SOLUTIONS TO SOME QUESTIONS OF WORKSHEET GIVEN ON TUESDAY 02-05-2017
1.
In a triangle ABC if A = (1, 2) and internal angle bisectors through B and C are
y = x and y = -2x. The in radius r of the ABC is
1
1
(a)
(b)*
(c)
2/3
(d)
none of these.
2
3
Solution: Image of A about y = x, y = -2x are A1 and A2 which lies on BC.
 11 2 
A1   2,1 , A2    ,  equation of BC is x – 7y + 5 = 0
 5 5
5
5
1
r  ID 


1  49 5 2
2
Let AB be any chord of the circle x2 + y2 – 4x – 4y + 4 = 0 which subtends an angle of 90 at the
point (2, 3), then the locus of the mid point of AB is circle whose centre is
(a)
(1, 5)
(b)*
(1, 5/2)
(c)
(1, 3/2)
(d)
(2, 5/2)
Solution: Let mid point of AB is M(h, k)
AB subtends 90 at (2, 3)
 AM = MB
3
=
 h  2  2   k  3 2
Also, CM2 + MB2 = CB2
 (h – 2)2 + (k – 2)2 + (h – 2)2 + (k – 3)2 = 4
 x2 + y2 – 2x – 5y + 17/2 = 0
7.
The conjugate hyperbola of the hyperbola 2 x 2  3 y 2  6 must be
(a)*
Solution:
y 2 x2

 1 (b)
2
3
y 2 x2

 1
2
3
The hyperbola is
y 2 x2

 1 (d)
3
2
y 2 x2

 1
3
2
x2 y 2

 1
3
2
 Conjugate hyperbola is
8.
(c)
x2 y 2
y 2 x2

 1 or

 1.
3
2
2
3
x2 y 2

 1 is
The number of rational points on the ellipse
16 9
(a)
none
(b)
m2  n2
.4
Solution: Take x  2
m  n2
two
(c)
four
2mn
, y 2
.3
m  n2
*(d)
Infinite.
2
2
 x  y
      1
4  3
For all m , n In particular for rational m & n.
9.
The foci of the ellipse x2 + 4x + 42y2 = 0 will lie on the line x = - 2, if
 1 1
 2 2
(a)    2, 2
*(b)     , 
(c)   1
(d) None of these.
Solution: The ellipse can be written as  x  2   4 2 y 2  4
2
or
 x  2
4
2

y2
1
 1 . Now the foci will lie on the line x + 2 = 0 if 4  2
1 2


 1 1
 2 2
or      , 
10.
If a circle cuts the rectangular hyperbola y 
1
at four distinct points
x
 xi , yi  ( i 1, 2,3, 4)
then necessarily
(a)
x1  x2 3  x4  0
(b)
y1  y2  y3  y4  0
(c)
1 1
1
1
  
1
x1 x2 x3 x4
(d)*
x1 x2  y3 y4 .
Solution: Let the circle be x2 + y2 + 2gx + 2fy + c = 0 then at intersection with y 
have x 2 
1
2f
 2 gx 
c 0
2
x
x
 x1 x2 x3 x4  1
12.
 x1 x2
 x 4  2 gx3  cx2 2 fx  1  0
1 1
 1  x1 x2  y3 y4
y3 y4
AB  p , BC  q , CD  r ,
Let ABCD be a cyclic quadrilateral with lengths of sides as
DA  s. Then
(a)
AC
must be equal to
BD
p2  q2
r 2  s2
(b)
1
we
x
pq
rs
(c)*
ps  qr
pq  rs
(d)
p+q
.
r+s
AC 2  p 2  q 2  2 pq cos B  r 2  s 2  2rs cos(180  B)
Solution:
 cos B 
p2  q2  r 2  s2
2( pq  rs)
 p 2  q 2  r 2  s 2  ( pr  qs)( ps  qr )

pq  rs
 2( pq  rs) 
Also AC 2  p 2  q 2  2 pq 
Similarly BD 2 
13.
( pq  rs)( pr  qs)
ps  qr
AC 2
( ps  qr ) 2

BD 2
( pq  rs)2

The minimum value of sin 3A + sin 3B + sin 3C in a triangle must be
(a)
-1
(b)
-2
(c)

3
2
(d)*
None of these.
Solution: Since thrice of an acute angle can cross 180 the given expression can become negative in a
triangle. Then the minimum value must be negative and must lie
three quantities sin 3A, sin 3B, sin 3C can not be
between -3 and 0. But all the
negative (for other wise A + B + C > 180)
 sin 3A + sin 3B + sin 3C can not be less than -2
It may become -2 when sin 3A = sin 3B = -1, sin 3C = 0
 A = B = 90, C = 60
Which is not possible in a triangle
 minimum value will be greater than -2
we can take A = B = 89, C = 1 to get
sin 3A + sin 3B + sin C  -1.897 which is less than all the choices given
 (d) is the correct answer
26.
If f ( x) and g ( x) are two twice differential functions then
 f ( x) g ( x) dx  f ( x) g ( x)  f ( x) g ( x) is equal to
(a)
 f ( x) g ( x) dx
(b)*
 f ( x) g ( x) dx
 f ( x) g ( x) dx
(c)
(d)
None of these
Sol: (b): On applying the formula for integration by parts ( taking f ( x) as second function
27.
 g ( x) f ( x)dx  g ( x) f ( x)   g ( x) f ( x)dx
 g ( x) f ( x)   g ( x) f ( x)   g ( x) f ( x)dx 
 g ( x) f ( x)  f ( x) g ( x)   g ( x) f ( x)dx
 (b) is correct .
If n is a positive integer. Let U n 
U n  Vn
(a)
Sol: ( b) U n 

π
0
(b)*
nx
2 dx
2 x

2sin
2
 2
2sin

π
0
2
π / 2 sin ny
1  cos nx
dx , Vn  
dy then
0
1  cos nx
sin 2 y
U n  2Vn
1
U n  Vn
2
(c)
put x  2 y then U n 

π/2
0
(d)
None of these
sin 2 ny
2dy  2Vn  (b) is
sin 2 y
correct.
28.
The value of the integral
π
 3
3
(a)
Sol: (b):

π
0
1  2cos dx 

π
0
(b)
2π
3
0

1  2cos x dx is
π
2 3
3
(1  2cos x) dx 
(c)

π
2π
3
π
4 3
3
 (1  2cos x) dx 
(d)
2π
4 3
3
π
2 3
3
 (a) is correct .
29.
Let f ( x ) be a function such that , f (0)  f (0)  0, f ( x)  sec4 x  4, then the function is .
(a)
1
l n(sin x)  tan 3 x  x 2
3
(b)*
2
1
ln(sec x)  tan 2 x  2 x 2
3
6
(c)
1
x2
2
ln (cos x)  cos x 
6
5
(c)
None of these .
Sol: (b) f ( x)  (1  tan 2 x)sec2 x  4

f ( x)  tan x 
tan 3 x
 4x  C
3
1
3
On putting x = 0 we get C  0 . Thus f ( x)  tan x  tan x (sec 2 x  1)  4 x
2
1
f ( x)  log sec x  tan 2 x  2 x 2  C. C , again turns out to be zero
3
6

 (b) is correct .

30.

0
f


x 2  1  x dx is equal to

1
) f ( x)dx
x2
(a)

(c)*
1 
1
(1  2 ) f ( x) dx

2 1
x
(1 
1
(b)
1 
1
(1  2 ) f ( x)dx

1
2
x
(d)
None of these
x2  1  x  t
Sol: (c): Put
x2  1  t  x

 dx 
31.
x 1  t  x  2  x
2
2
Sol: (a) .
2 2
3a 4
I 

a

t 2  1 dx
2t.2t  2(t 2  1)
x
,

2t dt
4t 2
1
1
1  2  dt when x  0 , t  1 when x   , t    (c) is correct .
2 t 
The value of the integral
(a)*
2
2

dx
a
x4 a2  x2
(b)
2 2
3a 4
dx
x (a x 2  1)1/ 2
5

is
(c)
2 1
2a 4
(d)
None of these
Put
32.
a2
 1  t 2 . when x  a , t  2 . When x   , t  1
2
x
If a,  0, b  0
(a)
lim
x 0
a
b
x b 
a  x 
(b)*
Sol: (b) lim it  lim
x 0
If sin x  x 
Then lim
x 0
(a)
b
a
(c)
x  b b 
 
a  x  x  
 b x b  b
 lim      
x 0
 a a  x  a
33.
(  x  denotes greatest integer  x is equal to )
( using
(  x  0 and
zero
(d)
1
a
 x  x  x
b 
  is finite ).
x
x3 x5 x 7
   ..... for all x and f ( x)  x sin(sin x)  sin 2 x
3! 5! 7!
f ( x)
is non zero and finite if m is
xm
4
(b)
-4


x3 x5
  ...... 
Sol: (d) f ( x)  x sin  x 
3! 5!


2
(c)
2


x3 x5
  x    ......... 
3! 5!


(d)*
6
2
3
5


 1
 1

x3 x5
x3 x5
x3 x5
= x  x    .....    x    ........    x    ......   ....
3! 5!
3! 5!
3! 5!


 3! 
 5! 

( x 
x3 x5
  ......)2
3! 5!
Now there are no terms of x0 , x1 , x3 and x5 . Coeff of x 2  1  1  0
Coeff of x 4  
1 1  2
     =0 If can be observed that coeff of x 6 is non zero
3! 3!  3! 
 x sin(sin x)  sin 2 x  Ax6  Bx7  ........( A  0)
x sin(sin x)  sin 2 x
A  Bx  Cx 2  ........ 
6
x

34.
 m = 6.
Using the results
1 1 1
π2



.........


, and
12 22 32
6

1
0
Sol:
lim it  A
log(1  x)  x 
x 2 x3
  ........
2 3
log x log(1  x)dx is equal to
(a)*
2  π2 / 6
(a):


x 2 x3 x 4
log x log(1  x)   log x  x     ......  .
2 3 4


Let I m 

1
0
(b)
1 π2 / 6
(c)
1 π2 / 6
(d)
none of these
x m log xdx
1
m 1
11 x
x m 1
1
 .
dx  0 
Then on integrating I m  log x.
0
m 1 0
x m 1
(m  1) 2


1
0
log x log(1  x)dx
1
1
1
1
1
 1


 .....
  2 

 .......   2 
2
2
2
2
 1.2 2.3 3.4
 1.2 2.3 3.4
1
1
1
1 1  1  1 1  1  1 1  1
 1 1 1

             ..... 


 ....   2  2  2  ...... 
1.2 2.3 3.4
1 2  2  2 3  3  3 4  4
2 3 4

 π2 1 
π2
1 1 1 1 1
= 1       ....  
 2  2
6
2 2 3 3 4
 6 1 
36.
The solution of y 5 x  y  x
dy
 0 is
dx
5
(a)
x4 1  x 
   C
4 5 y 
 (a) is correct .
4
(b)*
x5 1  x 
   C
5 4 y
5
 x  x4
C
 
 y 4
(c)
x5
( xy )   C
5
4
(d)
The given differential equation can be written as y5 xdx  ydx  xdy  0
Sol: (b):
Multiplying by
x3  ydx  xdy 
x3
4
we
have
x
dx

,

  0 . On Integrating ,
y3 
y2
y3

4
x5  1   x 
we get
    C .
5  4  y 
37.
Sol: (b):
f ( x) 


1
1
d
then
lim f ( x) must be equal to

da x a
sin x  sin a ( x  a) cos a
(a)
1
 sec3 a  sec a
2
(b)*
1
sec3 a  sec a
2
(c)
1
sec3 a  sec a
2
(d)
sec3 a
( x  a) cos a  sin x  sin a
( x  a) cos a (sin x  sin a)
f ( x) 
Since lim
xa

( x  a) cos a  sin x  sin a
xa
.
2
( x  a) cos a
sin x  sin a
( x  a) cos a  sin x  sin a
xa
1

we try to find lim
x

a
sin x  sin a
cos a
( x  a)2 cos a
If it exists then our separating becomes justified . The limit by LH rule
lim
x a


cos a  cos x
2( x  a) cos a
original limit =
f (a) 
=
lim
xa
1
1
1
tan a

2
cos a 2
cos x  cos a
1
.
xa
2 cos a

1
1
sec a sec 2 a  sec a tan 2 a
2
2
1
sec a tan a
2
  ( sin a )
1
2 cos a

1 3
1
sec a  sec a(sec 2 a  1)
2
2
1
 sec3 a  sec a
2
 (b) is correct
n
39.
 
1
 f x n

  must be equal to
Let f be a positive differentiable function on (0,  ) then , lim 
n  
f ( x) 




(a)
1
(b)
f ( x )
f ( x)
Solution: Value of given limit  e
40.
(c)*
1


 f ( x ) 
n 1

 f ( x)

lim 


n  
if x 1
0
log 2 x if x 1
Let f(x) = 
e
f ( x )
f ( x)
(d)
e.
n
e
 1
f  x   f ( x )
 n
n
lim
f ( x)
n 
e
lim
n 
f ( x h ) f ( x ) 1
.
f ( x)
f ( x)
Let f 2(x) = f( f (x)), f 3(x) = f ( f (2) (x))
and in general f ( n1) ( x)  f ( f ( n ) ( x)).
If T (n)  min{n  1: f ( n ) ( x)  0} then T (425262 ) must be equal to
(a)
7


Solution: f ( f ( x))  
f

(b)
f (0)
6
(c)*
5
x 1
x

 log  x  1
2


0
x1

 0
1 x  2

x
log  log  x  2
2
 
Again ff ( f ( x ))
 0

 
x
log  log 2 

 
 0



x 

log  log  log 2  




x2
if x  2
x  22
if x  22
(d)
4.
e
f ( x )
f ( x)
.

0

f ( f ( f ( x)))   


x 

log  log  log  log  log    
2 

   

0

f ( f ( f ( f ( f ( x)))))   


x 

log  log  log  log  log    
2 



 
since 216< 425262  22
16
x  22
2
x  22
2
x  22
22
x  22
22
Therefore the smallest n such that for which
f ( n ) ( x)  0 for x  425262 is 5
whence T(425262 ) = 5.