Download Algebra – Olympiad problems

ALGEBRA
TIPS 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=3abc
iii)
a4+b4+c4=2(b2c2+c2a2+a2b2) =1/2.(a2+b2+c2)2.
5.
For any three real numbers a,b,c
𝑎2 + 𝑏 2 + 𝑐 2 ≥ 𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎
𝑎4 + 𝑏 4 + 𝑐 4 ≥ 𝑎2 𝑏 2 + 𝑏 2 𝑐 2 + 𝑐 2 𝑎2
𝑎2 𝑏 2 + 𝑏 2 𝑐 2 + 𝑐 2 𝑎2 ≥ 𝑎𝑏𝑐(𝑎 + 𝑏 + 𝑐)
𝑎4 + 𝑏 4 + 𝑐 4 ≥ 𝑎𝑏𝑐(𝑎 + 𝑏 + 𝑐)
1.
2.
3.
4.
5.
6.
𝑎2 +𝑏 2
𝑎+𝑏
𝑎4 +𝑏 4
𝑎2 +𝑏 2
1
≥ 2 (𝑎 + 𝑏)
1
≥ 2 (𝑎2 + 𝑏 2 )
𝑎+𝑏+𝑐 2
1
7.
(
8.
√𝑎2 + 𝑏 2 + √𝑥 2 + 𝑦 2 ≥ √(𝑥 + 𝑎)2 + (𝑦 + 𝑏)2
3
) > 3 (𝑎𝑏 + 𝑏𝑐 + 𝑐𝑎)
𝑨𝑴 ≥ 𝑮𝑴 ≥ 𝑯𝑴
6.
𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛
𝑛
≥ (𝑎1 . 𝑎2 . … 𝑎𝑛 )1/𝑛 ≥ (
)
1
1
1
𝑛
+
+
⋯
…
+
𝑎1 𝑎2
𝑎𝑛
If 𝑎1 , 𝑎2 , … . , 𝑎𝑛 are n positive distinct real numbers then
7.
1.
2.
𝑚
𝑎1𝑚 +𝑎2𝑚 +⋯+𝑎𝑛
𝑛
𝑚
𝑎1𝑚 +𝑎2𝑚 +⋯+𝑎𝑛
𝑛
>(
<(
𝑎1 +𝑎2 +⋯+𝑎𝑛 𝑚
𝑛
)
𝑎1 +𝑎2 +⋯+𝑎𝑛 𝑚
𝑛
)
𝑖𝑓 𝑚 < 0 𝑜𝑟 𝑚 > 1
𝑖𝑓 0 < 𝑚 < 1
3.
𝑚
𝑏1 𝑎1𝑚 +𝑏2 𝑎2𝑚 +⋯+𝑏𝑛 𝑎𝑛
𝑏1 +𝑏2 +⋯+𝑏𝑛
𝑏1 𝑎1 +𝑏2 𝑎2 +⋯+𝑏𝑛 𝑎𝑛 𝑚
>(
)
𝑏1 +𝑏2 +⋯+𝑏𝑛
𝑖𝑓 𝑚 < 0 𝑜𝑟 𝑚 > 1 where
𝑎1 , 𝑎2 , … . , 𝑎𝑛 and 𝑏1 , 𝑏2 , … , 𝑏𝑛 𝑎𝑛𝑑 𝑚 𝑎𝑟𝑒 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠.
4.
8.
𝑚
𝑏1 𝑎1𝑚 +𝑏2 𝑎2𝑚 +⋯+𝑏𝑛 𝑎𝑛
𝑏1 +𝑏2 +⋯+𝑏𝑛
)
𝑏1 +𝑏2 +⋯+𝑏𝑛
𝑖𝑓 0 < 𝑚 < 1
If 𝑎1 , 𝑎2 , … . , 𝑎𝑛 are n positive distinct real numbers then
𝑝+𝑞+𝑟
𝑎1
𝑝+𝑞+𝑟
+𝑎2
𝑛
9.
𝑏1 𝑎1 +𝑏2 𝑎2 +⋯+𝑏𝑛 𝑎𝑛 𝑚
<(
𝑝+𝑞+𝑟
+⋯+𝑎𝑛
𝑝
>
𝑝
𝑝
𝑞
𝑞
𝑞
𝑟
𝑎1 +𝑎2 +⋯+𝑎𝑛 𝑎1 +𝑎2 +⋯+𝑎𝑛 𝑎1𝑟 +𝑎2𝑟 +⋯+𝑎𝑛
𝑛
.
𝑛
.
𝑛
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 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑎1 ≤ 𝑎2 ≤ … . ≤ 𝑎𝑛
and 𝑏1 ≤ 𝑏2 ≤ ⋯ ≤ 𝑏𝑛 𝑡ℎ𝑒𝑛
n(𝑏1 𝑎1 + 𝑏2 𝑎2 + ⋯ + 𝑏𝑛 𝑎𝑛 ) ≥ (𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 )(𝑏1 + 𝑏2 + ⋯ + 𝑏𝑛 )
(or)
𝑏1 𝑎1 + 𝑏2 𝑎2 + ⋯ + 𝑏𝑛 𝑎𝑛
𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 𝑏1 + 𝑏2 + ⋯ + 𝑏𝑛
≥
.
𝑛
𝑛
𝑛
11.
12.
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 + 𝑏 2 + 𝑐 2 )(𝑥 2 + 𝑦 2 + 𝑧 2 )
with equality iff a:b:c::x:y:z
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.
14.
If the rational number
𝑝
𝑞
(where p,q are integers q≠ 0, (𝑝, 𝑞) = 1, 𝑖. 𝑒; 𝑐𝑜𝑝𝑟𝑖𝑚𝑒𝑠 )
is a root of the equation 𝑎0 𝑥 𝑛 + 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
FUNCTIONAL EQUATIONS:
An equation involving an unknow3n function is called a functional equation.
1.
2.
3.
−𝑏
𝑐
If 𝛼, 𝛽 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 , 𝑡ℎ𝑒𝑛 𝛼 + 𝛽 = 𝑎 , 𝛼𝛽 = 𝑎
If 𝛼, 𝛽, 𝛾 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 = 0 , 𝑡ℎ𝑒𝑛
−𝑏
𝑐
−𝑑
𝛼+𝛽+𝛾 =
,
∑ 𝛼𝛽 =
𝛼𝛽𝛾 =
𝑎
𝑎
𝑎
If 𝛼, 𝛽, 𝛾, 𝛿 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑟𝑜𝑜𝑡𝑠 𝑜𝑓 𝑎𝑥 4 + 𝑏𝑥 3 + 𝑐𝑥 2 + 𝑑𝑥 + 𝑒 = 0 , 𝑡ℎ𝑒𝑛
−𝑏
𝑐
−𝑑
−𝑑
𝛼+𝛽+𝛾+𝛿 =
,
∑ 𝛼𝛽 = ; ∑ 𝛼𝛽𝛾 =
𝛼𝛽𝛾 =
𝑎
𝑎
𝑎
𝑎
MATHS OLYMPIAD QUESTIONS AND SOLUTIONS:
ALGEBRA
Q.NO
1.
QUESTIONS/SOLUTIONS
Find the maximum number of positive and negative real roots of the equation
𝑥 4 + 𝑥 3 + 𝑥 2 − 𝑥 − 1 = 0.
Solution:
Let f(x)= 𝑥 4 + 𝑥 3 + 𝑥 2 − 𝑥 − 1
The 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 − 1
The signs are +, -, +, +, There are three changes in sign
∴ three negative roots.
2.
If 𝛼, 𝛽, 𝛾, 𝛿, be the roots of the equation 𝑥 4 + 𝑝𝑥 3 + 𝑞𝑥 2 + 𝑟𝑥 − 𝑠 = 0. Show that
(1+𝛼 2 )(1 + 𝛽 2 )(1 + 𝛾 2 )(1 + 𝛿 2 ) =(1-q+s)2 +(p-r)2.
Solution:
Since 𝛼, 𝛽, 𝛾, 𝛿 𝑎𝑟𝑒 the roots of the equation 𝑥 4 + 𝑝𝑥 3 + 𝑞𝑥 2 + 𝑟𝑥 − 𝑠 = 0,
𝑥 4 + 𝑝𝑥 3 + 𝑞𝑥 2 + 𝑟𝑥 − 𝑠 = (𝑥 − 𝛼)(𝑥 − 𝛽)(𝑥 − 𝛾)(𝑥 − 𝛿)
If x=i ,
𝑖 4 + 𝑝𝑖 3 + 𝑞𝑖 2 + 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,
(−𝑖)4 + 𝑝(−𝑖)3 + 𝑞(−𝑖)2 + 𝑟(−𝑖) − 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)………..(ii)
(i)x(ii)=>
(1-q+s)2+(p-r)2=(𝑖 2 − 𝛼 2 ) (𝑖 2 − 𝛽 2 ) (𝑖 2 − 𝛾 2 )(𝑖 2 − 𝛿 2 )
=(1 + 𝛼 2 ) (1 + 𝛽 2 ) (1 + 𝛾 2 )(1 + 𝛿 2 ).
3.
Show that (𝑥 − 1)2 is a factor of 𝑥 𝑛 − 𝑛𝑥 + 𝑛 − 1.
Solution:
Let f(x)= 𝑥 𝑛 − 𝑛𝑥 + 𝑛 − 1.
f (1)=1-n+n-1=0
∴ ( x-1) is a factor.
f(x)= 𝑥 𝑛 − 𝑛𝑥 + 𝑛 − 1.
= 𝑥 𝑛 − 1 − 𝑛𝑥 + 𝑛.
=𝑥 𝑛 − 1 − 𝑛(𝑥 − 1).
=(x-1){𝑥 𝑛−1 + 𝑥 𝑛−2 + ⋯ + 1 − 𝑛}
=(x-1)g(x)………………(i)
𝑛−1
Now g(x)= 𝑥
+ 𝑥 𝑛−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 + cos 𝜃 = 1, (sin 𝜃 > 0, cos 𝜃 > 0).
Show that (1+cosec𝜃)(1 + 𝑠𝑒𝑐𝜃) ≥ 9.
Solution:
We know that AM ≥ GM.
sin 𝜃+cos 𝜃
2
≥ √sin 𝜃 cos 𝜃.
1
≥ √sin 𝜃 cos 𝜃.
1
≥ sin 𝜃 cos 𝜃
4
Let If sin 𝜃 = 𝑥, cos 𝜃 =y
2
1
1
(1+cosec𝜃)(1 + 𝑠𝑒𝑐𝜃)=(1+ 𝑥 ) (1 + 𝑦)
(𝑥+1)(𝑦+1)
=
𝑥𝑦
(𝑥𝑦+𝑥+𝑦+1)
=
𝑥𝑦
𝑥𝑦+2
=
𝑥𝑦
.
≥ 𝑝(𝑠𝑎𝑦)
𝑥𝑦 + 2 ≥ 𝑝𝑥𝑦
2≥ (𝑝 − 1)𝑥𝑦
2
𝑝−1
≥ 𝑥𝑦
1
but
∴
4
2
≥ 𝑥𝑦
𝑝−1
=
1
P=9.
5.
4x
9xy
If x  y  8 and 5 y  243 find the value of x-y.
3
2
4x
8
2xy
SOLUTION:

(2 2 ) x
8
2xy
22x
 xy  8
2

2 2 x x  y  8
.
4
6.
A po

2 x y  (2)3

xy3
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,
k
x  3 kx  k  3
kx  3k
2k  3
Remainder = 2k + 3
But,
2k + 3 = 5
 2k = 2
k=1
= 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
0
x  4x  3 x  2
00
x2
2
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b
cd

2
2

ab =
And
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.
𝑎
𝑏
𝑐
𝑑
If a,b,c,d be any four positive real numbers ,then prove that 𝑏 + 𝑐 +𝑑 + 𝑒 ≥ 4
SOLUTION: Applying the inequality of the means A.M ≥ G.M to the four possible
numbers
𝑎
𝑏
𝑎
𝑏
𝑐
𝑏 𝑐
𝑑
𝑎
𝑏
𝑐
𝑑
𝑎
𝑏
𝑐
𝑑
, 𝑐 ,𝑑 , 𝑒 we have ¼( 𝑏 + 𝑐 +𝑑 + 𝑒 ) ≥( 𝑏 + 𝑐 +𝑑 + 𝑒 )1/4
𝑑
→𝑏 + 𝑐 +𝑑 + 𝑒 ≥4Type equation here.
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+𝑎
2
1+𝑏
2
1+𝑐
2
≥(1.a)1/2
≥(1.b)1/2
≥(1.c)1/2
1+𝑑
2
≥(1.d)1/2
Multiplying the inequalities
(1+a)(1+b)(1+c)(1+d
16
≥ (abcd)1/2
(1+a)(1+b)(1+c)(1+d)≥16
10.
n+1 2n
Prove that nn (
n
) ˃(n!)3
Solution:
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,
n+1 2n
nn (
n
) ˃(n!)3
1
˃( (n!)3 )n
11. If x , y, z are three real numbers such that x + y + z =4 and x2 + y2 +z2 = 6 ,
2

then show thtat each of x , y ,z lies in the  , 2 . Can x attains
3

2
extreme value ?
3
Solution:
We have y + z = 4 - x and y2 +z2 = 6 - x2
From Cauchy’s Schwarz inequality we get y2 +z2 
1
( y + z )2
2
1
( 4-x)2 .This simplifies to (3x - 2) (x – 2)  0.
2
2
Hence we have
 x  2. Suppose x =2 then y +z =2 , y2 +z2 = 2 which
3
has solution
y = z = 1. Since the given relations are symmetric in x , y , z similar
assertions hold for y and z.
Hence 6 - x2 
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=c2y
y4+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=0
now 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
=0
Hence 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)
K
x  2 kx  k  2
kx  2k
k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,
x 2  3x  2  x  3 i.e. the remainder is (-x+3) only = -x+3
a x  bx  cx 0
15. If
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

a2 + b2 + c2 + 2ab – 2bc – 2ca – 2x (a+b-c) + x2 = 4 {ab – x(a+b)
ax bx
+ x2}

(a + b + c)2 + 2x (a + b + c ) – 3x2 = 4 (ab + bc + ca)

16.
(a + b + c + 3x) (a + b + c - x) = 4 (bc + ca + ab)
Let a+b+c =1 and ab  bc  ca 
1
3
a, b, c are real numbers.
Find the value of
(i)
a
b
c
 
b
c
a
(ii)
a
b
c


ba
ca
a 1
Solution :
a  b2
 b  c 2  c  a 2
=
2  a 2  b 2  c 2  ab  bc  ca 
=
2 a  b  c2  3 ab  bc  ca 


=
1

2 1  3  
3

=
0

So, a = b,

b = c, and c = a
So, a+b+c=1, gives a  b  c 
So,
a
b
c
 
b
c
a
=
1+1+1
1
3
=
3
and
a
b
c


b 1 c  a
a 1
=
1/ 3
1/ 3
1/ 3


4/3
4/3
4/3
=
3
4
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)
(a)18. If f(x) = ax7 + bx5 + cx3 – 6, and f(-9) = 3, find f(9).
(b)
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 =
68
2
 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
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
(i)
 a + 2b = 16
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
(ii)
 2b = 16 – 4
 b = 16 – 4
b=
12
2
=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 find
1 1 1
𝑎
+ 𝑏 +𝑐
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
−13+1
So,abc=
3
=-4
1 1 1
𝑎
+ 𝑏 +𝑐 =
𝑏𝑐+𝑐𝑎+𝑎𝑏
𝑎𝑏𝑐
=(-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,  ,  then
α +β+    = 20
(α +β)(    )+ α β+  =k
(α +β)  +(    ) α β =-590
α β  =-1992
α β=24
 =-1992/24
=-83
if α +β=x and    =y,
x+y=20
-83x+24y=-590
By solving x=10,y=10
K=(α +β)(    )+ α β+ 
=10.10+24-83
=41
22. Prove that 1< 1 + 1 + 1 +…+ 1
1001 1002
1003
3001
Solution:
Consider the numbers 1001,1002,..,3000,3001
A.M=
1001+1002+⋯+3001
2001
2001
=
=
1
(1001+3001)2001
2
4002
2
=2001
1
1
1
1
HM = (1001 +1002 + 1003 +…+3001
) -12001
2001
=
1
1
1
1
+
+
+⋯+
1001 1002 1003
3001
Since AM>HM,
2001
2001>
1
1
1
1
+
+
+⋯+
1001 1002 1003
3001
1
1>
1
1
1
1
+
+
+⋯+
1001 1002 1003
3001
1
1
1
1
+
+ 1003 +…+3001 > 1
1001 1002
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
4𝑥 2 +8𝑥+13
6(1+𝑥)
for x≥0
Solution:
4𝑥 2 +8𝑥+13
6(1+𝑥)
=
4(𝑥 2 +2𝑥+1)+9
6(1+𝑥)
=
=
4(𝑥+1)2 +9
6(1+𝑥)
4(𝑥+1)
=

6
9
+6(𝑥+1)
2(𝑥+1)
3
2(𝑥+1)
3
3
+2(𝑥+1)
3
+2(𝑥+1) ≥ 2
1
Since it is of the form a+𝑎 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:
19
361
Consider (13)2 = 169 > 2
19
∴ ( 13)8 > 24 > 13.
Thus 198 > 139
(198 )11 > (139 )11
1988 > 1399
∴ 1993 > 1399 .
26. If2𝑥 + 3𝑦 = 7, and 𝑥 ≥ 0, 𝑦 ≥ 0,then find the greatest value of 𝑥 3 𝑦 4
Solution:
2𝑥 3 3𝑦 4 3 3 4 4
Let Z=𝑥 3 𝑦 4 =( 3 ) ( 4 ) (2) (3) …………….(i)
2𝑥 3 3𝑦 4
∴ Z will have maximum value when ( 3 ) ( 4 ) is maximum.
2𝑥 3 3𝑦 4
But ( 3 ) ( 4 ) is the product of 3+4=7 factors,
2𝑥
3𝑦
the sum of which =3( 3 ) + 4 ( 4 ) = 2𝑥 + 3𝑦 = 7(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
2𝑥 3 3𝑦 4
∴ ( 3 ) ( 4 ) will be maximum if all the factors are equal,
i.e if
2𝑥
3
=
3𝑦
4
=
2𝑥+3𝑦
3+4
7
=7=1
2𝑥 3 3𝑦 4
∴ from (i) the maximum value of Z=( 3 ) ( 4 ) (1)3 (1)4
27
=8 𝑋
32
256
81
=3
27. Find the minimum value of bcx+cay+abz when xyz=abc
Solution:
We have xyz=abc
(bcx)(cay)(abz) =𝑎3 𝑏 3 𝑐 3 =Z
n =3
1
minimum value of bcx+cay+abz=n(𝑧)𝑛
=3(𝑎3 𝑏 3 𝑐 3 )1/3
=3abc.
BY CHENNAI REGION: