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ALGEBRA TIPS FOR MATHEMATICS OLYMPIAD 1. Every polynomial equation of degree nโฅ1 has exactly n roots. 2. If a c Complex polynomial equation with real coefficients has a complex root p+iq (p,q real numbers,qโ 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 ๐2 + ๐ 2 + ๐ 2 โฅ ๐๐ + ๐๐ + ๐๐ ๐4 + ๐ 4 + ๐ 4 โฅ ๐2 ๐ 2 + ๐ 2 ๐ 2 + ๐ 2 ๐2 ๐2 ๐ 2 + ๐ 2 ๐ 2 + ๐ 2 ๐2 โฅ ๐๐๐(๐ + ๐ + ๐) ๐4 + ๐ 4 + ๐ 4 โฅ ๐๐๐(๐ + ๐ + ๐) 1. 2. 3. 4. 5. 6. ๐2 +๐ 2 ๐+๐ ๐4 +๐ 4 ๐2 +๐ 2 1 โฅ 2 (๐ + ๐) 1 โฅ 2 (๐2 + ๐ 2 ) ๐+๐+๐ 2 1 7. ( 8. โ๐2 + ๐ 2 + โ๐ฅ 2 + ๐ฆ 2 โฅ โ(๐ฅ + ๐)2 + (๐ฆ + ๐)2 3 ) > 3 (๐๐ + ๐๐ + ๐๐) ๐จ๐ด โฅ ๐ฎ๐ด โฅ ๐ฏ๐ด 6. ๐1 + ๐2 + โฏ + ๐๐ ๐ โฅ (๐1 . ๐2 . โฆ ๐๐ )1/๐ โฅ ( ) 1 1 1 ๐ + + โฏ โฆ + ๐1 ๐2 ๐๐ If ๐1 , ๐2 , โฆ . , ๐๐ are n positive distinct real numbers then 7. 1. 2. ๐ ๐1๐ +๐2๐ +โฏ+๐๐ ๐ ๐ ๐1๐ +๐2๐ +โฏ+๐๐ ๐ >( <( ๐1 +๐2 +โฏ+๐๐ ๐ ๐ ) ๐1 +๐2 +โฏ+๐๐ ๐ ๐ ) ๐๐ ๐ < 0 ๐๐ ๐ > 1 ๐๐ 0 < ๐ < 1 3. ๐ ๐1 ๐1๐ +๐2 ๐2๐ +โฏ+๐๐ ๐๐ ๐1 +๐2 +โฏ+๐๐ ๐1 ๐1 +๐2 ๐2 +โฏ+๐๐ ๐๐ ๐ >( ) ๐1 +๐2 +โฏ+๐๐ ๐๐ ๐ < 0 ๐๐ ๐ > 1 where ๐1 , ๐2 , โฆ . , ๐๐ and ๐1 , ๐2 , โฆ , ๐๐ ๐๐๐ ๐ ๐๐๐ ๐๐๐ก๐๐๐๐๐ ๐๐ข๐๐๐๐๐ . 4. 8. ๐ ๐1 ๐1๐ +๐2 ๐2๐ +โฏ+๐๐ ๐๐ ๐1 +๐2 +โฏ+๐๐ ) ๐1 +๐2 +โฏ+๐๐ ๐๐ 0 < ๐ < 1 If ๐1 , ๐2 , โฆ . , ๐๐ are n positive distinct real numbers then ๐+๐+๐ ๐1 ๐+๐+๐ +๐2 ๐ 9. ๐1 ๐1 +๐2 ๐2 +โฏ+๐๐ ๐๐ ๐ <( ๐+๐+๐ +โฏ+๐๐ ๐ > ๐ ๐ ๐ ๐ ๐ ๐ ๐1 +๐2 +โฏ+๐๐ ๐1 +๐2 +โฏ+๐๐ ๐1๐ +๐2๐ +โฏ+๐๐ ๐ . ๐ . ๐ Weierstrass Inequality : 1. If ๐1 , ๐2 , โฆ . , ๐๐ are n positive real numbers, then for n โฅ 2 (1 + ๐1 ) (1 + ๐2 ) (1 + ๐3 ) โฆ..(1 + ๐๐ ) > 1 + ๐1 + ๐2 + โฏ + ๐๐ 2. If ๐1 , ๐2 , โฆ . , ๐๐ are n positive real numbers, then for n < 1 (1 โ ๐1 ) (1 โ ๐2 ) (1 โ ๐3 ) โฆ..(1 โ ๐๐ ) > 1 โ ๐1 โ ๐2 โฆ โ ๐๐ 10. Tchebychefโs Inequality: If ๐1 , ๐2 , โฆ . , ๐๐ and ๐1 , ๐2 , โฆ , ๐๐ ๐๐๐๐ ๐๐ข๐๐๐๐๐ ๐ ๐ข๐โ ๐กโ๐๐ก ๐1 โค ๐2 โค โฆ . โค ๐๐ and ๐1 โค ๐2 โค โฏ โค ๐๐ ๐กโ๐๐ n(๐1 ๐1 + ๐2 ๐2 + โฏ + ๐๐ ๐๐ ) โฅ (๐1 + ๐2 + โฏ + ๐๐ )(๐1 + ๐2 + โฏ + ๐๐ ) (or) ๐1 ๐1 + ๐2 ๐2 + โฏ + ๐๐ ๐๐ ๐1 + ๐2 + โฏ + ๐๐ ๐1 + ๐2 + โฏ + ๐๐ โฅ . ๐ ๐ ๐ 11. 12. CAUCHY-SCHWARZ Inequality: If a,b,c and x,y,z are any real numbers ( positive, negative or zero) then (๐๐ฅ + ๐๐ฆ + ๐๐ง)2 โค (๐2 + ๐ 2 + ๐ 2 )(๐ฅ 2 + ๐ฆ 2 + ๐ง 2 ) with equality iff a:b:c::x:y:z DESCARTEโS 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 ๐ ๐ (where p,q are integers qโ 0, (๐, ๐) = 1, ๐. ๐; ๐๐๐๐๐๐๐๐ ) is a root of the equation ๐0 ๐ฅ ๐ + ๐1 ๐ฅ ๐โ1 + โฏ + ๐๐ = 0, where ๐1 , ๐2 , โฆ . , ๐๐ are integers and ๐๐ โ ๐, then , p is a divisor of ๐๐ and q is a divisor of ๐0 FUNCTIONAL EQUATIONS: An equation involving an unknow3n function is called a functional equation. 1. 2. 3. โ๐ ๐ If ๐ผ, ๐ฝ ๐๐๐ ๐กโ๐ ๐๐๐๐ก๐ ๐๐ ๐๐ฅ 2 + ๐๐ฅ + ๐ = 0 , ๐กโ๐๐ ๐ผ + ๐ฝ = ๐ , ๐ผ๐ฝ = ๐ If ๐ผ, ๐ฝ, ๐พ ๐๐๐ ๐กโ๐ ๐๐๐๐ก๐ ๐๐ ๐๐ฅ 3 + ๐๐ฅ 2 + ๐๐ฅ + ๐ = 0 , ๐กโ๐๐ โ๐ ๐ โ๐ ๐ผ+๐ฝ+๐พ = , โ ๐ผ๐ฝ = ๐ผ๐ฝ๐พ = ๐ ๐ ๐ If ๐ผ, ๐ฝ, ๐พ, ๐ฟ ๐๐๐ ๐กโ๐ ๐๐๐๐ก๐ ๐๐ ๐๐ฅ 4 + ๐๐ฅ 3 + ๐๐ฅ 2 + ๐๐ฅ + ๐ = 0 , ๐กโ๐๐ โ๐ ๐ โ๐ โ๐ ๐ผ+๐ฝ+๐พ+๐ฟ = , โ ๐ผ๐ฝ = ; โ ๐ผ๐ฝ๐พ = ๐ผ๐ฝ๐พ = ๐ ๐ ๐ ๐ MATHS OLYMPIAD QUESTIONS AND SOLUTIONS: ALGEBRA Q.NO 1. QUESTIONS/SOLUTIONS Find the maximum number of positive and negative real roots of the equation ๐ฅ 4 + ๐ฅ 3 + ๐ฅ 2 โ ๐ฅ โ 1 = 0. Solution: Let f(x)= ๐ฅ 4 + ๐ฅ 3 + ๐ฅ 2 โ ๐ฅ โ 1 The signs are +,+,+,-,-, There is only one change of sign from positive to negative. โด there is only one positive root. F(-x)=(- ๐ฅ)4 + (โ๐ฅ)3 + (โ๐ฅ)2 โ (โ๐ฅ) โ 1 = ๐ฅ4 โ ๐ฅ3 + ๐ฅ2 + ๐ฅ โ 1 The signs are +, -, +, +, There are three changes in sign โด three negative roots. 2. If ๐ผ, ๐ฝ, ๐พ, ๐ฟ, be the roots of the equation ๐ฅ 4 + ๐๐ฅ 3 + ๐๐ฅ 2 + ๐๐ฅ โ ๐ = 0. Show that (1+๐ผ 2 )(1 + ๐ฝ 2 )(1 + ๐พ 2 )(1 + ๐ฟ 2 ) =(1-q+s)2 +(p-r)2. Solution: Since ๐ผ, ๐ฝ, ๐พ, ๐ฟ ๐๐๐ the roots of the equation ๐ฅ 4 + ๐๐ฅ 3 + ๐๐ฅ 2 + ๐๐ฅ โ ๐ = 0, ๐ฅ 4 + ๐๐ฅ 3 + ๐๐ฅ 2 + ๐๐ฅ โ ๐ = (๐ฅ โ ๐ผ)(๐ฅ โ ๐ฝ)(๐ฅ โ ๐พ)(๐ฅ โ ๐ฟ) If x=i , ๐ 4 + ๐๐ 3 + ๐๐ 2 + ๐๐ โ ๐ = (๐ โ ๐ผ)(๐ โ ๐ฝ)(๐ โ ๐พ)(๐ โ ๐ฟ) 1-pi-q+ri+s=(๐ โ ๐ผ)(๐ โ ๐ฝ)(๐ โ ๐พ)(๐ โ ๐ฟ) ( 1-q+s)-i(p-r)= (๐ โ ๐ผ)(๐ โ ๐ฝ)(๐ โ ๐พ)(๐ โ ๐ฟ)โฆโฆโฆ..(i) If x=-I, (โ๐)4 + ๐(โ๐)3 + ๐(โ๐)2 + ๐(โ๐) โ ๐ = (๐ + ๐ผ)(๐ + ๐ฝ)(๐ + ๐พ)(๐ + ๐ฟ) 1+pi-q-ri+s=(๐ + ๐ผ)(๐ + ๐ฝ)(๐ + ๐พ)(๐ + ๐ฟ) ( 1-q+s)+i(p-r) = (๐ + ๐ผ)(๐ + ๐ฝ)(๐ + ๐พ)(๐ + ๐ฟ)โฆโฆโฆ..(ii) (i)x(ii)=> (1-q+s)2+(p-r)2=(๐ 2 โ ๐ผ 2 ) (๐ 2 โ ๐ฝ 2 ) (๐ 2 โ ๐พ 2 )(๐ 2 โ ๐ฟ 2 ) =(1 + ๐ผ 2 ) (1 + ๐ฝ 2 ) (1 + ๐พ 2 )(1 + ๐ฟ 2 ). 3. Show that (๐ฅ โ 1)2 is a factor of ๐ฅ ๐ โ ๐๐ฅ + ๐ โ 1. Solution: Let f(x)= ๐ฅ ๐ โ ๐๐ฅ + ๐ โ 1. f (1)=1-n+n-1=0 โด ( x-1) is a factor. f(x)= ๐ฅ ๐ โ ๐๐ฅ + ๐ โ 1. = ๐ฅ ๐ โ 1 โ ๐๐ฅ + ๐. =๐ฅ ๐ โ 1 โ ๐(๐ฅ โ 1). =(x-1){๐ฅ ๐โ1 + ๐ฅ ๐โ2 + โฏ + 1 โ ๐} =(x-1)g(x)โฆโฆโฆโฆโฆโฆ(i) ๐โ1 Now g(x)= ๐ฅ + ๐ฅ ๐โ2 + โฏ + 1 โ ๐ g (1)=1+1+1+โฆ.1-n =0. โด ( x-1) is a factor of g(x) โด ( x-1)2 is a factor of f(x) {by (i)} 4. If sin ๐ + cos ๐ = 1, (sin ๐ > 0, cos ๐ > 0). Show that (1+cosec๐)(1 + ๐ ๐๐๐) โฅ 9. Solution: We know that AM โฅ GM. sin ๐+cos ๐ 2 โฅ โsin ๐ cos ๐. 1 โฅ โsin ๐ cos ๐. 1 โฅ sin ๐ cos ๐ 4 Let If sin ๐ = ๐ฅ, cos ๐ =y 2 1 1 (1+cosec๐)(1 + ๐ ๐๐๐)=(1+ ๐ฅ ) (1 + ๐ฆ) (๐ฅ+1)(๐ฆ+1) = ๐ฅ๐ฆ (๐ฅ๐ฆ+๐ฅ+๐ฆ+1) = ๐ฅ๐ฆ ๐ฅ๐ฆ+2 = ๐ฅ๐ฆ . โฅ ๐(๐ ๐๐ฆ) ๐ฅ๐ฆ + 2 โฅ ๐๐ฅ๐ฆ 2โฅ (๐ โ 1)๐ฅ๐ฆ 2 ๐โ1 โฅ ๐ฅ๐ฆ 1 but โด 4 2 โฅ ๐ฅ๐ฆ ๐โ1 = 1 P=9. 5. 4x 9x๏ซy If x ๏ซ y ๏ฝ 8 and 5 y ๏ฝ 243 find the value of x-y. 3 2 4x ๏ฝ8 2x๏ซy SOLUTION: ๏ (2 2 ) x ๏ฝ8 2x๏ซy 22x ๏ x๏ซy ๏ฝ 8 2 ๏ 2 2 x ๏ญx ๏ญ y ๏ฝ 8 . 4 6. A po ๏ 2 x ๏ญy ๏ฝ (2)3 ๏ x๏ญy๏ฝ3 The Polynomial p (x) leaves a remainder 3 when divided by x โ 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 โ1. Let, p (x) = k(x-1) + 3 = kx โ k + 3 Now, k x ๏ญ 3 kx ๏ญ k ๏ซ 3 kx ๏ญ 3k 2k ๏ซ 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 โ 4x + 3 Dividing p(x) by x2 โ 4x + 3 0 x ๏ญ 4x ๏ซ 3 x ๏ซ 2 0๏ซ0 x๏ซ2 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 โ ab + b2 = c2 โ cd + d2 ๏ (a+b)2 โ 3ab = (c + d)2 โ 3cd ๏ -3ab = -3cd ๏ ab = cd ๏ ab = cd Given a + b = c + d a๏ซb c๏ซd ๏ฝ 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. ๐ ๐ ๐ ๐ If a,b,c,d be any four positive real numbers ,then prove that ๐ + ๐ +๐ + ๐ โฅ 4 SOLUTION: Applying the inequality of the means A.M โฅ G.M to the four possible numbers ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ ๐ , ๐ ,๐ , ๐ we have ¼( ๐ + ๐ +๐ + ๐ ) โฅ( ๐ + ๐ +๐ + ๐ )1/4 ๐ โ๐ + ๐ +๐ + ๐ โฅ4Type 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)โฅ16 SOLUTION: Since A.MโฅG.M 1+๐ 2 1+๐ 2 1+๐ 2 โฅ(1.a)1/2 โฅ(1.b)1/2 โฅ(1.c)1/2 1+๐ 2 โฅ(1.d)1/2 Multiplying the inequalities (1+a)(1+b)(1+c)(1+d 16 โฅ (abcd)1/2 (1+a)(1+b)(1+c)(1+d)โฅ16 10. n+1 2n Prove that nn ( n ) ห(n!)3 Solution: Consider positive numbers 13,23,โฆ,n3 Since AMหGM, (13+23+โฆ+n3 )/nห(13.23โฆn3)1/n n(n + 1) 2 ( ) 2 n By simplifying, n+1 2n nn ( n ) ห(n!)3 1 ห( (n!)3 )n 11. If x , y, z are three real numbers such that x + y + z =4 and x2 + y2 +z2 = 6 , ๏ฉ2 ๏น then show thtat each of x , y ,z lies in the ๏ช , 2๏บ . Can x attains ๏ซ3 ๏ป 2 extreme value ? 3 Solution: We have y + z = 4 - x and y2 +z2 = 6 - x2 From Cauchyโs Schwarz inequality we get y2 +z2 ๏ณ 1 ( y + z )2 2 1 ( 4-x)2 .This simplifies to (3x - 2) (x โ 2) ๏ฃ 0. 2 2 Hence we have ๏ฃ x ๏ฃ 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 ๏ณ 12. If squares of the roots of x4+bx2+cx+d=0 are ฮฑ,ฮฒ,ฮณ,ฮด, then prove that: 64ฮฑฮฒฮณฮดโ[4โฮฑฮฒโ(โฮฑ)2]2=0. Solution: x4+bx2+cx+d=0......(1) Let y be the root of transformed equation. y=x^2 x4+bx2+d=โcx 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+(2bdโc2)y+d2=0...(2) It is the equation whose roots are squares of roots of x4+bx2+cx+d=0 now for (2)ฮฑ,ฮฒ,ฮณ,ฮด are the roots. โฮฑ =-2b โฮฑฮฒ=b2+2d โฮฑฮฒฮณ =-(2bd-c^2) ฮฑฮฒฮณฮด = d2 Now plug the values in required function. 64ฮฑฮฒฮณฮดโ[4โฮฑฮฒโ(โฮฑ)2]2 = 64d2โ(4(b2+2d)โ(โ2b)2)2 =64d2โ(4b2+8dโ4b2)2 =64d2โ(8d)2 =64d2โ64d2 =0 Hence it is proved. 13. Prove that 3a4โ4a3b+b4โฅ0 for all real numbers a and b. Solution: 3a4โ4a3b+b4 =a4+2a4+b4โ4a3b =a4+b4+2a4โ4a3b =(a2)2+(b2)2+2a4โ4a3b = (a2)2+(b2)2โ2a2b2+2a2b2+2a4โ4a3b = (a2โb2)2+2a2b2+2a4โ4a3b = (a2โb2)2+2a2(b2+a2โ2ab) = (a2โb2)2+2a2(aโb)2โฅ0 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 โ k + 2 Now, dividing p(x) by (x-2) K x ๏ญ 2 kx ๏ญ k ๏ซ 2 kx ๏ญ 2k k๏ฝ2 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 โ 3x + 2, x 2 ๏ญ 3x ๏ซ 2 ๏ญ x ๏ซ 3 i.e. the remainder is (-x+3) only = -x+3 a ๏ญx ๏ซ b๏ญx ๏ซ c๏ญx ๏ฝ0 15. If Prove that (a + b + c + 3x) (a + b + c - x) = 4(bc + ca + ab) Solution: a ๏ญx ๏ซ b๏ญx ๏ซ c๏ญx ๏ฝ0 ๏ a๏ญx ๏ซ b๏ญx ๏ฝ ๏ญ c๏ญx ๏ (a ๏ซ b ) ๏ซ ( b ๏ญ x ) ๏ซ 2 a ๏ญ x ๏ญ b ๏ญ x ๏ฝ c ๏ญ x ๏ a+bโcโx=-2 ๏ a2 + b2 + c2 + 2ab โ 2bc โ 2ca โ 2x (a+b-c) + x2 = 4 {ab โ x(a+b) a๏ญx b๏ญx + x2} ๏ (a + b + c)2 + 2x (a + b + c ) โ 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 ๏ซ bc ๏ซ ca ๏ฝ 1 3 a, b, c are real numbers. Find the value of (i) a b c ๏ซ ๏ซ b c a (ii) a b c ๏ซ ๏ซ b๏ซa c๏ซa a ๏ซ1 Solution : ๏จa ๏ญ b๏ฉ2 ๏ซ ๏จb ๏ญ c ๏ฉ2 ๏ซ ๏จc ๏ญ a ๏ฉ2 = 2 ๏ฆ๏ง a 2 ๏ซ b 2 ๏ซ c 2 ๏ญ ab ๏ญ bc ๏ญ ca ๏ถ๏ท = 2 ๏ฉ๏ช๏จa ๏ซ b ๏ซ c๏ฉ2 ๏ญ 3 ๏จab ๏ซ bc ๏ซ ca ๏ฉ๏น๏บ ๏ซ ๏ป = 1๏น ๏ฉ 2 ๏ช1 ๏ญ 3 ๏ ๏บ 3๏ป ๏ซ = 0 ๏จ So, a = b, ๏ธ b = c, and c = a So, a+b+c=1, gives a ๏ฝ b ๏ฝ c ๏ฝ So, a b c ๏ซ ๏ซ b c a = 1+1+1 1 3 = 3 and a b c ๏ซ ๏ซ b ๏ซ1 c ๏ซ a a ๏ซ1 = 1/ 3 1/ 3 1/ 3 ๏ซ ๏ซ 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 โ x + x โ 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 โ 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 + z2โ2zx+yz+2x-x2 โ xy) (y-z)5 + (z-x)5 + (x-y)5 = 5 (y-z) (z-x) (x-y) (x2+ y2+z2 -yz -zx โ xy) (a)18. If f(x) = ax7 + bx5 + cx3 โ 6, and f(-9) = 3, find f(9). (b) Find the value of (2002) 3 ๏ญ (1002) 3 ๏ญ (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 โ 6 ๏a (-9)7 + b (-9)5 + c (-9)3 โ 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 โ 6 = -9-6 = -15 Ans. (b) (2002)3 ๏ญ (1002)3 ๏ญ (1000)3 3 x (1002) x (1000) = (2002 ๏ญ1002) [( 2002) 2 ๏ซ (2002) (1002) ๏ซ (1002)2 ] ๏ญ (1000)3 3 x (1002) x (1000) 1000 [( 2002 ๏ญ1002 ) 2 ๏ซ 3 (2002 ) (1002 )] ๏ญ (1000 )3 = 3 x (1002 ) x (1000 ) (1000 ) 2 ๏ซ 3 (2002 ) (1002 ) ๏ญ (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 โ x2 โ y2 \ x2 โ xy + y2 = 49 ๏ x2 โ (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 โ 34 = -15 \ (x+ y)2 = x2 + y2 + 2xy = 34 + 2x (-15) = 34 โ 30 =4 ๏x+y=4 (x-y)2 = x2 + y2 โ 2xy = 34 โ 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 โ 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 โ c = 30 a + 2b = 16 9a + 2c = 12 6b โ 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 โ 4 ๏ b = 16 โ 4 ๏b= 12 2 =6 9 x 4 + 2c = 12 ๏ 2c = 12 โ 36 ๏c= ๏ญ 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 ๐ + ๐ +๐ 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 โ13+1 So,abc= 3 =-4 1 1 1 ๐ + ๐ +๐ = ๐๐+๐๐+๐๐ ๐๐๐ =(-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 ฮฑ,ฮฒ, ๏ง , ๏ค then ฮฑ +ฮฒ+ ๏ง ๏ซ ๏ค = 20 (ฮฑ +ฮฒ)( ๏ง ๏ซ ๏ค )+ ฮฑ ฮฒ+ ๏ง๏ค =k (ฮฑ +ฮฒ) ๏ง๏ค +( ๏ง ๏ซ ๏ค ) ฮฑ ฮฒ =-590 ฮฑ ฮฒ ๏ง๏ค =-1992 ฮฑ ฮฒ=24 ๏ง๏ค =-1992/24 =-83 if ฮฑ +ฮฒ=x and ๏ง ๏ซ ๏ค =y, x+y=20 -83x+24y=-590 By solving x=10,y=10 K=(ฮฑ +ฮฒ)( ๏ง ๏ซ ๏ค )+ ฮฑ ฮฒ+ ๏ง๏ค =10.10+24-83 =41 22. Prove that 1< 1 + 1 + 1 +โฆ+ 1 1001 1002 1003 3001 Solution: Consider the numbers 1001,1002,..,3000,3001 A.M= 1001+1002+โฏ+3001 2001 2001 = = 1 (1001+3001)2001 2 4002 2 =2001 1 1 1 1 HM = (1001 +1002 + 1003 +โฆ+3001 ) -12001 2001 = 1 1 1 1 + + +โฏ+ 1001 1002 1003 3001 Since AM>HM, 2001 2001> 1 1 1 1 + + +โฏ+ 1001 1002 1003 3001 1 1> 1 1 1 1 + + +โฏ+ 1001 1002 1003 3001 1 1 1 1 + + 1003 +โฆ+3001 > 1 1001 1002 Hence proved. 23. Let x be the set of positive integers โฅ 8.Let f:xโx be a function,such that f(x+y) =f(xy) for all xโฅ4,yโฅ4.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 4๐ฅ 2 +8๐ฅ+13 6(1+๐ฅ) for xโฅ0 Solution: 4๐ฅ 2 +8๐ฅ+13 6(1+๐ฅ) = 4(๐ฅ 2 +2๐ฅ+1)+9 6(1+๐ฅ) = = 4(๐ฅ+1)2 +9 6(1+๐ฅ) 4(๐ฅ+1) = ๏ 6 9 +6(๐ฅ+1) 2(๐ฅ+1) 3 2(๐ฅ+1) 3 3 +2(๐ฅ+1) 3 +2(๐ฅ+1) โฅ 2 1 Since it is of the form a+๐ where a>0 Since (a-1)2โฅ0 and a>0 (a-1)2/a โฅ0 (a2-2a+1)/aโฅ0 a-2+(1/a)โฅ0 a+1/aโฅ2 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 โด ( 13)8 > 24 > 13. Thus 198 > 139 (198 )11 > (139 )11 1988 > 1399 โด 1993 > 1399 . 26. If2๐ฅ + 3๐ฆ = 7, and ๐ฅ โฅ 0, ๐ฆ โฅ 0,then find the greatest value of ๐ฅ 3 ๐ฆ 4 Solution: 2๐ฅ 3 3๐ฆ 4 3 3 4 4 Let Z=๐ฅ 3 ๐ฆ 4 =( 3 ) ( 4 ) (2) (3) โฆโฆโฆโฆโฆ.(i) 2๐ฅ 3 3๐ฆ 4 โด Z will have maximum value when ( 3 ) ( 4 ) is maximum. 2๐ฅ 3 3๐ฆ 4 But ( 3 ) ( 4 ) is the product of 3+4=7 factors, 2๐ฅ 3๐ฆ the sum of which =3( 3 ) + 4 ( 4 ) = 2๐ฅ + 3๐ฆ = 7(๐๐๐๐ ๐ก๐๐๐ก) 2๐ฅ 3 3๐ฆ 4 โด ( 3 ) ( 4 ) will be maximum if all the factors are equal, i.e if 2๐ฅ 3 = 3๐ฆ 4 = 2๐ฅ+3๐ฆ 3+4 7 =7=1 2๐ฅ 3 3๐ฆ 4 โด from (i) the maximum value of Z=( 3 ) ( 4 ) (1)3 (1)4 27 =8 ๐ 32 256 81 =3 27. Find the minimum value of bcx+cay+abz when xyz=abc Solution: We have xyz=abc (bcx)(cay)(abz) =๐3 ๐ 3 ๐ 3 =Z n =3 1 minimum value of bcx+cay+abz=n(๐ง)๐ =3(๐3 ๐ 3 ๐ 3 )1/3 =3abc. BY CHENNAI REGION: