Solution Differentiation Test-3 - Continuity & Differentiability Class 12 Notes | EduRev

Class 12 : Solution Differentiation Test-3 - Continuity & Differentiability Class 12 Notes | EduRev

 Page 1


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Calculus : Continuity and Differentiability) 
 [ANSWERS] 
Topic: - Differentiability 
1. 
( )
lim lim
( ) 1
x x
f x K x
p p
- - ? ?
= +
( )
lim
0
1
h
K h p
?
= - + ? ?
? ?
1 Kp = + 
lim lim
( ) cos
x x
f x x
p p
+ +
? ?
=
lim lim
0 0
cos( h) cosh
h h
p
? ?
= - + = - cos0 1 = - = 
AT? 
1 1 Kp + = - 
2
K
p
- =
2.
2
2 cot y x =
( ) ( )
1
2 2
2
1
2. cot . cot
2
dy d
x x
dx dx
- =
2 2
2
1
cos .2
cot
ec x x
x
= - 2 2
2
2 .cos
cot
x ec x
x
- =
3.
2
sin cos y xy p + =
. diff
2sin cos sin ( .1) 0
dy dy
y y xy x y
dx dx
- + =
2sin .cos .sin .sin 0
dy dy
y y x xy y xy
dx dx
- - = 
(sin 2 .sin ) sin
dy
y x xy y xy
dx
- =
.sin
sin 2 .sin
dy y xy
dx y x xy
=
- 4. Let ,  v ,  w
x y x
u y x x = = =
b
u v w a + + =
Page 2


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Calculus : Continuity and Differentiability) 
 [ANSWERS] 
Topic: - Differentiability 
1. 
( )
lim lim
( ) 1
x x
f x K x
p p
- - ? ?
= +
( )
lim
0
1
h
K h p
?
= - + ? ?
? ?
1 Kp = + 
lim lim
( ) cos
x x
f x x
p p
+ +
? ?
=
lim lim
0 0
cos( h) cosh
h h
p
? ?
= - + = - cos0 1 = - = 
AT? 
1 1 Kp + = - 
2
K
p
- =
2.
2
2 cot y x =
( ) ( )
1
2 2
2
1
2. cot . cot
2
dy d
x x
dx dx
- =
2 2
2
1
cos .2
cot
ec x x
x
= - 2 2
2
2 .cos
cot
x ec x
x
- =
3.
2
sin cos y xy p + =
. diff
2sin cos sin ( .1) 0
dy dy
y y xy x y
dx dx
- + =
2sin .cos .sin .sin 0
dy dy
y y x xy y xy
dx dx
- - = 
(sin 2 .sin ) sin
dy
y x xy y xy
dx
- =
.sin
sin 2 .sin
dy y xy
dx y x xy
=
- 4. Let ,  v ,  w
x y x
u y x x = = =
b
u v w a + + =
Therefore 0
du dw dv
dx dx dx
+ + =
x
u y = 
Taking log both side 
log log
x
u y =
log .log u x y =
Differentiate both side w.r.t. to x 
1 1
. . . log .1
du dy
x y
u dx y dx
= + 
. log
du x dy
u y
dx y dx
? ?
= +
? ?
? ?
 
. log
x
du x dy
y y
dx y dx
? ?
= +
? ?
? ?
 
y
v x = 
Taking log both side 
log log
y
v x =
log .log v y x =
1 1
. . log .
dv dy
y x
v dx x dx
= +
log .
dv y dy
v x
dx x dx
? ?
= +
? ?
? ?
log .
y
dv y dy
x x
dx x dx
? ?
= +
? ?
? ?
x
w x = 
Taking log both side 
log log
x
w x =
log log w x x = 
1 1
. . log .1
dw
x x
w dx x
= + 
1
. 1 log
dw
x
w dx
= +
Page 3


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Calculus : Continuity and Differentiability) 
 [ANSWERS] 
Topic: - Differentiability 
1. 
( )
lim lim
( ) 1
x x
f x K x
p p
- - ? ?
= +
( )
lim
0
1
h
K h p
?
= - + ? ?
? ?
1 Kp = + 
lim lim
( ) cos
x x
f x x
p p
+ +
? ?
=
lim lim
0 0
cos( h) cosh
h h
p
? ?
= - + = - cos0 1 = - = 
AT? 
1 1 Kp + = - 
2
K
p
- =
2.
2
2 cot y x =
( ) ( )
1
2 2
2
1
2. cot . cot
2
dy d
x x
dx dx
- =
2 2
2
1
cos .2
cot
ec x x
x
= - 2 2
2
2 .cos
cot
x ec x
x
- =
3.
2
sin cos y xy p + =
. diff
2sin cos sin ( .1) 0
dy dy
y y xy x y
dx dx
- + =
2sin .cos .sin .sin 0
dy dy
y y x xy y xy
dx dx
- - = 
(sin 2 .sin ) sin
dy
y x xy y xy
dx
- =
.sin
sin 2 .sin
dy y xy
dx y x xy
=
- 4. Let ,  v ,  w
x y x
u y x x = = =
b
u v w a + + =
Therefore 0
du dw dv
dx dx dx
+ + =
x
u y = 
Taking log both side 
log log
x
u y =
log .log u x y =
Differentiate both side w.r.t. to x 
1 1
. . . log .1
du dy
x y
u dx y dx
= + 
. log
du x dy
u y
dx y dx
? ?
= +
? ?
? ?
 
. log
x
du x dy
y y
dx y dx
? ?
= +
? ?
? ?
 
y
v x = 
Taking log both side 
log log
y
v x =
log .log v y x =
1 1
. . log .
dv dy
y x
v dx x dx
= +
log .
dv y dy
v x
dx x dx
? ?
= +
? ?
? ?
log .
y
dv y dy
x x
dx x dx
? ?
= +
? ?
? ?
x
w x = 
Taking log both side 
log log
x
w x =
log log w x x = 
1 1
. . log .1
dw
x x
w dx x
= + 
1
. 1 log
dw
x
w dx
= +
(1 log )
dw
w x
dx
= + 
(1 log )
x
dw
x x
dx
= +
1
1
(1 log ) . log
. log .
x y x
x y
dw x x y x y y
dx x y x x
- - - + - - =
+
5. [ ] [ ] 1 cos , 0 cos
dx dy
a a
d d
? ?
? ?
= - = - sin
(1 cos )
dy a
d a
?
? ?
- =
- 2
2sin cos
2 2
tan
2
2sin
2
? ?
?
?
- = = - 
6.
2 3
3 2
x x
y e e = +
2 3
3 .2 2 .3
x x
dy
e e
dx
= + 
=
2
2 3
2
6 .2 6 .3
x x
d y
e e
dx
= + 
2 3
12 18
x x
e e = +
LHS
2
2
5 6
d y dy
y
dx dx
= - +
( ) ( )
2 3 2 3 2 3
12 18 5 6 6 6(3 2 )
x x x x x x
e e e e e e = + - + + +
2 3 2 3 2 3
12 18 30 30 18 12
x x x x x x
e e e e e e = + - - + + 
0 = 
7. 
1
cos x
a
y e
- =
1
cos
2
1
. .
1
x
a
dy
a e
dx
x
- = - - 1
cos
2
1 .
x
a
dy
x e a
dx
- - = - 1
cos
2
2
2
2 2
1( 2) ( 1)
1
2 1 1
x
a
d y dy
x ae a
dx dx
x x
- - - - + = - - -
Page 4


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Calculus : Continuity and Differentiability) 
 [ANSWERS] 
Topic: - Differentiability 
1. 
( )
lim lim
( ) 1
x x
f x K x
p p
- - ? ?
= +
( )
lim
0
1
h
K h p
?
= - + ? ?
? ?
1 Kp = + 
lim lim
( ) cos
x x
f x x
p p
+ +
? ?
=
lim lim
0 0
cos( h) cosh
h h
p
? ?
= - + = - cos0 1 = - = 
AT? 
1 1 Kp + = - 
2
K
p
- =
2.
2
2 cot y x =
( ) ( )
1
2 2
2
1
2. cot . cot
2
dy d
x x
dx dx
- =
2 2
2
1
cos .2
cot
ec x x
x
= - 2 2
2
2 .cos
cot
x ec x
x
- =
3.
2
sin cos y xy p + =
. diff
2sin cos sin ( .1) 0
dy dy
y y xy x y
dx dx
- + =
2sin .cos .sin .sin 0
dy dy
y y x xy y xy
dx dx
- - = 
(sin 2 .sin ) sin
dy
y x xy y xy
dx
- =
.sin
sin 2 .sin
dy y xy
dx y x xy
=
- 4. Let ,  v ,  w
x y x
u y x x = = =
b
u v w a + + =
Therefore 0
du dw dv
dx dx dx
+ + =
x
u y = 
Taking log both side 
log log
x
u y =
log .log u x y =
Differentiate both side w.r.t. to x 
1 1
. . . log .1
du dy
x y
u dx y dx
= + 
. log
du x dy
u y
dx y dx
? ?
= +
? ?
? ?
 
. log
x
du x dy
y y
dx y dx
? ?
= +
? ?
? ?
 
y
v x = 
Taking log both side 
log log
y
v x =
log .log v y x =
1 1
. . log .
dv dy
y x
v dx x dx
= +
log .
dv y dy
v x
dx x dx
? ?
= +
? ?
? ?
log .
y
dv y dy
x x
dx x dx
? ?
= +
? ?
? ?
x
w x = 
Taking log both side 
log log
x
w x =
log log w x x = 
1 1
. . log .1
dw
x x
w dx x
= + 
1
. 1 log
dw
x
w dx
= +
(1 log )
dw
w x
dx
= + 
(1 log )
x
dw
x x
dx
= +
1
1
(1 log ) . log
. log .
x y x
x y
dw x x y x y y
dx x y x x
- - - + - - =
+
5. [ ] [ ] 1 cos , 0 cos
dx dy
a a
d d
? ?
? ?
= - = - sin
(1 cos )
dy a
d a
?
? ?
- =
- 2
2sin cos
2 2
tan
2
2sin
2
? ?
?
?
- = = - 
6.
2 3
3 2
x x
y e e = +
2 3
3 .2 2 .3
x x
dy
e e
dx
= + 
=
2
2 3
2
6 .2 6 .3
x x
d y
e e
dx
= + 
2 3
12 18
x x
e e = +
LHS
2
2
5 6
d y dy
y
dx dx
= - +
( ) ( )
2 3 2 3 2 3
12 18 5 6 6 6(3 2 )
x x x x x x
e e e e e e = + - + + +
2 3 2 3 2 3
12 18 30 30 18 12
x x x x x x
e e e e e e = + - - + + 
0 = 
7. 
1
cos x
a
y e
- =
1
cos
2
1
. .
1
x
a
dy
a e
dx
x
- = - - 1
cos
2
1 .
x
a
dy
x e a
dx
- - = - 1
cos
2
2
2
2 2
1( 2) ( 1)
1
2 1 1
x
a
d y dy
x ae a
dx dx
x x
- - - - + = - - - 1
cos
2 2
2
2
2 2
1
1 .
1 1
x
a
d y dy a e
x
dx dx
x x
- - - =
- - ( )
1
cos
2
2 2
2
1
x
a
d y dy
x a e
dx dx
- - - =
( )
2
2 2
2
1 0
d y dy
x a y
dx dx
- - - = 
8.
2 2 2
( ) ( )   ( ) (1) x a y b C Given - + - = - - - - - - - Diff. both side w.r.t. to x
1
2( ) ( ) 0 x a y b y - + - =
1
( ) ( ) 0 (2) x a y b y - + - = - - - - - - - - - - - - - Again diff. both side
2 1 1
(1 0) ( ) . 0 y b y y y - + - + =
2
2 1
1 ( ) 0 y b y y + - + =
2
1
2
1 y
y b
y
- - - =
Put (y-b) in equation (1) 
2
1
1
2
1
( ) 0
y
x a y
y
? ? +
- - =
? ?
? ?
 
2
1
1
2
1
.
y
x a y
y
? ? +
- =
? ?
? ?
Put the value of (x-a) and (y-b) in equation (1) 
2 2
2 2
2 1 1
1
2 2
1 1 y y
y C
y y
? ? ? ? ? ? + +
+ =
? ? ? ? ? ?
? ? ? ? ? ?
( ) ( )
2 2
2 2 2
1 1 1
2
2 2
2 2
1 1 y y y
C
y y
+ +
+ =
( )( )
2 2
1 1
2
2
2
1 1 y y
C
y
+ +
=
( )
3
2
1
2
2
1 y
C
y
+
± =
Page 5


CBSE TEST PAPER-03 
CLASS - XII MATHEMATICS (Calculus : Continuity and Differentiability) 
 [ANSWERS] 
Topic: - Differentiability 
1. 
( )
lim lim
( ) 1
x x
f x K x
p p
- - ? ?
= +
( )
lim
0
1
h
K h p
?
= - + ? ?
? ?
1 Kp = + 
lim lim
( ) cos
x x
f x x
p p
+ +
? ?
=
lim lim
0 0
cos( h) cosh
h h
p
? ?
= - + = - cos0 1 = - = 
AT? 
1 1 Kp + = - 
2
K
p
- =
2.
2
2 cot y x =
( ) ( )
1
2 2
2
1
2. cot . cot
2
dy d
x x
dx dx
- =
2 2
2
1
cos .2
cot
ec x x
x
= - 2 2
2
2 .cos
cot
x ec x
x
- =
3.
2
sin cos y xy p + =
. diff
2sin cos sin ( .1) 0
dy dy
y y xy x y
dx dx
- + =
2sin .cos .sin .sin 0
dy dy
y y x xy y xy
dx dx
- - = 
(sin 2 .sin ) sin
dy
y x xy y xy
dx
- =
.sin
sin 2 .sin
dy y xy
dx y x xy
=
- 4. Let ,  v ,  w
x y x
u y x x = = =
b
u v w a + + =
Therefore 0
du dw dv
dx dx dx
+ + =
x
u y = 
Taking log both side 
log log
x
u y =
log .log u x y =
Differentiate both side w.r.t. to x 
1 1
. . . log .1
du dy
x y
u dx y dx
= + 
. log
du x dy
u y
dx y dx
? ?
= +
? ?
? ?
 
. log
x
du x dy
y y
dx y dx
? ?
= +
? ?
? ?
 
y
v x = 
Taking log both side 
log log
y
v x =
log .log v y x =
1 1
. . log .
dv dy
y x
v dx x dx
= +
log .
dv y dy
v x
dx x dx
? ?
= +
? ?
? ?
log .
y
dv y dy
x x
dx x dx
? ?
= +
? ?
? ?
x
w x = 
Taking log both side 
log log
x
w x =
log log w x x = 
1 1
. . log .1
dw
x x
w dx x
= + 
1
. 1 log
dw
x
w dx
= +
(1 log )
dw
w x
dx
= + 
(1 log )
x
dw
x x
dx
= +
1
1
(1 log ) . log
. log .
x y x
x y
dw x x y x y y
dx x y x x
- - - + - - =
+
5. [ ] [ ] 1 cos , 0 cos
dx dy
a a
d d
? ?
? ?
= - = - sin
(1 cos )
dy a
d a
?
? ?
- =
- 2
2sin cos
2 2
tan
2
2sin
2
? ?
?
?
- = = - 
6.
2 3
3 2
x x
y e e = +
2 3
3 .2 2 .3
x x
dy
e e
dx
= + 
=
2
2 3
2
6 .2 6 .3
x x
d y
e e
dx
= + 
2 3
12 18
x x
e e = +
LHS
2
2
5 6
d y dy
y
dx dx
= - +
( ) ( )
2 3 2 3 2 3
12 18 5 6 6 6(3 2 )
x x x x x x
e e e e e e = + - + + +
2 3 2 3 2 3
12 18 30 30 18 12
x x x x x x
e e e e e e = + - - + + 
0 = 
7. 
1
cos x
a
y e
- =
1
cos
2
1
. .
1
x
a
dy
a e
dx
x
- = - - 1
cos
2
1 .
x
a
dy
x e a
dx
- - = - 1
cos
2
2
2
2 2
1( 2) ( 1)
1
2 1 1
x
a
d y dy
x ae a
dx dx
x x
- - - - + = - - - 1
cos
2 2
2
2
2 2
1
1 .
1 1
x
a
d y dy a e
x
dx dx
x x
- - - =
- - ( )
1
cos
2
2 2
2
1
x
a
d y dy
x a e
dx dx
- - - =
( )
2
2 2
2
1 0
d y dy
x a y
dx dx
- - - = 
8.
2 2 2
( ) ( )   ( ) (1) x a y b C Given - + - = - - - - - - - Diff. both side w.r.t. to x
1
2( ) ( ) 0 x a y b y - + - =
1
( ) ( ) 0 (2) x a y b y - + - = - - - - - - - - - - - - - Again diff. both side
2 1 1
(1 0) ( ) . 0 y b y y y - + - + =
2
2 1
1 ( ) 0 y b y y + - + =
2
1
2
1 y
y b
y
- - - =
Put (y-b) in equation (1) 
2
1
1
2
1
( ) 0
y
x a y
y
? ? +
- - =
? ?
? ?
 
2
1
1
2
1
.
y
x a y
y
? ? +
- =
? ?
? ?
Put the value of (x-a) and (y-b) in equation (1) 
2 2
2 2
2 1 1
1
2 2
1 1 y y
y C
y y
? ? ? ? ? ? + +
+ =
? ? ? ? ? ?
? ? ? ? ? ?
( ) ( )
2 2
2 2 2
1 1 1
2
2 2
2 2
1 1 y y y
C
y y
+ +
+ =
( )( )
2 2
1 1
2
2
2
1 1 y y
C
y
+ +
=
( )
3
2
1
2
2
1 y
C
y
+
± =
( )
3
2
3
2
2
1
2
2
1 y
C
y
+
± =
3
2
2
2
2
1
dy
dx
C
d y
dx
? ?
? ?
+
? ?
? ?
? ?
? ?
? ?
=
Hence prove 
9.
1 1 2
sin sin 1 y x x
- - = + - Differentiate both side w.r.t. x 
( )
( )
2
2 2
2
1 1
1
1
1 1
dy d
x
dx dx
x
x
= + - - - - ( )
2 2 2
1 2 1 1
1 1 (1 ). 2 1
x
x x x
- +
- - - - ( )
2 2 2
2
1 1
.
1 1
1 1
x
x x
x
? ? - = +
? ?
- - ? ?
- - 
2 2 2
1 1
1 1
x
x x x
? ? - = +
? ?
- - ? ?
2 2
1 1
1 1
x
x
x x
? ? - = +
? ?
- - ? ?
2 2
1 1
0
1 1 x x
= - =
- - 
10.
sin cos
(sin cos )
x x
y x x
- = - Taking log both side 
sin cos
log log(sin cos )
x x
y x x
- = - log (sin cos ).log(sin cos ) y x x x x = - - Differentiate both side w.r.t. x 
1 1
. (sin cos ). (cos sin ) log(sin cos ).(cos sin )
(sin cos )
dy
x x x x x x x x
y dx x x
= - + + - +
- 
( ) ( ) ( ) (cos sin ) log (sin cos ) . cos sin
dy
y x x x x x x
dx
= + + - + ? ?
? ?
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shortcuts and tricks

;