
? ?
+ ? ?
? ? ? ?
     12. I = 
0
( x)
dx
1 sin sin ( x)
p
p -
+ a p -
?
          1
2I = 
0
dx
1 sin sin x
p
p
+ a
?
= 
2
0
dx
2
1 sin sin x
p/
p
+ a
?
? I = 
/2
0
2
dx
x
2 tan
2
1 sin
x
1 tan
2
p
p
+ a
+
?
       1
?
?
?
?
?
?
?
?
?
?
?
1
65/2/1/F (3)
65/2/1/F
I = 
1
2
0
2dt
1 t 2t sin
p
+ + a
?
              Put tan 
x
t
2
=
  1
2
? I = 
1
2 2
0
dt
2
(t sin ) cos
p
+ a + a
?
      1
= 
1
1
0
2 t sin
tan
cos cos
-
p ? + a ? ? ?
? ? ? ?
a a
? ? ? ?
       ? I = 
cos 2
p p ? ?
- a
? ?
a
? ?
    1
2
13. I = 
2
(2x 5) 10 4x 3x dx + - -
?
        = 
2 2
1 11
( 4 6x) 10 4x 3x dx 10 4x 3x dx
3 3
- - - - - + - -
? ?
                 1
= 
3
2
2
2
2
2 11 3 34 2
(10 4x 3x ) x dx
9 3 3 3
? ?
? ?
- - - + - -
? ?
? ?
? ?
? ?
? ?
?
              1 + 1
= 
3
2
2
2
2 1
2 34 2
x x
3 3 3 2 11 3 17 3x 2
(10 4x 3x ) sin C
9 3 2 9 34
-
? ?
? ?
? ? ? ?
? ?
- - - -
? ?
? ? ? ?
? ?
? ? -
? ? ? ?
- - - + + +
? ?
? ?
                 1
OR
x
2
 = y (say)
1
2
(y 1) (y 4)
(y 3) (y 5)
+ +
+ -
           = 
A B
1
y 3 y 5
+ +
+ -
        1
2
using partial fraction we get A = 
1 27
, B
4 4
=
   1
2 2
2 2
(x 1) (x 4)
dx
(x 3) (x 5)
+ +
+ -
?
            = 
2 2
1 dx 27 dx
1.dx
4 4
x 3 x 5
+ +
+ -
? ? ?
          1
                          = 
1
1 x 27 x 5
x tan log C
4 3 3 8 5 x 5
-
-
+ + +
+
              1
14. I = 
1
2
x sin x
dx
1 x
-
-
?
     put sin
–1
 x = t ? 
2
dx
1 x -
   = dt
1 1
2 2
+
  = 
t sin t dt ·
?
  1
= – t cos t + sin t + c
1
1
2
= 
2 1
1 x sin x x c
-
- - + +
        1
2
65/2/1/F (4)
65/2/1/F