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

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

 Page 1


CBSE TEST PAPER-01 
CLASS - XII MATHEMATICS (Calculus : Continuity & differentiability) 
Topic: - differentiation 
1. 
lim lim
2 2
cos
( )
2 x x
K x
f x
x
p p
p
- - ? ?
=
- lim
0
cos
2
2
2
h
K h
h
p
p
p
?
? ?
- ? ?
? ?
=
? ?
- - ? ?
? ?
lim
0
sin h
2 2
2
h
K
h
p
p
?
=
- +
lim
0
sin h
2
h
k
h
?
=
lim
0
sin h
.
2
h
K
h
?
? ?
=
? ?
? ?
2
K
=
lim lim
2 2
( ) 3
x x
f x
p p
+
? ?
= 
lim
0
3 3
3
2 1
h
K
AT?
?
= =
=
2. Let y = u + v
When u = x 
sinx
, v = (sinx)
cosx
sin
(1)
x
dy du dv
dx dx dx
u x
= + - - - - =
Taking log both side
K = 6 
Page 2


CBSE TEST PAPER-01 
CLASS - XII MATHEMATICS (Calculus : Continuity & differentiability) 
Topic: - differentiation 
1. 
lim lim
2 2
cos
( )
2 x x
K x
f x
x
p p
p
- - ? ?
=
- lim
0
cos
2
2
2
h
K h
h
p
p
p
?
? ?
- ? ?
? ?
=
? ?
- - ? ?
? ?
lim
0
sin h
2 2
2
h
K
h
p
p
?
=
- +
lim
0
sin h
2
h
k
h
?
=
lim
0
sin h
.
2
h
K
h
?
? ?
=
? ?
? ?
2
K
=
lim lim
2 2
( ) 3
x x
f x
p p
+
? ?
= 
lim
0
3 3
3
2 1
h
K
AT?
?
= =
=
2. Let y = u + v
When u = x 
sinx
, v = (sinx)
cosx
sin
(1)
x
dy du dv
dx dx dx
u x
= + - - - - =
Taking log both side
K = 6 
log u = log x
sinx 
 
log u = sinx . logx 
diff. both side w.r. to x 
1 1
sin . log .cos
sin
log .cos
du
x x x
u dx x
du x
u x x
dx x
= +
? ?
= +
? ?
? ?
sin
cos
sin log .cos
(sin )
x
x
du x x x x
x
dx x
v x
+ ? ?
=
? ?
? ?
=
Taking log both side 
log v = log (sinx)
cosx 
log cos .log(sin ) v x x =
Differentiation both side w.r. to x 
[ ]
1 1
. cos . (cos ) log(sin )( sin )
sin
cot .cos log(sin ).sin
dv
x x x x
v dx x
dv
v x x x x
dx
= + - = - [ ]
cos
(sin ) cot .cos log(sin ).sin
x
dv
x x x x x
dx
= - Hence 
[ ]
sin cos
sin log .cos
(sin ) cot cos log(sin ).sin
x x
dy x x x x
x x x x x x
dx x
+ ? ?
= + - ? ?
? ?
3.
1
sin t
x a
- =
Square both side 
1
2 sin t
x a
- =
Differentiation 
1
sin
2
1
2 .log (1)
1
t
dx
x a a
dt
t
- = - - - - - - 1
1
cos
2 cos
t
t
y a
y a
- - =
? =
Page 3


CBSE TEST PAPER-01 
CLASS - XII MATHEMATICS (Calculus : Continuity & differentiability) 
Topic: - differentiation 
1. 
lim lim
2 2
cos
( )
2 x x
K x
f x
x
p p
p
- - ? ?
=
- lim
0
cos
2
2
2
h
K h
h
p
p
p
?
? ?
- ? ?
? ?
=
? ?
- - ? ?
? ?
lim
0
sin h
2 2
2
h
K
h
p
p
?
=
- +
lim
0
sin h
2
h
k
h
?
=
lim
0
sin h
.
2
h
K
h
?
? ?
=
? ?
? ?
2
K
=
lim lim
2 2
( ) 3
x x
f x
p p
+
? ?
= 
lim
0
3 3
3
2 1
h
K
AT?
?
= =
=
2. Let y = u + v
When u = x 
sinx
, v = (sinx)
cosx
sin
(1)
x
dy du dv
dx dx dx
u x
= + - - - - =
Taking log both side
K = 6 
log u = log x
sinx 
 
log u = sinx . logx 
diff. both side w.r. to x 
1 1
sin . log .cos
sin
log .cos
du
x x x
u dx x
du x
u x x
dx x
= +
? ?
= +
? ?
? ?
sin
cos
sin log .cos
(sin )
x
x
du x x x x
x
dx x
v x
+ ? ?
=
? ?
? ?
=
Taking log both side 
log v = log (sinx)
cosx 
log cos .log(sin ) v x x =
Differentiation both side w.r. to x 
[ ]
1 1
. cos . (cos ) log(sin )( sin )
sin
cot .cos log(sin ).sin
dv
x x x x
v dx x
dv
v x x x x
dx
= + - = - [ ]
cos
(sin ) cot .cos log(sin ).sin
x
dv
x x x x x
dx
= - Hence 
[ ]
sin cos
sin log .cos
(sin ) cot cos log(sin ).sin
x x
dy x x x x
x x x x x x
dx x
+ ? ?
= + - ? ?
? ?
3.
1
sin t
x a
- =
Square both side 
1
2 sin t
x a
- =
Differentiation 
1
sin
2
1
2 .log (1)
1
t
dx
x a a
dt
t
- = - - - - - - 1
1
cos
2 cos
t
t
y a
y a
- - =
? =
1
cos
2
1
2 log (2)
1
t
dy
y a a
dt
t
- - = - - - - - - 
Dividing (2) and (1) 
1
1
cos
2
sin
2
1
log
2
1
1
2 log
1
t
t
dy
a a
y
t dt
dx
x a a
dt
t
- - - - =
- 1
1
cos
sin
.
t
t
y dy a
x dx
a
- - = - 1
1
cos 2
2
2
sin 2
.
t
t
a y
y dy y
x dx x
a x
- - ? ?
=
= - ? ?
? ? =
? ?
?
dy y
dx x
- =
4. y = (tan
-1
 x)
2
 (given)
Differentiation both side w.r. to x
1
1 2
2 1
1
1
2 tan .
1
(1 ) 2 tan
y x
x
x y x
- - =
+
+ =
Again differentiation both side w.r. to 
2
2 1 2
2 2
2 1
1
(1 ) .(2 ) 2.
1
(1 ) 2 ( 1) 2
x y y x
x
x y x x y
+ + =
+
+ + + =
5. y = x
2
 + 2 is continuous in [-2, 2] and differentiable in (-2, 2). Also f (-2) = f(2) = 6
Hence all the condition of Rolle’s Theorem are verified hence their exist value c such
that
f ' (c) = 0
0 = 2c.  
C = 0  
Hence prove. 
Page 4


CBSE TEST PAPER-01 
CLASS - XII MATHEMATICS (Calculus : Continuity & differentiability) 
Topic: - differentiation 
1. 
lim lim
2 2
cos
( )
2 x x
K x
f x
x
p p
p
- - ? ?
=
- lim
0
cos
2
2
2
h
K h
h
p
p
p
?
? ?
- ? ?
? ?
=
? ?
- - ? ?
? ?
lim
0
sin h
2 2
2
h
K
h
p
p
?
=
- +
lim
0
sin h
2
h
k
h
?
=
lim
0
sin h
.
2
h
K
h
?
? ?
=
? ?
? ?
2
K
=
lim lim
2 2
( ) 3
x x
f x
p p
+
? ?
= 
lim
0
3 3
3
2 1
h
K
AT?
?
= =
=
2. Let y = u + v
When u = x 
sinx
, v = (sinx)
cosx
sin
(1)
x
dy du dv
dx dx dx
u x
= + - - - - =
Taking log both side
K = 6 
log u = log x
sinx 
 
log u = sinx . logx 
diff. both side w.r. to x 
1 1
sin . log .cos
sin
log .cos
du
x x x
u dx x
du x
u x x
dx x
= +
? ?
= +
? ?
? ?
sin
cos
sin log .cos
(sin )
x
x
du x x x x
x
dx x
v x
+ ? ?
=
? ?
? ?
=
Taking log both side 
log v = log (sinx)
cosx 
log cos .log(sin ) v x x =
Differentiation both side w.r. to x 
[ ]
1 1
. cos . (cos ) log(sin )( sin )
sin
cot .cos log(sin ).sin
dv
x x x x
v dx x
dv
v x x x x
dx
= + - = - [ ]
cos
(sin ) cot .cos log(sin ).sin
x
dv
x x x x x
dx
= - Hence 
[ ]
sin cos
sin log .cos
(sin ) cot cos log(sin ).sin
x x
dy x x x x
x x x x x x
dx x
+ ? ?
= + - ? ?
? ?
3.
1
sin t
x a
- =
Square both side 
1
2 sin t
x a
- =
Differentiation 
1
sin
2
1
2 .log (1)
1
t
dx
x a a
dt
t
- = - - - - - - 1
1
cos
2 cos
t
t
y a
y a
- - =
? =
1
cos
2
1
2 log (2)
1
t
dy
y a a
dt
t
- - = - - - - - - 
Dividing (2) and (1) 
1
1
cos
2
sin
2
1
log
2
1
1
2 log
1
t
t
dy
a a
y
t dt
dx
x a a
dt
t
- - - - =
- 1
1
cos
sin
.
t
t
y dy a
x dx
a
- - = - 1
1
cos 2
2
2
sin 2
.
t
t
a y
y dy y
x dx x
a x
- - ? ?
=
= - ? ?
? ? =
? ?
?
dy y
dx x
- =
4. y = (tan
-1
 x)
2
 (given)
Differentiation both side w.r. to x
1
1 2
2 1
1
1
2 tan .
1
(1 ) 2 tan
y x
x
x y x
- - =
+
+ =
Again differentiation both side w.r. to 
2
2 1 2
2 2
2 1
1
(1 ) .(2 ) 2.
1
(1 ) 2 ( 1) 2
x y y x
x
x y x x y
+ + =
+
+ + + =
5. y = x
2
 + 2 is continuous in [-2, 2] and differentiable in (-2, 2). Also f (-2) = f(2) = 6
Hence all the condition of Rolle’s Theorem are verified hence their exist value c such
that
f ' (c) = 0
0 = 2c.  
C = 0  
Hence prove. 
6. 
1
1
2
sin
1 4
x
x
y
+
- ? ?
=
? ?
+
? ?
 
1
2
2 .2
sin
1 (2 )
x
x
- ? ?
=
? ?
+
? ?
1
2
 2 tan
2 tan
sin
1 tan
x
Put ?
?
?
- =
? ?
=
? ?
+
? ?
 
1
sin (sin 2 )
2
?
?
- =
=
1
2
2.tan 2
1
2. . (2 )
1 (2 )
x
x
x
y
dy d
dx dx
- =
=
+
2
.2 .log 2
1 4
x
x
=
+
7.
2
sin u x =
cos
2sin .cos
x
du
x x
dx
v e
=
=
( )
cos
cos
sin
2sin .cos
.sin
x
x
dv
e x
dx
du x x
dv e x
= - =
- cos
2cos
x
x
e
=
8. 1 1 0 x y y x + + + =
1 1 x y y x + = - +
Square both side
Page 5


CBSE TEST PAPER-01 
CLASS - XII MATHEMATICS (Calculus : Continuity & differentiability) 
Topic: - differentiation 
1. 
lim lim
2 2
cos
( )
2 x x
K x
f x
x
p p
p
- - ? ?
=
- lim
0
cos
2
2
2
h
K h
h
p
p
p
?
? ?
- ? ?
? ?
=
? ?
- - ? ?
? ?
lim
0
sin h
2 2
2
h
K
h
p
p
?
=
- +
lim
0
sin h
2
h
k
h
?
=
lim
0
sin h
.
2
h
K
h
?
? ?
=
? ?
? ?
2
K
=
lim lim
2 2
( ) 3
x x
f x
p p
+
? ?
= 
lim
0
3 3
3
2 1
h
K
AT?
?
= =
=
2. Let y = u + v
When u = x 
sinx
, v = (sinx)
cosx
sin
(1)
x
dy du dv
dx dx dx
u x
= + - - - - =
Taking log both side
K = 6 
log u = log x
sinx 
 
log u = sinx . logx 
diff. both side w.r. to x 
1 1
sin . log .cos
sin
log .cos
du
x x x
u dx x
du x
u x x
dx x
= +
? ?
= +
? ?
? ?
sin
cos
sin log .cos
(sin )
x
x
du x x x x
x
dx x
v x
+ ? ?
=
? ?
? ?
=
Taking log both side 
log v = log (sinx)
cosx 
log cos .log(sin ) v x x =
Differentiation both side w.r. to x 
[ ]
1 1
. cos . (cos ) log(sin )( sin )
sin
cot .cos log(sin ).sin
dv
x x x x
v dx x
dv
v x x x x
dx
= + - = - [ ]
cos
(sin ) cot .cos log(sin ).sin
x
dv
x x x x x
dx
= - Hence 
[ ]
sin cos
sin log .cos
(sin ) cot cos log(sin ).sin
x x
dy x x x x
x x x x x x
dx x
+ ? ?
= + - ? ?
? ?
3.
1
sin t
x a
- =
Square both side 
1
2 sin t
x a
- =
Differentiation 
1
sin
2
1
2 .log (1)
1
t
dx
x a a
dt
t
- = - - - - - - 1
1
cos
2 cos
t
t
y a
y a
- - =
? =
1
cos
2
1
2 log (2)
1
t
dy
y a a
dt
t
- - = - - - - - - 
Dividing (2) and (1) 
1
1
cos
2
sin
2
1
log
2
1
1
2 log
1
t
t
dy
a a
y
t dt
dx
x a a
dt
t
- - - - =
- 1
1
cos
sin
.
t
t
y dy a
x dx
a
- - = - 1
1
cos 2
2
2
sin 2
.
t
t
a y
y dy y
x dx x
a x
- - ? ?
=
= - ? ?
? ? =
? ?
?
dy y
dx x
- =
4. y = (tan
-1
 x)
2
 (given)
Differentiation both side w.r. to x
1
1 2
2 1
1
1
2 tan .
1
(1 ) 2 tan
y x
x
x y x
- - =
+
+ =
Again differentiation both side w.r. to 
2
2 1 2
2 2
2 1
1
(1 ) .(2 ) 2.
1
(1 ) 2 ( 1) 2
x y y x
x
x y x x y
+ + =
+
+ + + =
5. y = x
2
 + 2 is continuous in [-2, 2] and differentiable in (-2, 2). Also f (-2) = f(2) = 6
Hence all the condition of Rolle’s Theorem are verified hence their exist value c such
that
f ' (c) = 0
0 = 2c.  
C = 0  
Hence prove. 
6. 
1
1
2
sin
1 4
x
x
y
+
- ? ?
=
? ?
+
? ?
 
1
2
2 .2
sin
1 (2 )
x
x
- ? ?
=
? ?
+
? ?
1
2
 2 tan
2 tan
sin
1 tan
x
Put ?
?
?
- =
? ?
=
? ?
+
? ?
 
1
sin (sin 2 )
2
?
?
- =
=
1
2
2.tan 2
1
2. . (2 )
1 (2 )
x
x
x
y
dy d
dx dx
- =
=
+
2
.2 .log 2
1 4
x
x
=
+
7.
2
sin u x =
cos
2sin .cos
x
du
x x
dx
v e
=
=
( )
cos
cos
sin
2sin .cos
.sin
x
x
dv
e x
dx
du x x
dv e x
= - =
- cos
2cos
x
x
e
=
8. 1 1 0 x y y x + + + =
1 1 x y y x + = - +
Square both side
2 2
2 2 2 2
2 2 2 2
(1 ) (1 )
0
x y y x
x x y y xy
x y x y xy
+ = +
+ = +
- + - =
( )( ) ( ) 0
( )[ ] 0
x y x y xy x y
x y x y xy
- + + - =
- + + =
0
(1 )
x y xy
y x x
+ + =
+ = - 1
x
y
x
- =
+
2
(1 )(1) ( )(1)
(1 )
dy x x
dx x
? ? + - = - ? ?
+
? ?
2
2
1
(1 )
1
(1 )
x x
x
dy
dx x
? ? + - = - ? ?
+
? ?
- =
+
9. cos .cos( )    ( ) y x a y given = +
cos
cos( )
y
x
a y
=
+
2
cos( )( sin ) cos .( sin( ))
cos ( )
dx a y y y a y
dy a y
+ - - - +
=
+
2
cos .sin( ) sin cos( )
cos ( )
dx y a y y a y
dy a y
+ - +
=
+
2
2
sin( )
cos ( )
cos ( )
sin
dx a y y
dy a y
dy a y
dx a
+ - =
+
+
=
10. (cos .sin ) x a t t t = +
[ ] sin .cos sin .1
[ .cos ] (1)
dx
a t t t t
dt
a t t
= - + +
= - - - - -
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