MATHEMATICS Max Marks: 66

SECTION-I
(INTEGER ANSWER TYPE)
• This section contains SIX (06) questions.
• The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 TO 9, BOTH INCLUSIVE.
• For each question, enter the correct integer corresponding to the answer using the mouse and the on-screen virtual Numeric keypad in the place
designated to enter the answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks: +3 If ONLY the correct integer is entered;
Zero Marks: 0 If the question is unanswered;
Negative Marks: -1 In all other cases
37. Each of the 2001 students at school studies either Spanish or French and some study both.
The number who study Spanish is between 80% and 85% of the school population and
the number who study French is between 30% and 40% Let ‘m’ be the smallest number
of students who could study both languages and let ‘M’ be the largest number of students

who could study both languages then

IR
M m
149
=______

38. There are 6 boys on a merry-go-round(An amusement park ride with seats often in the
S
form of animals (such as horses)revolving about a fixed centre). The number of ways
K
that they can again sit such that no boy follows the originally he followed is K then 
6
M

________
39. If f : 1, 2,3, 4  1,2,3, 4 , y  f  x  be a function such that f      1 for
P

F
for   1, 2,3,4 , total number of such functions is F then  _______
4

40. Given N  11.22.33.44...........100100 , the number of zeros at the end of N is z then

z-1300= _______
41. Number of ways in which 32 identical blankets can be distributed among 4 persons so

that each of them gets number of blankets which are divisible by 2 but not by 4 is K C3
Then K=______
42. Number of ways in which 9 different toys be distributed among 4 children belonging
to different age groups in such way that distribution among the three elder children is
2N
even and the youngest one is to receive one toy more is N then  _______
7!
 SECTION – II
(ONE OR MORE CORRECT ANSWER TYPE)
• This section contains SIX (06) questions.
• Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is(are) correct answer(s).
• For each question, choose the option(s) corresponding to (all) the correct answer(s).
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks: +4 If only (all) the correct option(s) is(are) chosen; Partial Marks +3 If all the four options are correct but ONLY three options are chosen;
Partial Marks: +2 If three or more options are correct but ONLY two options are chosen, both of which are correct;
Partial Marks: +1 If two or more options are correct but ONLY one option is chosen and it is a correct option;
Zero Marks: 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks: -2 In all other cases.
43. If the number of ways in which 5 balls can be selected from a bag containing 5 identical
N
and 5 different balls is N then is greater than
4
A) 2 B) 4 C) 6 D) 8
44. Which of the following is /are true
A) The number of functions from the set 1, 2,3....m to the set 1, 2,3....n where n  m

such that f  i   f  j  whenever i  j is nCm

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B) The number of onto function from a set with 5 elements to a set with 4 elements is 240
C) The number of bijective functions f : A  A where A  u, v, w, x such that

f  u   w, f  v   x, f  w   v and f  x   u is 9
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D) Number of monotonic function from set 1, 2,3 to the set 1, 2,3, 4,5 is 65

45. If 100 C50 can be prime factorized as 2 .3 .5 . ........ where  ,  ,  ,  ...... are non
M

negative integers then correct relation(s) is/are

A)    B)   
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C)         1 D)     0
46. There are 6 different balls and 6 different boxes of the colour same as of the colour of
ball then choose correct statement (s)
A) The number of ways in which no ball goes in the box of the own colour is 265
B) The number of ways in which atleast 4 balls goes into their own boxes is 16
C) The number of ways in which atmost 2 balls goes into their own boxes is 664
D) The number of ways in which exactly 5 balls goes into their own boxes is 0
47. 16 players P1, P2 , P3......P16 take part in a tennis tournament. Lower suffix player is better
than any higher suffix player. These players are divided into 4 groups each comprising of
4 players and the best from each group is selected for semifinals then
A) Number of ways in which 16 players can be divided into four equal groups is

35 8
  2r  1
27 r 1

B) Number of ways in which they can be divided into four equal groups if players
11!
P1, P2 , P3 and P4 are in different groups is
108
C) Number of ways in which these 16 players can be divided into four equal groups such
12!
that when the best player is selected from each group, P6 is one among them is 20.
 4!3

35 8
  2r  1
54 r 1
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D) Number of ways in which 16 players can be divided into four equal groups is
S
48. There are 350 farmers in a large region. 260 farm beetroot, 100 farm yams, 70 farm
Radish, 40 farm beetroot and radish, 40 farm yams and radish and 30 farm beet root and
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yams then
A) Number of farmers that farm beet root only is 220
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B) Number of farmers that farm yams only is 60

C) Number of farmers that farm radish only is 20
D) Number of farmers that farm beetroot, yams and radish is 10
SECTION – III
(NUMERICAL VALUE TYPE)
• This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE.
• For each question, enter the correct numerical value of the answer using the mouse and the on-screen virtual
Numeric keypad in the place designated to enter the answer. If the numerical value has more than two decimal places, truncate/round-off the value to
TWO decimal places.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks: +4 If ONLY the correct numerical value is entered;
Zero Marks: 0 In all other cases..
49. Call a number “prime looking” if it is composite but not divisible by 2,3, or 5. The three
smallest prime looking numbers are 49, 77 and 91. There are 168 prime numbers less
K
than 1000. If K is number of prime looking numbers less than 1000 then  ______
200
50. Number of ways you can put 7 letters into their respective envelopes such that exactly
3 go into the right envelope is________
51. Let X  1, 2,3, 4 the number of one functions f : X  X satisfying f  f  i    i, for all

1  i  4 is ________
52. The highest power of 12 that divides 54! is _______
53. Number of homogeneous products of degree 4 from 5 Variables is equal to ______
54. The streets of city are arranged like the lines of chess board. There are 5 streets running
North to South and 4 streets running East to West. The number of ways in which a man
can travel from NW (North West) to SE (South East) corner going the shortest possible
distance is _______
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MATHEMATICS MAX.MARKS: 66
SECTION – 1 (Maximum Marks: 18)
This section contains SIX (06) questions.
Each question has FOUR options for correct answer(s). ONLY ONE of these four option is the correct answer.
For each question, choose the correct option corresponding to the correct answer.
Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If only the correct option is chosen.
Zero Marks: 0 If none of the option is chosen.(i.e the question is un answered)
Negative Marks: -1 In all other cases.
1. The number of ways of forming an arrangement of 5 letters from the letters of the word
“IITJEE” is

A) 60 B) 96 C) 120 D) 180

2. The number of times the digit 3 will be written when listing the integers from 1 to 1000,
is

A) 269 B) 300 C) 271 D) 302

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3. The coefficient of x 50 in the expansion of

1  x   2 x 1  x   3 x 2 1  x   ......  1001x1000 is
1000 999 998
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A) 1000
C50 B) 1001
C50 C) 1002
C50 D) 21001

4. A flight of stairs has 10 steps. A person can go up the steps one at a time, two at a time
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or any combination of 1s and 2s. The total number of ways in which the person can go
up the stairs is
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A) 75 B) 79 C) 85 D) 89

5. Five boys and four girls are to be seated in a row so that two particular girls will never
sit adjacent to a particular boy and all girls are separated. If the number of ways in which
N
they can be seated is N , then value of is-
480
A) 32 B) 36 C) 40 D) 48

6. The number of three digit numbers with three distinct digits such that one of the digits is
the arithmetic mean of the other two is

A) 120 B) 180 C) 112 D) 104

 SECTION - 2 (Maximum Marks : 24)
This section contains SIX (06) questions.
Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four option(s) is
(are) correct option(s).
For each question, choose the correct option(s) to answer the question.
Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If only (all) the correct option(s) is (are) chosen.
Partial Marks: +3 If all the four options are correct but ONLY three options are chosen.
Partial Marks: +2 If three or more options are correct but ONLY two options are chosen, both of which are
correct options.
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct
option.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
Negative Marks: -2 In all other cases.

Let f  x  be a real valued function satisfying f 3  x   6  f  x    11 f  x   6  0

2
7.

defined on  0,8 . If the function y  f  x  satisfies the condition that f  x  is

discontinuous at every integral value of x and continuous at every other x, then the
number of such functions is 2a  3b  7c (where a, b, c  N ), where

A) a  21 B) a  c  23
IR C) b  c  1 D) a  b  22

 7 
S
   14 
7
 k   r  14   n
8. The value of   14  r k  k   r   , where
n
 r  denotes C r , is
k 0     
  
 k 
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A) 67 B) greater than 76 C) 87 D) greater than 78

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9. The number of 4 digit natural numbers which contains

A) exactly 2 distinct digits is 567 B) atleast 2 distinct digits is 8090
C) atmost 2 distinct digits is 577 D) atleast 3 identical digits is 333

10. If 2n 1 Tn  2n C0  3  2n C 2   32  2n C 4   .....  3n  2n
C 2n  ,n  N(n  2), then

A) Tn ,Tn 1 ,Tn  2 are in AP B) Tn , 2Tn 1 ,Tn  2 are in AP

C) T3  52 D) T4  194

11. Let k be the number of 10 digit numbers, formed using the digits 0,1,2...9 (with
repetition allowed) ,such that the product of any two consecutive digits in the number is
a prime number ,then
 A) k is a perfect square of a natural number
k
B) is a perfect square of a natural number
2
C) 2k is a perfect cube of a natural number
D) 4k is a perfect cube of a natural number
12. Let p n denotes the number of permutations of n distinct things taken all at a time and

n 5  143   pn5 
xn  c4     ( where n  N ) . The possible value of n for which xn is
 96   pn3 
negative, can be
A) 1 B) 2 C) 3 D) 4
SECTION - 3 (Maximum Marks : 24)
This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE
For each question, enter the correct numerical value of the answer using the mouse and the on-screen virtual
numeric keypad in the place designated to enter answer. If the numerical value has more than two decimal
places truncate/round- off the value to TWO decimal places.

Zero Marks:
IR
Answer to each question will be evaluated according to the following marking scheme:
Full Marks: +4 If ONLY the correct numerical value is entered as answer.
0 In all other cases.
13. Let N be the number of positive integral divisors of 1998, then sum of digits of N is ___
S
14. The number of different seven digit numbers that can be written using only the three
digits 4,2,6 with the condition that digit 6 occurs twice in each number is ________
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15. Let X be the number of linear permutations of all the letters of the word SBISMART in
which the two S 's are always together, Y be the number of linear permutations of all the
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Y
letters of the word SBISMART in which the two S 's are never together, then is ________
X

16. Let P be ways in which 5 Boys and 5 Girls form a circle such that Boys and Girls are
alternate Q be ways in which 5 Boys and 5 Girls form a line such that Boys and Girls
Q
are alternate. Then the value of  10 is _________
P
17. Let S be the set of all 2  2 matrices formed by using the elements of the set
0,1, 1 .Then the number of singular matrices which belong to the set S is_____
18. Let P be the number of distinct ways of painting six sides of a cube with 6 distinct
colours, where you would regard two cube colourings the same if one can be obtained
from the other by rotation, then sum of digits of P is ________
 KEY SHEET
MATHEMATICS
1 D 2 B 3 C 4 D 5 B
6 C 7 BC 8 AB 9 AD 10 BCD
11 BC 12 ABC 13 7 14 672 15 3

16 0 17 33 18 3

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 SOLUTIONS
MATHEMATICS
2. Number of arrangements in which 2 are identical of one kind, two identical of another kind and
5!
one letter different from the remaining two letters is 2C1   60. Number of arrangements in
 2!
2

5!
which 2 are identical of one kind and the rest are different is 2C1   120
2!
3. Since 3 does not occur in 1000, we have to count the number of times 3 occurs when we list the
integers from 1 to 999. Any number between 1 and 999 is of the form xyz where 0  x, y, z  9. Let
us first count the number in which 3 occurs exactly once. Since 3 can occur at one place in 3C1 ways,
there are 3 C1 (9  9)  3  9 2 such numbers. Next, 3 can occur in exactly two places in

 3

C2  9   3  9 such numbers. Lastly, 3 can occur in all three digits in one number only.
Hence, the number of times 3 occurs is
1  3  9 2   2   3  9   3  1  300 .

4. Let

Subtract above equations

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Coefficient of in S = coefficient of in

1  , B  y1, y2 , y3
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6. A  x ,x ,x ,x ,x ,x ,x
2 3 4 5 6 7

f : A  B is onto  f  x   y
2
Exactly 3 elements x in is y . This can be done
2
In 7 ways
C
3
Remain A four elements in B 2 elements
4 2 4
 2  C  2  1  14
1
7
Total no.of onto functions  C 14
3
7
 Ck 7 14
  C7 7 14
r 14 
8.   14
Ck
 r
Ck . 14C r     k  k 14  k 
r k k r  k
. 
r 14  r 
k 0  r k  k 0  14
k
 7 14
 7 7
1
   7 Ck  14  k
Cr  k    7 Ck . 214 k  214  7 Ck  
k 0  r k  k 0 k 0 2
7
 1
 214 .  1    67  7 6
 2
 n
  a  b    a  b   2  2n C2r  a 2n 2r b 2r
2n 2n
10.
r 0

put a  1, b  3

     
n
2  2n C2r 
2r 2n 2n
3  3 1  3 1
r 0

 4  2 3  4  2 3
n n

 2   2  3    2  3  
n n
n

 
   C   3   2   2  3    2  3  
n 2r n n
2n n 1
2r
r 0  
 T   2  3    2  3  …….(i)
n n
n

 T   2  3    2  3   T   2  3  2  3  T 0

n 2
 n 1 n

 Tn  2  4Tn 1  Tn
 Tn ,2Tn 1 ,Tn  2 are in AP
T1  4
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   2  3
2 2
T2  2  3  14
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T3  4  T2  T1  4  14  4  52
T4  4  T3  T2  4  52  14  194
11. 
2! 8C2  5C2  3C2  30  k 
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Sum of all positive integral divisors of k is 72

Product of all positive integral divisors of k is k 4
12. Clearly digit 1 should occur at alternate places of the number and at the remaining 5 places
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either 2,3,5 or 7 should appear. If number starts with 1, then we have 45 such numbers and
when number starts with 2,3,5 or 7 number of such numbers = 45 so total number of required
numbers= 2  45
13. n 5  143   pn 5 
x n c4    
 96   pn 3 
given xn  0
n 5 143  n  5 !
hence c4  
96  n  3!
 n  5!  143   n  5 !
4! n  1! 96  n  3 n  2  n  1!
1 143 1
 
24 96  n  3 n  2 
  4  n  3 n  2   143  4  n 2  5n  6   143  0
 4n 2  20n  119  0
  2n  17  2n  7   0
 n  {1, 2,3}.
15. X  7!, Y  3*7!
16. P  4!5! Q  2  5!5!
17. One can work out to get P  1, 2
18. Colour one side with the ugliest colour, and put the cube on a table ugly side down. There
are 5 choices for the colour on top. For each of these choices, colour the side facing you with the
nicest remaining colour. The last three sides can be coloured in 3! ways, so the number of colourings
is (5)(3!).

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 IIT – ian’s P A C E

MATHS
41 B 42 D 43 C 44 C 45 B

46 B 47 D 48 B 49 C 50 A

51 C 52 D 53 B
IR 54 A 55 C

56 B 57 A 58 B 59 B 60 C
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 MATHS
n  m  m m! p! 
41.  n
C m 
 
 k 1 p  k p! m  p  ! k! p  k !  
 
m 1    
n  m m  m! 
  n Cm     m  k Cp  k  
 k 1 p  k
m 1    k! m  k ! 
n
 m 
  n Cm   2m  k m Ck 
m 1  k 1 

 
  n C m 1  2   2 m    n C m 3m  n C m 2 m 
n n
m

m 1 m 1

42. The given expression can be written as (1  x ) n .(1  y ) n .(1  1 ) n . The constant term is
xy
clearly C  C      C where cr  cr .
3
0
3
1
3
n
n

43. Given
1  x   C0  C1 x  c2 x 2  .....  Cn x n
n

Integrating both sides with respect to x , we get

1  x 
n 1
C0 x C1 x 2 C2 x3 C x n 1
    .....  n k IR
n 1 1 2 3 n 1
Putting x  0 ,
1
We get k 
n 1
S
1  x   1
n 1


n 1
C x C x 2 C x3 C x n 1
 0  1  2  .....  n
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1 2 3 n 1
Multiplying with x both sides
x 1  x 
n 1
x
C0 x 2 C1 x 3 C2 x 4 C x n 1
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   ......  n
n 1 1 2 3 n 1
Differentiating with respect to x
 n  1 x 1  x   1  x 
n 1
1
n

n 1
2C0 x 3C1 x 2 4C2 x 3  n  2  Cn x n1
    ... 
1 2 3 n 1
Now putting x  1 both sides, we get
2n 1   n  1 2n  1
n 1
2C 3C 4C  n  2  Cn
 0  1  2  ..... 
1 2 3 n 1
2  n  3  1 2C0 3C1 4C2
n
 n  2  Cn
    ........ 
n 1 1 2 3 n 1
44. f(x) =  x  a1  x  a 2   x  a 3    a1  x    a 2  x    a 3  x 
Now f(x)  –  as x  –  and f(x)  are x .
Again f(a1) = (a2 – a1) + (a3 – a1) > 0  a1  a 2  a 3 
 One root belongs to  , a1 
Also, f  a 3    a1  a 3    a 2  a 3   0
 One root belongs to  a1, a 3 
So f(x) = 0 has three distinct real roots.
45. Let the roots be 1 ,  2 ,....,  8
 1   2  ....  8  4
1
1 2 ..... 8  8
2
1     ...   8
 1 2 ..... 8    1 2
1/8

2 8
1
 AM=GM  all the roots are equal to .
2
7
1 1
 a1   C7     4
8

2 2
IR
6
1 7
a2  8C6     4
2 2
5
1
S
a3   C5   8

2
46. Putting 3  y, we have
x

 2a  4  y 2   2a  3  y  1  0
M

This equation must have real solution

  2a  3  4  2a  4   0
2
P

 4 a  20 a  25  0
2

  2a  5   0. This is true.
2

y 1
satisfies the equation
Since 3 is positive and 3x  30 , y  1
x

Product of the roots  1 y  1

1
 1
2a  4
5
 2a  4  1  a 
2
2a  3
Sum of the roots  1
2a  4

 2a  3    2a  4  0
2a  4
 1
 0a 2
2a  4
5
 2a
2
47. We have
n
Ck n  k 1
n

Ck 1 k
2
n
 n Ck 
 k  n 3

k 1  Ck 1 
 n  k 1
n 2

 
k 1
k3 
 k


n

 k  n  1  k 
2

k 1

 k  n  1 
n
  2  n  1 k  k 2
2

k 1
n
   k  n  1  2  n  1 k 2  k 3 
2

k 1

2
n
  n   n 
  n  1   k   2  n  1   k 2     k 3 
 k 1   k 1   k 1 
IR
 n  n  1  n  n  1 2n  1  n  n  1 
2

  n  1    2  n  1 
2

S
 2  6  2 
n  n  1 
2
2 n
  n  1   2n  1  
2  3 2
M

n  n  1 6  n  1  4  2n  1  3n
2

….. (i)  
2 6
n  n  1  n  2 
2
P


12
A, b c, d are in H.P
48.
2ac
b
ac
log 4 log 2
 1 x
 2 1
log(2  1) log(5.2 x  1)
log 2
1
log(5.2 x  1)
2 log 2 2 log 2
 1 x

log(2  1) [log 2  log(5.2 x  1)]
 10t  2  2 / t  1  10t 2  2t  2  t ( put 2 x  t )
 10t 2  5t  4t  2  0
  5t  2t  1  2(2t  1)  0
2 1
 t  , (rejected)
5 2
 x  1  log 2 5.
49. 2 log 9 (31 x  2)  1  log 3 (4(3x )  1)
 log 3 (31 x  2)  log 3 3(4(3x )  1)
 31 x  2  3[4(3x )  1]
3
Let y = 3x;  2  3(4 y  1)
y
 3  2 y  12 y 2  3 y
 12 y 2  5 y  3  0
 12 y 2  9 y  4 y  3  0
  4 y  3 3 y  1  0
3
 y  3x
4
3
 x  log 3    1  2 log 3 2
4
50.
p  1, q  1, 2,3, 4,5, 6  6
IR
p  2, q  1,3, 4,5, 6  3   2, 4  ,  2, 6  
S
p  3, q  1, 2, 4,5, 6  4   3, 6  
p  4, q  1,3,5, 6  3   4, 6  
p  5, q  1, 2,3, 4,6  5
M

p  6, q  1,5  2
51.
For x  4, the last digit of 1! 2! ....  x! is 3
P

For x  4, the given equation has only solutions

x  1, K  1 and x  3, K  3
  1,   3

a  b   
x 2 15 x 2 15
 a b  2a
 x 2  15  1  x  4,  14
1  4,2  4,3   14,4  14.
 1  2  3  4  123 4  0  16 14  224
1 1 1
  .......
52. a 2 4 8
a
x x x......  x and 49  20 6  5  2 6
x  3  0 and x  0
2
 x 3

   
a a a ...... x 2  x  3 x x x ...
5 2 6  52 6  10

   
x 2 3 x 2 3
 5 2 6  52 6  10

53. 1  x n  C0  C1x  C2 x 2  C3x 3  ....

Multiply both sides by x 2
x 2 1  x   C0 x 2  C1x 3  C2 x 4  C3 x 5  C4 x 6  C5 x 7  C6 x 8  C7 x 9  ....
n
…… (1)
Now put x = 1, ,  in (1) we get, 2

1.2n  C0  C1  C2  C3  C4  C5  C6  C7  ....
2 1    C02  C1  C2  C32  C4  C5  C62  C7  ...
n

 
n
 1  2  C0  C1  C22  C3  C4  C52  C6  C7  ...
Then add we get,
 
n
2n  2 1     1  2  3  C1  C4  C7  ...
n

4 i ni 2 i  ni
3  C1  C4  C7  ....  2n  e 3 .e 3 e IR
3 .e 3

 2 i 
1 3
     i e 3 
 2 2 
 4 i 
S
 1 3
2    i e 3 
 2 2 
 i 
 1
1    i
3
e3 
 
M

2 2
 i 
 1 3  
 1    2  i 2  e 
2 3
P

2 i ni 2 i ni

 2 e n 3 .e 3 e 3 e 3
 n 2 i  n 2 i
n2

 2 e n 3 e  3  2n  2 cos
3
1 n2 
C1  C 4  C7  ....   2 n  2 cos 
3 3 
54. 1  x   Cn
0  C1x  C2 x 2  C3 x 3  C4 x 4  ....
Put x = 1, , 2 and adding we get,
 
n
3  C0  C3  C6  ....  2n  2   
n

1    2  0 
 
1 3  
   i  cos  i sin
2 2 3 3
  
2  cos  i sin
3 3
1 n n n n  1  n 
C0  C3  C6  ....   2n  cos  i sin  cos  i sin    2 n  2 cos 
3 3 3 3 3  3 3
55. The natural numbers are written in rows
1 4 7 10 13
2 5 8 11 14
3 6 9 12 15
x  y is divisible by 3 if and only if and only if x  y is divisible by 3. The numbers x
3 3

and y are taken one from row 1 and other row 2 or both from row 3.
The desired number is 5C1  5C1  5C2  25  10  35
56. The numbers are written is rows
1 6 11
2 7 12
3 8 13
4 9 14
5 10 15
x  y  ( x  y )( x  y ) is divisible by 5  both x and y are from any of these rows or
2 2
IR
one from row 1 and the other from row 4 or one from row 2 and other from row 3 
desired number is
5 3 C2  2  3 C1   15  18  33
2
S
57. Given 8Ck 2  2 8Ck 3  8Ck  4  10C4
  8Ck  2  8Ck 3    8Ck 3  8Ck  4   10C4
M

 9Ck 3  9Ck  4  10C4

 10Ck  4  10C4 only 10 C5 10 C4  K  4  5  K  1
k  and  k  
P

Hence quadratic equation having roots  and  and  k and  k are identical and have
both roots common.
 m2
3 3
(B) For 1 x  or  x  , x  1  0
2 2
Therefore no solution is possible
For x  1, given equation is  2 x 2  5 x  3  x  1  0
 2 x 2  4 x  2  0  x 2  2 x  1  0  x  1.
 The equation has only one solution
 n  1.
1 1 1
(C ) Constant term C    ............ 
1.2 2.3 n  n  1
 1 1 1
tn   
n  n  1 n n  1
n
1
 C   tn  1   p 1
x 1 n 1
D)
 x  2a   1
2

 x  2b   1
2

x   1  2 a , x   1  2b
1  2a  1  2b  b  a  1
1  2a  1  2b  b  a  1
 a b 1
58. We first observe that n = 8
(A) r = 5
9r 1
(B) 8C .2r  8C 2r 1 r 1
 r6
r
r 2
But for r = 6, 8C6 .2  8C5 .2
6 5

8
 T6 and T7 are the terms whose coefficients have greatest value (equal to 7 (2 )).
IR
(C) Tr 1  nC .2r x r / 3 is an integer only if r = 0,3 &6. So the number of integral terms is 3
r

1
       
8 8 8 8
(D) x  R  2 1  1 2 1  2 1  2 1 1
2
S
 2  1
8
 I  f where f  1 

 R 1  f    2  1  2  1
8 8
1
M

59. (A) One Indian wife and one American wife can be selected in 2 C1  2C1 ways and
keeping an unmarried person in between these two wives the total number of linear
P

arrangements are 2 C1  2C1  7  2  40320

(B) Required number of ways  8  40320
(C) Required number of ways   7  1  2 2 C1 2 C1  5760
(D) Number of ways in which interviews can be arranged
 9 8 C2 6 C2 4 C2 2 C2  22680
60. (A) We have log 2 2  x 2  x   4log 2  x  1  log 2 x  1
 (log 2 ( x.  x  1  4.log 2  x  1 .log 2 x  1
2

  log 2 x  log 2  x  1   4.log 2  x  1 .log 2 x  1

2

 x 
 log2    1
 x 1 
 x 1
  2 or  x = 2 or -1
x 1 2
But x = -1 (rejected)
Hence x = 2
(8 x  3)
10 log10
7 log 7
ln 5log 5
(B) We have 2log e 2
 13
 8 x  3  13  8 x  16  x  2
(using a log N)
a N

 1 1 
(C) Let N =   
 log 2  log 6  
 N  log x 2  log 6
 N  log x 12 …….. (1)
As,   12  
2 3

 2  N  log 12  3
Hence, log 12 lies between 2 and 3
(D) We have l   log 3 4  log 2 9    log 3 4  log 2 9 
2 2

 4  (log3 4)   log 2 9  As,  a  b    a  b   4ab

2 2

 l  4
 log 4  log 9 

 log 3  log 2 
 log 2   log 3 
  4  
 log 3   log 2 
IR
 l  4  2  2  4  4  16
S
Also, m =  0.8  1  9log 8 
log 65 5
3

 8

    1   32 
 10 

log 3 8 log 65 5
M

8
 
log65 5
log382
    1 3
 10 
8 8
 1  64  65   5  m  4
P

log 5

10 10
Hence  l  m   16  4  20
 IIT – ian’s P A C E
(SINGLE CORRECT ANSWER TYPE)
This section contains 10 multiple choice questions. Each question has 4 options (A), (B), (C) and (D) for its
answer, out of which ONLY ONE option can be correct.
Marking scheme: +3 for correct answer, 0 if not attempted and -1 in all other cases.
n  m  m 
41.      n Cm  m Cp  p Ck   
m 1  k 1  p  k 

A) 3n  2n B) 4n  3n C) 3n  2n D) 4n  1

42. The term independent of x and y in the expansion of

1 2 1 2 1 2
[( x  ) ( y  )  ( xy  )  4]n is
x y xy

n
A) (  n cr ) 2 B)
n

 ( n cr ) 2
IR n
C) (  n cr )3 D)
n

( n
cr ) 3
r 0 r 0 r 0 r 0
S
n
43. If C0 , C1 , C2 ..... are binomial coefficients in the expansion Cr x r , then value of the
r 0
M

2C0 3C1 4C2 5C3

expression (series)     ......  is
1 2 3 4

2n  n  3  1 2n  n  2   1
P

A) 2  1 B) 2  1
n n
C) D)
n 1 n 1 n 1 n 1

i3 3
44. If f(x) =   x  a i    a i  3x , where ai< ai + 1, then f(x) = 0 has
i 1 i 1

A) only one real root B) three real roots of which two of them are equal

C) three distinct real roots D) three equal roots

45. The equation a8 x8  a7 x 7  a6 x 6  ...  a0  0 has all its roots positive and real

 where a 8  1, a7  4, a0  1/ 28  , then

1 1 7 7
A) a1  B) a1   C) a 2  D) a 2 
28 24 25 28
46. If both the roots of  2a  4  9 x   2a  3 3x  1  0 are non-negative, then

5 5
A) 0  a  2 B) 2  a  C) a  D) a  3
2 4
2
 n n  n  n  1  n  2 
2

47. If n  N and Ck  Ck , and  k  Ck

n 3
  then p is
k 1  nC  3p
 k 1 

A) 1 B) 2 C) 3 D) 4

48. If log 5 2x 1 2, log 21x 1 4, 1 (taken in order) are in harmonic progression then
    

A) x is a positive real B) x is a negative real

C) x is rational which is not integral D) x is a integer

49. If 1, log9(31 –x+ 2), log3(4(3x) – 1), (taken in order) are in arithmetic progression then x is
equal to IR
A) log34 B) 1 – log43 C) 1 – log34 D) log43
S
50. The number of distinct rational numbers of the form p/q, where p, q 1, 2,3, 4,5, 6 is

A) 23 B) 32 C) 36 D) 28
M

SECTION - II
(PARAGRAPH TYPE)
P

This section contains 3 Paragraph of questions. Each paragraph has 2 multiple choice questions based on
a paragraph. Each question has 4 choices A), B), C) and D) for its answer, out of which ONLY ONE IS correct.
Marking scheme: +3 for correct answer, 0 if not attempted and -1 in all other cases.
Paragraph for Question Nos. 51 & 52:
Let  a  b   
Q x  Q x  2
 a b  A Where   N , A  R and a 2  b  1

          
1 1
 a  b a  b 1 a  b  a  b and a  b  a  b

i.e.,  a  b    a  b  or  a  b 
1 1
51. If  ,  are the roots of the equation 1! 2! 3! ................   x  1! x !  k 2 and k I , Where
 and If 1 ,2 ,3 , 4 are the roots of the equation

   
x 2  1 2  3 2  4 3  5 4  x 2   5  
 
a b  a b  2a .

Where a 2  b  1 and . denotes G.I.F, then the value of 1  2  3  4  123 4 is

A) 216 B) 221 C) 224 D) 209

a a a ........
If   49  20 6    
x 2  x  3 x x x ............
 52 6  10 where a  x  3, then x is
2
52.
 

A)  2 B) 2 C) 2 D) 2

Paragraph for Question Nos. 53 & 54:

1  x n  C0  C1x  C2 x 2  ....  Cn x n , where x is complex number and C 0 , C1 ,...., C n are

constants. Then,

53. The value of C1  C 4  C7  .... will be

IR
1 n  1 n 
A)  2  2 cos  n  2   B)  2  2 cos  n  2  . 
S
3 4 3 3

1 n  1 n 
C)  2  2 cos  n  2   D)  2  2 cos  n  2  . 
3 4 3 3
M

54. The value of C0  C3  C6  .... will be

P

n n n n
A)  2n  2 cos  B)  2n  2 cos  C)  2n  2 cos  D)  2n  2 cos 
1 1 1 1
3 3  3 3  3 4  3 4 

Paragraph for Question Nos. 55 & 56:

Two numbers x and y are drawn without replacement from the set of the first 15 natural
numbers. The number of ways of drawing them such that
55. x 3  y 3 is divisible by 3
A) 21 B) 33 C) 35 D) 69
56. x 2  y 2 is divisible by 5
A) 21 B) 33 C) 35 D) 69
 SECTION - III
(Matching List Type)
This section contains four questions, each having two matching lists (List-1 & List-II). The options for the correct
match are provided as (A), (B),(C) and (D) out of which ONLY ONE is correct.
Marking scheme: +3 for correct answer, 0 if not attempted and -1 in all other cases.
57. For the following questions, match the items in column-I to one or more items in column-II

Column-I Column-II
If 8 Ck  2  2.8 Ck 3 8 Ck  4 10 C4, then the Quadratic
P) equations whose roots are  ,  and  k ,  k have m 1) 1
common roots, then m=
If the number of solutions of the equation
Q) 2) 2
2 x 2  5 x  3   x  1  0 is (are) n, then n=

If the constant term of the quadratic

expression   x 
R) n 1  1 3) 0
  x   as n   is p, then p=
k 1  k 1 k

S)
IR
The equation x 2  4a 2  1  4ax and x 2  4b 2  1  4bx have
only one root in common, then the value of a  b is 4) -1

5) -2
S
A) P-2; Q-1; R-1; S-1 B) P-3; Q-2; R-1; S-4
C) P-1; Q-4; R-2; S-1 D) P-4; Q-1; R-3; S-2
M

58. Consider the binomial expansion of (1+2x)n (n is a positive integer) in which the sum of
the coefficients is 6561. Let 1  2 x   R  I  f where I is the largest integer not exceeding
n

R and 0  f  1 .
P

Column-I Column-II
If r th term in the expansion is the greatest
P) 1) 3
term, then r cannot exceed
If ith term is having the greatest coefficient,
Q) 2) 4
then i can be
The number of integral terms in the
R) 3) 5
expansion when x  3 3 is less than
1
S) For x  , the value of R 1  f  is less than 4) 6
2
5) 7
A) P-1,2; Q-4,1; R-3,1; S-1,2 B) P-3,45; Q-4,5; R-2,3,4,5; S-1,2,3,4,5
C) P-2,1; Q-1,2; R-2,4; S-3,4 D) P-3,5; Q-1,2; R-2,1,3; S-1,3,4,5
59. There are 2 Indian couples, 2 American couples and one unmarried person
Column-I Column-II
The total number of ways in which they can
sit in a row such that an Indian wife and an
P) 1) 22680
American wife are always on either side of the
unmarried person, is
The total number of ways in which they can
Q) sit in a row such that the unmarried person 2) 5760
always occupy the middle position, is
The total number of ways in which they can
sit around a circular table such that an Indian
R) 3) 40320
wife and an American wife are always on
either side of the unmarried person, is
If all the nine persons are to be interviewed
one by one then the total number of ways of
S) 4) 24320
arranging their interviews such that no wife
gives interview before her husband, is
A) P -1,2,3; Q-4; R-3,4; S-4 B) P-3; Q-3; R-2; S-1
C) P-4,2; Q-2; R-3,4; S-3 D) P-2,3,4; Q-2; R-3,4; S-1 IR
60. Match the following questions:
Column-I Column-II
S
The Value(s) of x, which does not satisfy the equation
P) 2 1) 2
log 22 (x – x) – 4log2(x – 1) log2x = 1, is (are)
M

The value of x satisfying the equation

log10(8 x  3)
Q) 7 log 7 10 2) 3
ln 5log 5
log 2 e
2 = 13, is
P

The number
R)  1 1  3) 4
N    is less than
 log 2  log 6  
Let l   log 3 4  l o g 2 9    log 3 4  l o g 2 9 
2 2

S) 4) 5
 
log65 5
And m   0.8 1  91og
8
3
then l  m is divisible by

5) 6
A) P -1,2; Q-1; R -4, 5; S-1,2,3 B) P -1; Q-1; R -3, 4; S-2,4
C) P -2,3,4,5; Q-1; R-2,3,4,5; S -1,3,4 D) P -3,2; Q-2,4; R -1; S-3
MATHEMATICS
37 2 38 6 39 9 40 0 41 9 42 3

43 ABC 44 ABCD 45 ACD 46 ABCD 47 ABC 48 ABC

49 0.50 50 315 51 10 52 25 53 70 54 35

IR
S
M
P
 MATHEMATICS
37. Let S  Spanish, F  French, S  F  number of students who study both
80%  2001  1601  S  85%  2001  1700
 30%  2001  601  F   40%  2001  800 , by the principle of Inclusion-Exclusion
S  F  S  F  S  F  2001
For ' m '  S  F to be smallest, S and F must be minimized  1601+601-m=2001  m=201
For ' M '  S  F to be largest, S and F must be maximized  1700+800-M=2001  M=499
Qn  Dn 1  Dn  2  Dn 3  .......  1
n 3
38. Circular derangement D2
Q6  D5  D4  D3  D2  44  9  2  1  36
39.  f 1  1  1  1  f 1  1  0  f 1  2  f 1 can be 1 or 2
Similarly remaining, total no. of function=2.3.3.2=36
40.  power of 5in 5 .10 .15 ....100   5 
5 1  2  3  .....  20
5 10 15 100
.525  50  75 100  51300
41. Each one should get 4   2 blankets where  W
  4 1  2    4  2  2    4 3  2    4  4  2   32  1   2   3   4  6.
Number of ways C 9 6  4 1
4 1

42. Distribution 2,2,2 and 3 to the youngest. Now 3 toys for the youngest can be selected in

IR
C3 way, remaining 6 toys can be divided into three equal groups in 6! ways and can
9

 2!
3
3!
6! 9
be distributed in 3! ways  C3 . 
9

 21
3
3!.8
S
N
43. N  5C0  5 C1  5 C2  5 C3  5 C4  5 C5  2 5  32  8
4
Number of oneone function n Cm n
44. Number of strictly increasing function    Cm
M

m! m!
Number onto functions = 4  C1 .3  C2 .2  C3  240
5 4 5 4 5 4

Required number of bijections =number of derangement= 4! 1 1 1

P

  9
 2! 3! 4! 
Number of monotonic functions  2n  m 1
C m   n  2. C3  5  65
7

100!
45.
100
C50   2 .3.5.7........
50!5!
E2 100!  97, E2  50   47  E2  100 C50   3
Similarly E  3
100
C50   4, E 5  100 C50   0, E 7  100 C50   0   3,   4,   0,   0
1 1 1 1 1
46. 6!       265
 2! 3! 4! 5! 6! 
Either four balls or all balls go into correct boxes=16
Either all balls go incorrect or exactly one ball go into correct box or two balls go into
1 1 1 1
265  6 C1 .5!      6 C2 .9  664 correct.
 2! 3! 4! 5! 
28.8!  2r  1
8

16! 35 8
47. 1)Number of ways =  r 1
   2r  1
 4!  4!
4 5
4! 27 r 1
 p1 , p2 , p3 , p4
12! 11!
16  
 3!
4
108
12Others
2)
p1 , p2 , p3 , p4 , p5
3 4
16 p6 10! 
12!
 12 4
10  4!
3
3!7! .3!
p7 ..... p16 10  7 4
3)
10! 12!
To the ways=  .
3!7!  4! .3!
3

48.

49. To find “prime looking numbers” we can apply complementary counting.

We can split the number from 1 to 1000 into following groups IR
350  190  x    30  x   x   40  x    30  x    x  10  , x  30

1 ,{numbers divisible by 2= S },{numbers divisible by 3= S }, {numbers divisible by

S
2 3

5= S },{primes not including 2,3,5}, {prime looking numbers}

5

Hence prime looking numbers =1000-(168-3)-1- S  S  S 2 3 5

M

=1000-165-1-{500+333+200-166-66-100+33}=100
50. Number of ways in which 3 correct envelopes can be selected  C  35 T
3

Derangement of the remaining 4 envelopes and letters =91

P

 9  35  315
51. Case I: All are having self images : f(1)=,f(2)=2….,f(4)=4____(1)
f 1  1 f  3  3 4
Case II: Exactly two having self images:  C2 1 =(6)
f (2)  2 f  4  4 

Case III: Two are having images of each other and other two having images of each other
f 1 1 f  3  3 4 C2 .2 C2
   3
f (2) 2 f  4  4  26

 max power of 2 in 54! 

52. Max power of 2 2 in 54! =    25
 2 
Max power of 3 in 54!=26
Since the max power of 2 is less than the max power of 3=25
2

53. x a y b z c w d t e  a  b  c  d  e  4 4  5 1 C51  70

54.
 5  1   4  1!  7!
 35
 5  1!.  4  1! 4!.3!