MEET II - AUGUST 2016
MATHEMATICS - Hints and Solutions
1.

1
x sin ;
f(x) = 
x

0
;

x0
6.
x0
Lt f( x )  0 and f(0) = 0 since -1  sin
x 0
1
 1.
x
f(0  h )  f( 0)
 Lt
h
h 0

(0  h ) sin  
h
h
1
1
 since -1  sin
 1
h
h
 
dy
1
   
1

= tan tan     
dx
2
4 2 4 2
= Lt sin 
h 0
then limit oscillalate, between -1 1. Hence f(x) is not
differentiable at x = 0
2.
7.
x 1  y  y 1  x
f(x) = sin-1 (cos x) + cos-1 (sin x)




= sin-1 sin   x  + cos-1 cos   x 
2

2

x2(1 + y) = y2(1 + x)  x2 - y2 = xy2 - x2y
(x + y) (x - y) = xy (y - x) x  y if x = y do not satisfy
the given equation
 x + y = -xy y + xy = -x
=
dy
x
1
y(1 + x) = -x y = 1  x  dx  (1  x )2
3.


 
 sin  2  x  
cos x 

 
1
y  tan 
  tan 

1

sin
x


 1  cos  x  



2

1 

 
   
 2 sin  4  2  cos 4  2  




= tan 1 

x




2 cos 2   


4 2
 f(x) is continuous
RHS limit of f’(0) = hLt
0
C


 x   x    2x
2
2
 f' (x )  2
8.
xmyn = (x + y)m+n
m log x + n log y = (m + n) log(x + y)
Given x2 + y2 = t x4 + y4 = t2 +
m n dy m  n 
dy 
 x  y dx  x  y  1  dx 


1
t
1
1
2
2 2
2 2
2
2  (x + y ) - 2x y = t + 2
t
t
2

 1
  t    2x2 y2  t 2  2
t
t

m m  n  m  n n  dy
 

 
x x  y  x  y y  dx
2

4.
1
  
 2x2 y 2   t     t 2  2 
t 
 t 
my  nx my  nx dy
dy y



x( x  y ) ( x  y )y dx
dx x
f ( x)  f ( y )
We have f’(y) = xLt
 y (x  y)
2
=t 
|f ( x)  f ( y )|
( x  y )2
 Lt
x y |x  y|
x y |x  y|
1
t
2
 2  t2 
t2
 x 2 y 2  1  y 2 
|f '( y)| Lt
 Lt |x  y| 0
x y
Differentiating 2y
|f '( y)| 0  f'( y)  0  y  R
which shows that f’(x) = 0  x  R
 x3y
 f(x) = a constant  given f(0) = 0  f(1) = 0
5.
1
 2
1
x2
dy
2 2
  3   3
dx
x  x
dy
1
dx
sin y = x sin (a + y)
sin y
dx sin( a  y ) cos y  cos( a  y ). sin y
x = sin(a  y )  dy 
sin 2 ( a  y )
=
sin(a  y  y )
sin 2 ( a  y )

9.
y = tan-1
dy sin 2 ( a  y )
sin a


dx
sin a
sin 2 (a  y)


= tan
1

x 
 5  
a
= tan-1    2  
a 2  6x 2
x
 1  6  
 a  

5ax

3x 2 x


3x
2x
a
a
  tan 1    tan 1  
 1   3 x ,  2 x  
 a 
 a 
   

 a   a 

1 

dy

dx
1
 3x 
1 
 a 
2

3

a
1
 2x 
1

 a 
2

2
a
 sin  =
3
 sin (2) = 2 sin  cos 
5
 3 4  24

= 2. 
 5 5  25
3a
2a

a 2  9x 2 a2  4x 2
x = 2 cos t - cos 2t
=
10.
14.
dx
 2 sin t  2 sin 2 t  2(sin 2 t  sin t )
dt
y = 2 sin t - sin 2t


 6 


6
- 2 sin      sin ( 5   )
2



dx
 2 cos t  2 cos t  2 cos 2 t  2(cos 2 t  cos t )
dt
= 3[( cos)4  (  sin )4 ] - 2 {cos6  + sin6 }
= +3 (cos4  + sin4 ) - 2 (cos6 + sin6 )
= +3 ((sin2  + cos2  )2 - 2 (sin2  cos2 )
- 2 ((cos2 + sin2 )3 - 3 sin2  cos2 ) (sin2  +cos2 ))
= +3(1 - 2sin2  cos2 ) - 2 (1 - 3sin2 cos2 ) = 3 - 2 = 1
3t
t
sin
dy  2(cos 2 t  cos t )
 3t 
2
2


  tan  
3t
t
dx  2(sin 2 t  sin t )
2 
2 cos sin
2
2
 2 sin
3
3t
. sec 2
d y d
3t  dt
2
2


tan
.



2  dx  2(sin 2 t  sin t )
dx 2 dt 

2
15.
 3t 
sec 2  
8
 2

4 2 cos 3 t . sin t
2
2
=

16.
2
=
11.
cos A cos B sin A cos B  cos A sin B

cot A  cot B sin A sin A
sin A sin B


tan A  tan B sin A  sin B sin A cos B  cos A sin B
cos A cos B
cos A cos B
=
3
 3t 
t
, sec 2  . cos ec 
8
2
2
At = t =
 4  3


    sin 4 ( 3  )
3 sin 
 2



 d y 3

 3 
 . sec 3 
 - cosec  
,
2 dx 2 8
4
4


cos A cos B
 cot A cot B
sin A sin B
cot A  cot B
 tan A  tan B  cot A cot B  cot B cot C  cot C cot A
=1
 If A + B + C = 
then cot A cot B + cot B cot C + cot C cot A = 1
f’(x) = cos(log x)
y = f(2x + 3)
dy
d
 f' ( 2 x  3 )
( 2 x  3)
dx
dx
3
3
( 2 )3 ( 2 )  
8
2
d
( 2 x  3 )  2 f ' ( 2 x  3 )  2 cos log( 2 x  3 )
dx

2 


    and  
3
6
1

sin(  )       
2
6
sin(  )  1     
17.
Put x2 = cos (2)  2 = cos-1 x2   
1  x2  1  x2
2
1 x  1x
 2 
 5 
 tan  
 tan () . tan (2) = tan 
 3 
 6 

 



= tan     tan     =   tan 
3
6 
3


=
12.
3
1
3
1  tan 2 (7½)o
1  tan 2 (7½)o
=


  tan 
6

=
1
13.
 cos(2  7½)o 

1
3
1 1
3 1


 
2
2
2 2
2 2
25 cos2  + 5 cos  - 12 = 0
 (5 cos  - 3) (5 cos  + 4) = 0
 cos =
1  cos 2   1  cos 2 
1  cos 2   1  cos 2 
2 cos 2   2 sin 2 
2 cos 2   2 sin 2 
cos   sin  1  tan 



 tan   
cos   sin  1  tan 
4


  tan 1 tan     



4



 1
    cos  1 ( x 2 )
4
4 2
1
2x

Derivative =  2 .
1  x4
3
4
or cos  =
5
5

4
      in the 2nd quadrant  cos  =
2
5

 1  x2  1  x 2
 tan  1 
 1  x2  1  x2

cos15o = cos (45 - 30) = cos 45 cos 30 + sin45 sin 30
=
2
18.

x
1
u = tan 
2
1 1x




Put x = sin  = sin-1 x
2
1
cos-1 (x2)
2
x
1  x4
22.
 1  sin  

then u = tan 
 1  cos  
y = (x + 1) (x + 2) (x + 3) (x + 4) (x + 5)
dy
there will be 5 different terms in which there
dx
will be only one term which do not contains (x + 2).
All the other terms will be zero when we substitute
x = -2
In



 2 sin cos
2
2
= tan
 2 cos2 

2

1 





  tan 1 tan  
2

  2


 dy 


 dx ( x  2 ) = (-2 + 1) (-2 + 3) (-2 + 4) (-2 + 5)
1
V 1
1
1
sin1  x)


2
dx 2 1  x2 2 1  x2
= -1 × 1 × 2 × 3 = 6

1x
v = sin  2 cot 1

1x





23.
y = sin-1(xy)  sin y = xy  cos y
dy
dy
y
=x
dx
dx
put x = cos     cos1 x

1
v = sin  2 cot





 cos
1  cos  
1
2

 sin 2 cot

1  cos  
 sin 


2


dy
dy
y
 (cos y - x) dx  y  dx  cos y  x






24.
   


1
= sin  2 cot cot  = sin  2 ,   
2


  2 
 4x  3 1  x 2

5

y = cos-1 




5
5

4
= sin  =
1  cos2  =
du
1

 2x 
dx 2 1  x 2
3

1  4
4
Put x = cos   y  cos  cos   sin  
5
5
 cos  = &
1  x2
5
x
 y  cos 1 cos  cos   sin  sin  
1  x2
1
du
du dx
1
2 1  x2




x
dv dv
2x
2
dx
1x
19.
y=
where tan  =

2
 y  log x  y
20.
dy 1 dy
dy 1
 
 ( 2 y  1)

dx x dx
dx x
y = log ( x  x2  1 ) 
dy
1

dx x  x2  1
 x2  1  x 
1


 y1 


( x  x 2  1 ) 
x 2  1 
25.


 1  1  2x 


2 x2  1 

dy

dx
x  1 y2  y1 .
2 x2  1
1
(1  x )(nx n 1 )  (1  x n )( 1)
2
x 1
=
(1  x )2

nxn 1(1  x)  (1  xn )
2
 0  ( x  1)y 2  xy1  0
(1  x)2
1
 1 + 2x + 3x2 + .....  = (1  x )2
y
x
 ( x  1)y 2   xy 1  2  2
y1 x  1
since |x| < 1 and as n   xn, xn-1  
y = (x - 1) (x - 2) (x - 3) (x - 4) (x - 5)
Put u = x - 3
then y = (u + 2) (u + 1)u (u - 1) (u - 2)
= u (u2 - 1) (u2 - 4) = u (u4 - 5u2 + 4) = u5 - 5u3 + 4u
dy dy du

.
 ( 5u 4  15u 2  4 ).1
dx du dx
= 5(x - 3)4 - 15 ( x - 3)2 + 4
1  xn
1x
1 + 2x + 3x2 + ..... + (n - 1) xn-2
2
21.
1  x2
Differentiating both sides
Differentiating again
1  2x
1
1 + x + x2 + x3 + ...... + xn-1 =
 x 2  1 y1  1
2
3
4
3
 cos 1 cos(   )      cos 1 x  tan 1  
4
log x  y where y log x  log x  log x 
 2 y.
sin  =
26.
f(x) = |x|3
 x 3 for x  0
 x 3 for x  0
f(x) = 
3
3
5
 3 x 2 for x  0

f’(x) = 
=
1
1
3 1
sin(180  80 )  
 sin 80
4
4 2
4
=
1
3 1
 3
sin 80 
 sin 80 
4
8
4
8
  3 x 2 for x  0
 6 x for x  0
f”(x)= 
  6 x for x  0
 f”(4) = 6 × 4 = 24
27.
Given expression
2
2
2
2
2
 
3  
5

 1  cos   1  cos
  1  cos
4  
4  
4

=
2
2
2
 
 

 
 

 
 

2
7 
 
  1  cos

4 
 
2
 

 

 

= 2 cos
2
2  2  2  2  2 cos
31.
2
2

1
1  
1   1
1 3

   1 
    2  1   
 2  1 
4
2 2

2 
2  2



=
(C  D ) 
A B

CD
cos
  cos 180 
   cos

2
 2 


 2 
AB
 (C  D) 
 cos
  cos
0
2
2




32.
=
1
1
cos 20 sin 80  sin 80
2
4
2

2
 1  tan

 1  tan 2


2

2






2
 2 sec 
cos 
We have cos  + cos (120 + ) + cos (- 120) = 0
= 3 cos  . cos(120 + ) cos( - 120)
If a + b + c = 0 then a3 + b3 + c3 = 3abc
= 3 cos  ( cos ( + 120) cos ( - 120)
1
[cos 20  cos 60] sin 80
2
2 sin A sin B = cos(A-B) - cos (A + B)
1
1
[cos 20  ]sin 80
2
2
2
cos3  + cos3 (120 + ) + cos3( - 120)
1
(2 sin 40 sin 20) sin 80
2
=


1  tan
2
2

Given expression


1  tan
1  tan
2
2
2
AB
(C  D )
 180 
2
2
=


 2 cos
16
16
 




2 1  tan 2 
 1  tan    1  tan 
2
2
2

 
  
=
2 
2 
1  tan
1  tan
2
2
For thequadrilateral A, B, C, D  A + B + C + D = 360o
sin 20 sin 40 sin 80 =



 2 1  cos 
8
8

1  tan
A + B = 360 - (C + D )
29.
2.2 cos 2
=
2
2
2
2
1 
1  
1  
1  
1  
 1 









1


1


1

4 
2  
2  
2  
2  




8
2




2 
2 3 
2 5 
2 7 
   cos
   cos

=  cos    cos
8
8
8
8 

 
 
 
28.

1 



  2 1  cos   2.2 cos2
2  2  2 1 
4
8
2




30.
= 3 cos  ( cos2 - sin2 120)
cos (A + B) cos (A - B) = cos2 A - sin2 B
2

 3  

  sin 120  sin(90  30)
3 cos   cos 2   
 2  


 

= cos 30 =
=
1
1
.( 2 sin 80. cos 20 )  . sin 80
4
4
3
2
3
= 3 cos   cos 2   3   3 ( 4 cos   3 cos  )  3 cos( 3)

1
1
= [sin 100  sin 60 ]  . sin 80
4
4
33.
4
4
4
f(2) = 4 and f’(2) = 1
xf( 2 )  2 f(x )
f( 2 )  2f' (x )
 Lt
x 2
x 2
x2
1
Lt
1
1
1
= sin 100  sin 60  sin 80
4
4
4
L Hospital’s rule
4
= f(2) - 2f’(2)
34.
1
 m 2 ( x  x 2  1 )m 
=4-2×1=2
2
x 1
f(x + y) = f(x) + f(y)
Put y = 0, f(x) = f(x) f(0)  f(0) = 1
f’(x) = Lt
h 0
36.
f ( x  h )  f ( x)
h
sin x
 x 
x
 x 
cos . cos
.... cos  n  
3
 32 
 3  3 n sin  x 
 n
2 
Taking logarthm
f ( x )f ( h )  f ( x )
f’(x) = Lt
 f(x + h) = f(x) .f(h)
h 0
h
Lt f( x )
h 0
f( h )  1
f( h )  1
 f( x) Lt
h

0
h
h
 x
x
  log cos  2
3
 
3
log cos 
.....(A)
 x 
= log sin x - n log 3 = log sin  n 
2

f(h )  1
f( h )  1
 3  1 Lt
h 0
h 0
h
h
f’(0) = f(0). Lt
 Lt
h 0
f( h )  1
3
h
f' ( 5)  f(5). Lt
h 0
= 2 × 3= 6
35.
form
.....(A)
using
(B)
1
3
 x 
cos  n 
1
1
2 
cos x  n
=
sin x
2 sin  x 
 n
2 
1
1
+ b cos (log x).
x
x
 xy1 = - a sin (log x) + b cos (log x)

again differentitating
xy2 + y1 . 1 = -a cos (log x) .
=
38.
1
 x 
cot  n   cot x
2
2 
n
D(f3 + g3 + h3 - 3 f g h)
= 3f2 f1 + 3g2 g1 + 3h2h1 - 3 (f1gh + fg1h + fgh1)
If y = (x  x 2  1 )m
= 3 (f2h + 3g2f + 3h2g - 3
y 1  m( x  x 2  1 )m 1
=0
This shows f 3 +g3 + h3 - 3fgh is a constant function
 f3(2) + g3(2) + h3(2) - 3f(2) g(2) h(2)
= f3(0) + g3(0) + h3(0) - 3f(0) g(0) h(0)
x2  1
= 53 + 13 + (-1)3 - 3 × 5 × 1 × -1
 x 2  1 y 1  m( x  x 2  1 )m
Again differentiating
 x2  1 . y 2  y1.
(hgh + f. fh + fg. g)
= 3f2h + 3g2f + 3h2g - 3h2g - 3f2h - 3 f g2


 1  1  2x 

2 x 2  1 

m ( x  x 2  1 )m
=
1
x
1
1
 x 
 x 
tan  2 tan  2   .....  n tan  n 
3
3 3
3
3 
3 
1
1
- b sin (log x).
x
x
 x2y2 + xy1 = - (a cos (log x) + b sin (log x)) = -y
36.
 x 
 x 

 sin  2 
 sin  n 
 sin  
1
1
3
 
 3   ....
3 
,
n
   32
 x 
 x 
3
cos  
cos 2 
cos  n 
3
3 
3 
y = a cos (log x) + b sin (log x)
y1 = - a sin (log x) -
= 125 + 15 = 140
39.
f(x) = |x + 1| + |x - 2|
for x < -1, x + 1 is negative of x - 2 is negative
1  2x
2
2 x 1
 |x + 1| = -(x + 1) and |x - 2| = -(x - 2)
then f(x) = -(x + 1) - (x - 2) for x < -1
= -x -1 - x + 2

1  2 x 
2
2
m 1 
1
= m (x  x  1 )

2 x 2  1 

2

Differentiating
.....(B)
f( h )  1
h

 x 
  ......  log cos  n 

3 
2
 (x + 1) y2 + xy1 = m y
= -2x + 1 = 1 - 2x
f’(x) = -2  f’(-3) = -2
40.
u = log a x 
log e x
1
log e a
du
1 1
1

. 
dx log a x x log e a
v = logx a =
log e a
log e x
du
1
1
 log a
 log a.
. 
2
dx
(log x ) x x(log x )2
1
du
x log a
x(log x) 2
du dx
1





log a
dv dv
x log a
log a

2
dx
x(log x )
= 
(log x) 2
2
 log x 
  (log a x )2
 
2

log
a
(log a)


Version - D Hints and Solutions
D - 81, C - 101
D - 82, C - 102
D - 83, C - 1037
D - 84, C - 104
D - 85, C - 105
D - 86, C - 106
D - 87, C - 107
D - 88, C - 108
D - 89, C - 109
D - 90, C - 110
D - 91, C - 111
D - 92, C - 112
6
D - 93, C - 113
D - 94, C - 114
D - 95, C - 115
D - 96, C - 116
D - 97, C - 117
D - 98, C - 118
D - 99, C - 119
D - 100,C - 120
D - 101, C - 81
D - 102, C - 82
D - 103, C - 83
D - 104, C - 84
D - 105, C - 85
D - 106, C - 86
D - 107, C - 87
D - 108, C - 88
D - 109, C - 89
D - 110, C - 90
D - 111, C - 91
D - 112, C - 92
D - 113, C - 93
D - 114, C - 94
D - 115, C - 95
D - 116, C - 96
D - 117, C - 97
D - 118, C - 98
D - 119, C - 99
D - 120, C - 100