11 SPHS math T7
61.
There are four balls of different colours and four boxes of colours same as those of
the balls. The number of ways in which the balls, one in each box, could be placed
such that a ball does not go to box of its own colour is
(a) 8
(c)
(b) 7
Since
number
(c) 9
of
derangements
in
(d) None of these
such
a
problems
is
given
by

1
1
1
1
1
n ! 1 



 .......( 1)n

1
!
2
!
3
!
4
!
n
!

 Number of derangements are
1
1
1 


  12  4  1  9 .
2
!
3
!
4
!

= 4 !
62.
A five digit number divisible by 3 has to formed using the numerals 0, 1, 2, 3, 4 and 5
without repetition. The total number of ways in which this can be done is
(a) 216
(b) 240
(c) 600
(d) 3125
(a)
We know that a five digit number is divisible by 3, if and only if sum of its digits (= 15) is
divisible by 3, therefore we should not use 0 or 3 while forming the five digit numbers. Now, (i)
In case we do not use 0 the five digit number can be formed (from the digit 1, 2, 3, 4, 5) in 5 P5
ways.
(ii) In case we do not use 3, the five digit number can be formed (from the digit 0, 1, 2, 4, 5) in
5
P5 4 P4  5 !  4 !  120  24  96 ways.
 The total number of such 5 digit number
 5 P5  (5 P5 4 P4 )  120  96  216 .
63.
Ten persons, amongst whom are A, B and C to speak at a function. The number of
ways in which it can be done if A wants to speak before B and B wants to speak
before C is
(a)
(b) 3! 7!
(c)
(d) None of these
(a)
For A, B, C to speak in order of alphabets, 3 places out of 10 may be chosen first in 10 C 3
ways.
The remaining 7 persons can speak in 7 ! ways. Hence, the number of ways in which all the 10
person can speak is
64.
10
C3 . 7 ! 
10 !
10 !
.
.
3!
6
The number of ways in which an examiner can assign 30 marks to 8 questions,
awarding not less than 2 marks to any question is
(a)
(b)
(c)
(d) None of these
(a)
Since the minimum marks to any question is two, the maximum marks that can be
assigned to any questions is 16( 30  2  7), n1  n 2  ........  n8  30 . If n i are the marks assigned
11 SPHS math T7
to i th questions, then n1  n 2  .......... .  n8  30 with 2  ni  16 for i  1, 2,........, 8 . Thus the
required number of ways
= the coefficient of x 30 in (x 2  x 3  .......  x 16 )8
= the coefficient of x 30 in x 16 (1  x  .......... x 14 )8
= the coefficient of x
30
in x
16
 1  x 15

 1x





8
= the coefficient of x 14 in (1  x )8 . (1  x 15 )8
= the coefficient of x 14 in


8
8.9 2 8.9.10 3
x
x 
x  ...... (1 8 C1 x 15  ......)
1 
2!
3!
 1!

= the coefficient x 14 in 1 8 C1 x 9 C2 x 2 10 C3 x 3  .... 
since the second bracket has powers of x 0 , x 15 etc.
= 21 C14  21 C7 .
65.
Five balls of different colours are to be placed in three boxes of different sizes. Each
box can hold all five balls. In how many ways can we place the balls so that no box
remains empty
(a) 50
(b) 100
(c) 150
(d) 200
(c)
Let the boxes be marked as A, B, C . We have to ensure that no box remains empty and
in all five balls have to put in. There will be two possibilities.
(i) Any two containing one and 3 rd containing 3.
A
5
(1) B (1) C (3)
C1 . 4 C1 . 3 C 3  5 . 4 . 1  20 .
Since the box containing 3 balls could be any of the three boxes A, B, C .
Hence the required number is = 20  3  60 .
(ii) Any two containing 2 each and 3 rd containing 1.
A
5
(2) B (2) C (1)
C 2 . 3 C 2 .1 C1  10  3  1  30
Since the box containing 1 ball could be any of the three boxes A, B, C .
Hence the required number is = 30  3  90 .
Hence total number of ways are = 60  90  150 .
11 SPHS math T7
66.
The number of ways in which an arrangement of 4 letters of the word
‘PROPORTION’ can be made is
(a) 700
(c)
(b) 750
(c) 758
(d) 800
We have got 2 P , 2 R , 3O , 1I, 1T , 1 N i.e. 6 types of letters. We have to form words of 4
s
s
s
letters. We consider four cases
(i) All 4 different : Selection 6 C 4  15
Arrangement  15 . 4 !  15  25  360
(ii) Two different and two alike :
Ps, Rs
and O s in 3 C1  3 ways. Having chosen one pair we have to choose 2 different letters
out of the remaining 5 different letters in 5 C 2  10 ways. Hence the number of selections is
10  3  30 .
Each of the above 30 selections has 4 letters out of which 2 are alike and they can
be arranged in
4!
 12
2!
ways.
Hence number of arrangements is 12  30  360 .
(iii) 2 like of one kind and 2 of other :
Out of these sets of three like letters we can choose 2 sets in 3 C 2  3 ways. Each such
selection will consist of 4 letters out of which 2 are alike of one kind, 2 of the other. They can
be arranged in
4!
6
2!2!
ways.
Hence the number of arrangements is 3  6  18 .
(iv) 3 alike and 1 different :
There is only one set consisting of 3 like letters and it can be chosen in 1 way. The remaining
one letter can be chosen out of the remaining 5 types of letters in 5 ways.
Hence the number of selection  5  1 . Each consists of 4 letters out of which 3 are alike and
each of them can be arranged in
4!
4
3!
ways.
Hence the number of arrangements is 5  4  20 .
From (i), (ii), (iii) and (iv), we get
Number of selections  15  30  3  5  53
Number of arrangements
 360  360  18  20  758 .
67.
Let Z and W be two complex numbers such that | |
|
| |
̅̅̅|
. Then z is equal to
(a) 1 or
(b) or
(c) Let z  a  ib,| z |  1  a  b  1
2
2
and w  c  id,| w |  1  c 2  d 2  1
| z  iw| | a  ib  i(c  id)|  2
 (a  d)2  (b  c)2  4
| z  iw | | a  ib  i(c  id)|
......(i)
(c) 1 or 1
| |
and
(d) or 1
11 SPHS math T7
 (a  d)2  (b  c)2  4
......(ii)
From (i) and (ii), we get bc  0
 Either b  0 or c  0
If b  0 , then a 2  1 . Then, only possibility is a  1 or 1 .
68.
The maximum distance from the origin of coordinates to the point z satisfying the
1
equation z   a is
z
1
( a 2  1  a)
2
1
(c) ( a 2  4  a)
2
(a)
(b)
1
( a 2  2  a)
2
(d) None of these
(c) Let z  r (cos  i sin ) .
Then z 
 r2 
1
1
a z
z
z
2
 a2
1
 2 cos 2  a 2
r2
……(i)
Differentiating w.r.t.  we get
2r
dr
2 dr
 3
4
d r d
Putting
sin 2

dr
 0, we get   0,
d
2
r is maximum for  
r2 
69.
0

2
,
therefore from (i)
1
1
a  a2  4
 2  a2  r   a  r 
2
r
2
r
 z  z1  
  ,
is a complex number such that amp
 z  z2  4
then the value of | z  7  9i | is equal to
If z1  10  6i, z2  4  6i and
(a)
z
(b) 2 2
2
(c) 3 2
(c) Given numbers are
z1  10  6i, z 2  4  6i and z  x  iy
zz 

 (x  10)  i (y  6) 

1 
 amp
   amp  (x  4 )  i (y  6)   4


 z  z2  4
 (x  4 )(y  6)  (y  6)(x 210)  1
(x  4 )(x  10)  (y  6)
 12y  y 2  72  6 y  x 2  14 x  40
Now | z  7  9i| | (x  7)  i(y  9)|
.....(i)
(d) 2 3
11 SPHS math T7
 (x  7)2  (y  9)2
....(ii)
From (i), (x 2  14 x  49)  (y 2  18 y  81)  18
 (x  7)2  (y  9)2  18
or [(x  7)2  (y  9)2 ]1 / 2  [18]1 / 2  3 2
 | (x  7)  i(y  9)|  3 2 or | z  7  9i|  3 2 .
70.
If | z  25i | 15 , then | max .amp( z)  min .amp( z) |
 3
(a) cos 1  
5
 3
(b)   2 cos 1  
5
(c)

3
 cos 1  
2
5
 3
 3
(d) sin 1    cos 1  
5
5
(b) We have
max amp(z)=amp (z 2 ), min amp (z)=amp (z 1 )
Y
15
15
1
Z2
Z1
25
2
1
X
O
Now amp(z1 )  1  cos 1  15   cos 1  3 
 25 
amp(z 2 ) 

2
2 

2
5
 15  
1  3 
 sin1 
   sin  
25
2


5
 | max amp(z)  min amp(z)|


2


71.
2
 sin1


2
3
3
 cos 1
5
5
 cos 1
The value of
3
3
3
 cos 1    2 cos 1
5
5
5
8

  sin
r 1
(a)  1
2r
2r 
 i cos
 is
9
9 
(b) 1
(c) i
(d) We have
8

r 1
2r
2r 

 i cos
 sin

9
9 

8

i
r 1
2 r
i
e 9
8
8
 i  cos
r 1
 , when   e
i
r
r 1
2r
2r 
 i sin

9
9 
(2i / 9 )
(d)  i
11 SPHS math T7
 i
(1   8 )
(   9 )    1 
i
 i
  i
(1   )
1 
1  
( 9  e i2  cos 2  i sin 2  1 )
72.
The value of the expression 1.(2  )(2   2 )  2.(3   )(3   2 )  .......
....  (n  1).(n  )(n   2 ),
where  is an imaginary cube root of unity, is
1
(n  1)n(n 2  3n  4)
2
1
(c) (n  1)n(n 2  3n  4)
2
(a)
1
(n  1)n(n 2  3n  4)
4
(b)
(d)
1
(n  1)n(n 2  3n  4)
4
(b) r th term of the given series
= r[(r  1)  ][(r  1)   2 ]
= r[(r  1)2  (   2 )(r  1)   3 ]
= r[(r  1)2  (1)(r  1)  1]
= r[(r 2  3r  3]  r 3  3r 2  3r
(n  1)
Thus sum of the given series 
 (r
3
 3r 2  3r)
r 1
73.

1
1
1
(n  1)2 n 2  3. (n  1)(n)(2n  1)  3. (n  1)n
4
6
2

1
(n  1)n(n 2  3n  4 )
4
 1 i 3

If i   1, then 4  5  

2
2


334
(b)  1  i 3
(a) 1  i 3
√
(
)
(c) i 3
is equal to
(d)  i 3
(c) Given equation is
 1
3 
4  5   i
 2
2 

334
 1
3 
 3   i
 2
2 

2
2 

 4  5  cos
 i sin

3
3 

334
365
2
2 

 3 cos
 i sin

3
3 

365
668
668 

 4  5 cos
  i sin

3
3 

730
730 

3 cos
  i sin

3
3 

 
2 
2 

 4  5 cos 222 
  i sin 222 

3 
3 

 
11 SPHS math T7
 

 

 3 cos 243    i sin 243  
3
3 

 
2
2 




 4  5  cos
 i sin
  3  cos  i sin 
3
3 
3
3


 1
 1
3 
3 
 4  5   i
 3   i
 2
 2
2 
2 


 4  4  2i
74.
3
i 3
2
.
If a  cos(2 / 7)  i sin(2 / 7), then the quadratic equation whose roots are
  a  a 2  a 4 and   a 3  a 5  a 6 is
(a) x 2  x  2  0
(b) x 2  x  2  0
(c) x 2  x  2  0
(d) x 2  x  2  0
(d) a  cos(2 / 7)  i sin(2 / 7)
a 7  [cos(2 / 7)  i sin(2 / 7)]7
 cos 2  i sin 2  1
.....(i)
S      (a  a 2  a 4 )  (a 3  a 5  a 6 )
S  a  a 2  a 3  a 4  a5  a6 
S
a(1  a 6 )
1a
a  a7 a  1

 1
1a
1a
.....(ii)
P    (a  a 2  a 4 )(a 3  a 5  a 6 )
 a 4  a6  a7  a5  a7  a8  a7  a9  a10
n
 a 4  a 6  1  a 5  1  a  1  a 2  a 3 (From eq (i)]
 3  (a  a 2  a 3  a 4  a 5  a 6 )  3  S
 3 1  2
[From (ii)]
Required equation is, x 2  Sx  P  0
 x2  x  2  0 .
75.
The number of integral values of m, for which the x-co-ordinate of the point of
intersection of the lines 3x  4 y  9 and y  mx  1is also an integer is
(a) 2
(b) 0
(c) 4
5
(a) Solving 3 x  4 y  9, y  mx  1 we get x 
3  4m
x is an integer if 3  4 m  1,  1, 5,  5
 m
2 4 2 8
,
, ,
. So, m has two integral values
4 4 4 4
(d) 1
11 SPHS math T7
The graph of the function cos x cos( x  2)  cos 2 ( x  1) is
76.
(a) A straight line passing through (0,  sin 2 1) with slope 2
(b) A straight line passing through (0, 0)
(c) A parabola with vertex
sin


(d) A straight line passing through the point  , sin 2 1 and parallel to the x–axis
2

(d)
y  cos(x  1  1) cos(x  1  1)  cos 2 (x  1)
 cos 2 (x  1)  sin2 1  cos 2 (x  1)   sin2 1 ,
which represents a straight line parallel to x-axis with y   sin2 1 for all x and so also for x   / 2 .
77.
A line 4 x  y  1 passes through the point A(2,  7) meets the line BC whose
equation is 3x  4 y  1  0 at the point B. The equation to the line AC so that AB = AC,
is
(a) 52 x  89 y  519  0
(c) 89 x  52 y  519  0
(a)
(b)
(d) 89 x  52 y  519  0
Slopes of AB and BC are – 4 and
3
19
4
tan  

3 8
1  4 
4
3
4
respectively. If  be the angle between AB and BC , then
4 
.....(i)
A
4x+y = 1
Since AB  AC
 ABC  ACB  
C


3x–4y+1=0
B
Thus the line AC also makes an angle  with BC. If m be the slope of the line AC, then its equation is y  7  m(x  2)
.....(ii)
3

 m4
Now tan    
1  m. 3

4
 m  4 or –


19
4m  3


8
4  3m


52
.
89
But slope of AB is – 4, so slope of AC is 
52
.
89
Therefore the equation of line AC given by (ii) is 52 x  89 y  519  0 .
78.
The sides AB, BC , CD and DA of a quadrilateral are x  2 y  3, x  1, x  3 y  4,
5x  y  12  0 respectively. The angle between diagonals AC and BD is
(a) 45o
(c)
(b) 60°
(c) 90°
(d) 30°
The four vertices on solving are A(3, 3) , B(1, 1), C(1,  1) and D(2,2) . m 1 = slope of AC  1 , m 2 = slope
of BD  1 ;  m 1 m 2  1 .
Hence the angle between diagonals AC and BD is 90 .
11 SPHS math T7
79.
Given the four lines with equations x  2 y  3, 3x  4 y  7, 2 x  3 y  4 and
4 x  5 y  6, then these lines are
(a) Concurrent
(c) The sides of a rectangle
(b) Perpendicular
(d) None of these
(d)
These lines cannot be the sides of a rectangle as none of these are parallel nor they are perpendicular.
Now check concurrency
1
2
3
3 4
 7  1(16  21)  2(2)  3(1)  0
2 3
4
Hence neither concurrent.
80.
The figure formed by the lines x 2  4 xy  y 2  0 and x  y  4, is
(a) A right angled triangle
(c) An equilateral triangle
(c) S 1 
1
 2  4 1
S2 

1
 2  4 1
 13  tan 1
1
 ( 3  2)
2 3

1
2 3
 ( 3  2)  1
1  ( 3  2)
(b) An isosceles triangle
(d) None of these
 ( 3  2)
 tan 1
and S 3  1 .
 ( 3  3)
 ( 3  1)
 tan 1 ( 3 )  60 .
 23  tan 1
3  2 1
1 3 2
 tan 1
3 3
3 1
 tan 1 ( 3 )  60 .
81.
The angle between the lines joining the points of intersection of line y  3x  2 and
the curve x 2  2 xy  3 y 2  4 x  8 y  11  0 to the origin, is
 3 
(a) tan 1 

2 2
(b)
 
 2 
(b) tan 1 

2 2
 2 
(d) tan 1 

2 2
(c) tan 1 3
Finding the equation of lines represented by the points of intersection of curve and line
2
y  3x 
 y  3x 
  11
 0
 2 
 2 
with origin, we get x 2  2 xy  3 y 2  (4 x  8 y)
 x 2  2 xy  3 y 2  (2 xy  6 x 2  4 y 2  12 xy )

11 2 99 2 33
y 
x 
xy  0
4
4
2
2
Proceed and find the angle between the lines represented by it using   tan 1 2 h  ab .
ab
11 SPHS math T7
82.
The area bounded by the angle bisectors of the lines x 2  y 2  2 y  1 and the line
x  y  3 , is
(a) 2
(b) 3
(c) 4
x 2  y 2  2y  1
(a)
The angle bisectors of the lines given by
Therefore, required area Y
1
22  2 .
2
=
(d) 6
are x  0 , y  1 .
x + y =3
2
y =1
2
x=0
X
O
83.
3
    , then cosec 2  2 cot  is equal to
4
(a) 1 cot 
(b) 1 cot 
(c)  1 cot 
If
(c)
(d)  1 cot 
cosec 2  2 cot   1  cot 2   2 cot   | 1  cot  |
3
     cot   1  1  cot   0
4
But
Hence, | 1  cot  |  (1  cot ) .
84.
The value of sin
(a)
(d) sin

14
sin
1
8
(b)

14
 sin
3
5
7
9
11
13
is equal to
sin
sin
sin
sin
sin
14
14
14
14
14
14
sin

14
1
16
(c)
1
32
(d)
1
64
3
5
7
9
11
13
sin
sin
sin
sin
sin
14
14
14
14
14
14
sin
3
5
sin
1
14
14
5  
3  
 

 sin 
 sin 
 sin 

14
14
14

 
 

2

3
5
7 
1

 sin sin
sin
sin  
14
14
14 
64
 14
85.
.
tan   2 tan 2  4 tan 4  8 cot 8 
(a) tan 
(b) tan 2
(c) tan   2 tan 2  4 tan 4  8 cot 8
 sin 4
cos 8 
 tan   2 tan 2  4 
2

sin8 
 cos 4
(c) cot 
(d) cot 2
11 SPHS math T7
 tan   2 tan 2 
 cos 4 cos 8  sin 4 sin8  cos 4 cos 8 
4

sin8 cos 4


 cos 4  cos 4 cos 8 
 tan   2 tan 2  4 

sin8 cos 4


 cos 4 a(1  cos 8 ) 
 tan   2 tan 2  4 

 cos 4 sin8 
 2 cos 2 4 
 tan   2 tan 2  4 

 2 sin 4 cos 4 
 tan   2(tan 2  2 cot 4 )
 tan   2 tan 2  4 cot 4
 sin 2
cos 4 
 tan   2 
2

sin 4 
 cos 2
 cos 2 (1  cos 4 ) 
 tan   2 

 sin 4 cos 2 
 tan   2 cot 2 
86.
sin 2 cos 2

cos 
sin 2

cos   cos  cos 2
sin 2 cos 

1  cos 2
2 cos 2 

 cot 
sin 2
2 sin cos 
.
a2
If a sin x  b cos x  c, b sin y  a cos y  d and a tan x  b tan y, then 2 is equal
b
to
2
2
(b  c) (d  b)
( a  d ) (c  a )
(a)
2
(b)
2
( a  d ) (c  a )
(b  c) (d  b)
(b) a sin2 x  b cos 2 x  c  (b  a)cos 2 x  c  a
 (b  a)  (c  a)(1  tan 2 x )
b sin2 y  a cos 2 y  d  (a  b) cos 2 y  d  b
 (a  b)  (d  b)(1  tan 2 y)
 tan 2 x 

tan 2 x
tan 2 y

b c
ad
, tan 2 y 
ca
d b
(b  c)(d  b)
(c  a)(a  d )
…..(i)
tan x b
But a tan x  b tan y, i.e.,

tan y
From (i) and (ii),

a
b 2 (b  c)(d  b)

a 2 (c  a)(a  d )
a 2 (c  a)(a  d )
.

b 2 (b  c)(d  b)
…..(ii)
(c)
( d  a ) (c  a )
(b  c) (d  b)
(d)
(b  c) (b  d )
(a  c) (a  d )
11 SPHS math T7
87.
sin 4

4
1
2
(a)
(c) sin4
3
5
7
 sin 4
 sin 4

8
8
8
 sin 4
1
4
(b)

8
 sin4
(c)
3
5
7
 sin4
 sin4
8
8
8

3
2
(d)
3
4
2
2
= 1  2 sin2     2 sin2 3  
4 

8
8  


1
4

2
2

 
3  
 2 sin2    2 sin2
 
8
8  


2
2

= 1 1  cos    1  cos 3  
4 
4


4  
1
4
2
2

 
3  
1  cos   1  cos
 
4
4  




 
 
 
 
= 1 1  1   1  1    1 1  1   1  1  
4 
2 
2   4 
2 
2 
2
2

=
88.
2

1
1
3
(3)  (3) 
4
4
2
2


.
 1 1 1  1 1 1 
If a, b, c are in H.P., then the value of         , is
 b c a  c a b 
(a)
2
1
 2
bc b
(b)
1 1 1
, ,
a b c
(c) a, b, c are in H.P., then

3
2

2
c
ca
(c)
3
2

2
b
ab
are in A.P.
1 1 1 1
  
b a c b
Now,  1  1  1   1  1  1 
b
c
a  c
3
2
 3 2  1 
      2 
ab
 b a  b  b
89.
b
a
.
If a, b, c, d be in H.P., then
(a) a 2  c 2  b 2  d 2
(c) ac  bd  b 2  c 2
(b) a 2  d 2  b 2  c 2
(d) ac  bd  b 2  d 2
(d) None of these
11 SPHS math T7
1 1 1 1
, , ,
will be in A.P.
a b c d
1 1 1 1 1 1
2ac
     b 
b a c b d c
ac
(c) a, b, c, d be in H.P., then
Therefore
G.M. between a and c = ac .
Now as G.M  H.M ., so here ac  b or ac  b 2 .
Similarly bd  c or bd  c 2
Adding, ac  bd  b 2  c 2
90.
Suppose a, b, c are in A.P. and a 2 , b 2 , c 2 are in G.P. If
3
2
a < b < c and a  b  c  ,
then the value of a is
(a)
1
(b)
2 2
1
2 3
(d) b  a  d, c  a  2d , where d  0
Now a 2 , (a  d )2 , (a  2d)2 are in G.P.
 (a  d )4  a 2 (a  2d )2
or (a  d)2  a(a  2d)
or a 2  d 2  2ad   (a 2  2ad)
Taking (+) sign, d = 0 (not possible as a  b  c)
Taking (–) sign,
2a 2  4 ad  d 2  0 ,
3
1

 a  b  c  2 ,  a  d  2 


2
1
 1

2a 2  4 a  a     a   0
2
2

 

a 
or 4 a 2  4 a  1  0
1
1
1
1

.Here d   a  0 . So, a  .
2
2
2
2
Hence a 
1
1

2
2
.
(c)
1 1

2
3
(d)
1 1

2
2