ALGEBRATIPS FOR MATHEMATICS OLYMPIAD

1. Every polynomial equation of degree n≥1 has exactly n roots.

2.If a c Complex polynomial equation with real coefficients has a complex root p+iq (p,q real numbers,q≠0) then it also has a complex root p-iq.

3. Pigeonhole Principle:If more than n objects are distributed in n boxes, then atleast one box has more than one object in it.

4. If a+b+c=0 then i) a2+b2+c2=-2(ab+bc+ca)

ii) a3+b3+c3=3abciii) a4+b4+c4=2(b2c2+c2a2+a2b2) =1/2.(a2+b2+c2)2.

5. For any three real numbers a,b,c 1. a2+b2+c2≥ab+bc+ca2. a4+b4+c4≥a2b2+b2c2+c2a2

3. a2b2+b2c2+c2a2≥abc (a+b+c)4. a4+b4+c4≥ abc (a+b+c )

5. a2+b2

a+b≥ 1

2(a+b )

6. a4+b4

a2+b2 ≥12

(a2+b2)

7. ( a+b+c3 )

2

> 13(ab+bc+ca)

8. √a2+b2 + √ x2+ y2 ≥√(x+a)2+( y+b)2

6. AM≥GM ≥HMa1+a2+…+an

n≥(a1 . a2 .…an)

1 /n≥( n1a1

+1a2

+……+1an )

7. If a1, a2 ,…. , an are n positive distinct real numbers then

1.a1m+a2

m+…+anm

n>( a1+a2+…+an

n )m

if m<0∨m>1

2.a1m+a2

m+…+anm

n<( a1+a2+…+an

n )m

if 0<m<1

3.b1a1

m+b2a2m+…+bnan

m

b1+b2+…+bn>( b1a1+b2a2+…+bnan

b1+b2+…+bn)m

if m<0∨m>1 where a1 , a2 ,…. , an and

b1, b2 ,…,bn∧mare rational numbers .

4.b1a1

m+b2a2m+…+bnan

m

b1+b2+…+bn<( b1a1+b2a2+…+bnan

b1+b2+…+bn)m

if 0<m<1

8. If a1, a2 ,…. , an are n positive distinct real numbers then

a1p+q+ r+a2

p+q+ r+…+anp+q+r

n>a1p+a2

p+…+anp

n.a1

q+a2q+…+an

q

n.a1r+a2

r+…+anr

n

9. Weierstrass Inequality :

1. If a1, a2 ,…. , an are n positive real numbers, then for n≥2 (1+a1) (1+a2 ) (1+a3 ) …..(1+an )>1+¿ a1+a2+…+an

2. If a1 , a2 ,…. , an are n positive real numbers, then for n¿1 (1−a1 ) (1−a2 ) (1−a3 ) …..(1−an )>1−¿ a1−a2…−an

10. Tchebychef’s Inequality: If a1 , a2 ,…. , an and b1 , b2 ,…,bn realnumbers such that a1≤a2≤….≤an and b1≤b2≤…≤bn then n(b1a1+b2a2+…+bnan )≥ (a1+a2+…+an ) (b1+b2+…+bn )

(or)b1a1+b2a2+…+bnan

n≥a1+a2+…+an

n.b1+b2+…+bn

n

11. CAUCHY-SCHWARZ Inequality:If a,b,c and x,y,z are any real numbers ( positive, negative or zero) then

(ax+by+cz )2≤ (a2+b2+c2 ) (x2+ y2+z2 ) with equality iff a:b:c::x:y:z

12. DESCARTE’S RULE OF SIGNS: Suppose P(x) is a polynomial whose terms are arranged in descending powers of x of the variable. Thus , the number of positive real zeros of P(x) is the same as the number of changes in sign of the coefficients of the terms or less than this by an even number.

The number of negative real zeros of P(x) is the same as the number of changes in sign

of the coefficients of the terms of P(-x) or is less than this number by an even number.13. If the rational number

pq (where p,q are integers q≠0 , (p ,q )=1 , i . e ;coprimes ¿ is a

root of the equation a0 xn+a1 x

n−1+…+an=0, where a1 , a2 ,…. , an are integers and an ≠o , then , p is a divisor of an and q is a divisor of a0

14. FUNCTIONAL EQUATIONS: An equation involving an unknow3n function is called a functional equation.

1. If α ,β are the rootsof a x2+bx+c=0 , thenα+ β=−ba

,αβ= ca

2. If α ,β , γ are the rootsof a x3+b x2+cx+d=0 , then

α+β+γ=−ba

,∑ αβ= caαβγ=−d

a3. If α ,β , γ , δ are theroots of ax 4+b x3+c x2+dx+e=0 , then

α+β+γ+δ=−ba

,∑ αβ= ca;∑ αβγ=−d

aαβγ=−d

a

MATHS OLYMPIAD QUESTIONS AND SOLUTIONS:ALGEBRA

Q.NO QUESTIONS/SOLUTIONS1. Find the maximum number of positive and negative real roots of the equation

x4+x3+x2−x−1=0.

Solution:

Let f(x)=x4+x3+x2−x−1The signs are +,+,+,-,-,There is only one change of sign from positive to negative.∴ there is only one positive root. F(-x)=(- x¿¿4+(−x)3+(−x)2−(−x)−1 =x4−x3+x2+x−1The signs are +, -, +, +, -There are three changes in sign∴ three negative roots.

2. If α ,β , γ , δ , be the roots of the equation x4+ p x3+q x2+rx−s=0. Show that

(1+α 2¿ (1+β2 ) (1+γ 2 ) (1+δ2 )=¿(1-q+s)2 +(p-r)2.

Solution:Since α ,β , γ , δ are the roots of the equation x4+ p x3+q x2+rx−s=0,

x4+ p x3+q x2+rx−s=(x−α )( x−β)(x−γ )(x−δ )If x=i , i4+ p i3+q i2+ri−s=(i−α )(i−β)(i−γ)(i−δ )

1-pi-q+ri+s=(i−α )(i−β)(i−γ )(i−δ ) ( 1-q+s)-i(p-r)=(i−α )(i−β)(i−γ)(i−δ)………..(i)

If x=-I, ¿ 1+pi-q-ri+s=(i+α )(i+β )(i+γ )(i+δ ) ( 1-q+s)+i(p-r)¿( i+α )(i+ β)(i+γ )( i+δ)………..(ii)(i)x(ii)¿>¿ (1-q+s)2+(p-r)2=(i2−α 2) (i2−β2) (i2−γ2)(i2−δ 2) =(1+α 2) (1+β2) (1+γ2)(1+δ2).

3. Show that (x−1)2 is a factor of xn−nx+n−1.

Solution:

Let f(x)= xn−nx+n−1. f (1)=1-n+n-1=0∴ ( x-1) is a factor. f(x)= xn−nx+n−1. = xn−1−nx+n. =xn−1−n(x−1). =(x-1){xn−1+ xn−2+…+1−n } =(x-1)g(x)………………(i)Now g(x)=xn−1+ xn−2+…+1−n g (1)=1+1+1+….1-n =0.∴ ( x-1) is a factor of g(x)∴ ( x-1)2 is a factor of f(x) {by (i)}

4. If sin θ+cosθ=1 ,(sin θ>0 ,cos θ>0). Show that (1+cosecθ ¿(1+secθ)≥9.

Solution: We know that AM≥GM.

sin θ+cosθ

2≥√sin θ cosθ.

12≥√sin θ cosθ.

14≥ sinθ cosθ

Let If sin θ=x , cosθ=¿y

(1+cosecθ ¿¿)=(1+1x¿(1+ 1

y)

=(x+1)( y+1)

xy

=(xy+x+ y+1)

xy.

=xy+2xy

≥ p (say )

xy+2≥ pxy 2≥ (p−1 ) xy

2

p−1≥ xy

but 14≥xy

∴ 2

p−1=¿

14 .

P=9.

5.

If

4x

2x+ y=8and

9x+ y

35 y =243find the value of x-y.

SOLUTION:

4x

2x+ y=8

(22 )x

2x+ y =8

22 x

2x+ y=8

22 x−x− y=8

2x− y=(2 )3

x− y=3

6. A po The Polynomial p (x) leaves a remainder 3 when divided by x – 1 and a remainder

5 when divided by x-3. Find the remainder when p(x) is divided by (x-1) (x-3).

SOLUTION∵ The polynomial gives a remainder 3 on division by

x –1.

Let, p (x) = k(x-1) + 3

= kx – k + 3

Now,

¿Remainder = 2k + 3

But,

2k + 3 = 5

2k = 2

k = 1

P(x) = k (x-1) + 3

= 1 (x-1) + 3

=x-1 + 3

= x +2

Now ,

(x-1) (x-3)

= x2 – 4x + 3

Dividing p(x) by x2 – 4x + 3

¿Hence, the required remainder = x + 2.

7. Show that there do not exist any distinct natural numbers a, b, c, d such that

a3 + b3 = c3 + d3 and a + b = c + d

SOLUTION: a3 + b3 = c3 + d3 and a + b = c + d

a3 + b3 = c3 + d3

(a+b) (a2-ab+b2) = (c+d) (c2 - cd + d2)

a2 – ab + b2 = c2 – cd + d2

(a+b)2 – 3ab = (c + d)2 – 3cd

-3ab = -3cd

ab = cd

√ab = √cd

Given a + b = c + d

∴

a+b2

= c+d2

And √ab = √cd

Here, A.M. of a, b is equal to AM of c, d

And G.M. of a, b is equal to G.M. of c, d

This is only possible when a,b and c,d are equal

But, they must be distinct

∴ It is not possible

8. If a,b,c,d be any four positive real numbers ,then prove that ab +

bc +

cd +

de ≥ 4

SOLUTION: Applying the inequality of the means A.M ≥ G.M to the four possible

numbers ab ,

bc ,

cd ,

de we have ¼(

ab +

bc +

cd +

de¿ ≥(

ab +

bc +

cd +

de¿1/4

→ab +

bc +

cd +

de≥4Typeequationhere .

9. If a,b,c,d are four positive real numbers such that abcd=1 Prove that (1+a)(1+b)(1+c)

(1+d)≥16

SOLUTION: Since A.M≥G.M

1+a2 ≥(1.a)1/2

1+b2 ≥(1.b)1/2

1+c2 ≥(1.c)1/2

1+d2 ≥(1.d)1/2

Multiplying the inequalities

(1+a)(1+b)(1+c)¿¿ ≥ (abcd)1/2

(1+a)(1+b)(1+c)(1+d)≥16

10.

Solution:

Prove that nn ( n+1n )

2n

˃(n! )3

Consider positive numbers 13,23,…,n3

Since AM˃GM,(13+23+…+n3 )/n˃(13.23…n3)1/n

( n(n+1)2 )

2

n˃ ¿

By simplifying,

nn ( n+1n )

2n

˃(n! )3

11. If x , y, z are three real numbers such that x + y + z =4 and x2 + y2 +z2 = 6 ,

then show thtat each of x , y ,z lies in the [23, 2] . Can x attains

extreme value

23 ?

Solution:

We have y + z = 4 - x and y2 +z2 = 6 - x2

From Cauchy’s Schwarz inequality we get y2 +z2 ¿

12 ( y + z )2

Hence 6 - x2 ¿

12 ( 4-x)2 .This simplifies to (3x - 2) (x – 2) ≤ 0.

Hence we have

23 ≤ x≤ 2. Suppose x =2 then y +z =2 , y2 +z2 = 2 which

has solution y = z = 1. Since the given relations are symmetric in x , y , z similar assertions hold for y and z.

12. If squares of the roots of x4+bx2+cx+d=0 are α,β,γ,δ, then prove that:64αβγδ−[4∑αβ−(∑α)2]2=0.

Solution:x4+bx2+cx+d=0......(1)Let y be the root of transformed equation. y=x^2

x4+bx2+d=−cx since bring all even powers one side and odd powers other side.Squaring on both sides, we get(x4+bx2+d)2=c2x2

(y2+by+d)2=c2y since y=x2

y4+b2y2+d2+2by3+2bdy+2dy2=c2yy4+2by3+(b2+2d)y2+(2bd−c2)y+d2=0...(2)It is the equation whose roots are squares of roots of x4+bx2+cx+d=0now for (2)α,β,γ,δ are the roots.∑α =-2b∑αβ=b2+2d∑αβγ =-(2bd-c^2)αβγδ = d2

Now plug the values in required function.64αβγδ−[4∑αβ−(∑α)2]2 = 64d2−(4(b2+2d)−(−2b)2)2

=64d2−(4b2+8d−4b2)2

=64d2−(8d)2

=64d2−64d2

= 0Hence it is proved.

13. Prove that 3a4−4a3b+b4≥0 for all real numbers a and b. Solution:

3a4−4a3b+b4 =a4+2a4+b4−4a3b =a4+b4+2a4−4a3b =(a2)2+(b2)2+2a4−4a3b = (a2)2+(b2)2−2a2b2+2a2b2+2a4−4a3b = (a2−b2)2+2a2b2+2a4−4a3b = (a2−b2)2+2a2(b2+a2−2ab) = (a2−b2)2+2a2(a−b)2≥0 for all a, b As square is always positive and sum of the squares also positive.Hence proved.

14. If a polynomial is divided by x-1 and x-2, we obtain remainder 2 and 1

respectively. Find the remainder if it is divided by (x-1) (x-2).

Solution:

As, on dividing by (x-1) the polynomial leaves a remainder of 2.

So, the polynomial must be, p(x) = k (x-1) + 2 for a real K.

p(x) = kx – k + 2

Now, dividing p(x) by (x-2)

¿ i.e. remainder = k + 2 But, remainder must be 1

∴k+2 = 1 k = -1

∴p(x) = (-1) (x-1) + 2

= - x + 3

Now (x-1) (x-2)

= x2 - 3x + 2

Dividing p(x) by x2 – 3x + 2,

x2−3x+2√−x+3 i.e. the remainder is (-x+3) only = -x+3

15.If √a−x+√b−x+√c−x=0

Prove that (a + b + c + 3x) (a + b + c - x) = 4(bc + ca + ab)

Solution:

√a−x+√b−x+√c−x=0

√a−x+√b−x=−√c−x

(a+b)+(b−x )+2√a−x−√b−x=c−x

a + b – c – x = - 2 √a−x √b−x

a2 + b2 + c2 + 2ab – 2bc – 2ca – 2x (a+b-c) + x2 = 4 {ab – x(a+b)

+ x2}

(a + b + c)2 + 2x (a + b + c ) – 3x2 = 4 (ab + bc + ca)

(a + b + c + 3x) (a + b + c - x) = 4 (bc + ca + ab)

16.

Let a+b+c =1 and ab + bc + ca = 1

3

a, b, c are real numbers.

Find the value of

(i)

ab

+ bc

+ ca (ii)

ab+a

+ bc+a

+ ca+1

Solution :

(a−b )2 + (b−c )2 + (c−a )2

= 2 (a2 + b2 + c2 −ab − bc − ca )

=2 [ ( a + b + c )2 −3 (ab + bc + ca ) ]

=2 [1 −3 ⋅ 1

3 ]= 0

So, a = b, b = c, and c = a

So, a+b+c=1, gives a=b=c=1

3

So,

ab

+ bc

+ ca

= 1+1+1

= 3

and

ab+1

+ bc+a

+ ca+1

=

1 /34 /3

+ 1 /34 /3

+ 1 /34 /3

=

34

17. Factorize

(y-z)5 + (z-x)5 + (x-y)5

Solution:

Putting a, b, c for y-z, z-x and x-y respectively.

The expression reduces to

a5 + b5 + c5

Now a + b + c = y-z + z – x + x – y = 0 a + b = - c

a2 + b2 + 2ab = c2

- c5 = (a + b)5

= a5 + b5 + 5a4b + 10a3b2 + 10a2b2 + 5ab4

- (a5 + b5 + c5) = 5ab (a3 + b3) + 10a2b2 (a+b)

= 5ab (a + b) {(a2 – ab + b2) + 2ab}

= 5ab (-c) {a2 + b2 + ab)

= - 5abc (a2 + b2 + ab)

putting back the values of a, b and the expression

= 5 (y-z) (z-x) (x-y) {(y-z)2 + (z-x)2 + (y-z) (z-x)}

= 5 (y-z) (z-x) (x-y) (y2+z2 -2yz + z2–2zx+yz+2x-x2 – xy)

(y-z)5 + (z-x)5 + (x-y)5 = 5 (y-z) (z-x) (x-y) (x2+ y2+z2 -yz -zx – xy)

18. If f(x) = ax7 + bx5 + cx3 – 6, and f(-9) = 3, find f(9).

Find the value of

(2002)3−(1002)3−(1000 )3

3 x (1002) x (1000 )

Solution:

(a) F (x) = ax7 + bx5 + cx3 - 6

f (-9) = 3

∴ f (-9) = a (-9)7 + b (-9)5 + c (-9)3 – 6

a (-9)7 + b (-9)5 + c (-9)3 – 6 = 3

-a (9)7 + b (-9)5 + -c (9)3 = 9

-a (9)7 - b (9)5 - c (9)3 = 9

a (9)7 + b (9)5 + c (9)3 = - 9

∴ f (9) = a (9)7 + b (9)5 + c (9)3 – 6

= - 9 - 6

= -15 Ans.

(b)

(2002)3 − (1002 )3 − (1000)3

3 x (1002 ) x (1000)

=

(2002 − 1002) [(2002)2 + (2002 ) (1002) + (1002)2 ] −(1000 )3

3 x (1002) x (1000 )

=

1000 [(2002 − 1002)2 + 3 (2002 ) (1002) ]−(1000)3

3 x (1002 ) x (1000 )

=

(1000)2 + 3 (2002) (1002) −(1000 )2

3 x (1002)

=

3 (2002) (1002)3 x (1002 )

= 2002 Ans.

19. (a) Solve : x2 + xy + y2 = 19

x2 - xy + y2 = 49

(b) The quadratic polynomials p(x) = a (x-3)2 + bx + 1 and q (x) = 2x2 + c

(x-2) + 13 are equal for all values of x. Find the values of a, b and c.

Solution:

(a) x2 + xy + y2 = 19

xy = 19 – x2 – y2

\ x2 – xy + y2 = 49

x2 – (19-x2-y2) + y2 = 49

2x2 + 2y2 = 49 + 19

x2 + y2 =

682

x2 + y2 = 34

\ x2 + xy + y2 = 19

34 + xy = 19

xy = 19 – 34

= -15

\ (x+ y)2 = x2 + y2 + 2xy

= 34 + 2x (-15)

= 34 – 30

= 4

x + y = 4

(x-y)2 = x2 + y2 – 2xy

= 34 – 2 (-15)

= 34 + 30

= 64

x - y = 8

x + y = 4

x - y = 8

2x = 12

x = 6

6+y = 4

y = -2

x = 6

y = -2

3(b) p (x) = a(x-3)2 + bx + 1

q (x) = 2x2 + c(x-2) + 13

p(3) = a (3-3)2 + 3b + 1

= 3b + 1

q(3) = 2 x 32 +c(3-2) + 13

= 18 + c + 13

= 31 + c

3b + 1 = 31 + c

3b – c = 30 (i)

p(2) = a (2-3)2 + 2b + 1

= a+2b + 1

q(2) = 2 x 22 + c (2-2) + 13

= 4 + 13

=17

a + 2b + 1 = 17

a + 2b = 16 (ii)

p(o) = a (0-3)2 + 6(0) + 1

= 9a +1

q(o) = 2x02 + 6(0-2) + 13

= -2c + 13

9a + 1 = -2c + 13

9a + 2c = 12 (iii)

We have the following three eqn :

3b – c = 30

a + 2b = 16

9a + 2c = 12

6b – 2c = 60

+ 9a + 2c = 12

9a + 6b = 72

3a + 2b = 24

3a + 2b = 24

a + 2b = 16

2a = 8

a = 4

4 + 2b = 16

2b = 16 – 4

b = 16 – 4

b =

122 = 6

9 x 4 + 2c = 12

2c = 12 – 36

c = −24

2 = -12

a = 4

b = 6

c = -12

20. If a+b+c=1 ,a2+b2+c2=9, a3+b3+c3=1 find1a+

1b+

1c

SOLUTION:

Since (a+b+c)2=a2+b2+c2+2(ab+bc+ca)

ab+bc+ca =(1-9)/2 =-4

a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)

=1(9-(-4))

=13

So,abc=−13+1

3 =-4

1a+

1b+

1c =

bc+ca+ababc

=(-4)/(-4)

=1

21. The product of two of the four roots of x4-20x3+kx2+590x-1992 = 0 find k.

Solution:

If the roots be α,β,γ , δ then

α +β+γ+δ = 20

(α +β)(γ+δ )+ α β+γδ =k

(α +β)γδ +(γ+δ ) α β =-590

α βγδ =-1992

α β=24

γδ =-1992/24

=-83

if α +β=x and γ+δ =y,

x+y=20

-83x+24y=-590

By solving x=10,y=10

K=(α +β)(γ+δ )+ α β+γδ

=10.10+24-83

=41

22. Prove that 1< 1

1001 +1

1002+ 1

1003 +…+1

3001

Solution:

Consider the numbers 1001,1002,..,3000,3001

A.M= 1001+1002+…+3001

2001

=2001

2 (1001+3001)1

2001

=4002

2

=2001

HM = ¿ +1

1002+ 1

1003 +…+1

3001 ) -12001

=2001

11001

+ 11002

+ 11003

+…+ 13001

Since AM>HM,

2001>2001

11001

+ 11002

+ 11003

+…+ 13001

1>1

11001

+ 11002

+ 11003

+…+ 13001

11001 +

11002

+ 11003 +…+

13001

>1

Hence proved.

23. Let x be the set of positive integers ≥ 8.Let f:x→x be a function,such that

f(x+y) =f(xy) for all x≥4,y≥4.If f(8)=9,determine f(9)

Solution:

f(9)=f(4+5)

=f(20)

=f(16+4)

=f(64)

=f(8.8)

=f(8+8)

=f(16)

=f(4.4)

=f(4+4)

=f(8)

=9

Thus f(9)=9.

24. Find the smallest value of the expression

4 x2+8 x+136 (1+x )

for x≥0

Solution:

4 x2+8 x+136 (1+x )

¿ 4 (x¿¿2+2x+1)+96 (1+x )

¿

¿4 (x+1)2+9

6 (1+x )

=4 (x+1)

6 +

96(x+1)

¿2 ( x+1 )

3 +

32 ( x+1 )

2 ( x+1 )3

+3

2 ( x+1 ) ≥2

Since it is of the form a+1a where a>0

Since (a-1)2≥0 and a>0

(a-1)2/a ≥0

(a2-2a+1)/a≥0

a-2+(1/a)≥0

a+1/a≥2

Thus smallest value of the expression is 2.

25. Prove that without using tables or calculators,1993>1399.

Solution:Consider ¿

∴¿ 1913

¿¿8>24 ¿13.

Thus 198>139

¿ 1988>1399

∴1993>1399.

26. If2 x+3 y=7 , and x≥0 , y ≥0,then find the greatest value of x3 y 4

Solution:

Let Z=x3 y 4=( 2x3 )

3

( 3 y4 )

4

(32 )

3

( 43 )

4

…………….(i)

∴ Z will have maximum value when ( 2x3 )

3

( 3 y4 )

4

is maximum.

But ( 2x3 )

3

( 3 y4 )

4

is the product of 3+4=7 factors,

the sum of which =3( 2x3 )+4( 3 y

4 )=2x+3 y=7 (constant )

∴( 2x3 )

3

( 3 y4 )

4

will be maximum if all the factors are equal,

i.e if 2x3

=3 y4

=2 x+3 y3+4

=77=1

∴ from (i) the maximum value of Z=( 2x3 )

3

( 3 y4 )

4

(1)3 (1 )4

=278X 256

81

=323

27. Find the minimum value of bcx+cay+abz when xyz=abcSolution:We have xyz=abc (bcx)(cay)(abz) =a3b3c3=Z n =3 minimum value of bcx+cay+abz=n¿ =3¿)1/3 =3abc.

BY CHENNAI REGION: