Solution Applications of Derivatives Test- 3 Class 12 Notes | EduRev

Class 12 : Solution Applications of Derivatives Test- 3 Class 12 Notes | EduRev

 Page 1


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Calculus: Application of Derivatives) 
 [ANSWERS] 
Topic: - application of derivatives 
1. 
3
12 /  ( )
dv
cm s Given
dt
= 
2
1
  ( )
6
1
3
h r Given
v r h p
=
=
( )
2
3
1 1
. . 6 . 
3 6
1
. .36.
3
v h h h r
v h
p
p
? ?
= =
? ?
? ?
=
?
3
3
v=12
12 .3 .
h
dv dh
h
dt dt
p
p =
( ) [ ]
2
12 12 .3 4 . 4
12
12 .3.16
dh
h cm
dt
dh
dt
p
p
= =
=
1
/
48
dh
cm s
dt p
=
2. ( ) ( )
26 26
d
MR R x x
dx
= = +
]
17
25 17 26
442
x
MR
=
= × +
=
Page 2


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Calculus: Application of Derivatives) 
 [ANSWERS] 
Topic: - application of derivatives 
1. 
3
12 /  ( )
dv
cm s Given
dt
= 
2
1
  ( )
6
1
3
h r Given
v r h p
=
=
( )
2
3
1 1
. . 6 . 
3 6
1
. .36.
3
v h h h r
v h
p
p
? ?
= =
? ?
? ?
=
?
3
3
v=12
12 .3 .
h
dv dh
h
dt dt
p
p =
( ) [ ]
2
12 12 .3 4 . 4
12
12 .3.16
dh
h cm
dt
dh
dt
p
p
= =
=
1
/
48
dh
cm s
dt p
=
2. ( ) ( )
26 26
d
MR R x x
dx
= = +
]
17
25 17 26
442
x
MR
=
= × +
=
3. 
4sin
2 cos
y
?
?
?
= - +
( )( )
( )
2
2
2 cos 4cos 4sin
'( ) 1
2 cos
f
? ? ?
?
?
+ +
= - +
 
( )
( )
( )
2
2
2
8cos 4
1
2 cos
8cos 4 4 cos 4cos
2 cos
?
?
? ? ?
?
+
= - +
+ - + +
=
+
( )
( )
cos 4 cos
'( )
2 cos
cos 4 cos
'( ) 0 0,
2 cos 2
f
f
? ?
?
?
? ? p
? ?
?
- =
+
- ? ?
? = ? ?
? ?
+
? ?
>
4. 
1
'( ) .cos
sin
f x x
x
=
'( ) cot
'( ) 0 0,
2
f x x
f x x
p
=
? ?
? ?
? ?
? ?
>
 
and 
'( ) 0 ,
2
f x x
p
p
? ?
? ?
? ?
? ?
< 
Hence f(x) = log sinx is strictly increasing on 0,
2
p
? ?
? ?
? ?
and decreasing on ,
2
p
p
? ?
? ?
? ?
 
5.
2
( 2) (1) y x = - - - - - - - Slope of tangent to curve 
2( 2)
dy
x
dx
= - Slope of chord  
2 1
2 1
4 0
4 2
y y
m
x x
? ? - - = =
? ?
- - ? ?
 
  2( 2) 2
3
AT x
x
? - =
=
Put x = 3 in equation (1) 
1 y = 
Points (3, 1) 
Page 3


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Calculus: Application of Derivatives) 
 [ANSWERS] 
Topic: - application of derivatives 
1. 
3
12 /  ( )
dv
cm s Given
dt
= 
2
1
  ( )
6
1
3
h r Given
v r h p
=
=
( )
2
3
1 1
. . 6 . 
3 6
1
. .36.
3
v h h h r
v h
p
p
? ?
= =
? ?
? ?
=
?
3
3
v=12
12 .3 .
h
dv dh
h
dt dt
p
p =
( ) [ ]
2
12 12 .3 4 . 4
12
12 .3.16
dh
h cm
dt
dh
dt
p
p
= =
=
1
/
48
dh
cm s
dt p
=
2. ( ) ( )
26 26
d
MR R x x
dx
= = +
]
17
25 17 26
442
x
MR
=
= × +
=
3. 
4sin
2 cos
y
?
?
?
= - +
( )( )
( )
2
2
2 cos 4cos 4sin
'( ) 1
2 cos
f
? ? ?
?
?
+ +
= - +
 
( )
( )
( )
2
2
2
8cos 4
1
2 cos
8cos 4 4 cos 4cos
2 cos
?
?
? ? ?
?
+
= - +
+ - + +
=
+
( )
( )
cos 4 cos
'( )
2 cos
cos 4 cos
'( ) 0 0,
2 cos 2
f
f
? ?
?
?
? ? p
? ?
?
- =
+
- ? ?
? = ? ?
? ?
+
? ?
>
4. 
1
'( ) .cos
sin
f x x
x
=
'( ) cot
'( ) 0 0,
2
f x x
f x x
p
=
? ?
? ?
? ?
? ?
>
 
and 
'( ) 0 ,
2
f x x
p
p
? ?
? ?
? ?
? ?
< 
Hence f(x) = log sinx is strictly increasing on 0,
2
p
? ?
? ?
? ?
and decreasing on ,
2
p
p
? ?
? ?
? ?
 
5.
2
( 2) (1) y x = - - - - - - - Slope of tangent to curve 
2( 2)
dy
x
dx
= - Slope of chord  
2 1
2 1
4 0
4 2
y y
m
x x
? ? - - = =
? ?
- - ? ?
 
  2( 2) 2
3
AT x
x
? - =
=
Put x = 3 in equation (1) 
1 y = 
Points (3, 1) 
6. 
2
3 sin .cos
dx
a t tdt
dt
=
2
3 cos sin
cot
dy
b t tdt
dt
dy b
t
dx a
= - - =
2
0 0
t
dy b
dx a p
=
- ?
= × =
?
?
When 
2
, 9,  0 t x and y
p
= - = 
Equation of tangent 
( )
1 1
, 0 0( )
0
dy
y y x x y x a
dx
y
- = - - = - =
7. 3, 0.02 x x = ? =
( )
( ) ( ) ( )
2
( ) '( )
3 5 3 6 5 0.02
f x x f x f x x
f x x x x x
+ ? = + ?
+ ? = + + + + ×
Put 3, 0.02 x x = ? =
(3.02) 45.46 f =
8.
1 5
y x =
( )
1 5
32, 0.15 x x
y y x x
= ? =
+ ? = + ?
( )
( )
1 5
1 5
1 5
y x x y
x x x
? = + ? - = + ? - ( )
( ) ( ) ( )
1 5
1 5
4
1 5 1 5
5
.
1
. 32.15 32
5
dy
x x x x
dx
x x
- ? ?
? = + ? - ? ?
? ?
? = - ( )
1 5
32.15 2.0018 = 
Page 4


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Calculus: Application of Derivatives) 
 [ANSWERS] 
Topic: - application of derivatives 
1. 
3
12 /  ( )
dv
cm s Given
dt
= 
2
1
  ( )
6
1
3
h r Given
v r h p
=
=
( )
2
3
1 1
. . 6 . 
3 6
1
. .36.
3
v h h h r
v h
p
p
? ?
= =
? ?
? ?
=
?
3
3
v=12
12 .3 .
h
dv dh
h
dt dt
p
p =
( ) [ ]
2
12 12 .3 4 . 4
12
12 .3.16
dh
h cm
dt
dh
dt
p
p
= =
=
1
/
48
dh
cm s
dt p
=
2. ( ) ( )
26 26
d
MR R x x
dx
= = +
]
17
25 17 26
442
x
MR
=
= × +
=
3. 
4sin
2 cos
y
?
?
?
= - +
( )( )
( )
2
2
2 cos 4cos 4sin
'( ) 1
2 cos
f
? ? ?
?
?
+ +
= - +
 
( )
( )
( )
2
2
2
8cos 4
1
2 cos
8cos 4 4 cos 4cos
2 cos
?
?
? ? ?
?
+
= - +
+ - + +
=
+
( )
( )
cos 4 cos
'( )
2 cos
cos 4 cos
'( ) 0 0,
2 cos 2
f
f
? ?
?
?
? ? p
? ?
?
- =
+
- ? ?
? = ? ?
? ?
+
? ?
>
4. 
1
'( ) .cos
sin
f x x
x
=
'( ) cot
'( ) 0 0,
2
f x x
f x x
p
=
? ?
? ?
? ?
? ?
>
 
and 
'( ) 0 ,
2
f x x
p
p
? ?
? ?
? ?
? ?
< 
Hence f(x) = log sinx is strictly increasing on 0,
2
p
? ?
? ?
? ?
and decreasing on ,
2
p
p
? ?
? ?
? ?
 
5.
2
( 2) (1) y x = - - - - - - - Slope of tangent to curve 
2( 2)
dy
x
dx
= - Slope of chord  
2 1
2 1
4 0
4 2
y y
m
x x
? ? - - = =
? ?
- - ? ?
 
  2( 2) 2
3
AT x
x
? - =
=
Put x = 3 in equation (1) 
1 y = 
Points (3, 1) 
6. 
2
3 sin .cos
dx
a t tdt
dt
=
2
3 cos sin
cot
dy
b t tdt
dt
dy b
t
dx a
= - - =
2
0 0
t
dy b
dx a p
=
- ?
= × =
?
?
When 
2
, 9,  0 t x and y
p
= - = 
Equation of tangent 
( )
1 1
, 0 0( )
0
dy
y y x x y x a
dx
y
- = - - = - =
7. 3, 0.02 x x = ? =
( )
( ) ( ) ( )
2
( ) '( )
3 5 3 6 5 0.02
f x x f x f x x
f x x x x x
+ ? = + ?
+ ? = + + + + ×
Put 3, 0.02 x x = ? =
(3.02) 45.46 f =
8.
1 5
y x =
( )
1 5
32, 0.15 x x
y y x x
= ? =
+ ? = + ?
( )
( )
1 5
1 5
1 5
y x x y
x x x
? = + ? - = + ? - ( )
( ) ( ) ( )
1 5
1 5
4
1 5 1 5
5
.
1
. 32.15 32
5
dy
x x x x
dx
x x
- ? ?
? = + ? - ? ?
? ?
? = - ( )
1 5
32.15 2.0018 = 
9. (18 2 ) l x cm = - (18 2 ) b x cm
h xcm
v l b h
= - =
= × ×
2
(18 2 ) (18 2 )
(18 2 )
v x x x
v x x
= - × - ×
= - ×
( ) ( ) ( )
[ ]
2
(18 2 ) . 1 .2(18 2 ) 2
(18 2 ) (18 2 ) 4
dv
x x x
dx
dv
x x x
dx
= - + - - = - - - For maximum/minimum
0 (18 2 )(18 6 )
9  ( )
3
x x
x neglect
x
= - - =
=
( ) ( )
2
2
2
2
3
(18 2 ) 6 (18 6 ) 2
12 6
x
d v
x x
dx
d v
dx
=
= - - + - - ?
= - ×
?
?
72 maximum
18 - 2 3 12 l cm
= - = × =
12
3
b cm
h cm
=
=
10. 
2
2 2 s rh r p p = +
2
2
2
2
s r
h
r
v r h
p
p
p
- =
=
Page 5


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Calculus: Application of Derivatives) 
 [ANSWERS] 
Topic: - application of derivatives 
1. 
3
12 /  ( )
dv
cm s Given
dt
= 
2
1
  ( )
6
1
3
h r Given
v r h p
=
=
( )
2
3
1 1
. . 6 . 
3 6
1
. .36.
3
v h h h r
v h
p
p
? ?
= =
? ?
? ?
=
?
3
3
v=12
12 .3 .
h
dv dh
h
dt dt
p
p =
( ) [ ]
2
12 12 .3 4 . 4
12
12 .3.16
dh
h cm
dt
dh
dt
p
p
= =
=
1
/
48
dh
cm s
dt p
=
2. ( ) ( )
26 26
d
MR R x x
dx
= = +
]
17
25 17 26
442
x
MR
=
= × +
=
3. 
4sin
2 cos
y
?
?
?
= - +
( )( )
( )
2
2
2 cos 4cos 4sin
'( ) 1
2 cos
f
? ? ?
?
?
+ +
= - +
 
( )
( )
( )
2
2
2
8cos 4
1
2 cos
8cos 4 4 cos 4cos
2 cos
?
?
? ? ?
?
+
= - +
+ - + +
=
+
( )
( )
cos 4 cos
'( )
2 cos
cos 4 cos
'( ) 0 0,
2 cos 2
f
f
? ?
?
?
? ? p
? ?
?
- =
+
- ? ?
? = ? ?
? ?
+
? ?
>
4. 
1
'( ) .cos
sin
f x x
x
=
'( ) cot
'( ) 0 0,
2
f x x
f x x
p
=
? ?
? ?
? ?
? ?
>
 
and 
'( ) 0 ,
2
f x x
p
p
? ?
? ?
? ?
? ?
< 
Hence f(x) = log sinx is strictly increasing on 0,
2
p
? ?
? ?
? ?
and decreasing on ,
2
p
p
? ?
? ?
? ?
 
5.
2
( 2) (1) y x = - - - - - - - Slope of tangent to curve 
2( 2)
dy
x
dx
= - Slope of chord  
2 1
2 1
4 0
4 2
y y
m
x x
? ? - - = =
? ?
- - ? ?
 
  2( 2) 2
3
AT x
x
? - =
=
Put x = 3 in equation (1) 
1 y = 
Points (3, 1) 
6. 
2
3 sin .cos
dx
a t tdt
dt
=
2
3 cos sin
cot
dy
b t tdt
dt
dy b
t
dx a
= - - =
2
0 0
t
dy b
dx a p
=
- ?
= × =
?
?
When 
2
, 9,  0 t x and y
p
= - = 
Equation of tangent 
( )
1 1
, 0 0( )
0
dy
y y x x y x a
dx
y
- = - - = - =
7. 3, 0.02 x x = ? =
( )
( ) ( ) ( )
2
( ) '( )
3 5 3 6 5 0.02
f x x f x f x x
f x x x x x
+ ? = + ?
+ ? = + + + + ×
Put 3, 0.02 x x = ? =
(3.02) 45.46 f =
8.
1 5
y x =
( )
1 5
32, 0.15 x x
y y x x
= ? =
+ ? = + ?
( )
( )
1 5
1 5
1 5
y x x y
x x x
? = + ? - = + ? - ( )
( ) ( ) ( )
1 5
1 5
4
1 5 1 5
5
.
1
. 32.15 32
5
dy
x x x x
dx
x x
- ? ?
? = + ? - ? ?
? ?
? = - ( )
1 5
32.15 2.0018 = 
9. (18 2 ) l x cm = - (18 2 ) b x cm
h xcm
v l b h
= - =
= × ×
2
(18 2 ) (18 2 )
(18 2 )
v x x x
v x x
= - × - ×
= - ×
( ) ( ) ( )
[ ]
2
(18 2 ) . 1 .2(18 2 ) 2
(18 2 ) (18 2 ) 4
dv
x x x
dx
dv
x x x
dx
= - + - - = - - - For maximum/minimum
0 (18 2 )(18 6 )
9  ( )
3
x x
x neglect
x
= - - =
=
( ) ( )
2
2
2
2
3
(18 2 ) 6 (18 6 ) 2
12 6
x
d v
x x
dx
d v
dx
=
= - - + - - ?
= - ×
?
?
72 maximum
18 - 2 3 12 l cm
= - = × =
12
3
b cm
h cm
=
=
10. 
2
2 2 s rh r p p = +
2
2
2
2
s r
h
r
v r h
p
p
p
- =
=
2
2
3
2
.
2
1
2
2
s r
v r
r
v sr r
p
p
p
p
? ? - =
? ?
? ?
? ? = - ? ?
2
1
6
2
dv
s r
dr
p ? ? = - ? ?
[ ]
2
2
1
0 12
2
d v
r
dr
p = - For maximum/minimum 
2
2
2
2
6
6
1
12 .
2 6 s
r
s r
d v s
s
dr
p
p
p
p
=
=
?
? ?
= - ?
? ?
? ?
?
2
tive maximum
2 2 s rh r p p
= - = +
2
2 2
6
2 2 6
s r
rh r r
p
p p p
=
+ =
2
2 4
2 4
2
rh r
h r
h r
p p =
=
=
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