∴ (a − d) + (a + d) = 12 ⇒ a = 6 . Also, (a − d) a = 24 ⇒ 6−d = 24 =4 ⇒ d=2
6
∴ First term = a − d = 6 − 2 = 4
Example: 10 If Sr denotes the sum of the first r terms of an A.P., then S3r − Sr −1 is equal to
Solution: (b) S2r − S2r −1
Example: 11
Solution: (c) (a) 2r – 1 (b) 2r + 1 (c) 4r + 1 (d) 2r + 3
Example: 12
Solution: (a) 3r {2a + (3r − 1)d} − (r − 1) {2a + (r − 1 − 1) d} (2r + 1)a + d {3r(3r − 1) − (r − 1)(r − 2)}
Example: 13 2 2 = 2
Solution: (c) S3r − Sr −1 =
S2r − S2r −1 T2r a + (2r − 1)d
Example: 14
(2r + 1)a + d {8r 2 − 2} (2r + 1)a + d(4r 2 − 1) 2r +1
Solution: (a) 2 a + (2r − 1)d
= a + (2r − 1)d = =
If the sum of the first 2n terms of 2, 5, 8…. is equal to the sum of the first n terms of 57, 59, 61…., then n is equal to
[IIT Screening 2001]
(a) 10 (b) 12 (c) 11 (d) 13
We have, 2n {2 × 2 + (2n − 1)3} = n {2 × 57 + (n − 1)2} ⇒ 6n + 1 = n + 56 ⇒ n = 11
2 2
If the sum of the 10 terms of an A.P. is 4 times to the sum of its 5 terms, then the ratio of first term and common difference
is [Rajasthan PET 1986]
(a) 1 : 2 (b) 2 : 1 (c) 2 : 3 (d) 3 : 2
Let a be the first term and d the common difference
Then, 10 {{a + (10 − 1)d} = 4 × 5 {2a + (5 − 1)d} ⇒ 2a + 9d = 4a + 8d ⇒ d = 2a ⇒ a = 1 ,∴a:d=1:2
2 2 d 2
150 workers were engaged to finish a piece of work in a certain number of days. 4 workers dropped the second day, 4
more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in
which the work was completed is [Kurukshetra CEE 1996]
(a) 15 (b) 20 (c) 25 (d) 30
Let the work was to be finished in x days. ∴ Work of 1 worker in a day = 1
150x
Now the work will be finished in (x + 8) days. ∴ Work done = Sum of the fraction of work done
1 = 1 ×150 + 1 (150 − 4) + 1 (150 − 8) + ....... to (x + 8) terms
150x 150x 150 x
⇒ 1= x+8 2 × 150 + (x + 8 − 1) −4 ⇒ 150x = (x + 8){150 − 2(x + 7)} ⇒ (x + 8)(x + 7) − 600 =0
2 150x 150x
⇒ (x + 8)(x + 7) = 25 × 24 , ∴ x + 8 = 25
Hence work completed in 25 days.
If the sum of first p terms, first q terms and first r terms of an A.P. be x, y and z respectively, then
x (q − r) + y (r − p) + z (p − q) is
p q r
(a) 0 (b) 2 (c) pqr (d) 8 xyz
pqr
We have a, the first term and d, the common difference, x = {2a + (p − 1) d} p ⇒ x = a + (p − 1) d
2 p 2
Similarly, y = a + (q − 1) d and z = a + (r − 1) d
q 2 r 2
∴ x (q − r) + y (r − p) + z (p − q) = a + (p − 1) d (q − r) + a + (q − 1) d (r − p) + a + (r − 1) d (p − q)
p q r 2 2 2
= a{(q − r) + (r − p) + (p − q)} + d {(p − 1)(q − r) + (q − 1)(r − p) + (r − 1)(p − q)}
2
= a.. 0 + d [{pq − pr + rq − pq + pr − qr − {(q − r) + (r − p) + (p − q)} = 0 + d {0 − 0} = 0
2 2
Example: 15 The sum of all odd numbers of two digits is [Roorkee 1993]
Solution: (a)
(a) 2475 (b) 2530 (c) 4905 (d) 5049
Required sum, S = 11 + 13 + 15 + ....... + 99
Let the number of odd terms be n, then 99 = 11 + (n − 1)2 ⇒ n = 45
∴ S = 45 (11 + 99) = 45 × 55 = 2475 S = n (a + l)
2 2
Example: 16 If sum of n terms of an A.P. is 3n2 + 5n and Tm = 164 , then m = [Rajasthan PET 1991, 95; DCE 1999]
Solution: (b)
Example: 17 (a) 26 (b) 27 (c) 28 (d) None of these
Solution: (d) Tm = Sm − Sm−1 ⇒ 164 = (3m2 + 5m) − {3(m − 1)2 + 5(m − 1)} ⇒ 164 = 3(2m − 1) + 5 ⇒ m = 27
The sum of n terms of the series 1 3 + 1 5+ 1 7 + ....... is [UPSEAT 2002]
1+ 3+ 5+
(a) 2n + 1 (b) 1 2n + 1 (c) 2n − 1 (d) 1 ( 2n + 1 − 1)
2 2
Sn = 1 3 + 1 + 1 + ...... + 1 2n + 1
1+ 3+ 5 5+ 7 2n − 1 +
=( 3− 1 + 5− 3+ 7− 5 + ..... + 2n + 1 − 2n − 1
− 1)( 3 2 2 2
3 + 1)
= 1 [ 3 −1+ 5− 3+ 7− 5 + ..... + ( 2n + 1 − 2n − 1)] = 1 [ 2n + 1 − 1]
2 2
Example: 18 If a1, a2 ,......, an+1 are in A.P., then 1 + 1 + ..... + 1 is [AMU 2002]
Solution: (d) a1a 2 a2a3 a n a n+1
(a) n−1 (b) 1 (c) n+1 (d) n
a1 a n+1 a1 a n+1 a1 a n+1 a1 a n+1
1 1 1 1 − 1 1 − 1 1 − 1
a1a2 a2a3 anan+1 a1 a2 a2 a3 an an+1
S = + + .... + = + + ...... +
(a2 − a1) (a3 − a2) (an+1 − an)
As a1, a2, a3,......., an, an+1 are in A.P., i.e. a2 − a1 = a3 − a2 = ......... = an+1 − an = d (say)
∴ S= 1 1 − 1 + 1 − 1 + ...... + 1 − 1 = 1 1 − 1 = an+1 − a1 = [a1 + (n + 1 − 1) d] − a1
d a1 a2 a2 a3 an an+1 d d . a1 . an+1 d . a1. an+1
a1 an+1
S = nd = n
d a1 an+1 a1 an+1
Arithmetic Mean.
(1) Definitions
(i) If three quantities are in A.P. then the middle quantity is called Arithmetic mean (A.M.) between the other two.
If a, A, b are in A.P., then A is called A.M. between a and b.
(ii) If a, A1, A2, A3 ,....., An,b are in A.P., then A1, A2 , A3 ,......, An are called n A.M.’s between a and b.
(2) Insertion of arithmetic means
(i) Single A.M. between a and b : If a and b are two real numbers then single A.M. between a and b = a +b
2
(ii) n A.M.’s between a and b : If A1, A2, A3 ,......., An are n A.M.’s between a and b, then
A1 = a + d = a + b − a , A2 = a + 2d = a + 2 b − a , A3 = a + 3d = a + 3 b − a , ……., An = a + nd = a + n b− a
n + 1 n + 1 n + 1 n+ 1
Important Tips
• Sum of n A.M.’s between a and b is equal to n times the single A.M. between a and b.
i.e. A1 + A2 + A3 + .......... + An = n a +b
2
• If A1 and A2 are two A.M.’s between two numbers a and b, then A1 = 1 (2a + b), A2 = 1 (a + 2b) .
3 3
• Between two numbers, Sum of m A.M.' s = m .
Sum of n A.M.' s n
• If number of terms in any series is odd, then only one middle term exists which is n + 1 th term.
2
• If number of terms in any series is even then there are two middle terms, which are given by n th and n + 1th term.
2 2
Example: 19 After inserting n A.M.’s between 2 and 38, the sum of the resulting progression is 200. The value of n is [MP PET 2001]
Solution: (b)
(a) 10 (b) 8 (c) 9 (d) None of these
There will be (n + 2) terms in the resulting A.P. 2, A1, A2,......, An, 38
Sum of the progression = n + 2 (2 + 38) ⇒ 200 = (n + 2)× 20 ⇒n = 8
2
Example: 20 3 A.M.’s between 3 and 19 are
Solution: (a)
(a) 7, 11, 15 (b) 4, 6, 10 (c) 6, 10, 14 (d) None of these
Let A1, A2, A3 be three A.M.’s. Then 3, A1, A2, A3,19 are in A.P.
⇒ common difference d = 19 − 3 = 4 .Therefore A1 = 3 + d = 7 , A2 = 3 + 2d = 11 , A3 = 3 + 3d = 15
3+1
Example: 21 If a, b, c, d, e, f are A.M.’s between 2 and 12, then a + b + c + d + e + f is equal to
(a) 14 (b) 42 (c) 84 (d) None of these
Solution: (b) Since, a, b, c, d, e, f are six A.M.’s between 2 and 12
Therefore, a+b+c+d+e + f = 6 (a + f) = 6 (2 + 12) = 42
2 2
Properties of A.P..
(1) If a1, a2, a3 ..... are in A.P. whose common difference is d, then for fixed non-zero number K ∈ R.
(i) a1 ± K, a2 ± K, a3 ± K,..... will be in A.P., whose common difference will be d.
(ii) Ka1, Ka2, Ka3 ........ will be in A.P. with common difference = Kd.
(iii) a1 , a2 , a3 ...... will be in A.P. with common difference = d/K.
K K K
(2) The sum of terms of an A.P. equidistant from the beginning and the end is constant and is equal to sum of
first and last term. i.e. a1 + an = a2 + an−1 = a3 + an−2 = ....
(3) Any term (except the first term) of an A.P. is equal to half of the sum of terms equidistant from the term i.e.
an = 1 (an−k + an+k ) , k < n.
2
(4) If number of terms of any A.P. is odd, then sum of the terms is equal to product of middle term and
number of terms.
(5) If number of terms of any A.P. is even then A.M. of middle two terms is A.M. of first and last term.
(6) If the number of terms of an A.P. is odd then its middle term is A.M. of first and last term.
(7) If a1, a2 ,......an and b1,b2 ,......bn are the two A.P.’s. Then a1 ± b1, a2 ± b2 ,......an ± bn are also A.P.’s with
common difference d1 ≠ d2 , where d1 and d2 are the common difference of the given A.P.’s.
(8) Three numbers a, b, c are in A.P. iff 2b = a + c .
(9) If Tn , Tn+1 and Tn+2 are three consecutive terms of an A.P., then 2Tn+1 = Tn + Tn+2 .
(10) If the terms of an A.P. are chosen at regular intervals, then they form an A.P.
Example: 22 If a1, a2, a3, ....., a24 are in arithmetic progression and a1 + a5 + a10 + a15 + a20 + a24 = 225 , then a1 + a2 + a3 + .....
Solution: (d)
+a23 + a24 = [MP PET 1999; AMU 1997]
(a) 909 (b) 75 (c) 750 (d) 900
a1 + a5 + a10 + a15 + a20 + a24 = 225 ⇒ (a1 + a24 ) + (a5 + a20) + (a10 + a15) = 225 ⇒ 3(a1 + a24 ) = 225 ⇒ a1 + a24 = 75
(∵ In an A.P. the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of first and
last term)
a1 + a2 + ..... + a24 = 24 (a1 + a24 ) = 12 × 75 = 900
2
Example: 23 If a, b, c are in A.P., then 1 , 1 , 1 will be in [DCE 2002; MP PET 1985; Roorkee 1975]
Solution: (a) bc ca ab
(a) A.P. (b) G.P. (c) H.P. (d) None of these
a, b, c are in A.P., ⇒ 1 , 1 , 1 will be in A.P. [Dividing each term by abc]
bc ca ab
Example: 24 If log 2, log(2n − 1) and log(2n + 3) are in A.P., then n = [MP PET 1998; Karnataka CET 2000]
(a) 5/2 (b) log2 5 (c) log3 5 (d) 3
2
Solution: (b) As, log 2, log(2n − 1) and log(2n + 3) are in A.P. Therefore
2 log(2n − 1) = log 2 + log(2n + 3) ⇒ (2n − 5)(2n + 1) = 0
As 2n cannot be negative, hence 2n − 5 = 0 ⇒ 2n = 5 or n = log 2 5
Geometric progression(G.P.)
Definition.
A progression is called a G.P. if the ratio of its each term to its previous term is always constant. This constant
ratio is called its common ratio and it is generally denoted by r.
Example: The sequence 4, 12, 36, 108, ….. is a G.P., because 12 = 36 = 108 = ..... = 3, which is constant.
4 12 36
Clearly, this sequence is a G.P. with first term 4 and common ratio 3.
The sequence 1 , − 1 , 3 , − 9 ,.... is a G.P. with first term 1 and common ratio − 1 1 = − 3
3 2 4 8 3 2 3 2
General Term of a G.P..
(1) We know that, a, ar, ar 2, ar 3 ,.....ar n−1 is a sequence of G.P.
Here, the first term is ‘a’ and the common ratio is ‘r’.
The general term or nth term of a G.P. is Tn = ar n−1
It should be noted that,
r = T2 = T3 = ......
T1 T2
(2) pth term from the end of a finite G.P. : If G.P. consists of ‘n’ terms, pth term from the end = (n − p + 1)th
term from the beginning = ar n−p .
1 n−1
r
Also, the pth term from the end of a G.P. with last term l and common ratio r is l
Important Tips
• If a, b, c are in G.P. ⇒ b = c or b2 = ac
a b
• If Tk and Tp of any G.P. are given, then formula for obtaining Tn is
11
Tp
Tn n−k = Tk p−k
Tk
• If a, b, c are in G.P. then
⇒ b = c ⇒ a+b = b + c or a−b = a or a+b = a
a b a−b b − c b−c b b+c b
• Let the first term of a G.P be positive, then if r > 1, then it is an increasing G.P., but if r is positive and less than 1, i.e. 0< r < 1,
then it is a decreasing G.P.
• Let the first term of a G.P. be negative, then if r > 1, then it is a decreasing G.P., but if 0< r < 1, then it is an increasing G.P.
• If a, b, c, d,… are in G.P., then they are also in continued proportion i.e. a = b = c = .... = 1
b c d r
Example: 25 The numbers ( 2 + 1),1, ( 2 − 1) will be in [AMU 1983]
Solution: (b)
Example: 26 (a) A.P. (b) G.P. (c) H.P. (d) None of these
Solution: (b)
Clearly (1)2 = ( 2 + 1).( 2 − 1)
Example: 27
Solution: (c) ∴ 2 + 1,1, 2 − 1 are in G.P.
Example: 28
Solution: (d) If the pth, qth and rth term of a G.P. are a, b, c respectively, then aq−r ⋅ br−p ⋅ c p−q is equal to
[Roorkee 1955, 63, 73; Pb. CET 1991, 95]
(a) 0 (b) 1 (c) abc (d) pqr
Let x, xy, xy2, xy3,.... be a G.P.
∴ a = xy p−1, b = xyq−1, c = xyr−1
Now, aq−r . br − p. c p−q = (xy p−1)q−r (xyq−1)r − p(xyr −1)p−q = x(q−r)+(r − p)+(p−q). y(p−1)(q−r)+(q−1)(r − p)+(r −1)(p−q)
= x0. y p(q−r)+q(r − p)+r(p−q)−(q−r +r − p+ p−q) = x0. y0−0 = (xy)0 = 1
If the third term of a G.P. is 4 then the product of its first 5 terms is [IIT 1982; Rajasthan PET 1991]
(a) 43 (b) 44 (c) 45 (d) None of these
Given that ar 2 = 4
Then product of first 5 terms = a(ar)(ar 2)(ar3)(ar4 ) = a5r10 = [ar 2]5 = 45
If x, 2x + 2, 3x + 3 are in G.P., then the fourth term is [MNR 1980, 81]
(a) 27 (b) – 27 (c) 13.5 (d) – 13.5
Given that x, 2x + 2, 3x + 3 are in G.P.
Therefore, (2x + 2)2 = x(3x + 3) ⇒ x2 + 5x + 4 = 0 ⇒ (x + 4)(x + 1) = 0 ⇒ x = −1, − 4
Now first term a = x, second term ar = 2(x + 1)
2(x + 1) ar 3 2(x + 1) 3 8 (x + 1)3
x x x2
⇒ r = , then 4th term = = x =
Putting x = −4 , we get
T4 = 8 (−3)3 = − 27 = −13.5
16 2
Sum of First ‘n’ Terms of a G.P..
If a be the first term, r the common ratio, then sum Sn of first n terms of a G.P. is given by
Sn = a(1 − r n ) , |r|< 1
1−r
Sn = a(r n − 1) , |r|> 1
r −1
Sn = na , r=1
Selection of Terms in a G.P..
(1) When the product is given, the following way is adopted in selecting certain number of terms :
Number of terms Terms to be taken
3
a , a, ar
4 r
5 a , a , ar, ar 3
r3 r
a , a , a, ar, ar 2
r2 r
(2) When the product is not given, then the following way is adopted in selection of terms
Number of terms Terms to be taken
3 a, ar, ar 2
a, ar, ar 2 , ar 3
4 a, ar, ar 2 , ar 3 , ar 4
5
100 100
∑ ∑Let an be the nth term of the G.P. of positive numbers. Let
Example: 29 a2n = α and a2n−1 = β , such that α ≠ β, then the
Solution: (a)
n=1 n=1
common ratio is [IIT 1992]
(a) α (b) β (c) α (d) β
β α β α
Let x be the first term and y, the common ratio of the G.P.
100 100
∑ ∑Then, α = a2n = a2 + a4 + a6 + .... + a200 and β = a2n−1 = a1 + a3 + a5 + ...... + a199
n=1 n=1
⇒ α = xy + xy3 + xy5 + ..... + xy199 = xy 1 − (y2)100 = xy 1 − y200
1− y2 1− y2
β = x + xy2 + xy4 + ..... + xy198 = x ⋅ 1 − (y 2 )100 = x ⋅ 1 − y200
1− y2 1− y2
∴ α = y . Thus, common ratio = α
ββ
Example: 30 The sum of first two terms of a G.P. is 1 and every term of this series is twice of its previous term, then the first term will be
Solution: (b)
[Rajasthan PET 1988]
(a) 1 (b) 1 (c) 2 (d) 3
4 3 3 4
We have, common ratio r = 2; an = 2
an−1
Let a be the first term, then a + ar =1 ⇒ a(1 + r) = 1 ⇒ a= 1 = 1 = 1
1+r 1+ 2 3
Sum of Infinite Terms of a G.P..
(1) When |r|< 1, (or − 1 < r < 1)
S∞ = a
1−r
(2) If r ≥ 1, then S∞ doesn’t exist
Example: 31 The first term of an infinite geometric progression is x and its sum is 5. Then [IIT Screening 2004]
Solution: (b)
Example: 32 (a) 0 ≤ x ≤ 10 (b) 0 < x < 10 (c) −10 < x < 0 (d) x > 10
Solution: (b)
According to the given conditions, 5 = x , r being the common ratio ⇒ r =1− x
Example: 33 1−r 5
Now, |r|< 1 i.e. −1 < r < 1 ⇒ −1 < 1− x < 1 ⇒ −2< − x <0 ⇒ 2> x > 0 i.e. 0 < x < 2, ∴ 0 < x < 10
5 5 5 5
∑lim n 1 e r is [AIEEE 2004]
r =1 n n
n→∞
(a) e + 1 (b) e – 1 (c) 1 – e (d) e
∑ ∑lim n 1 er / n 1 n er / n = lim 1 (e1 / n e2 / n e3 / n en / n) 1 ⋅ [e1 / n (e1 / n )2 (e1 / n )3 + ..... + (e1 / n)n]
r =1 n n r =1 n n
n→∞ n→∞
= lim ⋅ + + + ..... + = lim + +
n→∞ n→∞
= lim 1 e1 / n 1 − (e1 / n)n = lim 1 e1 / n 1−e = lim (1 − e)(e1 / n − 1 + 1) = lim (e − 1) + lim (e − 1)⋅ 1
n 1 − e1 / n n 1 − e1 / n n (1 − e1 / n) n e1 / n −
n→∞ n→∞ n→∞ n→∞ n→∞ n
1
Put 1 = h, we get h → 0
n
= 0 + (e − 1) lim h 0 form
eh −1 0
h→0
= (e − 1) lim 1 = (e − 1).1 = e − 1 .
eh
h→0
The value of .2 3• 4• .234 is [MNR 1986; UPSEAT 2000]
(a) 232 (b) 232 (c) 0.232 (d) 232
990 9990 990 9909
Solution: (a) .2 3• 4• = .234343434.... = 2 + 34 + 34 + 34 + ...... = 2 + 34 1 + 1 + 1 + .......
10 1000 100000 107 10 1000 100 (100)2
= 1 + 17 1 = 1 + 17 100 = 1 1 + 17 = 116 = 232
5 500 1 5 500 × 99 5 495 990
1 − 99
100
Example: 34 If a, b, c are in A.P. and |a|, |b|, |c| < 1, and
Solution: (c)
x = 1 + a + a2 + ....∞
y = 1 + b + b2 + ....∞
z = 1 + c + c2 + .....∞
Then x, y, z shall be in [Karnataka CET 1995]
(a) A.P. (b) G.P. (c) H.P. (d) None of these
x = 1+ a + a2 + ....∞ = 1 x, y, z are in H.P.
1−a
y = 1 + b + b2 + ....∞ = 1 1
−b
z = 1 + c + c2 + .....∞ = 1 1 c
−
Now, a, b, c are in A.P.
⇒ 1 – a, 1 – b, 1 –c are in A.P. ⇒ 1 , 1 1 b , 1 1 c are in H.P. Therefore
1−a − −
Geometric Mean.
(1) Definition : (i) If three quantities are in G.P., then the middle quantity is called geometric mean (G.M.)
between the other two. If a, G, b are in G.P., then G is called G.M. between a and b.
(ii) If a, G1, G2 , G3 ,.... Gn,b are in G.P. then G1, G2 , G3 ,.... Gn are called n G.M.’s between a and b.
(2) Insertion of geometric means : (i) Single G.M. between a and b : If a and b are two real numbers
then single G.M. between a and b = ab
(ii) n G.M.’s between a and b : If G1, G2, G3 ,......, Gn are n G.M.’s between a and b, then
1 23 n
b b b b
G1 = ar = a a n+1 , G2 = ar 2 = a a n+1 , G3 = ar 3 = a a n+1 , ……………….., Gn = ar n = a a n+1
Important Tips
• Product of n G.M.’s between a and b is equal to nth power of single geometric mean between a and b.
i.e. G1 G2 G3...... Gn = ( ab )n
• G.M. of a1 a2 a3...... an is (a1 a2 a3.....an)1 / n
• If G1 and G2 are two G.M.’s between two numbers a and b is G1 = (a2b)1 / 3, G2 = (ab2)1 / 3 .
• The product of n geometric means between a and 1 is 1.
a
b 1
a
• If n G.M.’s inserted between a and b then r = n+1
3.13 Properties of G.P..
(1) If all the terms of a G.P. be multiplied or divided by the same non-zero constant, then it remains a G.P.,
with the same common ratio.
(2) The reciprocal of the terms of a given G.P. form a G.P. with common ratio as reciprocal of the common
ratio of the original G.P.
(3) If each term of a G.P. with common ratio r be raised to the same power k, the resulting sequence also
forms a G.P. with common ratio r k .
(4) In a finite G.P., the product of terms equidistant from the beginning and the end is always the same and is
equal to the product of the first and last term.
i.e., if a1, a2 , a3 ,...... an be in G.P. Then a1 an = a2 an−1 = a3 an−2 = an an−3 = .......... = ar . an−r+1
(5) If the terms of a given G.P. are chosen at regular intervals, then the new sequence so formed also forms a
G.P.
(6) If a1, a2, a3 ,....., an...... is a G.P. of non-zero, non-negative terms, then log a1, log a2, log a3 ,..... log an,......
is an A.P. and vice-versa.
(7) Three non-zero numbers a, b, c are in G.P. iff b 2 = ac .
(8) Every term (except first term) of a G.P. is the square root of terms equidistant from it.
i.e. Tr = Tr−p ⋅ Tr+p ; [r > p]
(9) If first term of a G.P. of n terms is a and last term is l, then the product of all terms of the G.P. is (al)n / 2 .
(10) If there be n quantities in G.P. whose common ratio is r and Sm denotes the sum of the first m terms,
then the sum of their product taken two by two is r r 1 Sn Sn−1 .
+
Example: 35 The two geometric mean between the number 1 and 64 are [Kerala (Engg.) 2002]
Solution: (b)
(a) 1 and 64 (b) 4 and 16 (c) 2 and 16 (d) 8 and 16
Example: 36
Let G1 and G2 are two G.M.’s between the number a = 1 and b = 64 [DCE 2002]
11 11
G1 = (a2b)3 = (1.64)3 = 4 , G2 = (ab2)3 = (1.642)3 = 16
The G.M. of the numbers 3, 32, 33......3n is
2 n+1 n n−1
(a) 3n (b) 3 2 (c) 3 2 (d) 3 2
1+ 2+3+....+n n(n+1) n+1 (d) 1 : 2 : 4
Solution: (b) G.M. of (3. 32. 33......3n) = (3. 32. 33......3n)1 / n = (3) n = 3 2n = 3 2
Example: 37
If a, b, c are in A.P. b – a, c – b and a are in G.P., then a : b : c is
Solution: (a)
(a) 1 : 2 : 3 (b) 1 : 3 : 5 (c) 2 : 3 : 4
Given, a, b, c are in A.P. ⇒ 2b = a + c
b – a, c – b, a are in G.P. So (c − b)2 = a (b − a)
2b = a + c
c
⇒ (b − a)2 = (b − a) a ⇒ b + b = a +
⇒ b − a = c − b
⇒ b = 2a [∵ b ≠ a]
Put in 2b = a + c, we get c = 3a. Therefore a : b : c = 1 : 2 : 3
Harmonic progression(H.P.)
Definition.
A progression is called a harmonic progression (H.P.) if the reciprocals of its terms are in A.P.
Standard form : 1 + 1 + 1 + ....
a a+d a + 2d
Example: The sequence 1, 1 , 1 , 1 , 1 ,... is a H.P., because the sequence 1, 3, 5, 7, 9, ….. is an A.P.
3 5 7 9
General Term of an H.P..
If the H.P. be as 1 , a 1 d , a 1 ,.... then corresponding A.P. is a, a + d, a + 2d,.....
a + + 2d
Tn of A.P. is a + (n − 1)d
∴ Tn of H.P. is 1
a + (n − 1)d
In order to solve the question on H.P., we should form the corresponding A.P.
Thus, General term : Tn = 1 or Tn of H.P. = Tn 1
a + (n − 1)d of A.P.
Example: 38 The 4th term of a H.P. is 3 and 8th term is 1 then its 6th term is [MP PET 2003]
Solution: (b) 5 3 [Rajasthan PET 1997]
Example: 39 (a) 1 (b) 3 (c) 1 (d) 3
Solution: (c) 6 7 7 5
Example: 40 Let 1 , a 1 d , a 1 , ....... be an H.P.
Solution: (c) a + + 2d
∴ 4th term = 1 ⇒ 3 = 1
a + 3d 5 a + 3d
⇒ 5 = a + 3d …..(i)
3
Similarly, 3 = a + 7d …..(ii)
From (i) and (ii), d = 1 , a = 2
3 3
∴ 6th term = 1 = 1 = 3
a + 5d 7
2 + 5
3 3
If the roots of a(b − c) x 2 + b(c − a) x + c (a − b) = 0 be equal, then a, b, c are in
(a) A.P. (b) G.P. (c) H.P. (d) None of these
As the roots are equal, discriminate = 0
⇒ {b(c − a)}2 − 4a(b − c) c(a − b) = 0 ⇒ b2c2 + a2b2 − 2ab2c − 4a2bc + 4a2c2 + 4ab2c − 4abc2 = 0
⇒ (b 2c 2 + 2ab 2c + a 2b 2 ) = 4ac{ab + bc − ac} ⇒ (ab + bc)2 = 4ac (ab + bc − ac) ⇒ {b(a + c)}2 = 4abc(a + c) − 4a2c2
⇒ b2(a + c)2 − 2b(a + c)⋅ 2ac + (2ac)2 = 0 ⇒ [b(a + c) − 2ac]2 = 0
∴ b = 2ac
a+c
Thus, a, b, c are in H.P.
If the first two terms of an H.P. be 2 and 12 then the largest positive term of the progression is the
5 23
(a) 6th term (b) 7th term (c) 5th term (d) 8th term
For the corresponding A.P., the first two terms are 5 and 23 i.e. 30 and 23
2 12 12 12
Common difference = − 7
12
∴ The A.P. will be 30 , 23 , 16 , 9 , 2 , − 5 , ......
12 12 12 12 12 12
The smallest positive term is 2 , which is the 5th term. ∴ The largest positive term of the H.P. will be the 5th term.
12
Harmonic Mean.
(1) Definition : If three or more numbers are in H.P., then the numbers lying between the first and last are
called harmonic means (H.M.’s) between them. For example 1, 1/3, 1/5, 1/7, 1/9 are in H.P. So 1/3, 1/5 and 1/7
are three H.M.’s between 1 and 1/9.
Also, if a, H, b are in H.P., then H is called harmonic mean between a and b.
(2) Insertion of harmonic means :
(i) Single H.M. between a and b = 2ab
a+b
1 1 1 + ..... + 1
H a1 + a2 an
(ii) H, H.M. of n non-zero numbers a1, a2 , a3 ,...., an is given by = .
n
(iii) Let a, b be two given numbers. If n numbers H1, H 2,...... Hn are inserted between a and b such that the
sequence a, H1, H 2, H3 ...... Hn,b is an H.P., then H1, H 2,...... Hn are called n harmonic means between a and b.
Now, a, H1, H 2 ,...... Hn ,b are in H.P. ⇒ 1 , 1 , 1 , ...... 1 , 1 are in A.P.
a H1 H2 Hn b
Let D be the common difference of this A.P. Then,
1 = (n + 2)th term = Tn+ 2
b
1 = 1 + (n + 1) D ⇒ D = a−b
b a (n + 1)ab
Thus, if n harmonic means are inserted between two given numbers a and b, then the common difference of
the corresponding A.P. is given by D = a−b
(n + 1)ab
Also, 1 = 1 + D, 1 = 1 + 2D ,……., 1 = 1 + nD where D = a−b
H1 a H2 a Hn a (n + 1)ab
Important Tips
• If a, b, c are in H.P. then b = 2ac .
a+c
• If H1 and H2 are two H.M.’s between a and b, then H1 = 3ab and H2 = 3ab
a + 2b 2a + b
Properties of H.P..
(1) No term of H.P. can be zero.
(2) If a, b, c are in H.P., then a−b = a .
b−c c
(3) If H is the H.M. between a and b, then
(i) 11 11 (ii) (H − 2a)(H − 2b) = H 2 (iii) H + a + H +b =2
H−a + H−b =a+b H − a H −b
Example: 41 The harmonic mean of the roots of the equation (5 + 2)x2 − (4 + 3) x + 8 + 2 3 = 0 is [IIT 1999]
Solution: (b)
(a) 2 (b) 4 (c) 6 (d) 8
Example: 42
Solution: (c) Let α and β be the roots of the given equation
Example: 43 ∴ a + β = 4 + 3 , αβ = 8 + 2 3
Solution: (d) 5+ 2 5+ 2
= 2αβ 2 8+2 3 2(8 2 3) 4(4 + 3)
α+β 5+ 2 4 3 4+ 3
Hence, required harmonic mean = = + = = 4
4+ 3 +
5+ 2
If a, b, c are in H.P., then the value of 1 + 1 − 1 1 + 1 − 1 is [MP PET 1998; Pb. CET 2000]
b c a c a b
(a) 21 (b) 32 (c) 32 (d) None of these
bc + b2 c2 + ca b2 − ab
a, b, c are in H.P. ⇒ 1 , 1 , 1 are in A.P.
a b c
∴ 1 + 1 = 2
a c b
Now, 1 + 1 − 1 1 + 1 − 1 = 1 + 1 + 1 − 2 2 − 1 = 1 + 2 − 2 1 = 1 3 − 2 = 3 − 2
b c a c a b a c a b b b b a b b b a b2 ab
b
If a, b, c are in H.P., then which one of the following is true [MNR 1985]
(a) 1 + 1 = 1 (b) ac =b (c) b+a + b+c =1 (d) None of these
b−a b−c b a+c b−a b−c
a, b, c are in H.P. ⇒ b= 2ac , ∴ option (b) is false
a+c
b−a = 2ac − a = a (c − a) ⇒ b−c = c (a − c)
a+c c+a a+c
∴ 1 1 = a+c − 1 + 1 = a + c ⋅ a−c = a+c = a+c ⋅ 2 = 2 , ∴ option (a) is false
b−a + b−c a−c a a − c ac ac 2ac b
c
b+a + b+c = (c + a)(b + a) + (b + c)(a + c) = a+c − b + a + b+c = a+c b − b = a+c ⋅ (a − c) b
b−a b−c a (c − a) c (a − c) a−c a a−c c a a−c ac
c
= a+c ⋅b = a+c ⋅ 2b = 1 ⋅ 2b = 2
ac 2ac b
∴ option (c) is false.
Arithmetico-geometric progression(A.G.P.)
nth Term of A.G.P..
If a1, a2, a3, ......, an, ...... is an A.P. and b1, b2, ......, bn, ...... is a G.P., then the sequence a1b1, a2b2 , a3b3 ,
......, anbn,..... is said to be an arithmetico-geometric sequence.
Thus, the general form of an arithmetico geometric sequence is a,(a + d)r,(a + 2d)r 2,(a + 3d)r 3 ,.....
From the symmetry we obtain that the nth term of this sequence is [a + (n − 1)d]r n−1
Also, let a,(a + d)r,(a + 2d)r 2,(a + 3d)r 3 ,..... be an arithmetico-geometric sequence. Then, a + (a + d)r
+ (a + 2d)r 2 + (a + 3d)r 3 + ... is an arithmetico-geometric series.
Sum of A.G.P..
(1) Sum of n terms : The sum of n terms of an arithmetico-geometric sequence a,(a + d)r,(a + 2d)r 2,
(a + 3d)r 3 ,..... is given by
1n2a−[2ra++d(nr (1 − r n−1 ) {a + (n − 1) d}r n , when r ≠1
(1 − r)2 1 − r
−
Sn =
− 1)d], when r = 1
(2) Sum of infinite sequence : Let |r|< 1. Then r n,r n−1 → 0 as n → ∞ and it can also be shown that
n.r n → 0 as n → ∞. So, we obtain that Sn → a + dr , as n → ∞.
1−r (1 − r)2
In other words, when |r|< 1 the sum to infinity of an arithmetico-geometric series is S∞ = a dr
1−r + (1 − r)2
Method for Finding Sum.
This method is applicable for both sum of n terms and sum of infinite number of terms.
First suppose that sum of the series is S, then multiply it by common ratio of the G.P. and subtract. In this
way, we shall get a G.P., whose sum can be easily obtained.
Method of Difference.
If the differences of the successive terms of a series are in A.P. or G.P., we can find nth term of the series by
the following steps :
Step I: Denote the nth term by Tn and the sum of the series upto n terms by Sn .
Step II: Rewrite the given series with each term shifted by one place to the right.
Step III: By subtracting the later series from the former, find Tn .
Step IV: From Tn , Sn can be found by appropriate summation.
Example: 44 1+ 3 + 5 + 7 + ......∞ is equal to [DCE 1999]
2 22 23
(a) 3 (b) 6 (c) 9 (d) 12
S = 1+ 3 + 5 + 7 + .......∞
2 22 23
Solution: (b)
Example: 45 1 S = 1 + 3 + 5 + .......∞
Solution: (b) 2 2 22 23
Example: 46 1 2 2 2
Solution: (b) 2 S =1+ 2 + 22 + 23 + ........ (on subtracting)
Example: 47 ⇒ S =1+ 2 1 + 1 + 1 + ....∞ ⇒ S = 1+ 2 × 1 1 /2 2 = 3. Hence S=6
Solution: (c) 2 2 22 23 2 − 1/
Example: 48 Sum of the series 1 + 2.2 + 3.22 + 4.23 + ..... + 100.299 is [IIIT (Hydrabad) 2000; Kerala (Engg.) 2001]
(a) 100.2100 + 1 (b) 99.2100 + 1 (c) 99.2100 − 1 (d) 100.2100 − 1
…..(i)
Let S = 1 + 2.2 + 3.22 + 4.23 + .... + 100.299 …..(ii)
2S = 1.2 + 2.22 + 3.23 + ..... + 99.299 + 100.2100
Equation (i) – Equation (ii) gives,
−S = 1 + (1.2 + 1.22 + 1.23 + ..... upto 99 terms) − 100.2100 =1+ 2(299 − 1) − 100.2100
2−1
⇒ S = −1 − 2100 + 2 + 100.2100 = 1 + 99.2100
The sum of the series 3 + 33 + 333 + ….. + n terms is [Rajasthan PET 2000]
(a) 1 (10n+1 + 9n − 28) (b) 1 (10n+1 − 9n − 10) (c) 1 (10n+1 + 10n − 9) (d) None of these
27 27 27
S = 3 + 33 + 333 + ...... to n terms
S = 3 + 33 + .......
0 = 3 + 30 + 300 + ......... to n terms − Tn (on subtracting)
∴ Tn = 3(1 + 10 + 100 + ....... to n terms) = 3 × 1 ⋅ 10n −1 = 1 (10n − 1)
10 −1 3
n 1 1 n 1 n 1 10 10n −1 1
n=1 3 3 3 n=1 3 10 −1 3
10n
n=1
∑ ∑ ∑Sn (10n
= − 1) = − 1= ⋅ − n
S = 1 (10n+1 − 9n − 10)
27
The sum of n terms of the following series 1 + (1 + x) + (1 + x + x2) + .... will be [IIT 1962]
(a) 1− xn (b) x(1 − xn) (c) n (1 − x) − x (1 − xn) (d) None of these
1− x 1− x (1 − x)2
S = 1 + (1 + x) + (1 + x + x2) + ...... (on subtracting)
S = 1 + (1 + x) + .......
0 = (1 + x + x2 + ..... to n terms) − Tn
∴ Tn = 1− xn
1− x
n n 1− xn 1 n 1 n 1 1 1− xn
n=1 1− x 1− x n=1 − n=1 − − 1− x
∑ ∑ ∑ ∑Sn= Tn = = 1 − 1 x xn = 1 x ⋅ n − 1 x ⋅ x ⋅
n=1
= n − x (1 − xn) = n (1 − x) − x (1 − xn)
1− x (1 − x)2 (1 − x)2
The sum to n terms of the series 1 + 3 + 7 + 15 + 31 + ....... is [IIT 1963]
(a) 2n+1 − n (b) 2n+1 − n − 2 (c) 2n − n − 2 (d) None of these
Solution: (b) S = 1 + 3 + 7 + 15 + 31 + ......
S = 1 + 3 + 7 + 15 + .......
0 = (1 + 2 + 4 + 8 + 16 + ..... to n terms) − Tn (on subtracting)
∴ Tn = 1+ 2+ 4 + 8 + ........... to n terms = 1⋅ 2n − 1 = 2n −1
2−1
∑ ∑ ∑ ∑Snn Tn n (2n n 2n n 2 ⋅ 2n − 1 n 2n+1 −n−2
= n=1 = n=1 − 1) = n=1 − n=1 1= 2−1 − =
Miscellaneous series
Special Series.
There are some series in which nth term can be predicted easily just by looking at the series.
If Tn = α n3 + β n2 + γ n + δ
nn n n nn
∑ ∑ ∑ ∑ ∑ ∑Then Sn = Tn = (α n3 + β n2 + γ n + δ ) = α n3 + β n2 + γ n + δ 1
n=1 n=1 n=1 n=1 n=1 n=1
= α n(n + 1) 2 + β n(n + 1)(2n + 1) + γ n(n + 1) + δ n
2 6 2
∑Note = 12 22 32 n n(n + 1)(2n + 1)
: Sum of squares of first n natural numbers + + + ....... + n2 = = 6
r2
r =1
Sum of cubes of first n natural numbers = 13 + 23 + 33 + 43 + ....... + n3 n n(n + 1) 2
∑ = = 2
r3
r =1
Vn Method.
(1) To find the sum of the series 1 1 + ..... + 1
a1a2a3 .....ar + a2a3 .....ar+1 a n a n+1 .....a n+r −1
Let d be the common difference of A.P. Then an = a1 + (n − 1)d .
Let Sn and Tn denote the sum to n terms of the series and nth term respectively.
Sn = 1 + 1 + ..... + 1
a1a2 .....ar a2a3 .....ar+1 a n a n+1 .....a n+r −1
∴ Tn = 1
a n a n+1 .....a n+r −1
Let Vn = 1 ; Vn−1 = 1
an+1an+2 .....an+r−1 a n a n+1 .....a n+r − 2
⇒ Vn − Vn−1 = 1 1 = an − an+r−1
an+1an+2 .....an+r−1 − anan+1.....an+r−2 a n a n+1 .....a n+r −1
= [a1 + (n − 1)d] − [a1 + {(n + r − 1) − 1}d] = d (1 − r) Tn
a n a n+1 .....a n+r −1
Tn 1 {Vn−1 − Vn } , Sn n 1 (V0 − Vn )
∑∴ = d (r − 1) ∴ = = d (r − 1)
Tn
n=1
Sn = 1 1 − 1
− 1)(a2
(r − a1 ) a1a 2 ....ar −1 a n+1 a n+ 2 ......a n+r −1
Example: If a1, a2 ,.....an are in A.P., then 1 1 + ... + 1 = 1 1 − 1
a1a 2 a 3 + a2a3a4 a n a n+1 a n+ 2
2(a 2 − a1 ) a1a 2 a n+1 a n+ 2
(2) If Sn = a1a2 .....ar + a2a3 .....ar+1.... + anan+1...an+r−1
Tn = anan+1.....an+r−1
Let Vn = anan+1....an+r−1an+r , ∴ Vn−1 = an−1an+1......an+r−1
⇒ Vn − Vn−1 = anan+1an+2 .....an+r−1(an+r − an−1 ) = Tn {[a1 + (n + r − 1)d] − [a1 + (n − 2)d]} = Tn (r + 1)d
∴ Tn = Vn − Vn−1
(r + 1)d
n 1 n − Vn−1 ) = 1 (Vn − V0 ) = 1 {(a a ....an+r ) (a a1 ....ar )}
+ 1)d + 1) + 1)
∑ ∑Sn= Tn = (r (Vn (r d (r d n n+1 − 0
n=1 n=1
= (r 1 − a1 ) {anan+1....an+r − a0a1.....ar }
+ 1)(a2
Example: 1.2.3.4 + 2.3.4.5 + ...... + n(n + 1)(n + 2)(n + 3) = 1 {n(n + 1)(n + 2)(n + 3) − 0.1.2.3}
5.1
= 1 {n(n + 1)(n + 2)(n + 3)}
5
Example: 49 The sum of 13 + 23 + 33 + 43 + .... + 153 is [MP PET 2003]
(a) 22000 (b) 10000 (c) 14400 (d) 15000
S = 13 + 23 + 33 + ...... + 153 n (n + 1) 2 15 × 16 2
2 2
Solution: (c) ; For n = 15 , the value of = = 14400
Example: 50 A series whose nth term is n + y , the sum of r terms will be [UPSEAT 1999]
x
(a) r (r + 1) + ry (b) r (r − 1) (c) r (r − 1) − ry (d) r (r + 1) − rx
2x 2x 2x
2x
r r n + y 1 r r 1 r (r + 1) r (r + 1)
n=1 n=1 x x n=1 x 2 2x
n+y 1=
n=1
Solution: (a) ∑ ∑ ∑ ∑Sr= tn = = + yr = + ry
Example: 51
If (12 − t1) + (22 − t2) + .... + (n2 − tn) = 1 n (n2 − 1) , then tn is
3
(a) n (b) n − 1 (c) n + 1 (d) n
2
Solution: (d) 1 n (n2 − 1) = (12 + 22 + .... + n2) − (t1 + t2 + ..... + tn)
3
⇒ t1 + t2 + ..... + tn = 12 + 22 + 32 + ....... + n2 − 1 n (n2 − 1) = n (n + 1)(2n + 1) − 1 n (n2 − 1) = n (n + 1) [2n + 1 − (2n − 2)]
3 6 3 6
∴ t1 + t2 + t3 + ..... + tn = n (n + 1) ⇒ Sn = n (n + 1)
2 2
tn = Sn − Sn−1 = n (n + 1) − (n − 1) n = n
2 2
Example: 52 The sum of the series 1 + 7 1 + 1 + .... is [MNR 1984; UPSEAT 2000]
3×7 × 11 11 × 15
(a) 1 (b) 1 (c) 1 (d) 1
3 6 9 12
Solution: (d) S = 1 + 1 + 1 + .... = 1 1 − 1 + 1 − 1 + 1 − 1 + ....... 1 = 1 1 − 1 = 1 1 − 0 = 1
Example: 53 3×7 7 ×11 11 × 15 4 3 7 7 11 11 15 4 3 ∞ 4 3 12
∞
Solution: (c)
The sum of the series 1.2.3 + 2.3.4 + 3.4.5 + ….. to n terms is [Kurukshetra CEE 1998]
(a) n (n + 1)(n + 2) (b) (n + 1)(n + 2)(n + 3)
(c) 1 n (n + 1)(n + 2)(n + 3) (d) 1 (n + 1)(n + 2)(n + 3)
4 4
Tn = n (n + 1)(n + 2) = n3 + 3n2 + 2n
n nnn
∑ ∑ ∑ ∑∴ S = 1.2.3 + 2.3.4 + 3.4.5 + ...... to n terms = (n3 + 3n2 + 2n) = n3 + 3 n2 + 2 n
n=1 n=1 n=1 n=1
S = n (n + 1) 2 +3 n (n + 1)(2n + 1) + 2 n (n + 1) = 1 n (n + 1)[n (n + 1) + 2 (2n + 1) + 4]
2 6 2 4
= 1 n (n + 1)[n2 + 5n + 6] = 1 n (n + 1)(n + 2)(n + 3)
4 4
Properties of Arithmetic, Geometric and Harmonic means between Two given Numbers.
Let A, G and H be arithmetic, geometric and harmonic means of two numbers a and b. Then,
A = a +b ,G = ab and H = 2ab
2 a+b
These three means possess the following properties :
(1) A ≥ G ≥ H
A = a + b ,G = ab and H = 2ab
2 a+b
∴ A−G = a+b − ab = ( a− b)2 ≥0
2 2
⇒ A≥G …..(i)
G−H = ab − 2ab = ab a + b− 2 ab = ab ( a− b)2 ≥ 0
a+b a+ b +b
a
⇒ G≥H …..(ii)
From (i) and (ii), we get A ≥ G ≥ H
Note that the equality holds only when a = b
(2) A, G, H from a G.P., i.e. G2 = AH
AH = a +b × 2ab = ab = ( ab )2 = G2
2 a+b
Hence, G2 = AH
(3) The equation having a and b as its roots is x 2 − 2Ax + G2 = 0
The equation having a and b its roots is x 2 − (a + b)x + ab = 0
⇒ x 2 − 2Ax + G2 = 0 A = a +b and G = ab
2
The roots a, b are given by A ± A2 − G2
(4) If A, G, H are arithmetic, geometric and harmonic means between three given numbers a, b and c, then
the equation having a, b, c as its roots is x3 − 3 Ax 2 + 3G 3 x − G3 =0
H
111
a b c a+b+c
A= + 3 + ,G = (abc)1 / 3 and 1 = 3
H
⇒ a + b + c = 3A, abc = G3 and 3G 3 = ab + bc + ca
H
The equation having a, b, c as its roots is x 3 − (a + b + c)x 2 + (ab + bc + ca)x − abc = 0
⇒ x3 − 3 Ax 2 3G 3 x − G3 =0
+H
Relation between A.P., G.P. and H.P..
(1) If A, G, H be A.M., G.M., H.M. between a and b, then a n+1 + b n+1 A when n = 0
an + bn = G when n = −1 / 2
H when n = −1
(2) If A1, A2 be two A.M.’s; G1, G2 be two G.M.’s and H1, H 2 be two H.M.’s between two numbers a and b
then G1G2 = A1 + A2
H1 H 2 H1 + H2
(3) Recognization of A.P., G.P., H.P. : If a, b, c are three successive terms of a sequence.
Then if, a−b = a , then a, b, c are in A.P.
b−c a
If, a−b = a , then a, b, c are in G.P.
b−c b
If, a−b = a , then a, b, c are in H.P.
b−c c
(4) If number of terms of any A.P./G.P./H.P. is odd, then A.M./G.M./H.M. of first and last terms is middle term
of series.
(5) If number of terms of any A.P./G.P./H.P. is even, then A.M./G.M./H.M. of middle two terms is
A.M./G.M./H.M. of first and last terms respectively.
(6) If pth, qth and rth terms of a G.P. are in G.P. Then p, q, r are in A.P.
(7) If a, b, c are in A.P. as well as in G.P. then a = b = c .
(8) If a, b, c are in A.P., then x a , x b , x c will be in G.P. (x ≠ ±1)
Example: 54 If the A.M., G.M. and H.M. between two positive numbers a and b are equal, then [Rajasthan PET 2003]
(a) a = b (b) ab = 1 (c) a > b (d) a < b
Solution: (a) A.M. = G.M.
Example: 55 ⇒ a+b = ab ⇒ ( a )2 − 2 a b +( b )2 = 0 ⇒ ( a− b )2 = 0 ⇒ a=b
Solution: (b) 2 2 2
Example: 56
Solution: (a) G.M. = H.M.
Example: 57
Solution: (c) ⇒ ab = 2ab ⇒ a+b−2 ab = 0 ⇒ ( a − b )2 = 0 ⇒ a = b ∴ a=b
Example: 58 a+b
Solution: (a)
Thus A.M. =(G.M.) (H.M.) So a = b
Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic
equation [AIEEE 2004]
(a) x2 − 18x − 16 = 0 (b) x2 − 18x + 16 = 0 (c) x2 + 18x − 16 = 0 (d) x2 + 18x + 16 = 0
A = 9, G = 4 are respectively the A.M. and G.M. between two numbers, then the quadratic equation having its roots as
the two numbers, is given by x2 − 2Ax + G2 = 0 i.e. x2 − 18x + 16 = 0
If a , b , c are in H.P., then [UPSEAT 2002]
b c a
(a) a2b, c2a, b2c are in A.P. (b) a2b, b2c, c2a are in H.P.
(c) a2b, b2c, c2a are in G.P. (d) None of these
a , b , c are in H.P.
b c a
⇒ b , c , a are in A.P. ⇒ abc × b , abc × c , abc × a are in A.P. ⇒ b2c, ac2, a2b are in A.P.
a b c a b c
∴ a2b, c2a, b2c are in A.P.
If a, b, c are in G.P., then loga x, logb x, logc x are in [Rajasthan PET 2002]
(a) A.P. (b) G.P. (c) H.P. (d) None of these
a, b, c are in G.P.
⇒ log x a, log x b, log x c are in A.P. ⇒ 1 x , 1 x , 1 x are in A.P.
log a logb log c
∴ loga x, logb x, logc x are in H.P.
If A1, A2 ; G1, G2 and H1, H2 be two A.M.’s, G.M.’s and H.M.’s between two quantities, then the value of G1G2 is
H1H2
[Roorkee 1983; AMU 2000]
(a) A1 + A2 (b) A1 − A2 (c) A1 + A2 (d) A1 − A2
H1 + H2 H1 + H2 H1 − H2 H1 − H2
Let a and b be the two numbers
∴ A1 = a + b − a = 2a + b , A2 = a + 2 b− a = a + 2b
3 3 3 3
a b 1 / 3 = a2/3 b1/ 3 , b 1 / 3 2
a a
G1 = G2 = a = a1 / 3b2 / 3
H1 = 1 = 3 = 3ab , H2 = 3ab
a + 2b 2a + b
1 + 1 − 1 1 2 + 1
a b a 3 a b
∴ G1G2 = (a2 / 3b1 / 3)(a1 / 3b2 / 3) = (a + 2b)(2a + b)
H1H2 9ab
3ab ⋅ 3ab
a + 2b 2a + b
A1 + A2 = 2a + b + a + 2b = a+b
3 3
H1 + H2 = 3ab + 3ab = 3ab 2a + b + a + 2b = 9ab (a + b)
a + 2b 2a + b (a + 2b)(2a + b) (a + 2b)(2a + b)
∴ A1 + A2 = (a + 2b)(2a + b) = G1G2
H1 + H2 9ab H1H2
Example: 59 If the ratio of H.M. and G.M. of two quantities is 12 : 13, then the ratio of the numbers is [Rajasthan PET 1990]
Solution: (d)
(a) 1 : 2 (b) 2 : 3 (c) 3 : 4 (d) None of these
Let x and y be the numbers
∴ H.M. = 2xy , G.M. = xy
x+y
∴ H.M. = 2 xy = 2 x/y ⇒ 12 = 2r , ( r = x )⇒ 12r 2 − 26r + 12 = 0 ⇒ 6r 2 − 13r + 6 = 0
G.M. x+y 13 r2 +1 y
x +1
y
∴ r = 13 ± 132 − 4.6.6 = 13 ± 5 = 18 , 8 = 3 , 2
2×6 12 12 12 2 3
∴ Ratio of numbers = x = r2 :1 = 9 :1 or 4 :1= 9:4 or 4 : 9
y 4 9
Example: 60 If the A.M. of two numbers is greater than G.M. of the numbers by 2 and the ratio of the numbers is 4 : 1, then the
Solution: (c)
numbers are [Rajasthan PET 1988]
Example: 61
(a) 4, 1 (b) 12, 3 (c) 16, 4 (d) None of these
Let x and y be the numbers ∴ A.M. = G.M. + 2 ⇒ x+y = xy + 2
2
Also, x = 4 :1 ⇒ x = 4y
y
∴ 4y + y = 4y.y + 2 ⇒ 5y = 2y + 2 ⇒ y=4 ⇒ x = 4 × 4 = 16
2 2
∴ The numbers are 16, 4.
If the ratio of A.M. between two positive real numbers a and b to their H.M. is m : n, then a : b is
(a) m − n + n (b) n + m − n (c) m + m − n (d) None of these
m−n − n n − m−n m − m−n
m (a + b) / 2 m (a + b)2 a + 12 m a a 12 m a 1 a
n 2ab /(a + b) n 4ab b n b b n b b
Solution: (c) We have, = ⇒ = = ⇒ 4 = + ⇒ 2 = +
a
4 b
Let a = r2 , ∴ 2 m r = (1 + r 2) ⇒ 2 mr = n + n r2 ⇒ nr2 − 2 mr + n =0
b n
∴ r= 2 m± 4m − 4n = m ± m−n
2 n n
Considering +ve sign, r = m+ m−n ( m+ m − n)( m − m − n) = m − (m − n) = n
=
n n( m − m − n) n( m − m − n) m − m − n
∴ r2 = m+ m−n ⋅ n . Hence, a = m + m−n .
n m−n b m − m−n
m−
Applications of Progressions.
There are many applications of progressions is applied in science and engineering. Properties of progressions
are applied to solve problems of inequality and maximum or minimum values of some expression can be found by
the relation among A.M., G.M. and H.M.
Example: 62 If x = log5 3 + log7 5 + log9 7 then
Solution: (c)
(a) x ≥ 3 (b) x ≥ 1 (c) x ≥ 3 (d) None of these
Example: 63 2 32 32
x = log5 3 + log7 5 + log9 7
log 5 3 + log7 5 + log 9 7 ≥ (log 5 3. log7 5. log9 7)1 / 3 [A.M. ≥ G.M.]
3
x x ≥ 3(log9 91 / 2)1 / 3 1 1 / 3 3
3 2 32
⇒ ≥ (log9 3)1 / 3 ⇒ ⇒ x ≥ 3 . Hence x ≥
If a, b, c, d are four positive numbers then
(a) a + b c + d ≥ 4⋅ a (b) a + c b + d ≥ 4 ⋅ a
b c d e e b d c e e
(c) a + b + c + d + e ≥ 5 (d) b + c + d + e + a ≥ 1
b c d e a a b c d e 5
ab
b+c a b 1 / 2
2 b c
Solution: (a,b,c) We have ≥ ⋅ ; (A.M. ≥ G.M.)
⇒ a + b ≥ 2 a …..(i)
b c c
Similarly, c + d ≥2 c …..(ii)
d e e
Multiplying (i) by (ii),
a + b c + d ≥ 4 a c ⇒ a + b c + d ≥ 4 a , ∴ (a) is true
b c d e c e b c d e e
a c b d 2 a c 1 / 2 2 b d 1 / 2 a c b d ≥ a
b d c e b d c e b d c e e
Next, + + ≥ ⋅ ⋅ ⋅ ⇒ + + 4 , ∴ (b) is true
ab c de
+ + + + a b c d e 1 / 5 a b c d e
b c d e a b c d e a b c d e a
5 ≥ ⋅ ⋅ ⋅ ⋅ ⇒ + + + + ≥ 5 , ∴ (c) is true
bcdea 5 b c d e a 1 / 5 b c d e a
a+b+c +d+e a b c d e a b c d e
Now, ≥ ⋅ ⋅ ⋅ ⋅ ⇒ + + + + ≥5, ∴ (d) is false
Example: 64 Let an = product of first n natural numbers. Then for all n ∈ N
Solution: (a,b)
(a) nn ≥ an (b) n + 1 n ≥ n! (c) nn ≥ an + 1 (d) None of these
2
We have an = 1.2.3.............n = n ! , nn = n.n.n........ to n times
⇒ nn ≥ n.(n − 1)(n − 2)....{n − (n − 1)} ⇒ nn ≥ n.(n − 1)(n − 2).........2.1 ⇒ nn ≥ n ! ∴ nn ≥ an . So (a) is true
nn ≥/ (n + 1)! ⇒ nn ≥/ an + 1 . So (c) is false
1+ 2 + 3+ ..... + n ≥ (1.2.3......n)1 / n ⇒ n(n + 1) ≥ (n ! )1 / n ⇒ n +1 ≥ (n ! )1 / n . ∴ n + 1 n ≥ n! . So (b) is true.
n 2n 2 2
Example: 65 In the given square, a diagonal is drawn and parallel line segments joining points on the adjacent sides are drawn on both
sides of the diagonal. The length of the diagonal is n 2 cm. If the distance between consecutive line segments be 1 cm
2
then the sum of the lengths of all possible line segments and the diagonal is
(a) n (n + 1) 2 cm (b) n2 cm (c) n (n + 2) cm (d) n2 2 cm
Solution: (d) Let us consider the diagonal and an adjacent parallel line
Example: 66 Length of the line PQ = RS = AC – (AR + SC) = AC – 2AR ( AR = SC) Q
Solution: (b,c) C
= AC – 2.PR ( AR = PR)
=n 2 −2⋅ 1 =n 2− 2 = (n − 1) 2 cm P 1/√2 cm. S
2 45o
Length of line adjacent to PQ, other than AC, will be ((n − 1) − 1) 2 = (n − 2) 2 cm R
A
∴ Sum of the lengths of all possible line segments and the diagonal
= 2 × [n 2 + (n − 1) 2 + (n − 2) 2 + .....] − n 2 , n∈N
= 2× 2[n + (n − 1) + (n − 2) + ...... + 1] − n 2=2 2 × n(n + 1) − n 2 =n 2{n + 1 − 1} = n2 2 cm
2
Let f(x) = 1 − xn+1 and g(x) = 1 − 2 + 3 − .... + (−1)n n+1 . Then the constant term in f′(x)× g(x) is equal to
1− x x x2 xn
(a) n (n2 − 1) when n is even (b) n (n + 1) when n is odd (c) − n (n + 1) when n is even (d) − n (n − 1) when n is odd
6 2 2 2
f(x) = 1 − xn+1 = (1 − x)(1 + x + x2 + ..... + xn) =1+ x+ x2 + ....... + xn ; f′(x) = 1 + 2x + 3x2 + ...... + n xn−1
1− x (1 − x)
f ′(x).g(x) = (1 + 2x + 3x 2 + .... + n x n−1 ) × 1 − 2 + 3 − ...... + (−1)n n+ 1
x x2 xn
∴ constant term in f′(x)× g(x) is
c = 12 − 22 + 3 2 − 4 2 + ....... + n2 (−1)n−1 = [12 + 22 + 3 2 + 4 2 + ...... + n2 ] − 2[22 + 4 2 + 6 2 + ....]
when n is odd,
c = [12 + 22 + ..... + n2] − 2 [22 + 42 + 62 + ...... + (n − 1)2] = n(n + 1)(2n + 1) − 2.22[12 + 22 + 32 + ...... + n − 1 2
6 2
n − 1 ⋅ n − 1 + 1 2 n−1 + 1
2 2 2
= n (n + 1)(2n + 1) −8 = n (n + 1)(2n + 1) − n (n − 1)(n + 1)
6 6 6 3
= n (n + 1) (2n +1 − 2(n − 1)) = n (n + 1) × 3 = n (n + 1)
6 6 2
= [12 22 n2] − 2 [2 2 42 n2] = n (n + 1)(2n + 1) 2 2 22 n 2
6 1 2
when n is even, c + + ..... + + + ..... + − 2.2 + + .... +
n (n + 1)(2n + 1) n ⋅ n + 1 2 ⋅ n + 1 n (n + 1)(2n + 1) 1
6 2 2 2 6 3
= −8 6 = − n (n + 1)(n + 2)
= 1 n (n + 1)(2n + 1 − 2(n + 2)) = − 1 n (n + 1)
6 2
***
Quadratic-equation-&-inequation
Polynomial.
Algebraic expression containing many terms of the form cx n , n being a non-negative integer is called a
polynomial. i.e., f(x) = a0 + a1 x + a2 x 2 + a3 x 3 + ...... + an−1 x n−1 + an x n , where x is a variable, a0, a1, a2 .......an are
constants and an ≠ 0
Example : 4 x4 + 3x3 − 7x2 + 5x + 3 , 3x3 + x2 − 3x + 5 .
(1) Real polynomial : Let a0, a1, a2 .......an be real numbers and x is a real variable.
Then f(x) = a0 + a1 x + a2 x 2 + a3 x 3 + ...... + an x n is called real polynomial of real variable x with real
coefficients.
Example : 3x 3 − 4 x 2 + 5x − 4, x 2 − 2x + 1 etc. are real polynomials.
(2) Complex polynomial : If a0, a1, a2 .......an be complex numbers and x is a varying complex number.
Then f(x) = a0 + a1 x + a2 x 2 + a3 x 3 + ...... + an x n is called complex polynomial of complex variable x with
complex coefficients.
Example : 3x 2 − (2 + 4i)x + (5i − 4), x 3 − 5i x 2 + (1 + 2i)x + 4 etc. are complex polynomials.
(3) Degree of polynomial : Highest power of variable x in a polynomial is called degree of polynomial.
Example : f(x) = a0 + a1 x + a2 x 2 + ...... + an−1 x n−1 + an x n is a n degree polynomial.
f(x) = 4 x 3 + 3x 2 − 7x + 5 is a 3 degree polynomial.
f(x) = 3x − 4 is single degree polynomial or linear polynomial.
f(x) = bx is an odd linear polynomial.
A polynomial of second degree is generally called a quadratic polynomial. Polynomials of degree 3 and 4 are
known as cubic and biquadratic polynomials respectively.
(4) Polynomial equation : If f(x) is a polynomial, real or complex, then f(x) = 0 is called a polynomial
equation.
Types of Quadratic Equation.
A quadratic polynomial f(x) when equated to zero is called quadratic equation.
Example : 3x 2 + 7x + 5 = 0, − 9x 2 + 7x + 5 = 0, x 2 + 2x = 0, 2x 2 = 0
or
An equation in which the highest power of the unknown quantity is two is called quadratic equation.
Quadratic equations are of two types :
(1) Purely quadratic equation : A quadratic equation in which the term containing the first degree of the
unknown quantity is absent is called a purely quadratic equation.
i.e. ax 2 + c = 0 where a, c ∈ C and a ≠ 0
(2) Adfected quadratic equation : A quadratic equation which contains terms of first as well as second
degrees of the unknown quantity is called an adfected quadratic equation.
i.e. ax 2 + bx + c = 0 where a, b, c ∈ C and a ≠ 0, b ≠ 0.
(3) Roots of a quadratic equation : The values of variable x which satisfy the quadratic equation is called
roots of quadratic equation.
Important Tips
• An equation of degree n has n roots, real or imaginary.
• Surd and imaginary roots always occur in pairs in a polynomial equation with real coefficients i.e. if 2 – 3i is a root of an equation, then
2 + 3i is also its root. Similarly if 2 + 3 is a root of given equation, then 2 − 3 is also its root.
• An odd degree equation has at least one real root whose sign is opposite to that of its last term (constant term), provided that the
coefficient of highest degree term is positive.
• Every equation of an even degree whose constant term is negative and the coefficient of highest degree term is positive has at least two
real roots, one positive and one negative.
Solution of Quadratic Equation.
(1) Factorization method : Let ax2 + bx + c = a(x − α)(x − β ) = 0 . Then x = α and x = β will satisfy the
given equation.
Hence, factorize the equation and equating each factor to zero gives roots of the equation.
Example : 3x 2 − 2x + 1 = 0 ⇒ (x − 1)(3x + 1) = 0
x = 1, − 1 / 3
(2) Hindu method (Sri Dharacharya method) : By completing the perfect square as
ax 2 + bx + c =0 ⇒ x2 + b x+ c =0
a a
b 2 x + b 2 b 2 − 4ac which gives, x = − b ± b 2 − 4ac
2a 2a 2a
Adding and subtracting , − 4a 2 = 0
Hence the quadratic equation ax 2 + bx + c = 0 (a ≠ 0) has two roots, given by
α = −b+ b2 − 4ac , β = −b− b 2 − 4ac
2a 2a
Note : Every quadratic equation has two and only two roots.
Nature of Roots.
In quadratic equation ax 2 + bx + c = 0 , the term b 2 − 4ac is called discriminant of the equation, which plays
an important role in finding the nature of the roots. It is denoted by ∆ or D.
(1) If a, b, c ∈ R and a ≠ 0, then : (i) If D < 0, then equation ax 2 + bx + c = 0 has non-real complex roots.
(ii) If D > 0, then equation ax 2 + bx + c = 0 has real and distinct roots, namely α = − b+ D, β = − b− D
2a 2a
and then ax 2 + bx + c = a (x − α)(x − β ) …..(i)
(iii) If D = 0, then equation ax 2 + bx + c =0 has real and equal roots α = β = −b
2a
and then ax 2 + bx + c = a (x − α)2 …..(ii)
To represent the quadratic expression ax 2 + bx + c in form (i) and (ii), transform it into linear factors.
(iv) If D ≥ 0, then equation ax 2 + bx + c = 0 has real roots.
(2) If a, b, c ∈ Q, a ≠ 0, then : (i) If D > 0 and D is a perfect square ⇒ roots are unequal and rational.
(ii) If D > 0 and D is not a perfect square ⇒ roots are irrational and unequal.
(3) Conjugate roots : The irrational and complex roots of a quadratic equation always occur in pairs.
Therefore
(i) If one root be α + iβ then other root will be α − iβ .
(ii) If one root be α + β then other root will be α − β .
(4) If D1 and D2 be the discriminants of two quadratic equations,then
(i) If D1 + D2 ≥ 0 , then
(a) At least one of D1 and D2 ≥ 0 . (b) If D1 < 0 then D2 > 0
(ii) If D1 + D2 < 0 , then
(a) At least one of D1 and D2 < 0 . (b) If D1 > 0 then D2 < 0 .
Roots Under Particular Conditions.
For the quadratic equation ax 2 + bx + c = 0 .
(1) If b = 0 ⇒ roots are of equal magnitude but of opposite sign.
(2) If c = 0 ⇒ one root is zero, other is − b / a .
(3) If b = c = 0 ⇒ both roots are zero.
(4) If a = c ⇒ roots are reciprocal to each other.
(5) If a > 0 c < 0 ⇒ roots are of opposite signs.
a < 0 c > 0
(6) If a > 0 b>0 c > 0 ⇒ both roots are negative, provided D ≥ 0.
a < 0 b<0 c < 0
(7) If a > 0 b<0 c > 0 ⇒ both roots are positive, provided D ≥ 0.
a < 0 b>0 c < 0
(8) If sign of a = sign of b ≠ sign of c ⇒ greater root in magnitude, is negative.
(9) If sign of b = sign of c ≠ sign of a ⇒ greater root in magnitude, is positive.
(10) If a + b + c = 0 ⇒ one root is 1 and second root is c/a.
(11) If a = b = c = 0 , then equation will become an identity and will be satisfied by every value of x.
(12) If a = 1 and b, c ∈ I and the root of equation ax 2 + bx + c = 0 are rational numbers, then these roots
must be integers.
Important Tips
• If an equation has only one change of sign, it has one +ve root and no more.
• If all the terms of an equation are +ve and the equation involves no odd power of x, then all its roots are complex.
Example: 1 Both the roots of given equation (x − a)(x − b) + (x − b)(x − c) + (x − c)(x − a) = 0 are always
Solution: (c)
[MNR 1986; IIT 1980; Kurukshetra CEE 1998]
(a) Positive (b) Negative (c) Real (d) Imaginary
Given equation (x − a)(x − b) + (x − b)(x − c) + (x − c)(x − a) = 0 can be re-written as 3x2 − 2(a + b + c)x + (ab + bc + ca) = 0
D = 4[(a + b + c)2 − 3(ab + bc + ca)] = 4[a2 + b2 + c2 − ab − bc − ac] = 2[(a − b)2 + (b − c)2 + (c − a)2] ≥ 0
Hence both roots are always real.
Example: 2 If the roots of (b − c)x2 + (c − a)x + (a − b) = 0 are equal then a + c = [Kurukshetra CEE 1992]
Solution: (a)
(a) 2b (b) b2 (c) 3b (d) b
Example: 3
Solution: (a) b−c+c−a+a−b =0
Example: 4
Solution: (c) Hence one root is 1. Also as roots are equal, other root will also be equal to 1.
Example: 5 Also α.β = a −b ⇒ 1.1 = a −b ⇒ a−b =b−c ⇒ 2b = a + c
Solution: (b) b −c b −c
If the roots of equation 1 1 1 are equal in magnitude but opposite in sign, then (p + q) =
x+ p + x+q =r
(a) 2r (b) r (c) – 2r [Rajasthan PET 1999]
(d) None of these
Given equation can be written as x2 + (p + q − 2r)x + [pq − (p + q)r] = 0
Since the roots are equal and of opposite sign, ∴ Sum of roots = 0
⇒ −(p + q − 2r) = 0 ⇒ p + q = 2r
If 3 is a root of x2 + kx − 24 = 0 , it is also a root of [EAMCET 2002]
(a) x2 + 5x + k = 0 (b) x2 − 5x + k = 0 (c) x2 − kx + 6 = 0 (d) x2 + kx + 24 = 0
Equation x2 + kx − 24 = 0 has one root as 3,
⇒ 32 + 3k − 24 = 0 ⇒ k = 5
Put x = 3 and k = 5 in option
Only (c) gives the correct answer i.e. ⇒ 32 − 15 + 9 = 0 ⇒ 0 = 0
For what values of k will the equation x2 − 2(1 + 3k)x + 7(3 + 2k) = 0 have equal roots [MP PET 1997]
(a) 1, –10/9 (b) 2, –10/9 (c) 3, –10/9 (d) 4, –10/9
Since roots are equal then [−2(1 + 3k)]2 = 4.1.7(3 + 2k) ⇒ 1 + 9k2 + 6k = 21 + 14k ⇒ 9k2 − 8k − 20 = 0
Solving, we get k = 2, − 10 / 9
Relations between Roots and Coefficients.
(1) Relation between roots and coefficients of quadratic equation : If α and β are the roots of
quadratic equation ax 2 + bx + c = 0 , (a ≠ 0) then
Sum of roots = S =α + β = −b = − coefficient of x
a coefficient of x 2
Product of roots = P = α.β = c = constant term
a coefficient of x 2
If roots of quadratic equation ax 2 + bx + c = 0 (a ≠ 0) are α and β then
(i) (α − β ) = (α + β )2 − 4αβ = ± b2 − 4ac = ± D
a a
(ii) α2 + β2 = (α + β )2 − 2αβ = b2 − 2ac
a2
(iii) α 2 − β 2 = (α + β ) (α + β )2 − 4αβ = −b b 2 − 4ac = ± bD
a2 a2
(iv) α3 + β 3 = (α + β )3 − 3αβ (α + β) = − b(b 2 − 3ac)
a3
(v) α 3 − β 3 = (α − β )3 + 3αβ(α − β ) = (α + β )2 − 4αβ {(α + β )2 − αβ} = ± (b 2 − ac) b2 − 4ac
a3
= b 2 − 2ac 2 c 2
a2 a 2
(vi) α4 +β4 = {(α + β )2 − 2αβ}2 − 2α 2 β 2 − 2
(vii) α 4 − β 4 = (α 2 − β 2 )(α 2 + β 2) = ± b(b 2 − 2ac) b 2 − 4ac
a4
(viii) α2 + αβ + β 2 = (α + β )2 − αβ = b 2 − ac
a2
(ix) α +β = α2 +β2 = (α + β )2 − 2αβ = b2 − 2ac
β α αβ αβ ac
(x) α 2β + β 2α = αβ (α + β) = − bc
a2
(xi) α 2 β 2 a4 +β4 (α 2 + β 2)2 − 2α 2β 2 b 2 D + 2a 2c 2
β α α 2β 2 α 2β 2 a 2c 2
+ = = =
(2) Formation of an equation with given roots : A quadratic equation whose roots are α and β is given
by (x − α)(x − β ) = 0
∴ x 2 − (α + β )x + αβ = 0 i.e. x 2 − (sum of roots)x + (product of roots) = 0
∴ x 2 − Sx + P = 0
(3) Equation in terms of the roots of another equation : If α, β are roots of the equation
ax 2 + bx + c = 0 , then the equation whose roots are
(i) –α, –β ⇒ ax 2 − bx + c = 0 (Replace x by – x)
(ii) 1 / α,1 / β ⇒ cx 2 + bx + a = 0 (Replace x by 1/x)
(iii) α n , β n ; n ∈ N ⇒ a(x1/ n )2 + b(x1/ n ) + c = 0 (Replace x by x1/ n )
(iv) kα, kβ ⇒ ax 2 + kbx + k 2c = 0 (Replace x by x/k)
(v) k + α , k + β ⇒ a(x − k)2 + b(x − k) + c = 0 (Replace x by (x – k))
(vi) α , β ⇒ k 2 ax 2 + kbx + c = 0 (Replace x by kx)
k k
(vii) α 1/ n , β 1/ n ; n ∈ N ⇒ a(x n )2 + b(x n ) + c = 0 (Replace x by x n )
(4) Symmetric expressions : The symmetric expressions of the roots α, β of an equation are those
expressions in α and β, which do not change by interchanging α and β. To find the value of such an expression, we
generally express that in terms of α + β and αβ.
Some examples of symmetric expressions are :
(i) α 2 + β 2 (ii) α 2 + αβ + β 2 (iii) 1 + 1 (iv) α + β
αβ βα
(v) α 2 β + β 2α (vi) α 2 + β 2 (vii) α 3 + β 3 (viii) α 4 + β 4
β α
4.7 Biquadratic Equation.
If α, β, γ, δ are roots of the biquadratic equation ax 4 + bx 3 + cx 2 + dx + e = 0 , then
S1 =α + β +γ +δ = −b / a , S2 = α.β + α.γ + αδ + βγ + β .δ + γδ = (−1)2 c = c
a a
or S2 = (α + β )(γ + δ ) + αβ + γδ = c/a, S3 = αβγ + βγδ + γδα + αβδ = (−1)3 d = −d / a
a
or S3 = αβ(γ + δ ) + γδ (α + β) = −d / a and S4 = α.β.γ .δ = (−1)4 e = e
a a
Example: 6 If the difference between the corresponding roots of x2 + ax + b = 0 and x2 + bx + a = 0 is same and a ≠ b , then
Solution: (a)
Example: 7 [AIEEE 2002]
Solution: (c)
(a) a + b + 4 = 0 (b) a + b − 4 = 0 (c) a − b − 4 = 0 (d) a − b + 4 = 0
Example: 8
Solution: (a) α + β = −a , αβ = b ⇒ α − β = a2 − 4b and γ + δ = −b , γδ = a ⇒ γ − δ = b2 − 4a
Example: 9 According to question, α − β = γ − δ ⇒ a2 − 4b = b2 − 4a ⇒ a + b + 4 = 0
Solution: (b)
Example: 10 If the sum of the roots of the quadratic equation ax2 + bx + c = 0 is equal to the sum of the squares of their reciprocals,
Solution: (b)
then a / c, b / a, c / b are in [AIEEE 2003; DCE 2000]
(a) A.P. (b) G.P. (c) H.P. (d) None of these
As given, if α, β be the roots of the quadratic equation, then
⇒ α +β = 1 + 1 = (α + β )2 − 2αβ ⇒ − b = b2 / a2 − 2c / a = b2 − 2ac
α2 β2 α2β 2 a c2 / a2 c2
⇒ 2a = b2 + b = ab2 + bc2 ⇒ 2a2c = ab2 + bc2 ⇒ 2a b c
c c2 a ac 2 b =c+a
c , a , b are in A.P. ⇒ a , b , c are in H.P.
a b c c a b
Let α, β be the roots of x2 − x + p = 0 and γ, δ be root of x2 − 4 x + q = 0 . If α, β, γ, δ are in G.P., then the integral value
of p and q respectively are [IIT Screening 2001]
(a) – 2, – 32 (b) – 2, 3 (c) – 6, 3 (d) – 6, – 32
α + β = 1 , αβ = p , γ + δ = 4 , γδ = q
Since α, β, γ, δ are in G.P.
r = β /α =γ / β =δ /γ
α + αr = 1 ⇒ α(1 + r) = 1 , α(r 2 + r3) = 4 ⇒ α.r 2(1 + r) = 4
So r 2 = 4 ⇒ r = ±2
If r = 2 , α + 2α = 1 ⇒ α = 1 / 3 and r = −2 , α − 2α = 1 ⇒ α = −1
But p = αβ ∈ I ∴ r = −2,α = −1
∴ p = −2 , q = α 2r5 = 1(−2)5 = −32
If 1 – i is a root of the equation x2 + ax + b = 0 , then the values of a and b are [Tamil Nadu Engg. 2002]
(a) 2, 1 (b) – 2, 2 (c) 2, 2 (d) 2, – 2
Since 1 − i is a root of x2 + ax + b = 0 . ∴ 1 + i is also a root.
Sum of roots ⇒ 1 − i + 1 + i = −a ⇒ a = −2
Product of roots ⇒ (1 − i)(1 + i) = b ⇒ b = 2
Hence a = −2 , b = 2
If the roots of the equation x2 − 5x + 16 = 0 are α, β and the roots of equation x2 + px + q = 0 are α 2 + β 2 , αβ / 2 ,
then [MP PET 2001]
(a) p = 1, q = −56 (b) p = −1, q = −56 (c) p = 1, q = 56 (d) p = −1, q = 56
Since roots of the equation x2 − 5x + 16 = 0 are α, β .
⇒ α+β = 5,αβ = 16 and α 2 + β 2 + αβ = −p ⇒ (α + β )2 − 2αβ + αβ = −p ⇒ 25 − 2(16) + 16 = −p ⇒ p = −1
2 2 2
and (α 2 + β 2) αβ = q ⇒ [(α + β )2 − 2αβ ] αβ =q ⇒ (25 − 32)8 = q ⇒ q = −56
2 2
Example: 11 If α ≠ β , but α 2 = 5α − 3, β 2 = 5β − 3 , then the equation whose roots are α and β is
Solution: (b) βα
Example: 12 [EAMCET 1989; AIEEE 2002]
Solution: (c)
Example: 13 (a) x2 − 5x − 3 = 0 (b) 3x2 − 19x + 3 = 0 (c) 3x2 + 12x + 3 = 0 (d) None of these
Solution: (a)
Example: 14 S = α + β = α 2 + β 2 = 5α − 3 + 5β − 3 α 2 = 5α − 3
Solution: (d) β 2 = 5β − 3
Example: 15 β α αβ αβ
Solution: (d)
S = 5(α + β ) − 6 , p = α ⋅ β = 1 ⇒ p = 1 . α, β are roots of x2 − 5x + 3 = 0 . Therefore α + β = 5 , αβ = 3
Example: 16 αβ β α
S = 5(5) − 6 = 19
3 3
∵ x 2 − 19 x +1= 0 ⇒ 3x2 − 19x + 3 = 0
3
Let α, β be the roots of the equation (x − a)(x − b) = c , c ≠ 0 , then the roots of the equation (x − α)(x − β ) + c = 0 are
(a) a, c (b) b, c (c) a, b [IIT 1992; DCE 1998, 2000; Roorkee 2000]
(d) a, d
Since α, β are the roots of (x − a)(x − b) = c i.e. of x2 − (a + b)x + ab − c = 0
∴ α + β = a + b ⇒ a + b = α + β and αβ = ab − c ⇒ ab = αβ + c
∴ a, b are the roots of x2 − (α + β )x + αβ + c = 0 ⇒ (x − α)(x − β ) + c = 0
Hence (c) is the correct answer
If α and β are roots of the equation x2 − ax + b = 0 and Vn = α n + β n , then [Rajasthan PET 1995; Karnataka CET 2000]
(a) Vn+1 = a Vn − b Vn−1 (b) Vn+1 = b Vn − a Vn−1 (c) Vn+1 = a Vn + b Vn−1 (d) Vn+1 = b Vn + a Vn−1
Since α and β are roots of equation, x2 − ax + b = 0 , therefore α + β = a , αβ = b
Now, Vn+1 = α n+1 + β n+1 = (α + β )(α n + β n) − αβ(α n−1 + β n−1) ⇒ Vn+1 = a . Vn − b . Vn−1 [IIT Screening 2004]
If one root of the equation x2 + px + q = 0 is the square of the other, then
(a) p3 + q2 − q(3p + 1) = 0 (b) p3 + q2 + q(1 + 3p) = 0
(c) p3 + q2 + q(3p − 1) = 0 (d) p3 + q2 + q(1 − 3p) = 0
Let α and α 2 be the roots then α + α 2 = −p , α.α 2 = q
Now (α + α 2)3 = α 3 + α 6 + 3α 3(α + α 2) ⇒ − p3 = q + q2 − 3pq ⇒ p3 + q2 + q(1 − 3p) = 0
Let α and β be the roots of the equation x2 + x + 1 = 0 , the equation whose roots are α19, β 7 is [IIT 1994; Pb. CET 2001]
(a) x2 − x − 1 = 0 (b) x2 − x + 1 = 0 (c) x2 + x − 1 = 0 (d) x2 + x + 1 = 0
Roots of x2 + x +1 = 0 are x = −1± 1−4 ,= −1± 3i = ω,ω 2
2 2
Take α = ω, β = ω 2
∴ α19 = w19 = w, β 7 = (w2)7 = w14 = w2
∴ Required equation is x2 + x + 1 = 0
If one root of a quadratic equation is 1 , then the equation is [Rajasthan PET 1987]
2+ 5
(d) None of these
(a) x2 + 4 x + 1 = 0 (b) x2 + 4 x − 1 = 0 (c) x2 − 4 x + 1 = 0
Solution: (b) Given root = 1 5 = 2− 5 = −2 + 5 , ∴ other root = −2 − 5
2+ −1
Again, sum of roots = – 4 and product of roots = – 1. The required equation is x2 + 4 x − 1 = 0
Condition for Common Roots.
(1) Only one root is common : Let α be the common root of quadratic equations a1x2 + b1x + c1 = 0 and
a2x 2 + b2x + c2 = 0 .
∴ a1α 2 + b1α + c1 = 0 , a2α 2 + b2α + c2 = 0
By Crammer’s rule : α2 = α = 1 or α2 = α = 1
− c1 b1 a1 − c1 a1 b1 b1c2 − b2c1 a2c1 − a1c2 a1b2 − a2b1
− c2 b2
a2 − c2 a2 b2
∴ α = a2c1 − a1c2 = b1c2 − b2c1 , α ≠ 0
a1b2 − a2b1 a2c1 − a1c2
∴ The condition for only one root common is (c1a2 − c 2a1)2 = (b1c 2 − b2c1)(a1b2 − a2b1)
(2) Both roots are common: Then required condition is a1 = b1 = c1 .
a2 b2 c2
Important Tips
• To find the common root of two equations, make the coefficient of second degree term in the two equations equal and subtract. The
value of x obtained is the required common root.
• Two different quadratic equations with rational coefficient can not have single common root which is complex or irrational as imaginary
and surd roots always occur in pair.
Example: 17 If one of the roots of the equation x2 + ax + b = 0 and x2 + bx + a = 0 is coincident. Then the numerical value of (a + b) is
Solution: (b)
[IIT 1986; Rajasthan PET 1992; EAMCET 2002]
Example: 18
Solution: (a) (a) 0 (b) – 1 (c) (d) 5
If α is the coincident root, then α 2 + aα + b = 0 and α 2 + bα + a = 0
⇒ α2 = α a = 1 a
a2 − b2 b− b−
α 2 = −(a + b) , α = 1 ⇒ −(a + b) = 1 ⇒ (a + b) = −1
If a, b, c are in G.P. then the equations ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have a common root if d , e , f are in
a b c
[IIT 1985; Pb. CET 2000; DCE 2000]
(a) A.P. (b) G.P. (c) H.P. (d) None of these
As given, b2 = ac ⇒ ax2 + 2bx + c = 0 can be written as ax2 + 2 ac x + c = 0 ⇒ ( a x + c )2 = 0 ⇒ x = − c
a
This must be common root by hypothesis
So it must satisfy the equation, dx2 + 2ex + f =0 ⇒ d c − 2e c + f = 0
a a
d f 2e c 2e ⇒ d f 2e
a+c = c a= c. a a+c = b
Hence d , e , f are in A.P.
a b c
Properties of Quadratic Equation.
(1) If f(a) and f(b) are of opposite signs then at least one or in general odd number of roots of the equation
f(x) = 0 lie between a and b.
Y f(a) = +ve Y f(a) = +ve
x=b A x=b
X
O x=a X x=a B
f(b) = –ve f(b) = –ve
(2) If f(a) = f(b) then there exists a point c between a and b such that f ′(c) = 0 , a < c < b .
Y
Q
P
O a bX
A B
As is clear from the figure, in either case there is a point P or Q at x = c where tangent is parallel to x-axis
i.e. f ′(x) = 0 at x = c .
(3) If α is a root of the equation f(x) = 0 then the polynomial f(x) is exactly divisible by (x − α) or (x − α) is
factor of f(x) .
(4) If the roots of the quadratic equations ax 2 + bx + c = 0 , a2 x 2 + b2 x + c 2 = 0 are in the same ratio
i.e. α1 = α2 then b12 / b 2 = a1c1 / a2c2 .
β1 β2 2
(5) If one root is k times the other root of the quadratic equation ax 2 + bx + c = 0 then (k + 1)2 = b2 .
k ac
Example: 19 The value of ‘a’ for which one root of the quadratic equation (a2 − 5a + 3) x2 + (3a − 1)x + 2 = 0 is twice as large as the
other is [AIEEE 2003]
(a) 2/3 (b) – 2/3 (c) 1/3 (d) – 1/3
Solution: (a) Let the roots are α and 2α
Now, α + 2α = 1 − 3a 3 , α . 2α = a2 2 + 3 ⇒ 3α = 1 − 3a 3 , 2α 2 = a2 2 + 3
a2 − 5a + − 5a a2 − 5a + − 5a
⇒ 2 1 (1 − 3a)2 = a2 2 + 3 ⇒ (1 − 3a)2 = 9 ⇒ 9a2 − 45a + 27 = 1 + 9a2 − 6a ⇒ 39a = 26 ⇒ a= 2/3
9 (a2 − 5a + 3)2 − 5a a2 − 5a + 3
Quadratic Expression.
An expression of the form ax 2 + bx + c , where a, b, c ∈ R and a ≠ 0 is called a quadratic expression in x. So
in general, quadratic expression is represented by f(x) = ax 2 + bx + c or y = ax 2 + bx + c .
(1) Graph of a quadratic expression : We have y = ax 2 + bx + c = f(x)
a x b 2 D D a x b 2
2a 4a 2 4a 2a
y = + − ⇒ y + = +
Now, let y+ D =Y and X = x+ b
4a 2a
Y = a.X 2 ⇒ X2 = 1 Y
a
(i) The graph of the curve y = f(x) is parabolic.
(ii) The axis of parabola is X =0 or x + b = 0 i.e. (parallel to y-axis).
2a
(iii) (a) If a > 0, then the parabola opens upward.
(b) If a < 0, then the parabola opens downward.
a > 0, D < 0 x axis
x axis a < 0, D < 0
(iv) Intersection with axis
(a) x-axis: For x axis, y=0 ⇒ ax 2 + bx + c = 0 ⇒ x = −b± D
2a
For D > 0, parabola cuts x-axis in two real and distinct points i.e. x = − b± D .
2a
For D = 0, parabola touches x-axis in one point, x = −b / 2a .
a<0 D>0
x-axis
a>0 x-axis
D>0
a<0 D=0
x-axis
a>0 x-axis
D=0
For D < 0, parabola does not cut x-axis(i.e. imaginary value of x).
a<0 D<0
x-axis
a>0 D<0 x-axis
(b) y-axis : For y axis x = 0 , y = c
(2) Maximum and minimum values of quadratic expression : Maximum and minimum value of
quadratic expression can be found out by two methods :
(i) Discriminant method : In a quadratic expression ax 2 + bx + c .
(a) If a > 0, quadratic expression has least value at x = −b / 2a . This least value is given by 4ac − b2 = − D .
4a 4a
(b) If a < 0, quadratic expression has greatest value at x = −b / 2a . This greatest value is given by
4ac − b2 = − D .
4a 4a
(ii) Graphical method : Vertex of the parabola Y = aX 2 is X = 0 , Y = 0
i.e., x+ b = 0, y+ D = 0 ⇒ x = −b / 2a , y = −D / 4a
2a 4a
Hence, vertex of y = ax 2 + bx + c is (−b / 2a, − D / 4a)
(a) For a > 0, f(x) has least value at x = − b . This least value is given by
2a
a>0
f − b = − D .
2a 4a vertex
vertex
(b) For a < 0, f(x) has greatest value at x = −b / 2a . This greatest value is given by a<0
f − b = − D .
2a 4a
(3) Sign of quadratic expression : Let f(x) = ax 2 + bx + c or y = ax 2 + bx + c
Where a, b, c ∈ R and a ≠ 0, for some values of x, f(x) may be positive, negative or
zero. This gives the following cases :
(i) a > 0 and D < 0, so f(x) > 0 for all x ∈ R i.e., f(x) is positive for all real values of x.
(ii) a < 0 and D < 0, so f(x) < 0 for all x ∈ R i.e., f(x) is negative for all real values of x.
(iii) a > 0 and D = 0 so, f(x) ≥ 0 for all x ∈ R i.e., f(x) is positive for all real values of x except at vertex,
where f(x) = 0 .
(iv) a < 0 and D = 0 so, f(x) ≤ 0 for all x ∈ R i.e. f(x) is negative for all real values of x except at vertex,
where f(x) = 0 .
(v) a > 0 and D > 0
Let f(x) = 0 have two real roots α and β (α < β ), then f(x) > 0 for all x ∈ (−∞,α) ∪ (β, ∞) and f(x) < 0 for
all x ∈ (α, β ).
(vi) a < 0 and D > 0
Let f(x) = 0 have two real roots α and β (α < β ),
Then f(x) < 0 for all x ∈ (−∞,α) ∪ (β, ∞) and f(x) > 0 for all x ∈ (α, β )
Example: 20 If x be real, then the minimum value of x2 − 8x + 17 is [MNR 1980]
Solution: (c)
(a) – 1 (b) 0 (c) 1 (d) 2
Since a =1> 0 therefore its minimum value is = 4ac − b2 = 4(1)(17) − 64 = 4 =1
4a 4 4
Example: 21 If x is real, then greatest and least values of x2 − x +1 are [IIT 1968; Rajasthan PET 1988]
Solution: (b) x2 + x +1
(d) None of these
Example: 22 (a) 3, –1/2 (b) 3, 1/3 (c) – 3, –1/3
Solution: (b)
Let y = x2 − x +1
Example: 23 x2 + x +1
Solution: (b)
x2(y − 1) + (y + 1)x + (y − 1) = 0
∵ x is real, therefore b2 − 4ac ≥ 0
⇒ (y + 1)2 − 4(y − 1)(y − 1) ≥ 0 ⇒ 3y2 − 10y + 3 ≤ 0 ⇒ (3y − 1)(y − 3) ≤ 0 ⇒ y − 1 (y − 3) ≤ 0 ⇒ 1 ≤ y ≤ 3
3 3
Thus greatest and least values of expression are 3, 1/3 respectively.
If f(x) is quadratic expression which is positive for all real value of x and g(x) = f(x) + f ′(x) + f ′′(x) . Then for any real value
of x [IIT 1990]
(a) g(x) < 0 (b) g(x) > 0 (c) g(x) = 0 (d) g(x) ≥ 0
Let f(x) = ax2 + bx + c , then g(x) = ax2 + bx + c + 2ax + b + 2a = ax2 + (b + 2a)x + (b + c + 2a)
∵ f(x) > 0 . Therefore b2 − 4ac < 0 and a > 0
Now for g(x),
Discriminant = (b + 2a)2 − 4a(b + c + 2a) = b2 + 4a2 + 4ab − 4ab − 4ac − 8a2 = (b2 − 4ac) − 4a2 < 0 as b2 − 4ac < 0
Therefore sign of g(x) and a are same i.e. g(x) > 0 .
If α, β (α < β ) are roots of the equation x2 + bx + c = 0 where (c < 0 < b) then [IIT Screening 2000]
(a) 0 < α < β (b) α < 0 < β <|α | (c) α < β < 0 (d) α < 0 <|α |< β
Since f(0) = 0 + 0 + c = c < 0
∴ Roots will be of opposite sign, α + β = −b = −ve (b > 0)
It is given that α < β
So, α + β = −ve is possible only when |α | > β
⇒ α < 0, β > 0,|α | > β ⇒ α < 0 < β <|α |
Wavy Curve Method.
Let f(x) = (x − a1 )k1 (x − a2 )k2 (x − a3 )k3 .......( x − a )n−1 kn−1 (x − an )kn …..(i)
Where k1,k2,k3 ..., kn ∈ N and a1, a2, a3 ,......, an are fixed natural numbers satisfying the condition
a1 < a2 < a3 ..... < an−1 < an
First we mark the numbers a1, a2, a3 ,......, an on the real axis and the plus sign in the interval of the right of
the largest of these numbers, i.e. on the right of an . If kn is even then we put plus sign on the left of an and if kn is
odd then we put minus sign on the left of an . In the next interval we put a sign according to the following rule :
When passing through the point an−1 the polynomial f(x) changes sign if kn−1 is an odd number and the
polynomial f(x) has same sign if kn−1 is an even number. Then, we consider the next interval and put a sign in it using
the same rule. Thus, we consider all the intervals. The solution of f(x) > 0 is the union of all intervals in which we have
put the plus sign and the solution of f(x) < 0 is the union of all intervals in which we have put the minus sign.
Position of Roots of a Quadratic Equation.
Let f(x) = ax 2 + bx + c , where a, b, c ∈ R be a quadratic expression and k,k1,k2 be real numbers such that
k1 < k2 . Let α, β be the roots of the equation f(x) = 0 i.e. ax 2 + bx + c = 0. Then α = −b+ D, β = − b− D
2a 2a
where D is the discriminant of the equation.
(1) Condition for a number k (If both the roots of f(x) = 0 are less than k)
a>0 f(k) (–b/2a, –D/4a)
–b/2a k x-axis
α βk
x-axis α –b/2a β
a<0 f(k)
(–b/2a, –D/4a)
(i) D ≥ 0 (roots may be equal) (ii) a f(k) > 0 (iii) k > −b / 2a , where α ≤ β
(2) Condition for a number k (If both the roots of f(x) = 0 are greater than k)
a>0 (–b/2a, –D/4a)
f(k)
k x-axis
–b/2a α –b/2a β
k α β x-axis
f(k) a < 0
(–b/2a, –D/4a)
(i) D ≥ 0 (roots may be equal) (ii) a f(k) > 0 (iii) k < −b / 2a , where α ≤ β
(3) Condition for a number k (If k lies between the roots of f(x) = 0)
a>0 (–b/2a, –D/4a)
k β x-axis α k β x-axis
α a<0
(–b/2a, –D/4a)
(i) D > 0 (ii) a f(k) < 0 , where α < β
(4) Condition for numbers k1 and k2 (If exactly one root of f(x) = 0 lies in the interval (k1, k2))
f(k1) a > 0 k1 f(k2) β
k1 α f(k2) (k2) β x-axis x axis α (k2)
f(k1)
a<0
(i) D > 0 (ii) f(k1)f(k2 ) < 0 , where α < β .
(5) Condition for numbers k1 and k2 (If both roots of f(x) = 0 are confined between k1 and k2)
(–b/2a, –D/4a)
f(k1) a > 0 f(k2) k1 k2
α β
k1 α β k2 a<0
(–b/2a, –D/4a)
(i) D ≥ 0 (roots may be equal) (ii) a f(k1) > 0 (iii) a f(k2 ) > 0
(iv) k1 < −b / 2a < k2 , where α ≤ β and k1 < k2
(6) Condition for numbers k1 and k2 (If k1 and k2 lie between the roots of f(x) = 0)
a>0
α k1 k2 β x-axis
k1 k2 β x-axis a<0
α f(k1) f(k2)
(i) D > 0 (ii) a f(k1) < 0 (iii) a f(k2 ) < 0 , where α < β
Example: 24 If the roots of the equation x2 − 2ax + a2 + a − 3 = 0 are real and less than 3, then [IIT 1999; MP PET 2000]
(a) a < 2 (b) 2 ≤ a ≤ 3 (c) 3 < a ≤ 4 (d) a > 4
Solution: (a) Given equation is x2 − 2ax + a2 + a − 3 = 0
If roots are real, then D ≥ 0
⇒ 4a2 − 4(a2 + a − 3) ≥ 0 ⇒ −a + 3 ≥ 0 ⇒ a − 3 ≤ 0 ⇒ a ≤ 3
As roots are less than 3, hence f(3) > 0
9 − 6a + a2 + a − 3 > 0 ⇒ a2 − 5a + 6 > 0 ⇒ (a − 2)(a − 3) > 0 ⇒ a < 2, a > 3 . Hence a < 2 satisfy all the conditions.
Example: 25 The value of a for which 2x2 − 2(2a + 1)x + a(a + 1) = 0 may have one root less than a and another root greater than a are
Solution: (d)
given by (b) −1 < a < 0 (c) a ≥ 0 [UPSEAT 2001]
(a) 1 > a > 0
(d) a > 0 or a < −1
The given condition suggest that a lies between the roots. Let f(x) = 2x2 − 2(2a + 1)x + a(a + 1)
For ‘a’ to lie between the roots we must have Discriminant ≥ 0 and f(a) < 0
Now, Discriminant ≥ 0
4(2a + 1)2 − 8a(a + 1) ≥ 0 ⇒ 8(a2 + a + 1 / 2) ≥ 0 which is always true.
Also f(a) < 0 ⇒ 2a2 − 2a(2a + 1) + a(a + 1) < 0 ⇒ − a2 − a < 0 ⇒ a2 + a > 0 ⇒ a(1 + a) > 0 ⇒ a > 0 or a < −1
Descarte's Rule of Signs.
The maximum number of positive real roots of a polynomial equation f(x) = 0 is the number of changes of
sign from positive to negative and negative to positive in f(x).
The maximum number of negative real roots of a polynomial equation f(x) = 0 is the number of changes of
sign from positive to negative and negative to positive in f(−x) .
Example: 26 The maximum possible number of real roots of equation x5 − 6x2 − 4 x + 5 = 0 is [EAMCET 2002]
(a) 0 (b) 3 (c) 4 (d) 5
Solution: (b) f(x) = x5 − 6x2 − 4x + 5 = 0
+– – +
2 changes of sign ⇒ maximum two positive roots.
f(−x) = −x5 − 6x2 + 4 x + 5
–– ++
1 changes of sign ⇒ maximum one negative roots.
⇒ total maximum possible number of real roots = 2 + 1 = 3.
Rational Algebraic Inequations.
(1) Values of rational expression P(x)/Q(x) for real values of x, where P(x) and Q(x) are quadratic
expressions : To find the values attained by rational expression of the form a1 x 2 + b1 x + c1 for real values of x,
a2x 2 + b2x + c2
the following algorithm will explain the procedure :
Algorithm
Step I: Equate the given rational expression to y.
Step II: Obtain a quadratic equation in x by simplifying the expression in step I.
Step III: Obtain the discriminant of the quadratic equation in Step II.
Step IV: Put Discriminant ≥ 0 and solve the inequation for y. The values of y so obtained determines the set
of values attained by the given rational expression.
(2) Solution of rational algebraic inequation: If P(x) and Q(x) are polynomial in x, then the inequation
P(x) > 0, P(x) < 0, P(x) ≥ 0 and P(x) ≤ 0 are known as rational algebraic inequations.
Q(x) Q(x) Q(x) Q(x)
To solve these inequations we use the sign method as explained in the following algorithm.
Algorithm
Step I: Obtain P(x) and Q(x).
Step II: Factorize P(x) and Q(x) into linear factors.
Step III: Make the coefficient of x positive in all factors.
Step IV: Obtain critical points by equating all factors to zero.
Step V: Plot the critical points on the number line. If there are n critical points, they divide the number line
into (n + 1) regions.
Step VI: In the right most region the expression P(x) bears positive sign and in other regions the expression
Q(x)
bears positive and negative signs depending on the exponents of the factors.
Algebraic Interpretation of Rolle’s Theorem.
Let f(x) be a polynomial having α and β as its roots, such that α < β . Then, f(α) = f(β ) = 0 . Also a
polynomial function is everywhere continuous and differentiable. Thus f(x) satisfies all the three conditions of Rolle’s
theorem. Consequently there exists γ ∈ (α, β ) such that f ′(γ ) = 0 i.e. f ′(x) = 0 at x = γ . In other words x = γ is a
root of f ′(x) = 0 . Thus algebraically Rolle’s theorem can be interpreted as follows.
Between any two roots of polynomial f(x), there is always a root of its derivative f ′(x).
Lagrange’s theorem : Let f(x) be a function defined on [a b] such that
(i) f(x) is continuous on [a b] and
(ii) f(x) is differentiable on (a, b), then c ∈ (a, b), such that f ′(c) = f(b) − f(a)
b−a
Lagrange’s identity : If a1, a2 , a3 ,b1,b2 ,b3 ∈ R then :
(a12 + a 2 + a 2 )(b12 + b22 + b32 ) − (a1b1 + a2b2 + a3b3 )2 = (a1b2 − a2b1 )2 + (a2b3 − a3b2)2 + (a3b1 − a1b3 )2
2 3
Example: 27 If 2x 2 2x + 2 > x 1 , then [IIT 1987]
Solution: (c) + 5x +1
Example: 28 (a) −2 > x > −1 (b) −2 ≥ x ≥ −1 (c) −2 < x < −1 (d) −2 < x ≤ −1
Solution: (c)
Given 2x − 1 >0 ⇒ 2x 2 + 2x − 2x 2 − 5x − 2 > 0 ⇒ −3x − 2 > 0
Example: 29 + 5x x +1 (2x + 1)(x + 2)(x + 1) 1)(x + 2)(x
Solution: (d) 2x 2 + 2 (2x + + 1)
Example: 30 ⇒ (x + −3(x + 2 / 3) + 1) > 0 ⇒ (x + (x + 2 / 3) + 1) < 0
Solution: (c) 1)(x + 2)(2x 1)(x + 2)(2x
Equating each factor equal to 0, + ++
We get x = −2, − 1, − 2 / 3, − 1 / 2 –2 – –1 –2/3 – –1/2
∴ x ∈] − 2,−1[ ∪ ] − 2 / 3, − 1 / 2[ ⇒ −2 / 3 < x < −1 / 2 or −2 < x < −1
If for real values of x, x 2 − 3x + 2 > 0 and x 2 − 3x − 4 ≤ 0 , then [IIT 1983]
(a) −1 ≤ x < 1 (b) −1 ≤ x < 4 (c) −1 ≤ x < 1 or 2 < x ≤ 4 (d) 2 < x ≤ 4
x 2 − 3x + 2 > 0 or (x − 1)(x − 2) > 0
∴ x ∈ (−∞,1) ∪ (2,∞) …..(i) + – +
1 2
Again x 2 − 3x − 4 ≤ 0 or (x − 4)(x + 1) ≤ 0
∴ x ∈[−1, 4] …..(ii) ++
–
–1 4
From eq. (i) and (ii), x ∈[−1,1) ∪ (2 4] ⇒ −1 ≤ x < 1 or 2 < x ≤ 4
If b > a , then the equation (x − a)(x − b) − 1 = 0 , has [IIT Screening 2000]
(a) Both roots in [a b] (b) Both roots in (– ∞, a)
(c) Both roots in (b, ∞) (d) One root in (– ∞, a) and other in (b, +∞)
We have, (x − a)(x − b) − 1 = 0
(x − a)(x − b) = 1 > 0 ⇒ (x − a)(x − b) > 0 [∵ b > a]
++
a–b
x ∈] − ∞, a [ ∪ ] b ,+∞[ , i.e. (−∞, a) and (b, ∞).
The number of integral solution of x +1 > 1 is [Orissa JEE 2002]
x2 + 2 4
(d) None of these
(a) 1 (b) 2 (c) 5
x +1 1 0 x 2 − 4x − 2 0 ⇒ (x − (2 + 6))(x − (2 − 6)) < 0
x2 + 2 4 x2 + 2
− > ⇒ <
⇒ 2− 6 < x <2+ 6 + +
2–√6 – 2+√6
Approximately, −0.4 < x < 4.4
Hence, integral values of x are 0, 1, 2, 3, 4
Hence, number of integral solution = 5
Example: 31 If 2a + 3b + 6c = 0 then at least one root of the equation ax 2 + bx + c = 0 lies in the interval
[Kurukshetra CEE 2002; AIEEE 2002]
(a) (0, 1) (b) (1, 2) (c) (2, 3) (d) (3, 4)
Solution: (a) Let f ′(x) = ax 2 + bx + c
∫∴ f(x) = f′(x) dx = ax 3 + bx 2 + cx
3 2
Clearly f(0) = 0 , f(1) = a + b + c = 2a + 3b + 6c = 0 = 0
3 2 6 6
Since, f(0) = f(1) = 0 . Hence, there exists at least one point c in between 0 and 1, such that f ′(x) = 0 , by Rolle’s theorem.
Trick: Put the value of a = −3, b = 2, c = 0 in given equation
− 3x 2 + 2x = 0 ⇒ 3x 2 − 2x = 0 ⇒ x(3x − 2) = 0
x = 0, x = 2 / 3 , which lie in the interval (0, 1)
Equation and Inequation containing Absolute Value.
(1) Equations containing absolute values
By definition, | x |= x, if x ≥ 0 0
− x, if x <
Important forms containing absolute value :
Form I: The equation of the form | f(x) + g (x)|=| f(x)|+| g (x)| is equivalent of the system f(x).g (x) ≥ 0 .
Form II: The equation of the form | f1(x)| + | f2(x)| + | f3 (x)| +......| fn(x)|= g (x) …..(i)
Where f1(x), f2(x), f3 (x)...... fn(x), g(x) are functions of x and g(x) may be a constant.
Equations of this form can be solved by the method of interval. We first find all critical points of
f1(x), f2(x)..... fn(x). If coefficient of x is +ve, then graph starts with +ve sign and if it is negative, then graph starts
with negative sign. Then using the definition of the absolute value, we pass form equation (i) to a collection of
system which do not contain the absolute value symbols.
(2) Inequations containing absolute value
By definition, |x| < a ⇒ − a < x < a (a > 0), | x |≤ a ⇒ − a ≤ x ≤ a ,
|x| > a ⇒ x < −a or x > a and | x | ≥ a ⇒ x ≤ −a or x ≥ a
Example: 32 The roots of | x − 2|2 +| x − 2|− 6 = 0 are [UPSEAT 2003]
Solution: (a)
(a) 0, 4 (b) –1, 3 (c) 4, 2 (d) 5, 1
We have | x − 2|2 +| x − 2|− 6 = 0
Let | x − 2| = X
X2 + X −6 = 0
⇒ X = −1± 1 + 24 = 2, − 3 ⇒ X=2 and X = −3
2
∴ | x − 2| = 2 and | x − 2| = −3 , which is not possible.
Example: 33 ⇒ x − 2 = 2 or x − 2 = −2 [IIT Screening 2002]
Solution: (b) ∴ x = 4 or x = 0
The set of all real numbers x for which x2 −| x + 2| + x > 0 , is
Example: 34
Solution: (c) (a) (−∞, − 2) ∪ (2, ∞) (b) (−∞, − 2) ∪ ( 2, ∞) (c) (−∞, − 1) ∪ (1, ∞) (d) ( 2, ∞)
Example: 35
Solution: (b) Case I: If x + 2 ≥ 0 i.e. x ≥ −2 , we get
x2 − x − 2 + x > 0 ⇒ x2 − 2 > 0 ⇒ (x − 2)(x + 2) > 0
⇒ x ∈ (−∞, − 2) ∪ ( 2,∞) + +
But x ≥ −2 –√2 √2
∴ x ∈ [−2, − 2) ∪ ( 2 ∞) …..(i)
Case II: x + 2 < 0 i.e. x < −2 , then
x2 + x + 2 + x > 0 ⇒ x2 + 2x + 2 > 0 ⇒ (x + 1)2 + 1 > 0 . Which is true for all x
∴ x ∈ (−∞, − 2) …..(ii)
From (i) and (ii), we get, x ∈ (−∞, − 2) ∪ ( 2,∞)
Product of real roots of the equation t2x2 +| x | + 9 = 0 (t ≠ 0) [AIEEE 2002]
(a) Is always +ve (b) Is always –ve (c) Does not exist (d) None of these
Expression is always +ve, so t2x2 +| x | + 9 ≠ 0 . Hence roots of given equation does not exist.
The number of solution of log4(x − 1) = log2(x − 3) [IIT Screening 2001]
(a) 3 (b) 1 (c) 2 (d) 0
We have log4(x − 1) = log2(x − 3)
(x − 1) = (x − 3)2 ⇒ x − 1 = x2 + 9 − 6x ⇒ x2 − 7x + 10 = 0 ⇒ (x − 5)(x − 2) = 0
x = 5 or x = 2
But x − 3 < 0 , when x = 2 . ∴ Only solution is x = 5 .
Hence number of solution is one.
Rectangular-cartesian-co-ordinates
Introduction .
Co-ordinates of a point are the real variables associated in an order to a point to describe its location in some
space. Here the space is the two dimensional plane. The work of describing the Y Quadrant I
position of a point in a plane by an ordered pair of real numbers can be done in (+,+)
different ways. Quadrant II
(–,+)
The two lines XOX' and YOY' divide the plane in four quadrants. XOY, X' X
YOX', X' OY', Y'OX are respectively called the first, the second, the third and the Quadrant III O Quadrant IV
fourth quadrants. We assume the directions of OX, OY as positive while the (+,–)
directions of OX', OY' as negative. (–,–)
Y'
Quadrant x-coordinate y-coordinate point
First quadrant + + (+,+)
Second quadrant – + (–,+)
Third quadrant – – (–,–)
Fourth quadrant + – (+,–)
Cartesian Co-ordinates of a Point .
This is the most popular co-ordinate system.
Let us consider two intersecting lines XOX' and YOY', which are perpendicular to each other. Let P be any
point in the plane of lines. Draw the rectangle OLPM with its adjacent sides X′ Y
OL,OM along the lines XOX', YOY' respectively. The position of the point P can
be fixed in the plane provided the locations as well as the magnitudes of OL, OM M x P(x, y)
are known.
y
Axis of x : The line XOX' is called axis of x.
O LX
Axis of y : The line YOY' is called axis of y.
Co-ordinate axes : x axis and y axis together are called axis of Y′
co-ordinates or axis of reference.
Origin : The point ‘O’ is called the origin of co-ordinates or the origin.
Oblique axes : If both the axes are not perpendicular then they are called as oblique axes.
Let OL = x and OM = y which are respectively called the abscissa (or x-coordinate) and the ordinate (or y-
coordinate). The co-ordinate of P are (x, y).
Note : Co-ordinates of the origin is (0, 0).
The y co-ordinate of every point on x-axis is zero. Y P(r,θ)
The x co-ordinate of every point on y-axis is zero. X r X
Polar Co-ordinates . θ
Let OX be any fixed line which is usually called the initial line and O be a fixed point O
on it. If distance of any point P from the O is 'r' and ∠XOP = θ , then (r, θ ) are called the
polar co-ordinates of a point P.
If (x, y) are the cartesian co-ordinates of a point P, then
x = r cosθ ; y = r sinθ and r = x 2 + y 2 Y'
θ = tan −1 y
x
Distance Formula .
The distance between two points P(x1, y1) and Q(x 2 , y2 ) is given by
PQ = (PR)2 + (QR)2 = (x2 − x1)2 + (y2 − y1)2 YQ
Note : The distance of a point M(x0 , y0 ) from origin O (0, 0) P R
X′ O X
OM = (x 2 + y 2 ) .
0 0
If distance between two points is given then use ± sign. Y′
When the line PQ is parallel to the y-axis, the abscissa of point P and Q will be equal i.e, x1 = x2 ;
∴ PQ =| y2 − y1 |
When the segment PQ is parallel to the x-axis, the ordinate of the points P and Q will be equal i.e.,
y1 = y2 . Therefore PQ =| x2 − x1 |
(1) Distance between two points in polar co-ordinates : Let O be the pole and OX be the initial line.
Let P and Q be two given points whose polar co-ordinates are (r1,θ1) and (r2,θ 2 )respectively.
Then OP = r1, OQ = r2 P(r1,θ1)
∠POX = θ1 and ∠QOX = θ 2
then ∠POQ = (θ1 − θ2) Q(r2,θ2)
X
(OP)2 + (OQ)2 − (PQ)2 r1 (θ1-θ2)
2OP. OQ O
In ∆POQ, from cosine rule cos(θ 1 −θ2) = θ1 r2
θ2
∴ (PQ)2 = r12 + r22 − 2r1r2 cos(θ1 − θ2) M
∴ PQ = r12 + r22 − 2r1r2 cos(θ1 − θ2)
Note : Always taking θ1 and θ2 in radians.
Example: 1 If the point (x, y) be equidistant from the points (a + b,b − a) and (a − b,a + b) , then [MP PET 1983, 94]
(a) ax + by = 0 (b) ax − by = 0 (c) bx + ay = 0 (d) bx − ay = 0
Solution: (d) Let points P(x,y) , A (a + b, b − a), B(a − b, a + b) .
According to Question, PA = PB , i.e., PA2 = PB2
⇒ (a + b − x)2 + (b − a − y)2 = (a − b − x)2 + (a + b − y)2
⇒(a + b)2 + x 2 − 2x(a + b) + (b − a)2 + y 2 − 2y(b − a) = (a − b)2 + x2 − 2x(a − b) + (a + b)2 + y2 − 2y(a + b)
⇒ 2x(a − b − a − b) = 2y(b − a − a − b) ⇒ − 4bx = − 4ay ⇒ bx − ay = 0
Example: 2 If cartesian co-ordinates of any point are ( 3,1) , then its polar co-ordinates is
(a) (2,π / 3) (b) ( 2,π / 6) (c) (2,π / 6) (d) None of these
Solution: (c) We know that x = r cos θ , y = r sinθ
∴ 3 = r cosθ , 1 = r sinθ
r= ( 3)2 + (1)2 = 2 , θ = tan −1 1 = π / 6
3
Polar co-ordinates = (2,π / 6) .
Geometrical Conditions.
(1) Properties of triangles
(i) In any triangle ABC, AB + BC > AC and | AB − BC|< AC .
(ii) The ∆ABC is equilateral ⇔ AB = BC = CA .
(iii) The ∆ABC is a right angled triangle ⇔ AB2 = AC2 + BC2 or AC2 = AB2 + BC2 or BC2 = AB2 + AC2 .
(iv) The ∆ABC is isosceles ⇔ AB = BC or BC = CA or AB = AC .
(2) Properties of quadrilaterals D C
(i) The quadrilateral ABCD is a parallelogram if and only if θθ
(a) AB = DC, AD = BC , or (b) the middle points of BD and AC are the same, A B
In a parallelogram diagonals AC and BD are not equal and θ ≠ π .
2
(ii) The quadrilateral ABCD is a rectangle if and only if
(a) AB = CD, AD = BC and AC2 = AB2 + BC2 or, (b) AB = CD, AD = BC, AC = BD or, (c) the middle points
of AC and BD are the same and AC=BD. (θ ≠ π / 2 ) DC
θ
AB
(iii) The quadrilateral ABCD is a rhombus (but not a square) if and only if (a) AB = BC = CD = DA and
AC ≠ BD or, (b) the middle points of AC and BD are the same and AB = AD but AC ≠ BD . (θ = π / 2)
D C
θ
(iv) The quadrilateral ABCD is a square if and only if AB
(a) AB = BC = CD = DA and AC = BD or (b) the middle points of AC and BD are the same and AC = BD,
(θ = π / 2) , AB = AD . D C
Note : Diagonals of square, rhombus, rectangle and parallelogram always θ
bisect each other.
Diagonals of rhombus and square bisect each other at right angle. A B
Four given points are collinear, if area of quadrilateral is zero.
Example: 3 ABC is an isosceles triangle. If the co-ordinates of the base are B(1,3) and C (– 2,7) the co-ordinates of vertex A can be
[Orissa JEE 2002]
(a) (1, 6) (b) − 1 , 5 (c) 5 , 6 (d) None of these
2 6
Solution: (c) Let the vertex of triangle be A(x, y) .
Then the vertex A(x, y) is equidistant from B and C because ABC is an isosceles triangle, therefore
(x − 1)2 + (y − 3)2 = (x + 2)2 + (y − 7)2 ⇒ 6x − 8y + 43 = 0
Thus, any point lying on this line can be the vertex A except the mid point − 1 , 5 of BC. Hence vertex A is 5 , 6
2 6
Example: 4 The extremities of diagonal of parallelogram are the points (3, – 4) and (– 6,5) if third vertex is (– 2,1), then fourth vertex is
Solution: (b)
Example: 5 [Rajasthan PET 1987]
Solution: (c)
(a) (1,0) (b) (–1,0) (c) (1,1) (d) None of these
Example: 6
Solution: (a) Let A(3,−4) and C(−6, 5) be the ends of diagonal of parallelogram ABCD. Let B(−2,1) and D be (x, y), then mid points of
diagonal AC and BD coincide. So, x − 2 = −6 + 3 and y +1 = 5 −4
2 2 2 2
x = −1, y = 0 . ∴ Coordinates of D are (–1, 0)
The vertices A and D of square ABCD lie on positive side of x and y-axis respectively. If the vertex C is the point (12, 17),
then the coordinate of vertex B are (b) (15, 3) Y
(a) (14, 16) (d) (17, 12)
(c) (17, 5) M 12 C
Let the co-ordinate of B be (h, k) a
5 θ
a
Draw BL and CM perpendicular to x-axis and y-axis. D
∴ a cosθ = CM = OD = AL = 12 12 θ a aB
and a sinθ = DM = OA = BL = 5 O 5A
∴ k = BL = DM = OM − OD = 17 − 12 = 5 5
θ X
12 L
∴ h = OL = OA + AL = 5 + 12 = 17
Hence, Point B is (17, 5). [AIEEE 2002]
A triangle with vertices (4, 0); (–1, –1); (3, 5) is
(a) Isosceles and right angled (b) Isosceles but not right angled
(c) Right angled but not isosceles (d) Neither right angled nor isosceles
Let A (4,0); B(−1,−1); C(3,5) then
AB = 26, AC = 26 , BC = 52 ; i.e. AB = AC
So triangle is isosceles and also (BC)2 = (AB)2 + (AC)2 . Hence ∆ ABC is right angled isosceles triangle.
Section Formulae.
If P(x, y) divides the join of A(x1, y1) and B(x 2 , y2 ) in the ratio m1 : m2 (m1,m2 > 0)
(1) Internal division : If P(x, y)divides the segment AB internally in the ratio of m1 : m2
⇒ PA = m1 (x2,y2)
PB m2 B
The co-ordinates of P(x,y) are P (x, y)
x = m1 x 2 + m2 x1 and y = m1 y 2 + m2y1 (x1,y1)
m1 + m2 m1 + m2 A
(2) External division : If P(x,y) divides the segment AB externally in the ratio of m1 : m2
⇒ PA = m1 P(x, y)
PB m2
(x2, y2)
m1 x 2 − m2 x1 m1 y 2 − m2y1
The co-ordinates of P(x,y) are x = m1 − m2 and y = m1 − m2 (x1,y1) A B
Note : If P(x,y) divides the join of A(x1, y1 ) and B(x 2 , y2 ) in the ratio λ : 1(λ > 0) , then
x = λx2 ± x1 ; y = λy2 ± y1 . Positive sign is taken for internal division and negative sign is taken
λ ±1 λ ±1
for external division.
The mid point of AB is x1 + x2 , y1 + y2 [Here m1 : m2 :: 1 : 1 ]
2 2
For finding ratio, use ratio λ : 1 . If λ is positive, then divides internally and if λ is negative, then
divides externally.
Straight line ax + by + c = 0 divides the join of points A(x1, y1) and B(x2, y2) in the
ratio − ax1 + by1 + c .
ax 2 + by2 + c
If ratio is –ve then divides externally and if ratio is +ve then divides internally.
Example: 7 The co-ordinate of the point dividing internally the line joining the points (4, –2) and (8, 6) in the ratio 7: 5 will be
Solution: (c) [AMU 1979; MP PET 1984]
Example: 8
Solution: (b) (a) (16, 18) (b) (18, 16) (c) 19 , 8 (d) 8 , 19
Example: 9 3 3 3 3
Solution: (a)
Let point (x, y) divides the line internally.
Then x = m1 x 2 + m2 x1 = 7(8) + 5(4) = 19 , y = m1y2 + m2y1 = 7(6) + 5(−2) = 8 .
m1 + m2 12 3 m1 + m2 12 3
The line x + y = 4 divides the line joining the points (–1,1) and (5, 7) in the ratio [IIT 1965, UPSEAT 1999]
(a) 2 : 1 (b) 1 : 2 Internally (c) 1 : 2 Externally (d) None of these
Required ratio = − ax1 + by1 + c = – −1 + 1 − 4 = 4 = 1 (Internally)
ax2 + by2 + c 5+7− 4 8 2
The line joining points (2, –3) and (–5, 6)is divided by y-axis in the ratio [MP PET 1999]
(a) 2 : 5 (b) 2 : 3 (c) 3 : 5 (d) 1 : 2
Let ratio be k :1 and coordinate of y-axis are (0, b). Therefore, 0 = k(−5) + 1(2) ⇒ k = 2
k +1 5
Some points of a Triangle.
(1) Centroid of a triangle : The centroid of a triangle is the point of A(x1,y1)
intersection of its medians. The centroid divides the medians in the ratio 2 : 1
(Vertex : base) 2 E
11 2
If A(x1, y1 ), B(x2, y2) and C(x 3 , y3 ) are the vertices of a triangle. If G be the F 2 G1
centroid upon one of the median (say) AD, then AG : GD = 2 : 1 B C
(x2,y2) D (x3,y3)
x1 + x2 + x3 y1 + y2 + y3
⇒ Co-ordinate of G are 3 , 3
Example: 10 The centroid of a triangle is (2,7)and two of its vertices are (4, 8) and (–2, 6) the third vertex is [Kerala (Engg.) 2002]
(a) (0, 0) (b) (4, 7) (c) (7, 4) (d) (7, 7)
Solution: (b) Let the third vertex (x, y)
2= x + 4 − 2 , 7 = y + 8 + 6 , i.e. x = 4, y=7
3 3
Hence third vertex is (4, 7).
(2) Circumcentre : The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of
the sides of a triangle. It is the centre of the circle which passes through the
vertices of the triangle and so its distance from the vertices of the triangle is the A(x1,y1)
same and this distance is known as the circum-radius of the triangle.
Let vertices A, B, C of the triangle ABC be (x1, y1),(x2, y2) and (x 3 , y3 )and FE
let circumcentre be O(x, y) and then (x, y) can be found by solving O
(OA)2 = (OB)2 = (OC)2 (x2,y2)B D C(x3,y3)
i.e., (x − x1)2 + (y − y1)2 = (x − x2)2 + (y − y2)2 = (x − x3)2 + (y − y3)2
Note : If a triangle is right angle, then its circumcentre is the mid point of hypotenuse.
If angles of triangle i.e., A, B, C and vertices of triangle A(x1, y1), B(x2, y2) and C (x 3 , y3 ) are given,
then circumcentre of the triangle ABC is
x1 sin 2A + x 2 sin 2B +x 3 sin 2C , y1 sin 2A + y 2 sin 2B + y3 sin 2C
sin 2A + sin 2B + sin 2C sin 2A + sin 2B + sin 2C
Example: 11 If the vertices of a triangle be (2, 1); (5, 2) and (3,4) then its circumcentre is [IIT 1964]
Solution: (b)
(a) 13 , 9 (b) 13 , 9 (c) 9 , 13 (d) None of these
2 2 4 4 4 4
Let circumcentre be O (x, y) and given points are A(2,1); B(5, 2); C(3, 4) and OA2 = OB2 = OC2
∴ (x − 2)2 + (y − 1)2 = (x − 5)2 + (y − 2)2 .....(i)
and (x − 2)2 + (y − 1)2 = (x − 3)2 + (y − 4)2 .....(ii)
On solving (i) and (ii), we get x = 13 , y = 9
4 4
A
(3) Incentre : The incentre of a triangle is the point of intersection of internal F IE
bisector of the angles. Also it is a centre of a circle touching all the sides of a triangle. B DC
Co-ordinates of incentre ax1 + bx2 + cx3 , ay1 + by2 + cy3
a+b+c a+b+c
Where a, b, c are the sides of triangle ABC.
(4) Excircle : A circle touches one side outside the triangle and other two extended sides then circle is known
as excircle. Let ABC be a triagle then there are three excircles with three excentres.
Let I1, I 2 , I 3 opposite to vertices A,B and C respectively. If vertices of triangle are I3 A I2
A(x1, y1), B(x2, y2) and C (x3, y3) then
I1 ≡ − ax 1 + bx 2 + cx 3 , − ay1 + by2 + cy3 BC
− a+b+c −a+b+c I1
I2 ≡ ax 1 − bx 2 + cx 3 , ay1 − by2 + cy 3 , I3 ≡ ax1 + bx2 − cx3 , ay1 + by2 − cy3
a−b+c a−b+c a+b−c a+b−c
Note : Angle bisector divides the opposite sides in the ratio of remaining sides e.g. BD = AB = c
DC AC b
Incentre divides the angle bisectors in the ratio (b + c) : a,(c + a) : b and (a + b) : c
Excentre : Point of intersection of one internal angle bisector and other two external angle bisector
is called as excentre. There are three excentres in a triangle. Co-ordinate of each can be obtained by
changing the sign of a,b,c respectively in the formula of in-centre.
Example: 12 The incentre of the triangle with vertices (1, 3),(0,0) and (2, 0) is [IIT Screening 2000]
Solution: (d)
(a) 1, 3 (b) 2 , 1 (c) 2 , 3 (d) 1, 1
2 3 3 3 2 3
Here AB = BC = CA (A1,
∴ The triangle is equilateral .
So, the incentre is the same as the centroid.
∴ Incentre = 1 + 0 + 2 , 3 + 0 + 0 = 1, 1 . 2 2
3 3 3 C
(2
60o
B 2
(0,
(5) Orthocentre : It is the point of intersection of perpendiculars drawn from vertices on opposite sides (called
altitudes) of a triangle and can be obtained by solving the equation of any two
A (x1, y1)
altitudes.
Here O is the orthocentre since AE ⊥ BC , BF ⊥ AC and CD ⊥ AB D OF
then OE ⊥ BC, OF ⊥ AC, OD ⊥ AB
Solving any two we can get coordinate of O.
B C
(x2, y2) E (x3, y3)
Note : If a triangle is right angled triangle, then orthocentre is the point where right angle is formed.
• If the triangle is equilateral then centroid, incentre, orthocentre, circum-centre coincides.
• Orthocentre, centroid and circum-centre are always collinear and centroid divides the line
joining orthocentre and circum-centre in the ratio 2 : 1
• In an isosceles triangle centroid, orthocentre, incentre, circum-centre lie on the same line.
Example: 13 The vertices of triangle are (0, 3) (– 3, 0) and (3, 0). The co-ordinate of its orthocentre are [AMU 1991; DCE 1994]
Solution: (c)
(a) (0, – 2) (b) (0, 2) (c) (0, 3) (d) (0, –3)
Here AB ⊥ BC .
In a right angled triangle, orthocentre is the point where right angle is formed. B (0,3)
∴ Orthocentre is (0, 3)
C O A
(– 3, 0) (3, 0)
Example: 14 If the centroid and circumcentre of triangle are (3, 3); (6, 2), then the orthocentre is [DCE 2000]
Solution: (d)
(a) (9, 5) (b) (3, –1) (c) (– 3, 1) (d) (– 3, 5)
Let orthocentre be (α, β ) . We know that centroid divides the line joining orthocentre and circumcentre in the ratio 2 : 1
∴3 = 2(6) + 1(α ) ⇒ α = −3 , 3 = 2(2) + 1(β ) ⇒ β = 5
2 + 1 2 + 1
Hence orthocentre is (–3, 5).
Area of some Geometrical figures.
(1) Area of a triangle : The area of a triangle ABC with vertices A(x1,y1); B(x2,y2) and C(x3, y3) . The area
of triangle ABC is denoted by ‘∆’and is given as A (x1, y1)
1 x1 y1 1 1
2 x2 y2 2
∆ = x3 y3 1 = (x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)
1
In equilateral triangle B C
(x2, y2) (x3, y3)
(i) Having sides a, area is 3 a2 .
4
(ii) Having length of perpendicular as 'p' area is (p 2 ) .
3
Note : If a triangle has polar co-ordinates (r1,θ1), (r2,θ2) and (r3,θ3) then its area
∆ = 1 [r1r2 sin(θ 2 − θ1) + r2r3 sin(θ3 − θ2) + r3r1 sin(θ1 − θ3)]
2
If area is a rational number. Then the triangle cannot be equilateral.
(2) Collinear points : Three points A(x1, y1 ); B(x 2 , y2 ); C(x 3 , y3 ) are collinear. If area of triangle is zero,
i.e., (i) ∆=0 ⇒ 1 x1 y1 1 x1 y1 1
2 x2 y2 1 =0 ⇒ x2 y2 1 =0
x3 y3 1 x3 y3 1
(ii) AB + BC = AC or AC + BC = AB or AC + AB = BC
(3) Area of a quadrilateral : If (x1, y1);(x2, y2);(x3, y3) and (x 4 , y4 ) are vertices of a quadrilateral, then its
Area = 1 [(x 1 y 2 − x 2y1) + (x 2y3 − x3y2 ) + (x3y4 − x4 y3 ) + (x4 y1 − x1y4 )]
2
Note : If two opposite vertex of rectangle are (x1, y1) and (x2, y2) , then its area is
(y2 − y1)(x2 − x1) .
It two opposite vertex of a square are A(x1, y1) and C(x2, y2) , then its area is
= 1 AC2 = 1 [(x 2 − x1)2 + (y2 − y1 )2 ]
2 2
(4) Area of polygon : The area of polygon whose vertices are (x1, y1);(x 2 , y2 );(x 3 , y3 );....(xn,yn ) is
= 1 |{(x 1 y 2 − x 2y1) + (x 2y3 − x 3 y2 ) + .... + (xn y1 − x1yn )}|
2
Or Stair method : Repeat first co-ordinates one time in last for down arrow use positive sign and for up
arrow use negative sign.
x1 y1
x2 y2
∴ Area of polygon = 1 | x3 y3 |= 1 |{(x1y2 + x2y3 + .... + xny1) − (y1x2 + y2x3 + .... + yn x1 )}|
2 : : 2
Example: 15
Solution: (d) : :
Example: 16
xn yn
Solution: (a) x1 y1
Example: 17
Solution: (a) The area of the triangle formed by the points (a,b + c),(b,c + a),(c,a + b) is
Example: 18 [IIT 1963; EAMCET 1982; Rajasthan PET 2003]
Solution: (b,c)
(a) abc (b) a2 + b2 + c2 (c) ab + bc + ca (d) 0
a b+c 1a a+b+c 1a 1 1
1 c+a 1 b+c+a a+b+c 1 1 =0
Area = 2 b a+b 1 = 2 b c+a+b 1 , (Applying C2 → C1 + C2 ) = 2 b 1 1
c 1 c 1 c
Three points are A(6, 3), B(– 3, 5), C(4, – 2) and P (x, y) is a point, then the ratio of area of ∆PBC and ∆ABC is
[IIT 1983]
(a) x+y−2 (b) x−y+2 (c) x−y−2 (d) None of these
7 2 7
Area of ∆PBC 1 [− 3(−2 − y) + 4(y − 5) + x(5 + 2)] = 7x + 7y − 14 x+y−2
Area of ∆ABC 2 2) − 3(−2 − 3) + 4(3 − 5)] 49 7
= 1 =
2
[6(5 +
If the points (2K, K)(K, 2K) and (K, K) with K > 0 enclose triangle of area 18 square units then the centroid of triangle is
equal to
(a) (8, 8) (b) (4, 4) (c) (– 4, – 4) (d) (4 2, 4 2)
∆ = 1 2K K 1 K2 = 18 ⇒ K =± 6 . Consider K = +6 because K > 0, then the points (12, 6) (6,12) and (6,6).
2 K 2K 1 = 18 ⇒ 2
K K 1
Hence, centroid = 12 +6 +6,6 + 12 + 6 = (8, 8)
3 3
If the points (x+1, 2); (1,x + 2); x 1 1 , x 2 1 are collinear, then x is [Rajasthan PET 2002]
+ +
(a) 4 (b) 0 (c) – 4 (d) None of these
Let A = (x + 1, 2) ; B = (1, x + 2) ; C = x 1 1 , x 2 1 . A, B, C are collinear, if area of ∆ABC = 0
+ +
x +1 2 1 x −x 0
⇒ 1 x+2 1 =0⇒ 1 x+2 1 =0 (R1 → R1 − R2)
1 2 1 2
x +1 x +1 1 x +1 x +1 1
x 00
⇒ 1 x+3 1 =0 (C2 → C1 + C2) ⇒ x2(x + 4) = 0 ⇒ x = 0, − 4
1 3
x +1 x +1 1
Example: 19 The points (1, 1); (0, sec 2 θ ); (cosec 2θ ,0) are collinear for [Roorkee 1963]
Solution: (b)
(a) θ = nπ / 2 (b) θ ≠ nπ / 2 (c) θ = nπ (d) None of these
1 1 1 1
2 sec 2 θ 1 = 0 ⇒ 1(sec 2 θ ) + 1(cosec 2θ ) + 1(−cosec 2θ .sec 2 θ ) = 0
The given points are collinear, if Area of ∆ = 0
cosec 2θ 0 1
⇒ 1 1 1 =0 1 1 =0 ⇒0=0
cos2 θ + sin2 θ − sin2 θ . cos2 θ ⇒ sin2 θ .cos2 θ − sin2 θ .cos2 θ
Therefore the points are collinear for all value of θ , except only θ = nπ because at θ = nπ , sec2 θ =∞ (Not defined).
2 2
Example: 20 The points (0, 8/3) (1, 3) and (82, 30) are the vertices of [IIT 1983; Rajasthan PET 1988]
Solution: (d)
(a) An equilateral triangle (b) An isosceles triangle
(c) A right angled triangle (d) None of these
Here A = (0, 8 / 3), B = (1,3) and C = (82, 30)
AB = 1+1/9 = 10 / 9 , BC = (81)2 + (27)2 = 27 10 = 81 10 , AC = (82)2 + (30 − 8 / 3)2 = 82 10
9 9
Since AB + BC = (1 + 81) 10 = 82 10 = AC . ∴ Points A, B, C are collinear.
9 9
Transformation of Axes .
(1) Shifting of origin without rotation of axes : Let P ≡ (x, y) with respect to axes OX and OY.
Let O' ≡ (α, β ) with respect to axes OX and OY and let P ≡ (x' , y' ) with respect to axes O'X' and O'Y', where
OX and O'X' are parallel and OY and O'Y' are parallel. Y
Then x = x'+α, y = y' + β or x' = x − α, y' = y − β
Y′
Thus if origin is shifted to point (α, β ) without rotation of axes, then new
equation of curve can be obtained by putting x + α in place of x and y + β in (α,β ) P(x, y)
place of y. (x', y')
O′ y′
O
x′ X′
X
(2) Rotation of axes without changing the origin : Let O be the origin. Let P ≡ (x, y) with respect to axes
OX and OY and let P ≡ (x' , y' ) with respect to axes OX′ and OY′ where Y' Y
∠X' OX = ∠YOY' = θ
then x = x' cosθ − y' sinθ P (x,y)
y = x' sinθ + y' cosθ (yx′′,y′)
and x' = x cosθ + y sinθ θ y θ X'
y' = −x sinθ + y cosθ
x′
O xθ X
The above relation between (x, y) and (x' , y' ) can be easily obtained with the help of following table
x↓ y↓
x' → cos θ sinθ
y' → − sinθ cos θ
(3) Change of origin and rotation of axes : If origin is changed to Y
O'(α, β ) and axes are rotated about the new origin O' by an angle θ in the Y'
anticlock-wise sense such that the new co-ordinates of P(x, y) become θ
(x' , y' ) then the equations of transformation will be x = α + x' cosθ − y' sinθ O′ P(x, y)
O (x', y')
and y = β + x' sinθ + y' cosθ X'
θ
X
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