NCERT SOLUTIONS
SET THEORY
Solution :
PAGE 20 :( Proved numerically in NCERT )
Prove this result in this question theoretically
(i) To prove  A  B  '  A ' B '
We will show that  A  B  '  A ' B ' and  A ' B '   A  B  '
First let x   A  B  ' then x   A  B 
 x  A and x  B (Note that despite union we have ‘and’ word !)
 x  A ' and x  B '  x   A '  B '
(i)
  A  B  '  A ' B '
Again let x   A '  B ' then x  A ' and x  B '
 x  A and x  B  x  A  B
(Note !)
 x  A  B'
Thus  A ' B '   A  B  '
(ii)
From (i) and (ii) it follows that  A  B  '  A ' B '
This is called demorgan law. In words it states that compliment of union is
intersection of the compliments.
(ii) Another form of demorgan law. Proof is similar to the part (i). Do yourself.
(iii) Let x   A ' ' then x  A '  x  A   A ' '  A
Again if x  A then x  A '  x   A ' '  A   A ' '
Thus  A ' '  A
(iv) Let x  A   B  C  then x  A or x   B  C 
 x  Aor  x  B and x  C 
  x  A or x  B  and  x  A or x  C 
 x   A  B  and x   A  C 
 x  A  B   A  C   A   B  C    A  B   A  C 
By simply writing backwards we will arrive at  A  B    A  C   A   B  C 
Thus A   B  C    A  B    A  C 
SOLVED EXAMPLE 30 PAGE 25 :
If A and B are two subsets of same universal set and if A  B  A  B then A  B . Is
the converse true?
Solution :
Given A  B  A  B
We have to prove A  B
Let x  A then x  A  B
 x A B
( A B  A B )
 x  A and x  B  x  B  A  B
If x  B then x  A  B
 x A B  x A  B  A
Thus A  B
31
Conversely if A  B then
A B  A A  A
A B  A A  A
 A  B  A  B. Thus the converse is also true.
NOTE : It is interesting to note that converse is not true in case of probability
theory. Let a dice be thrown once. Let A and B denote the event of getting 2 and 3
respectively then probabilities of two sets A  2 , B  3 are equal but
P  A  B  2 / 6 , P  A  B  0
Solved Example 31 Page 25
Following the notation of previous example, prove that for any two sets A and B
P  A  B   P  A  P  B  .
Solution :
In order to prove that P  A  B   P  A  P  B  , it is enough to show that
P  A  B   P  A  P  B  and P  A  P  B   P  A  B  .
First let X  P  A  B 
 X  A  B  X  A and X  B
 X  P  A and X  P  B 
 X  P  A  P  B 
 P  A  B   P  A  P  B 
(i)
Again let X  P  A  P  B. Then,
 X  P  A and X  P  B 
 X  A and X  B  X  A  B
 X  P  A  B

P  A  P  B   P  A  B 
(ii)
From (i) and (ii), it follows that P  A  B   P  A  P  B  .
MISC.EXERCISE PAGE 26-27
3.
Given A  B  A  C On taking intersection with set C on both sides we get
 A  B  C   A  C   C
  AC B C  C
[Since  A  C   C  C ]
  A  B   B  C   C
[Since A  C  A  B ] (i)
Again, A  B  A  C   A  B   B   A  C   B
 B   A  B   C  B 
[Since  A  B   B  B ]
 B   A  B   B  C 
(ii)
BC .
From (i) and (ii), it follows that
10.
Let A be the set of persons living in Delhi and Agra, B be the set persons living
in Delhi and Lucknow, C be the set of persons living in Delhi and Amritsar
then A  B  A  C , But B  C
32
11
Given, A  X  B  X for some set X . As in previous example .
 A A X   AB  X 
 A   A  B   A  X 
[Since A   A  X   A ]
[Since A  X  φ (given)]
 A   A  B  φ
 A  A B
 A B
Again, A  X  B  X
 B  A X   B B  X 
…(i)
  B  A   B  X   B
[Since B   B  X   B ]
[Since B  X  φ (given)]
  B  A  φ  B
 B  A  B  A B  B
B A
…(ii)
From (i) and (ii), it follows that A  B .
12.
We may take A  1, 2 , B  2,3 , C  3,1
We have A  B  2 , B  C  3 A  C  1
15.
(i) N  H   25, N T   26, N  I   26
N  H  I   9, N  H  T   11
N T  I   8, N  H  T  I   3
The answer to the first part  N  H  T  I 
(ii)
16.
 N  H   N T   N  I   N  H  I   N  H  T   N T  I   N  H  T  I 
 25  26  26  9 11 8  3  52
Number of persons who read exactly one newspaper)
N  H  T  I   N  H  T   N T  I   N  H  I   2N  H T  I 
 52  9 11  8  6  30
NOTE:- Do the problem by x, y, z, t,.....
Let N ( S ) denote number of elements in a set S . We have
N  A  21, N  B   26, N C   29
N  A  B   14, N  C  A  12, N  B  C   14
N  A B C  8 .

We have to determine N C  A  B

From venn diagram N C  A  B


 N C   N  B  C   N  A  C   N  A  B  C 
 29 14 12  8  11
NOTE: We can generally do such a question by
taking mutually non intersecting zones in the
venn
diagram as x, y, z, t...... conveniently.
33
MATHEMATICAL INDUCTION
16.
1
1
1
1

RHS 

1.4 4
3 1  1 4
1
1
1
1
k


 .......
Let P  k  be true then

1.4 4.7 7.10
 3k  2 3k  1 3k  1
For n  1,
LHS 
(i)
To prove P  k  1 we must show that
1
1
1
1
1
k 1


 ...... 


1.4 4.7 7.10
 3k  23k  1 3k  13k  4  3  k  1  1
(Note that the extra term is being written by pattern i.e. adding 3 or by replacing k by
k 1 )
1
1
1
1
LHS 

 ..... 

1.4 4.7
 3k  2 3k  1 3k  13k  4 
k
1
from P  k 


3k  1  3k  1 3k  4 
18.

1 
1 
k

3k  1 
3k  4 

k 1
3k  4

1  3k 2  4k  1
3k  1  3k  4 

 3k  1 k  1
 3k  1 3k  4 
 RHS
1
9
2
 2 1  1 i.e. 1 
8
8
1
2
Let P  k  be true then 1  2  3  .....  k   2k  1
(i)
8
1
2
We must show that 1  2  3  ......  k   k  1   2k  3 (ii)
8
1
2
we have 1  2  3  .....  k   k  1   2k  1   k  1 by (iii)
8
1
1
2
2
From (iii) the statement (ii) follows if  2k  1   k  1   2k  3 or
8
8
2
2
2
2
 2k  1  8  k  1   2k  3 or 4k  12k  9  4k  12k  9 which is, indeed true
The result is true for n  1 since 1 
for all k .Thus P  k   P  k  1
19.
Let k  k  1 k  5  3m
We must show that  k  1 k  2 k  6 is also a multiple of 3 .
Now  k  1 k  2 k  6   k  1 k  2 k  6  k  1 k  2
(Avoid touching 6  k  1 k  2 ) since it is a divisible by 3 )
 k  k  1 k  5  3  3m  k  k  1 k  5  3k  k  1  3m
34
 3m  3k  k  1  3m which is divisible by 3 Thus P  k   P  k  1
21.
A polynomial f  x  is divisible by another polynomial g  x  if there exist a polynomial
h  x  such that f  x   g  x  h  x 
Let x2k  y 2k   x  y  h  x  where h  x  is a polynomial.
Now x 2 k 2  y 2 k 2  x 2 k .x 2  y 2 k 2  x 2  x  y  h  x   y 2 k   y 2 k  2
  x  y  x2 h  x   x 2 y 2k  y 2k 2  x 2  x  y  h  x   y 2 k  x 2  y 2 
 x2  x  y  h  x   y 2k  x  y  x  y  which is also divisible by x  y
Thus P  k 
22.
P  k  1

Let 32 k  2  8k  9  8m
Now 32k 4  8  k  1  9  32.32k 2  8  k  1  9
 32 8m  8k  9  8  k  1  9  72m  72k  81  8k  8  9
 72m  64k  64

P  k  1 is also true since the last expression is divisible by 64
23.
 IH 
Let 41k  14k  27 m
k 1
k 1
 41.41k  14.14k
Now 41  14
 41.27m  41.14k  14.14k
 41 27m  14k   14.14k
 27m  14k.  41 14 
 27 m  14k .27

P  k  1 is also true since the last expression is divisible by 27
24.
 2n  7    n  3 
2
(i)
The result is true for n  1 since 9  16
2
Let the result be true for n  k then 2k  7   k  3
(ii)
To prove the result for n  k 1 we must show that 2  k  1  7   k  4 
2
(iii)
Now 2  k  1  7   2k  7   2   k  3  2 it is now sufficient to show
2
 k  3
P k 
2
 2   k  4

2
or
6k 13  8k 16
or
2k  3  0 which is true thus
P  k  1
PERMUTATIONS & COMBINATIONS
EXERCISE 7.3
1.
There are 3 places. The first place can be filled in 9 ways, the second can be filled in 8
ways and the third can be filled in 7 ways. Thus required number of 3 digit numbers
which can be made by using the nine digits 1,2,3,----------9 when digit can not repeat
 8  9  7  504
35
2.
3.
4.
5.
Short answer :Ans must be 9 P3 since we have to find permutations of nine
distinct objects taken three at a time.
When digits are not specified they will be 0,1,2,3,........,9 . The required answer
 9  9  8  7  4536 .
Even numbers will either end with 2 or with 4 or with 6 (3 cases).
If a digit number ends with 2 then number of three digit even numbers  5  4  20
Required answer must be 20  3  60 .

The answer to the first part is clearly 5 P4 or 5  4  3  2 . For the second part we have to
fix the last place in 2 ways.
Chairman can be chosen in 8 ways. Corresponding to each of these ways vice chairman
can be chosen in 7 ways.
 Required number of ways  8  7  56
n 1
6.
n
P3 1

P4
9
(n  1)! 1

n!
9

(n  1)! (n  4)! 1


(n  4)
n!
9
(n  1)! 1

n(n  1)! 9
5!
6!
2  6!
5
6
(i)
Pr  2.
Pr 1 
2

(5  r )
(6  r  1)! (7  r )!
5!
6  5!
5!
on cancelling

 2
(5  r )!
(7  r )(6  r )(5  r )!
(5  r )!
12
we get 1 
(7  r )(6  r )

7.

(n  1)!
(n  1  3)! 1

n!
9
(n  4)!
(Cancelling (n  4)!) 


r 2  13r  30  0
(7  r )(6  r )  12


r  3 or r  10
But r  10 is not possible since 5 Pr is not defined at r  10 
(ii)
5
Pr 6 Pr 1

8.
10.
r 2  13r  36  0


1
(r  4)(r  9)  0
1 1
 
n 9
n 9.
(r  3)(r  10)  0
r 3
6
(7  r )((6  r )
 r  4 (why?)
Required answer must be permutations of 8 objects (E,Q,U,A,T,I,O,N) taken all at a
time
9.
5!
6!

(5  r )! (7  r )!


or 8 P8 
8!
0!
 40320
6
(i)
(ii) 6!
P4
(Now you can write reasoning since the same has been described in previous questions)
(iii) There are 2 vowels only in the word MONDAY
Required answer  2  5  4  3  2 1  240

The configuration of the word MISSI SSIPPI is SSSS;1,1,1,1;P;P;M
The number of arrangements taken all at a time 
36
11!
4!4!2!1!
Now we have to discard those cases in which four I’s are together. Treating four I’s
as one (or tying four I’ with a string) the number of such arrangement 
11!
8!
1110  9  8! 8!



4! 4! 2!
4!2!
4!4!2! 4!2!
8! 161 8  7  6  5 161
8! 1110  9 

 1



4!2! 
24
4!2! 4!
2! 4

 42  5  161  33810 .
Fix P and S in the beginning and in the end respectively after which we have to
10!
arrange E,R,M,U,T,T,A,I,O,S linearly ANS 
 1814400
2!

11.
Required answer 
(i)
(ii)
Tie all five vowels A,E,I,O,U with a string and treating them as one object the
required answer must be
(iii)
1.
2.
3.
8!
4!2!
8! 5!
 2419200
2!
P can be at first place then S has to be at the sixth place. Similarly P can be at
the second place then S has to be at the seventh place and so on.
Finally P can be at 7th place (7 ways). But S can also proceed P (2 ways). For
10!
each such case (14 cases) the remaining will permute in
way.
2!
10!
Required answer  14  25401600

2!
EXERCISE 7.4
If C8  C2 then either 8  2 or 8  2  n . Since 8  2 is not possible, 10  n or n  10 .
n
n
product of threecon sec utive int egers
2n(2n  1)(2n  2)

3!
6
n
(
n

1)(
n

2)
2
n
(2
n

1)(2
n
 2) 4(2n  1)
2n
Similarly n C3 
C3 : nC3 

6
n(n  1)(n  2)
n2
4(2n  1)
2n  1
But 2n C3 : nC3 is given as 12 
 12 
3
n2
n2
2n  1  3n  6 
n5

2n
C3 
(ii) same as part (i).
Selection of any two points from 21 points will give rise to a chord
 Total possible chords 21 C2 
21 20
 210
2
4.
3 boys
3 girls
From diagram, required number of selections 5 C2  4C3  4
37
5.
3
3
3
From diagram, required number of selections  6C3  5C3  5C3  2000
6.
(when exactly 1 ace is there, there should be 4 non aces selected from 48 non aces).
 Required number of selections = 4 C1  48C4
7.

8.
9.
Required number of choosing a cricket eleven 5 C4 12 C7  3960
Easy. Write the solution yourself by drawing a diagram.
The answer should clearly be 7 C3  35
MISCELLANEOUS EXERCISE
1.
2.
3.
Number of selections  3C2  5C3  30 each selection will make 5! words
Required answer  30  5!  3600

Tie up vowels E, U,A,I,O with one string and consonants Q,T,N with another string then
number of arrangements  2! 5! 3!
(i)
38
From diagram the required answer  4C3  9C4  504
(ii)
Following cases are possible from the above diagram
 Required answer  4C3  9C4  4C4  9C3  588
(iii)
ANS = N (0,7)  N (1,6)  N (2,5)
 4C0  7C7  4C1  7C6  4C2  7C5  4C3  7C4  1632
10!
2!2!2!
 AA, I I,NN are there)
4.
Fix E at first place and arrange remaining 10 letters in
5.
Fix 0 at the last place so that we ensure that the number is divisible by 10 after which
permute 1,3,5,7,9 at first five places
 Ans = 5!
Same as Q.No.1
Same as Q.3 (ii)
Same as Q.6 Ex.7.4.
There are nine places. Women have to occupy four even places in 4! ways and
corresponding to each such case men will occupy odd places in 5! ways. Therefore
required answer  5! 4!  2880
There are two cases only
(i) Either all three go in the party in this case we have to select 7 person only from
remaining 22 persons in 22 C7 ways.
(ii)
Or None of the three persons go for party. In this case we will have to select 10
persons from remaining 22 persons in 22 C10 ways.
Required answer 22 C7  22C10

We have S,S,S,S;A,A,A,I,I,N,N;T,I,O. Tie up four S with one string then we have to
arrange S; A,A,A;I,I;N,N;T,I,O linearly
 907200
6.
7.
8.
9.
10.
11.
 Required answer 
11!
 151200
3!2!2!
BINOMIAL THEOREM (EX.8.2)
1.
8 r
Tr 1  Cr x .3
8
r
we must have 
2.
Tr 1 12 Cr a12r  2b 
r 3
r

coeff. of x5 8 C3 33  56  27  1512
 12Cr a12r  2 br
r
We must have 12  r  5, r  7 . The first equation is satisfied for r  7 .
7
Term of a5b7 does exist  coeff. of a5b7  12C7  2  101376

39
9.
In the expansion of 1  x  coeff. of x p  l C p 
l
 l  m  n, b  m 
m n Cm

mn Cn
10.
 x  1
 1  x 
n

C   C  

coeff. of a m in the expansion of a m
= coeff. of a n in 1  a 
m n
n
coeff. of  r  1 th term  nCr 2 ,Coeff. of r th term n Cr 1 , coeff. of (r  1)th term
 nCr
n!
 r  2 ! n  r  2 ! 1
 ,
n!
3
 r  1 n  r  1!

n
Cr  2 1
 ,
n
Cr 1 3

Now given n Cr 2 : nCr 1 : nCr  1:3:5
n!
 r  1! n  r  1!
n
r ! n  r !
Now n  r  2   n  r  2  n  r  1! and

n
Cr 1 3

Cr
5
n
3
5
 r  1!   r  1 r  2! etc.
r 1
1
r
3
 ,

n  r  2 3 n  r 1 5
3r  3  n  r  2, 5r  3n  3r  3 
n  4r  5, 3n  8r  3

The last equations easily give n  7, r  3.
The above equations now become
11.
Coefficient of xn in the expansion of 1  x 
2n
Coefficient of xn in the expansion of 1  x 
It is sufficient to show 2n Cn  2. 2n1Cn1
is
2 n 1
2n
is
Cn
2 n 1
Cn1
(2n  1)!
(2n)(2n  1)! (2n)! 2 n

 Cn
(n  1)!(n)!
n(n  1)!n!
n!n!
m(m  1)

 6 etc.
2
LHS  2. 2n1Cn1  2.
12.
m
C2  6
MISCELLANEOUS EXERCISE
1.
 a  b
n
 C0 a b  C1an1b  nC2 a n2b2  .....
n
n 0
n
Now according to the given condition
n
n
n
C0 anb0  729
C1an1b  7290
C2 a
(*)
n2 2
b  30375
The system of equations (*) can be written as
a n  729
(**)
nan1b  7290
n(n  1) n2 2
a b  30375
2
Dividing second by first and IIIrd by second we get
nb 7290

 10
a
729
 n  1 b 30375
. 
2
a 7290
(***)
40
on dividing last two equations we get
 n  1
30375
125 5
n 1 5



2n
7290  10 300 12
n
6
n
6
6
Now a  729
a  729  3


2.


n6
a3
nb
b can be calculated from first relation at *** we have
 10
a
6b
b5


 10
3
In the expansion of  3  ax  9
Tr 1  9Cr 39r  ax 
 9Cr 39r ar xr
r
For coeff. of x2 , r  2
For coeff. of x 3 , r  3

9
93
C7 39  2 a 2  9 C3 3 a 3

36  37  a2  84  36  a3
(cancelling 36 )

3.
(cancelling 12)
3  3  a 2  7  a3
9
On cancelling a 2 , a 
(If a  0 then x is finished!)
7
6
7
7
6
1  2x  1  x   1  x  1  2x 

 1  7C1 x  7C2 x 2  7C3 x3  7C4 x 4  7C5 x5  7C6 x 6  7C7 x 7
1 
6

C1 (2 x)  C2 (2 x)  C3 (2 x)  C4 (2 x)  C5 (2 x)  6C6 (2 x) 6
6
2
6
3
6
4
6
5

 coeff. of x  1 C5 2  C C4 2  C2 . C3 2  C C2 .2  C4 .6 C1.2  7C5
 171
5
4.
6
5
7
6
1
4
7
6
3
7
6
3
2
7
Following the hint given in the question
a n  bn   a  b  b   b n
n
 nC0  a  b  b0  nC1  a  b 
n
  a  b  n C1  a  b 
n
n 1
n 1
b1  nC2  a  b 
b n C2  a  b 
n 2
n2
b2  ....  nCn1  a  b  bn1  nCn  a  b  bn  bn
1
0
b2  .... n Cn1  a  b  (Cancelling n Cn bn and bn )
Since  a  b  is common in the last expression the result follows.
5.
 x  y   x  y
6
6
 2  6 C1 x5 y1  6C3 x3 y 3  6C5 x1 y 3 
 2 6 x5 y  20 x3 y 3  6 xy 5 
 4 xy 3x 4  10 x 2 y 2  3 y 4 
  3  2
 4 6  27  60  12
2 3. 3   10  3   2   3 2  



the given expression
4
4 3
6.
 x  y
4
 99 
5
2

3 2
2
6
6
4
  x  y   2  4 C0 x 4 y 0  4C2 x 2 y 2  4C4 x 0 y 4 
4
Now x  a 2 ,
7.
Now x  3, y  2
 396 2
 2  x 4  6 x 2 y 2  y 4 
y  a2  1
1 

 1  .01  1 

 100 
5
5
1 5  1 
 1  C1
 C2 

100
 100 
5
41
5
(upto three term)
 1  .05  .001
8.
 .9510
Fifth term from beginning  T4 , Fifth term from end  Tn 5 2  Tn 3
(Or Tn 1 is Ist from end, Tn is second from end, Tn 1 is IIIrd from end , Tn  2 is IVth from
end and finally Tn 3 is fifth from end)
n
T
6
Now, As, is given 4 

Tn  4
1
2

n4
4
.31
21 3
n4

4n
4
1
1
2 4 .3
n 8 1

4
2


 6
4n
4
n

0
3
n 8
4
.3
n4
4
1 4
1 4
4
n4

6
1




 31 4 
3

1
4

n 8
4
 6 
6
n8
4
 61 2
n  10
x 2  
x   2 

 1  2  x    1  2    1  x  

 
 

4
2
14
Cn  4
C4  nCn4

 6
4
9.
n
  3 
 2  3 
C4 21 4
4
1
2
2
3
x  2
x  2
x  2
x  2 




 2
 4 C0 1       4 C1  1       4 C2 1       4 C3 1      4 C4   
2
x
2
x
2
x
2
x

 


 


 




 x
4
2
2
x 8
x  24 
x  32 
x  16

 1    1    2 1    1    4
x 2
x  2
x  2 x
 2
2
3
4
x
x 4 x 4 x
8
x
x 2 3 x3 
 1  4C1  4C2
 C3  C4
 1  3C1  3C2
 C2 
2
4
8
16 x 
2
4
8
24 
x 2  32  x  16
 2 1  x    3 1    4
x 
4  x  2 x
3
3
x
x4 8
24 24
32 16
 1  2 x  x 2     12  6 x  x 2  2 
6 3  2
2
2 16 x
x
x
x
x

x 4 x3 1 2
16 24 16
  x  4 x  5   2  4 (writing in descending powers of x )
16 2 2
x x
x
SEQUENCE & SERIES
EXERCISE PAGE 9.2
a b
ab

n 1
a b
2
n
2a  2b n  a n  ab n 1  ba n 1  b n

a n  ab n 1  b n  ba n 1  0


n
15.
n
n 1
  a  b   a n 1  b n 1   0

a  a n 1  b n 1   b  a n 1  b n 1   0
n 1
n 1
either a  b or a  b
42
4
If a  b then a n 1  b n 1 for all n . If a  b then a n 1  b n 1 is possible only when
ab
a n  bn
i.e. n  1 . Thus n 1 n1 can become
if either a  b or n  1
n 1  0
2
a b
(in case a  b )
22.
EXERCISE 9-3 PAGE 192/193
Let A be the first term and R be the common ratio of the G.P then
a  AR p 1 , b  AR q 1 , c  AR r 1
Now a q r br b c p q   AR p 1 
23.
q 1 r  p
r 1 p  q
p 1 q r  r  p q 1  p q r 1
 Aqr r  p pq .R          A0 R0  1
a  a, ar n1  b , P  a.ar.ar 2 .......ar n 1  a n r1 23.......n 1
 a .r
n
24.
 AR   AR 
q r
 n 1 n
2

P2  a2n .r
n n1
  a.ar n1    ab 
n
Sum S1 of n terms of a G.P a  ar  ar  .....  ar
2
n 1
n
is given by S1 
r 1
ar  r  1
n
Sum S 2 of n terms of G.P ar n  ar n 1  ....... is given by S2 
It is obvious that
25.
a  r n  1
n
r 1
S1 1

S2 r n
If a, b, c, d are G.P we may take a  a, b  ar , c  ar 2 , d  ar 3
Now LHS   a 2  b 2  c 2  b 2  c 2  d 2 
  a 2  a 2 r 2  a 2 r 4  a 2 r 2  a 2 r 4  a 2 r 6 
 a 2 1  r 2  r 4  a 2 r 2 1  r 2  r 4 
26.
3
28.
 a 4 r 2 1  r 2  r 4 
2
RHS   ab  bc  cd 
  a.ar  ar.ar 2  ar 2 .ar 3 
  a2r  a2r 3  a 2r 5 
 a 2 r 1  r 2  r 4 
2


2
2
 a 2 r 2 1  r 2  r 4 
2

LHS  RHS
Let the numbers inserted by a and b then 3, a, b,81 must be in G.P whose first term is
and fourth term is 81 . If r be the common ratio then 81  3.r 3  r  3
 a  ar  9 , b  ar 2  27
Let the numbers be a and b then a  b  6 ab *
We have to show
a 3 2 2

b 3 2 2
One method could be to divide equation * by b to get x 2  1  6 x where x 
a
then
b
our target becomes x 2
A tricky and fascinating method using componendo and divinendo is as follows
43
a  b  6 ab
 a  b   a  b
(*)
2
 4ab  36ab  4ab  32ab
Dividing (*) by ** we get
**
ab
3

a b 2 2
a b  a b
3 2 2

 a  b  a  b 3  2 2
Eliminating b we get a 
G2
 2A
a


a 2  2 Aa  G 2  0
2 A  4 A2  4G 2
 A  A2  G 2 , Similarly b  A  a2  G2
2
The signs (Plus or – at  ) chosen for a and b must be opposite.
 
 8,
Let the roots be  and  then
  5 or     16,   25
2
25
 16
Eliminating  we get  



 2  16  25  0
 is a root of x 2  16 x  25  0
Similarly  is a root of the equation x 2  16 x  25  0

Quadratic is x 2  16 x  25  0

32.
a  b  4 2. ab
a 3 2 2

b 3 2 2
ab
, G  ab
Let the positive numbers be a and b then A 
2

a  b  2 A, G 2  ab

29.

a
MISCELLANEOUS EXERCISE
1.
Tmn  a   m  n 1 d , Tmn  a   m  n  1 d
2.
Tmn  Tmn  2a   2m  2 d  2  a   m  1 d 
Let the three numbers in A.P. be a  d , a, a  d
 2.Tm which proves our assertion.
we have a  d  a  a  d  24
(i),
 a  d  a  a  d   440
From first equation 3a  24, a  8
On putting a  8 in (ii) we get 8  d  8 8  d   440
3.
64  d 2  55 
d2  9 

d  3  or  3

Numbers are 8  3,8,8  3 or 8  3,8,8  3 or 5,8,11:11,8,5
(Note that values of a and d are not first term and common difference)
n
S1   2a   n  1 d   na  n  n  1 d 2
2
2n
3n
 2a   2n  1 d   2na  n  2n  1 d , S3   2a   3n  1 d 
, S2 
2
2
44
(ii)
4.
n  1

We have show S3  3  S2  S1  , we have S2  S1  na  nd  2n  1 
2 

nd
3nd
 na   3n  1

3  S2  S1   3na   3n  1
2
2
3n
3nd
RHS  S3 
 3na   3n  1
 2a   3n  1 d 
. Thus LHS  RHS
2
2
The first number which is divisible by 7 (between 200 and 400) is 203. The last number
is 497. We have to find sum S  203  210  217  .....  407 .
Let the series contain n terms then 497  203   n  1 7
43
 2  203   7  1 7  etc.
2 
Let S1 be the number of numbers from 1 to 100 which are divisible by 2 , S 2 be the
number of
numbers from 1 to 100 which are divisible by 5, S 3 be the number of
numbers from 1 to 100
which are divisible by 10 then required
sum  S1  S2  S3 .Find S1 , S2 , S3 as in the last question.
The smallest two digit number which yields remainder 1 when divided by 4 is 13 , the
last is 97 . Let S  13 17  21 ....  97
22
 2 13   22  1 4 
Then 97  13   n 1 4  n  22  S 
 1210
2 

5.
6.
7.
7  n  1  294 

n  43
S
f  x  y   f  x  f  y  . Putting x  y  1 we get f 1  1  f 1 f 1   f 1 
Again f  2  1  f  2 f 1   f 1  f 1
2
Inductively we get f  n    f 1 
Now


  f 1 
3
2
 f  3   f 1 
3
n
 f  x   120  f 1  f  2  f 3  .....  f  n  120
f 1   f 1    f 1   ....   f 1   120
2
3
3  3  3  .....  3  120
2
3
n
5  2  1
n

3  3n  1
3 1
 120  3n  81  n  4
n
8.
a  5, r  2,
9.

n6 
a  1, ar 2  ar 4  90
10.
315 

2n  1  63
2 1
Last term  ar 5  5.26 etc.
r 4  r 2  90  0



2n  64  26
r
2
 10  r 2  9   0

r  3
Let the three numbers in GP be a, ar , ar 2 . Given a  ar  ar 2  56 (i)
The fresh numbers are a  1, ar  7, ar 2  21 . Since fresh numbers are in AP
2  ar  7   a 1  ar 2  21

a  r 2  2r  1  8
 2ar  a  ar 2  1  21  14 
(ii)
45
a  2r  1  r 2   8
1  r  r 2 56

7
r 2  2r  1 8
1  r  r 2  7r 2  14r  7
6r 2  15r  6  0


1
5  25  16
2r 2  5r  2  0
 2 or

 r
2
4
If r  2 then from (i) a  8 . Hence the numbers are 8,16,32 (Answer to other case is not
given)
Let the GP a  ar  ar 2  ar 3  .......  ar 2 n 1 contains 2n (even) terms.
Dividing (i) by (ii) we get
11.
It is given sum of all terms  5 ( sum of odd terms) 
12.
a  r 2 n  1
r 1
 5.


4a  6d  56 
2a  3d  28 . But a  11 is given  d  2
Now sum of last four terms.
 a   n 1 d  a  n  2 d  a   n  3 d  a   n  4 d (in the reverse order)
 44  d  4n 10
According to the question 44  2  4n 10  112 etc.
13.
Note that whenever y 2  xz then three numbers x, y, z are in GP since y 2  xz
y z


x y
a  bx b  cx

 ab  acx  b 2 x  bcx 2  ab  acx  b 2 x  bcx 2
Now
a  bx b  cx
2b 2 x  2acx 
b 2  ac

 x  0

14.
n
r 2 1
(Since common ratio of GP a  ar 2  ar 4 ..... will be r 2 )
1
5
r4



r  1  r  1 r  1
Let the number of terms in AP be n
It is given that sum of first four terms  56 
a   a  d    a  2d   a  3d  56
 4a  d  4n 10
a, b, c are in G.P
b  cx c  dx

Similarly
will given c 2  bd
b  cx c  dx
 b, c, d are in G.P . Combining the two facts we get a, b, c, d are G.P .
Let the G.P be a  ar  ar 2  .....  ar n 1 then S  a  ar  ar 2  .....  ar n 1
1 1
1
1
P  a.ar.ar 2 ........ar n 1 , R    2  ......  n 1 or
a ar ar
ar
n
1 1 
1

   
n  n 1
a   r  
1 rn 
a 1  r n 

n 1 2  3.....  n 1
n
2
a r

, R
S
, P  a .r
1
a 1  r  r n 1
1 r
1
r
46

a r 2  1
Now LHS  P R
2
15.
n
1  r 
n n
2 n n n 1
a r

a n 1  r n 
n
 S n  RHS
a 1  r  r
1  r 
Let A and D be the first term and common difference of the AP then
a  A   p  1 D ,
b  A   q  1 D ,
c  A   r  1 D
n
n
n n 1
n
LHS   q  r  a   r  p  b   p  q  c
  q  r   A   p  1 D    r  p   A   q  1 D    p  q   A   r  1 D 
 A  q  r  r  p  p  q   D  q  r  p  1   r  p  q  1   p  q  r  1 
16.
17.
18.
 A.0  D.0  0
a a b b c c
Given  ,  ,  are in AP adding 1 in all terms
b c c a a b
a a
b b
c c
  1;   1;   1 are in AP

b c
c a
a b
ac  ab  bc ba  bc  ca cb  ac  ab
;
;

are in AP
bc
ca
ab
1 1 1
abc abc abc
, ,
,
,

are in AP 
are in AP  a, b, c are in AP
bc ca ab
bc ca ab
2
Take a  a, b  ar , c  ar 2 , d  ar 3 . Now show  bn  c n    a n  bn  c n  d n 
Take a  a, b  ar , c  ar 2 , d  ar 3 (Since a, b, c, d are in G.P )
Since a , b are the roots of the equation x 2  3x  p  0  a  ar  3, a.ar  p
Similarly ar 2  ar 3  12, ar 2 .ar 3  q
The above four relations can be written as
a 1  r   3, a2r  p, ar 2 1  r   12, a2r 5  q
 r  2
q  p a 2 r 5  a 2 r r 4  1 17
Now



q  p a 2 r 5  a 2 r r 4  1 15
21.
(ii)
.6  .66  .666  .......
6 66 666
  2  3  .......
10 10 10

from first and third relation r 2  4
6  9 99 999

 2  3  ........

9 10 10 10

6 
1 
1 
1 
6 10  1 102  1
10n  1

 1    1  2   .....  1  n  
 


......

2
n 
9  10   10 
9  10
10
10 
 10  
n


11

1


6  1
1
1   6  10  10 
  6  n  1 .10n 10n  1 
  n    2  .....  n    n 

1
 9  9
9   10 10
10   9 

1


10


47
n  n  1
n  n  1 2n  1
 n  n  1 
S1 
, S2 
, S3  

2
6
2 

Show that two sides are equal.
 n  n  12 



  n  12
2
13  23  33  ....  n3
n2 n 1


Tn 

 

1  3  5  .....  2n  1
4 2 4
n2
4
1
1
1
1 n  n  1 2n  1 1 n  n  1 n

 .

 Tn  4  n2  2  n   4
4
6
2
2
4
n
n
 2n 2  9n  13


 n  1 2n  1  6  n  1  6 
24
24
2
24.
25.
n
26.
Sum of the series in the numerator   j  j  1
2
j 1
  j  j 2  2 j  1
   j3  2 j 2  j 
n
n
j 1
j 1
  j 3  2 j 2   j
n  n  1 2n  1 n  n  1
 n  n  1 


  2.
2 
6
2

n  n  1
n  n  1
3n2  11n  10

3n  n  1  4  2n  1  6  
12
12
2
n  n  1
 3n  1 n  5
12
Sum of the series in the denominator

 n  n  1  n  n  1 2n  1

  j  j  1   j   j
 
2 
6

n  n  1
n  n  1


3n 2  7n  2 
3n  n  1  2  2n  1 

12
12
n  n  1
3n  5

 3n  1 3n  2  whence the ratio of sums 
3n  2
12
6000 12  
5500 12 

Total amount paid back by farmer  6000   500 
   500 

100  
100 

5000 12 
500 12 


  500 
  ......   500 

100 
100 


2
2
27.
3
 12000 
29.
2
12
 6000  5500  .....  500   16680
100
(The AP has 12 terms)
Total amount spent after 8th set is mailed   4  42  43  .....  48 
48
1
2
4  48  1 1

  43690
4 1
2
30.
 10000  5 
Amount in 15th years  10000  
 14
 100 
Amount in 20 years 10000 
 7000
10000  5
 20  20000
100
t
31.
32.
r 

We may apply Pt  P0 1 
 , Here P0  15625, r  20, t  5 etc.
 100 
Let the “unit job” be x (i.e. the amount of work done by a worker in one day) and let n
be the number of days in which the job should have finished.
We must have 150x 146x 142x  ..... upto n terms  150 x  n  8
n
 2 150 x    n  1 4 x    150 x  n  8 
2
152n  2n 2  150n 1200
n 150  2n  2  150  n  8 ,

n 2  n  600  0

 n  25 n  24  0

n  25
LIMITS
Exercise 13.1 Page 301-303
23.
lim f  x   lim f  0  h 
x 0
h0
x 0
h0
lim f  x   lim f  0  h 

 lim  2  0  h   3  3
h 0
 lim3  0  h  1  3
h0
lim f  3  3
x 0
f  x  6
We can easily show lim
x 1
24.
lim f  x   lim f 1  h 
x 1
h0
lim f  x   lim f 1  h 
x 1

26.
h0
2
 lim 1  h   1  0

h0 
2
 lim  1  h   1  2

h0 
lim does not exist
x1
 x
1 if
  x x  0

We can write f  x   
or f  x   1 if
 x x0
 0 if

 x
lim1 f  x   lim f  0  h   lim  1  1
x 0
h0
h0
49
x0
x0
x  0( given)
lim f  x   lim  1  1  lim f  x  does not exist.
x 0
27.
h0
x 0
 x  5 if x  0
f  x  
 x  5 if x  0
lim5 f  x   lim f  5  h 
 lim  5  h  5
h0
x 5
h0
lim f  x   lim  5  h  5  0
x 5
h 0
lim f  x   0

h0
 lim  h   0
x 5
f  x  has been used)
(Note that only second formula for lim
x 5
28.
 lim  a  b 1  h    a  b
lim f  x   lim f 1  h 
f 1  4 ,
x 11
h0
h0
lim f  x   lim  b  a 1  h    b  a
x 11
h0
We must have lim1 f  x   lim f  x   4
x1
29.
x1

a  b  b  a  4 . On solving we get a  0, b  4
lim f  x    a1  a1  a1  a2  a1  a3  .......  a1  an   0
x a1
lim f  x    a  a1  a  a2  .....  a  an 
x a
30.
By direct substitutions.
 x  1 x  0

We can write f  x    0
x0
 x 1 x  0

If a  0 then lim f  x   lim   x  1  a  1
x a
x a
If a  0 then lim f  x   a 1 Both limit exists.
x a
But if a  0 then lim f  x   1, lim f  x   1
x 0
31.
x 0
lim f  x  does not exist. Thus lim f  x  exist if a  0 .

x 0
x a
Let lim f  x   L
x 1
f  x  2
 f  x   2 must be zero at x  1

x2  1
x 2  1  0 at x  1
Now lim
x 1


f 1  2  lim f  x   f 1  2
We are assuming that f  x  is continuous at x  1
32.
x 1
lim f  x  will exist if lim f  x  and lim f  x  exist and are equal.
x 0
lim f  x   lim f  0  h 
x 0
x 0
h0
 lim  n  0  h   m
h 0

h  0
x 0
m
lim f  x   lim f  0  h   lim f  h 
x 0
h0
h0
 lim m  0  h   n  n

h0 
2
Thus m  n
50
(i)
Again lim f  x   lim f 1  h 
h0
x 1
 lim  n 1  h   m  n  m
h 0
lim f  x   lim n 1  h   m  n  m

h0 

lim1 f  x   lim f  x  for all m and n
3
x 1
x1
x1
But lim exist for m  n
x 0
Thus both lim f  x  and lim f  x  exist if the integers m and n are equal.
x 0
x 1
STRAIGHT LINE
EXERCISE 10.2 PAGE 219-220
Q.8
Let the line cut x  axis at A and y  axis at
B . Let OA  a, OB  b . Then equation of
x y
line AB is   1
(i)
a b
OA
2
 sec30 
OF
3
2
10

OA  5.

3
3
OB
 cos ec30

Again
OB  5  2  10
OF
x
y
Thus equation of OF is
  1 or x 3  y  10 (ii)
10 10
3
Note:- At a later stage the answer (ii) is directly written by using x cos   y sin   p
Here   30, p  5 .
Now from OAF ,
9.
The mid point M of side PQ is
 2   2  1  3 
,

 or is  0, 2  Now slope
2
2 

52 3
RM 

40 4

Equation of median RM is
3
y  2   x  0  or 4 y  3x  8  0
4
10.
Slope of line joining A  2,5 and B  3, 6  
51
of median
65
1

3  2
5


11.
Slope of a line  to this line  5
Equation of line through  3,5 is y  5  5  x  3 or 5 x  y  20  0
Let R divide AB in 1: n then R is the point
 1 2  n  1 1 3  n  0   n  2 3 
.
,



n 1   n 1 n 1 
 n 1
30
3
Also slope AB 
2 1

12
13.
Equation of line R and  AB is
3
1
n2
y
  x

n 1
3
n 1 
Let the equation of the line which cuts equal intercepts on axes be
x y
  1 or x  y  a . Since it passes through  2,3
a a

23  a
a 5
x y
Let the equation of the line be   1
(i)
a b
2 2
Since (i) passes through  2, 2  we have   1 (ii)
a b
Also a  b  9
(iii)
2
2
1
Putting b  9  a in (ii) we get 
a 9a

2  9  a   2a  a  9  a  or a 2  9a  18  0

 a  6 a  3  0
If a  6 then b  3
a  6 or 3
 Equation of one such line is
Similarly the other line is
14.

x y
  1.
6 3
x y
 1
3 6
2
 tan120  tan 180  60   tan 60   3
3
Equation of first line is y  2   3  x  0  or y  x 3  2
Slope of the line  tan

If a line is parallel to it then its slope is same i.e. it is  3 but the second line
passes through  0, 2 

15.
Equation of second line is y  2   3  x  0  or y  x 3  2
The line passes through  2,9 
Slope OF =
90
9

2  0
2
52


16.
2
9
Equation of the line is
2
y  9   x  2
9
9 y  2 x  85
Slope of line 
The graph of the relation between L and C will be a line joining two points
A 124.942, 20 and B 125.134,110 . Hence equation of AB (or a relation between L
and C ) can be easily written.
18. Let the line cut x-axis at A ,y axis at B where OA  h , OB  k
then equation of required line is
x y
 1
h k
but P is the mid point of A(h,0) and B(0, k )
a
h0
0c
,b 
 h  2a , k  2b
2
2
x
y

1
2a 2b
x y
Let OA  a, OB  b then equation of line is   1
a b
2  a  1x0
2  0  1 b
3h
x
y
Now h 
,k
 a  , b  3k ,  Ans is

1
3
3
3h / 2 3k
2
Find the equation of line joining A(3,0) and B(2, 2) and show that the
equation of
line is satisfied by C (8,2)
Hence equation of the required line is
19.
20.
Exercise 10.3 Page 227
1.
(ii) Slope intercept form is y  mx  c
6x  3y  5  0 
3 y  6 x  5

y  2 x 
5
3
which is the required slope intercept form with m  2, c 
5
.
3
Parts (i) and (iii) are very simple.
2.
x y
 1
a b
3x  2 y  12  0
3 x  2 y  12 on dividing by 12

3x 2 y 12
x y


 1
or
12 12 12
4 6
Intercept form is
(i)
53
(ii)
(iii)
3.
(i)
4 x 3 y 6

 
6
6
6
4x  3y  6

Intercepts are
3
and 2
2
x
y

1
3 2 2
y
3y
 1 
1
2
2 3
Since the line is parallel to x  axis , there will be no x  intercept.
x  3 y  8   x  3 y  8 on dividing by
x  3y  8  0


3y  2  0
 3

3 y  2
1
3
y  4 which is the normal form with
 2 we get  x 
2
2
1
3
p  4 . Also cos    ,sin  
2
2
2

 
 lies in second quadrant
3
 1
10.
2

2
Slope of line joining A(h,3) and B(4,1) is
Slope of 7 x  9 y  19  0 is

3 1
2

h4 h4
7
. Since the lines are  m1m2  1
9
2 7
.  1  14  9h  36  h  22 / 9
h4 9
Miscellaneous Exercise
1.
(a) If line is parallel to x-axis x term should be absent k  3  0
(b)
If a line is parallel to y-axis  4  k 2  0  k  2
(c)
If it passes through origin k 2  7k  6  0
4.
Let any point on x  asix be ( ,0) . Its  distance from the line

 0 
   1
3 4 
2
1 1
   
3  4

2
 3
1
5
3 4

x y
 1  0
3 4
4  12
. But this is given as 4
5
4  12
 4  4  12  20  4  12  20 OR 4  12  20
5
Hence the points on x-axis are (8,0) and (2,0)

5.
Slope 
2cos
 
sin
 
cos
 
sin   sin 
2
2 
2

 
 
 
cos  cos 
2sin
sin
sin
2
2
2
54
   8 ,   2
 Equation of line is
y  sin   
sin

y sin
 
 sin  sin
 
 
cos
2
 
2
  x cos
 
2
2
2
 
 
  

x cos
 y sin
 cos   


2
2
2 

 
 
 
x cos
 y sin
 cos

2
2
2
 
Hence  distance from origin  cos
2
14.
( x  cos  )
 cos cos
 
2
(Note )
Let the segment AB cuts the line x  y  4 at P such that AP : PB  k :1 then
 5k  1 7k  1 
P is 
,
 . But P lies on the line x  y  4
 k 1 k 1 
5k  1 7k  1


 4 . Now determine k .
k 1 k 1
19.
The slopes of lines y  3x  1,2 y  x  3 and
y  mx  4
1
are 3, and m respectively
2
1
m
m3
2  2m
We must have

m
1  3m 1 
2m
2
m3 2m
m3
2m
OR
 either


1  3m 2  m
2m
2m
21.
Solve two quadratic equations .
One of them will give us two values of m.
Indeed there will be two lines bisecting the angle between y  3x  1 , 2 y  x  3.
The lines 9 x  6 y  7  0 and 3x  2 y  6  0 are parallel
The line EF should also be parallel to them. Let equation EF be 3x  2 y  k  0
 Distance between AB and EF


7
Now  0,  is a point on AB then
6



1
 Distance between AB and CD
2
(*)
7
7
k
3 0  2   6
1
6
6

2
2
2
2
3 2
3  22
3 0  2 
55
(*)
22.
7 1
7
25
 6 
3 2
3
6
7 25
7
25
or k   

k 
3 6
3
6
11
 k
6
11
Hence line is 3x  2 y   0 . The second value of k is not acceptable (why?)
6
20
2
30
3
Slope of incident line I is
slope of reflected line R is


1 h 1 h
5h 5h
Slope of I is also equal to tan(90   )   cot 

k

 cot  
2
1 h
Slope of R is also equal to tan(90   )  cot 

23.
cot  
3
you get h now
5h
by eliminating cot 
For convenience let
a 2  b2  c
Now length of  from (c,0) to the line
p1 
c
cos   1
a
cos 2  sin 2 
 2
a2
b
Similarly p2 
x
y
cos  sin   1 is given by
a
b
c
 cos   1
a
cos 2  sin 2 
 2
a2
b
c2
cos 2 
2
a
( x  y) ( x  y)  x2  y 2
 p1 p2 
cos 2  sin 2 
 2
a2
b
2
 1

 1 a 2  b2

c
b 2  2  2 2 cos 2  
b 2  2  2 2 cos 2  
ab
ab
b
  b


2
2
2
2
cos  sin 
cos  sin 
 2
 2
2
a
b
a2
b
2
2
2



1
cos

cos

sin

cos 2  
2
b2  2 

b


 2

b
b2
a2 
a2 
 b
 b2
 

cos 2  sin 2 
cos 2  sin 2 
 2
 2
a2
b
a2
b
1

56
