Solution Applications of Derivatives Test- 2 Class 12 Notes | EduRev

Class 12 : Solution Applications of Derivatives Test- 2 Class 12 Notes | EduRev

 Page 1


CBSE TEST PAPER-02 
CLASS - XII MATHEMATICS (Calculus: Application of Derivatives) 
 [ANSWERS] 
Topic: - application of derivatives 
1. 2 /
dx
cm s
dt
= 
2 2 2
0.02 / sec
5
dx
m
dt
x y
=
+ =
2 2 0
dx dy
x y
dt dt
+ =
When 4 x = 
2 2
5 4
3
y = - =
( ) 2 4 0.02 2 3 0
2 4 0.02
2 3
dy
dt
dy
dt
× + × =
× ×
=
×
0.08 100
3 100
8
/ sec
3
cm
= - ×
- 2.
3
6 2 (1) y x = + - - - - - - 2
2
6 3
6 8 3 8
dy dx
x
dt dt
dx dx dy dx
x
dt dt dt dt
=
? ?
× = =
? ?
? ?
?
2
16
4
x
x
=
= ±
Put the value of x in equation (1) 
Page 2


CBSE TEST PAPER-02 
CLASS - XII MATHEMATICS (Calculus: Application of Derivatives) 
 [ANSWERS] 
Topic: - application of derivatives 
1. 2 /
dx
cm s
dt
= 
2 2 2
0.02 / sec
5
dx
m
dt
x y
=
+ =
2 2 0
dx dy
x y
dt dt
+ =
When 4 x = 
2 2
5 4
3
y = - =
( ) 2 4 0.02 2 3 0
2 4 0.02
2 3
dy
dt
dy
dt
× + × =
× ×
=
×
0.08 100
3 100
8
/ sec
3
cm
= - ×
- 2.
3
6 2 (1) y x = + - - - - - - 2
2
6 3
6 8 3 8
dy dx
x
dt dt
dx dx dy dx
x
dt dt dt dt
=
? ?
× = =
? ?
? ?
?
2
16
4
x
x
=
= ±
Put the value of x in equation (1) 
3
 4
6 (4) 2
when x
y
=
= +
64 2
68
6
y
= +
=
( ) 4,11
 4 when x = - 3
6 ( 4) 2
64 2
y = - +
= - +
62
6
62
4,
6
y
- =
- ? ?
- ? ?
? ?
31
4,
3
- ? ?
- ? ?
? ?
 
3. ( ) sin 3 f x x =
'( ) 3cos3
'( ) 0
f x x
f x
=
=
cos3 0
3
2
x
x
p
=
=
6
x
p
=
int. Sign of f’(x) Result 
0,
6
p ? ?
?
?
? ?
+tive increase 
,
6 2
p p ? ?
?
?
? ?
-tive Decrease 
Page 3


CBSE TEST PAPER-02 
CLASS - XII MATHEMATICS (Calculus: Application of Derivatives) 
 [ANSWERS] 
Topic: - application of derivatives 
1. 2 /
dx
cm s
dt
= 
2 2 2
0.02 / sec
5
dx
m
dt
x y
=
+ =
2 2 0
dx dy
x y
dt dt
+ =
When 4 x = 
2 2
5 4
3
y = - =
( ) 2 4 0.02 2 3 0
2 4 0.02
2 3
dy
dt
dy
dt
× + × =
× ×
=
×
0.08 100
3 100
8
/ sec
3
cm
= - ×
- 2.
3
6 2 (1) y x = + - - - - - - 2
2
6 3
6 8 3 8
dy dx
x
dt dt
dx dx dy dx
x
dt dt dt dt
=
? ?
× = =
? ?
? ?
?
2
16
4
x
x
=
= ±
Put the value of x in equation (1) 
3
 4
6 (4) 2
when x
y
=
= +
64 2
68
6
y
= +
=
( ) 4,11
 4 when x = - 3
6 ( 4) 2
64 2
y = - +
= - +
62
6
62
4,
6
y
- =
- ? ?
- ? ?
? ?
31
4,
3
- ? ?
- ? ?
? ?
 
3. ( ) sin 3 f x x =
'( ) 3cos3
'( ) 0
f x x
f x
=
=
cos3 0
3
2
x
x
p
=
=
6
x
p
=
int. Sign of f’(x) Result 
0,
6
p ? ?
?
?
? ?
+tive increase 
,
6 2
p p ? ?
?
?
? ?
-tive Decrease 
Hence, f(x) is increasing on 0,
6
p
? ?
? ?
? ?
 and decreasing on ,
6 2
p p
? ?
? ?
? ?
4. ( ) sin cos f x x x = +
'( ) cos sin
'( ) 0
f x x x
f x
= - =
sin 0
sin cos
cos cos
cox x
x x
x x
- =
=
tan 1
5
, 
4 4
x
x as
p p
=
=
0 2 x p = = 
int Singh of f’(x) Result 
0,
4
p
? ?
? ?
? ?
+tive Increase 
5
,
4 4
p p
? ?
? ?
? ?
 
-tive Decrease 
5
,2
4
p
p
? ?
? ?
? ?
+tive increase 
5. Let (x, y) be the point a
(a) y = x
2
 – 2x + 7  ---------- (1)
1
,
2 2
2 2
x y
dy
x
dx
dy
x
dx
= - ?
= - ?
?
 Slope of line  = 2 
1
1
 2 2 2
2
AT x
x
? - =
=
Page 4


CBSE TEST PAPER-02 
CLASS - XII MATHEMATICS (Calculus: Application of Derivatives) 
 [ANSWERS] 
Topic: - application of derivatives 
1. 2 /
dx
cm s
dt
= 
2 2 2
0.02 / sec
5
dx
m
dt
x y
=
+ =
2 2 0
dx dy
x y
dt dt
+ =
When 4 x = 
2 2
5 4
3
y = - =
( ) 2 4 0.02 2 3 0
2 4 0.02
2 3
dy
dt
dy
dt
× + × =
× ×
=
×
0.08 100
3 100
8
/ sec
3
cm
= - ×
- 2.
3
6 2 (1) y x = + - - - - - - 2
2
6 3
6 8 3 8
dy dx
x
dt dt
dx dx dy dx
x
dt dt dt dt
=
? ?
× = =
? ?
? ?
?
2
16
4
x
x
=
= ±
Put the value of x in equation (1) 
3
 4
6 (4) 2
when x
y
=
= +
64 2
68
6
y
= +
=
( ) 4,11
 4 when x = - 3
6 ( 4) 2
64 2
y = - +
= - +
62
6
62
4,
6
y
- =
- ? ?
- ? ?
? ?
31
4,
3
- ? ?
- ? ?
? ?
 
3. ( ) sin 3 f x x =
'( ) 3cos3
'( ) 0
f x x
f x
=
=
cos3 0
3
2
x
x
p
=
=
6
x
p
=
int. Sign of f’(x) Result 
0,
6
p ? ?
?
?
? ?
+tive increase 
,
6 2
p p ? ?
?
?
? ?
-tive Decrease 
Hence, f(x) is increasing on 0,
6
p
? ?
? ?
? ?
 and decreasing on ,
6 2
p p
? ?
? ?
? ?
4. ( ) sin cos f x x x = +
'( ) cos sin
'( ) 0
f x x x
f x
= - =
sin 0
sin cos
cos cos
cox x
x x
x x
- =
=
tan 1
5
, 
4 4
x
x as
p p
=
=
0 2 x p = = 
int Singh of f’(x) Result 
0,
4
p
? ?
? ?
? ?
+tive Increase 
5
,
4 4
p p
? ?
? ?
? ?
 
-tive Decrease 
5
,2
4
p
p
? ?
? ?
? ?
+tive increase 
5. Let (x, y) be the point a
(a) y = x
2
 – 2x + 7  ---------- (1)
1
,
2 2
2 2
x y
dy
x
dx
dy
x
dx
= - ?
= - ?
?
 Slope of line  = 2 
1
1
 2 2 2
2
AT x
x
? - =
=
[ ]
2
1 1 1
1
2 7      (1)
4 4 7
7
y x x from
y
= - +
= - +
=
Equation of tangent 
( )
1
,
7 2 4
dy
y y x x
dx
y x
- = - - = - 2 3 0 x y - + = 
(b) 5 15 13 y x - = 
 Slope of Line = 
( 15)
3
5
- - = 
( )
5
6
 2 , 2 3 1,  AT x x ? - × = - =
Put x
1
 in equation (1) 
2
5 5
2 7
6 6
25 10
7
36 6
y
? ? ? ?
= - +
? ? ? ?
? ? ? ?
= - +
25 60 7 36
36
35 252
36
- + ×
=
- +
=
217
36
=
Equation of tangent 
( )
1 1
217 1 5
36 3 6
dy
y y x x
dx
y x
- = - - ? ?
- = - ? ?
? ?
36 217 1 5
36 3 1 6
36 217 1 6 5
36 3 6
12 36 217 0
y x
y x
x y
- - ? ?
= - ? ?
? ?
- - - ? ?
=
? ?
? ?
+ - =
Page 5


CBSE TEST PAPER-02 
CLASS - XII MATHEMATICS (Calculus: Application of Derivatives) 
 [ANSWERS] 
Topic: - application of derivatives 
1. 2 /
dx
cm s
dt
= 
2 2 2
0.02 / sec
5
dx
m
dt
x y
=
+ =
2 2 0
dx dy
x y
dt dt
+ =
When 4 x = 
2 2
5 4
3
y = - =
( ) 2 4 0.02 2 3 0
2 4 0.02
2 3
dy
dt
dy
dt
× + × =
× ×
=
×
0.08 100
3 100
8
/ sec
3
cm
= - ×
- 2.
3
6 2 (1) y x = + - - - - - - 2
2
6 3
6 8 3 8
dy dx
x
dt dt
dx dx dy dx
x
dt dt dt dt
=
? ?
× = =
? ?
? ?
?
2
16
4
x
x
=
= ±
Put the value of x in equation (1) 
3
 4
6 (4) 2
when x
y
=
= +
64 2
68
6
y
= +
=
( ) 4,11
 4 when x = - 3
6 ( 4) 2
64 2
y = - +
= - +
62
6
62
4,
6
y
- =
- ? ?
- ? ?
? ?
31
4,
3
- ? ?
- ? ?
? ?
 
3. ( ) sin 3 f x x =
'( ) 3cos3
'( ) 0
f x x
f x
=
=
cos3 0
3
2
x
x
p
=
=
6
x
p
=
int. Sign of f’(x) Result 
0,
6
p ? ?
?
?
? ?
+tive increase 
,
6 2
p p ? ?
?
?
? ?
-tive Decrease 
Hence, f(x) is increasing on 0,
6
p
? ?
? ?
? ?
 and decreasing on ,
6 2
p p
? ?
? ?
? ?
4. ( ) sin cos f x x x = +
'( ) cos sin
'( ) 0
f x x x
f x
= - =
sin 0
sin cos
cos cos
cox x
x x
x x
- =
=
tan 1
5
, 
4 4
x
x as
p p
=
=
0 2 x p = = 
int Singh of f’(x) Result 
0,
4
p
? ?
? ?
? ?
+tive Increase 
5
,
4 4
p p
? ?
? ?
? ?
 
-tive Decrease 
5
,2
4
p
p
? ?
? ?
? ?
+tive increase 
5. Let (x, y) be the point a
(a) y = x
2
 – 2x + 7  ---------- (1)
1
,
2 2
2 2
x y
dy
x
dx
dy
x
dx
= - ?
= - ?
?
 Slope of line  = 2 
1
1
 2 2 2
2
AT x
x
? - =
=
[ ]
2
1 1 1
1
2 7      (1)
4 4 7
7
y x x from
y
= - +
= - +
=
Equation of tangent 
( )
1
,
7 2 4
dy
y y x x
dx
y x
- = - - = - 2 3 0 x y - + = 
(b) 5 15 13 y x - = 
 Slope of Line = 
( 15)
3
5
- - = 
( )
5
6
 2 , 2 3 1,  AT x x ? - × = - =
Put x
1
 in equation (1) 
2
5 5
2 7
6 6
25 10
7
36 6
y
? ? ? ?
= - +
? ? ? ?
? ? ? ?
= - +
25 60 7 36
36
35 252
36
- + ×
=
- +
=
217
36
=
Equation of tangent 
( )
1 1
217 1 5
36 3 6
dy
y y x x
dx
y x
- = - - ? ?
- = - ? ?
? ?
36 217 1 5
36 3 1 6
36 217 1 6 5
36 3 6
12 36 217 0
y x
y x
x y
- - ? ?
= - ? ?
? ?
- - - ? ?
=
? ?
? ?
+ - =
6. 
2 2
2 2
1
x y
a b
- = 
2 2
2 2
0
x y dy
a b dx
- =
2
2
2
2
x
dy
a
y
dx
b
- =
- 2
2
2
2
x b
a y
- = ×
- 2
2
2
2
.
.
o o
x y
o
o
dy x b
dx y a
x b
y a
?
=
?
?
=
Equation 
( )
2
0
1 1 0 0 2
0
2 2 2 2 2 2
( )
o o o o
x dy b
y y x x y y x x
dx y a
yy a y a xx b x b
- = - ? - = - - = - 2 2 2 2 2 2
o o o o
x b y a xx b yy a - = - Dividing by a
2
b
2
 
2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2
2 2 2 2
o o o o
o o o o
x b y a xx b yy a
a b a b a b a b
x y xx yy
a b a b
- = - - = - 2 2
1
o o
xx yy
a
a b
= - From (1) 
7. Let
1
2
y x =
( )
1
2
0.0036, 0.0001 x x
y y x x
= ? =
+ ? = + ?
( )
( ) ( )
1
2
1 1
2 2
.
y x x y
dy
x x x x
dx
? = + ? - ? ?
? = + ? - ? ?
? ?
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