[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.

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1 [ 11 ] Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural umbers. If a 0, the we say that f ( x ) is a polyomial of degree r. Example x x x x 1 is a polyomial of degree 4 ad 1 is a zero of the polyomial as Also, x ix ix 1 0 is a polyomial of degree ad i is a zero of his polyomial as i i. i i. i 1 i i Agai, x ( ) x 6 is a polyomial of degree ad is a zero of this polyomial as ( ) ( ) Note : The above defiitio ad examples refer to polyomial fuctios i oe variable. similarly polyomials i,,..., variables ca be defied, the domai for polyomial i variables beig set of (ordered) tuples of complex umbers ad the rage is the set of complex umbers. Example : f ( x, y, z) x xy z 5 is a polyomial i x, y, z of degree as both degree each. k k k x ad xy have Note : I a polyomial i variables say x1, x,..., x, a geeral term is x 1 1, x,..., x where degree is k1 k... k where ki 0, i 1,,...,. The degree of a polyomial i variables is the maximum of the degrees of its terms. Divisio i Polyomials If P( x ) ad ( x) are ay two polyomials the we ca fid polyomials Q( x ) ad R( x ) such hat P( x) ( x) Q( x) R( x) where the degree of R( x) degree of Q( x ). Q( x ) is called the quotiet ad R( x ), the remaider. 11

2 [ 1 ] I particular if P( x ) is a polyomial with complex coefficiets ad a is a complex umber the there exists a polyomial Q( x ) of degree 1 less tha P( x ) ad a complex umber R, such that P( x) ( x a) Q( x) R. Example : x ( x a)( x ax a x a x a ) a Here 5 P( x) x, 4 4 Q( x) x ax a x a x a ad R a Remaider Theorem ad Factor Theorem Remider Theorem : If a polyomial f ( x ) is divided by ( x a) the the remaider is equal to f ( a ). Proof : f ( x) ( x a) Q( x) R (1) ad so f ( a) ( a a) Q( a) R R If R 0 the f ( x) ( x a) Q( x) ad hece ( x a) is a factor of f ( x ). Further f ( a) 0 ad thus a is a zero of the polyomial f ( x ). This leads to the factor theorem. Factor Theorem : ( x a) is a factor of polyomial f ( x ) if ad oly if f ( a) 0. Fudametal theorem of algebra : Every polyomial fuctio of degree 1 has at least oe zero i the complex umbers. I other words if we have f ( x) a x a x... a x a with 1, the there exists at least oe h C such that, a h a h... a h a 0 From this it is easy to deduce that a polyomial fuctio of degree has exactly zeroes. Example 1. Fid the polyomial fuctio of lowest degree with itegral coefficiet s with 5 as oe of its zeroes. Solutio : Sice the order of the surd 5 is, you ca expect the polyomial of the lowest degree to be a polyomial of degree. Let P( x) ax bx c; a, b, c Q P( 5) 5a 5b c 0 But 5 is a zero, so 5a c 0 ad 5b 0 c 5a ad b 0. So the required polyomial fuctio is P( x) ax 5 x. You ca fid the other zero of this polyomial to i.e., 5. 1

3 Example. [ 1 ] If x, y, z be positive umbers, show that ( x y z) 7 xyz. Solutio : Sice A.M. (arithmetic mea) G.M. (geometric me), therefore x y z 1/ ( xyz). Cubig both sides ad multiplyig throughout by 7, we have ( x y z) 7 xyz. Example. If a1,..., a, b1,..., b be real umbers ad oe of the b i s be zero, the prove that Solutio : ( a... a )( b... b ) ( a / b... a / b ) Applyig Cauchy-Schwarz iequality to the umbers a,..., a, b,..., b, we have or b 1 b... (.. )(... ), a a a a b b a1 a ( a1... a )( 1... )... b b. b1 b Example 4. (Triagle Iequality). If x1, x, y1, y, be ay real umbers, the show that {( x x ) ( y y ) } ( x y ) ( x y ), Solutio : where the sig deotes the positive square root ( x x ) ( y y ) ( x y ) ( x y ) ( x x y y ). (i) By Cauchy-Schwarz iequality, ( x x y y ) ( x y )( x y ), i.e., From (i) ad (ii), we have x x y y ( x y ) ( x y ) (ii) ( x x ) ( y y ) ( x y ) ( x y ) 1 1 x y ) ( x y ), or ( x x ) ( y y ) { ( x y ) ( x y )} Takig positive square roots, we have {( x x ) ( y y ) } ( x y ) ( x y ). Remark : Geometrically iterpreted, the above iequality expresses the fact the sum of two sides of a triagle ca ever be less tha the third side ad this is precisely the reaso for the ame triagle iequality. 1

4 [ 14 ] Example 5. If c1,..., c be positive real umbers, show that 1 1 ( c... c ) ( c... c ). Whe does the iequality reduce to equality? Solutio : If a1,..., a, b1,..., b, be real umbers, the by Cauchy-Schwarz iequality, ( a b... a b ) ( a... a )9 b... b ). (1) / 1/ i i i i Puttig a c, b c, ( i 1,,..., ) i the above iequality, we have ( c... c ) ( c.. c )( c... c ). () Agai, puttig a c, b 1, ( i 1,,..., ) i (1), we have i i i 1 1 ( c... c ) ( c... c ). () Squarig both sides of () ad usig (), we immediately have 1 1 ( c... c ) ( c... c ). The above iequality reduces to a equality iff each of the iequalities () ad () reduces to a equality, i.e., iff / / 1/ 1/ 1 1 c :..: c :: c :...: c, ad c1 :...: c 1:...:1, i.e., iff c1 c... c. Example 6. If x, y, z be positive real umbers such that Solutio : x y z 81. Applyig Cauchy-Schwarz iequality to the two sets of umbers we have / / / 1/ 1/ 1/ x, y, z ; x, y, z x y z 7, the show that ( x y z ) ( x y z )( x y z). (i) Agai, applyig Cauchy-Schwarz iequality to the two sets of umbers we have Squarig both sides of (i), we have x, y, z ;1,1,1 ( x y z) ( x y z ) (ii) 4 ( x y z ) ( x y z ) ( x y z ) (iii) i.e., Sice x y z 7, we have from (iii) 4 ( x y z ) ( x y z ) ( x y z ) ( x y z ) (81). 14

5 Takig positive square roots, we have Tcheby Chef s Iequality [ 15 ] x y z 81. Example 7. If a1, a, a, b1, b, b are ay real umbers such that a1 a a, b1 b b, the show that ( a b a b a b ) ( a a a ) ( b b b ) Solutio : Sice a1 a, b1 b, therefore, a1 a, b1 b are of the same sig or at least oe of them is zero, so that Theorem : ( a a ) ( b b ) 0, ad therefore 1 1 a1b 1 ab a1b ab 1. Similarly, ab ab ab ab, (ii) ad ab a1b 1 ab1 a1b. (iii) Addig (i), (ii) ad (iii) ad the addig a1b1 ab ab to both sides of resultig iequality, we have ( a b a b a b ) ( a a a ) ( b b b ) If a1,..., a ad b 1,..., b are ay real umbers, such that (i) a1... a, b1... b, the ( a b... a b ) ( a... a )( b... b ) (ii) a1... a, b1... b, the ( a b... a b ) ( a.. a )( b... b ) Proof : (i) For every pair of distict suffixes p ad q, the differeces a p aq ad bp bq are of the same sig or at least oe of them is zero. Hece, ( a a )( b b ) 0 p q p q i.e., d b a b a b a b. p p q q p q q p There are 1 ( 1) iequalities of the above type (for there are 1 ( 1) pairs of distict suffixes p, q), Addig the correspodig sides of all such iequalities, we obtai ( 1)( a b... a b ) ( a... a )( b... b ) ( a b.. a b ) i.e., ( a1b 1... ab ) ( a1... a)( b1... b ). (ii) For every pair of distict suffixes p ad q, a p aq ad bp bq are of opposite sigs or at least oe of them is zero. Hece, 15

6 [ 16 ] ( a a )( b b ) 0 p q p q i.e., a b a b a b a b. p p q q p q q p Addig the correspodig sides of all the 1 ( 1) iequalities of the above type, we obtai ( 1)( a b... a b ) ( a... a )( b... b ) ( a b... a b ), i.e., ( a1b 1... ab ) ( a1... a)( b1... b ). Remark : The iequality above ca be put i the followig symmetric form : a1b 1... ab a1... a b1... b.. This form suggests the followig geeralisatio which we state without proof. If 1,..., ; 1,..., ;...; 1,..., a a b b k k are real umbers such that a... a, b... b,..., k... k, a the, 1b1... k1... ab... k a 1... a b 1... b k 1... k.... We shall refer to this iequality as Geeralised Tchebychef's Iequality. Example 8. Show that : (a) ( 1) 1... ; (b) / ( 1) Solutio : (a) Applyig Tchebychef's iequality to the sets of umbers 1,..., ; 1,...,, we have 1/4 ( ) ( 1... ), or or (1... ) ( 1... ), ( 1) ( 1... ). ( 1) Therefore, (b) Applyig Tchebychef's iequality to the sets of umbers 1, 1,..., 1 ;1, 1,..., 1, we obtai 16

7 [ 17 ] Takig positive square roots of both sides, we have , ( 1) , ( 1) Agai, applyig Tchebychef's iequality to the sets of umbers 1,,..., :1,,...,, we have From (i) ad (ii), we have Therefore Example ( 1) ( 1) If a, b, c are all positive ad o two of them are equal, the prove that (a) (b) ( a b c) a b c abc a b c abc( a b c). 1/4 (i) (ii) Solutio : (a) Without ay loss of geerality we may assume that a b c. By applyig the geeralised Tchebychef's iequality to three sets of umbers each of which is the same as a, b, c, we obtai i.e., a b c a b c. a b c. a b c, ( a b c) a b c (i) 9 Agai, sice the arithmetic mea exceeds the geometric mea 17

8 [ 18 ] a b c abc From (i) ad (ii), we obtai the iequalities (ii) ( a b c) a b c abc. (a) 9 (b) As i (a), without ay loss of geerality we may assume that a b c. Sice a b c, therefore, a b c. Applyig Tchebychef's iequality to the sets of umbers Also, from (a) From (iii) ad (iv), we have a, b, c; a, b, c, we obtai a b b a b c. a b c (iii) a b c Example 10. If a, b, c are positive ad uequal, show that abc. (iv) a b c abc( a b c) ( a b c )( a b c ) ( a b c )( a b c ), Solutio : ( a b c )( a b c ) ( a b c )( a b c ), ( a b a b a b a b ), 5 5 a b ( a b a b a b ), a b ( a b )( a b ). The differeces a b, a b are both of the same sig, ad therefore, ( a b )( a b ) is positive. Similarly, the other two terms i the above sum are also positive. Therefore, ( a b c )( a b c ) ( a b c )( a b c ) 0. Example 11. If a, b, c are positive ad if p, q, r are ratioal umbers such that p q r( 0) ad r( 0) have the same sig, the show that p p p q q q pr pr pr qr qr qr ( a b c )( a b c ) ( a b c )( a b c ). Show that if either (i) a b c, or (ii) p q r, or (iii) r 0, the equality holds. p p p q q q pr pr pr qr qr qr Solutio : ( a b c )( a b c ) ( a b c )( a b c ). q p q p pr qr qr pr ( b a a b a b a c ), q q pq pq pqr r r pqr a b ( a b a b a b ), 18

9 [ 19 ] q q pqr pqr r r a b ( a b )( a b ). Sice p q r ad r have the same sig, the differeces a the same sig or are both zero. Therefore, q q pqr pqr r r a b ( a b )( a b ) 0, pqr pqr b ad a r r b have ad similarly each of the other two terms i the above sum is also o-egative, so that the sum is o-egative. This proves the iequality. Also, if ay of the give coditios is satisfied, the at least oe of the factors i each term q q pqr pqr r r i a b ( d b )( a b ) vaishes ad therefore the sum is zero. This proves that the equality holds. 1. Idetities : IMPORTANT TERMS AND RESULTS IN ALGEBRA (a) (b) (c) If If If a b c 0, a b c ( bc ca ab ) a b c 0, a b c abc a b c 0, a b c ( b c c a a b ) 1 ( ) a b c. Periodic fuctio : A fuctio f is said to be periodic, with period k..if. f ( x k) f ( x) x. Pigeo Hole Priciple (PHP) : If more tha objects are distributed i boxes, the at least oe box has more tha oe object i it. 4. Polyomials : (a) A fuctio f defied by f ( x) a x a x... a where a0 0, is a positive iteger or zero ad a ( i 0,1,,..., ) are fixed complex umbers, is called a polyomial of degree i x. The umbers a0, a1, a,..., a are called the coefficiets of f. If be a complex umber such that f ( ) 0, the is said to be a zero of the polyomial f. (b) If a polyomial f ( x ) is divided by x h, where h is ay complex umber, the remaider is equal to f(h). (c) If h is a zero of a polyomial f ( x ), the ( x h) is a factor of f ( x ) ad coversely. i 19

10 [ 0 ] (d) Every polyomial equatio of degree 1 has exactly roots. (e) If a polyomial equatio wish real coefficiets has a complex root p iq (p, q real umbers, q 0 ) the it also has a complex root p iq. (f) If a polyomial equatio with ratioal coefficiets has a irratioal root p q (p, q ratioal, q > 0, q ot the square of a ratioal umber), the it also has a irratioal root p q. (g) If the ratioal umber p q (a fractio i its lowest terms so that p, q are itegers, prime to each other, q 0 ) is a root of the equatio 1 a0x a1 x... a 0 where a0, a1,..., a are itegers ad a 0, the p is a divisor of a ad q, is a divisor of a 0. (h) A umber is a commo root of the polyomial equatios f ( x) 0 ad g( x) 0 iff it is a root of h( x) 0, where h( x ) is the G.C.D. of f ( x ) ad g( x ). (i) A umber is a repeated root of a polyomial equatio f ( x) 0 iff it is a commo root of f ( x) 0 ad f ( x) Fuctioal equatio : A equatio ivolvig a ukow fuctio is called a fuctioal equatio. 6. (a) If, be the roots of the equatio (b) If,, be the roots of the equatio ax bx c 0 the b ad a ax bx cx d 0 the, b c d ; ;, a a a (c) If,,, be the roots of the equatios 4 ax bx cx dx e 0 the, b ; a d a c a. c a e a 4 Questio 1. The product of two roots of the equatio 4x 4x 1x 6x 8 0 is 1, fid all the roots. Solutio : Suppose the roots are,,, ad 1. 4 Now, 1 ( ) ( ) 6 (1) 4 0

11 [ 1 ] ( )( ) ( ) ( ) 1 () 4 4 ( ga ) ( ) ( ) ( ) () 4 (4) From Eq. () ad Eq. (4), we get 5 ( ) ( ) 4 (5) From Eq. () ad Eq. (4), we get ( ) ( ) (6) From Eq. (1) ad Eq. (6), we get ( ) 15 or Questio. If,, are the roots of (i) (ii) 5 x px q 0, the prove that Solutio : (i) Sice,, are the roots of we have, x px q 0, (1) 1

12 [ ] p q 0 p q 0 p q 0 () From (), p( ) q 0 But 0, from Eq. (1) q ( ) 0 p ( p ) p (4) Multiplyig (1) by x, we get 5 ad,, are three roots of Eq. (5). So x px qx 0 (5) 5 p q 0 5 p q 0 5 p q 0 (6) From Eq. (6), 5 p q 0 or ( p q ) 5 [ p( q) q( p)] (7) pq pq 5 pq pq (8) 5 Multiplyig Eq. (1) by x, we get 4 x px qx 0 (9)

13 [ ] ad hece 4 p q 0 4 p ( 0) Agai multiplyig Eq. (1) by x 4, we get x px qx 0 (10) ad hece p qga 0 or p q p 5 pq q( p ) 5 p q p q Questio. Solutio : or or Fid the commo roots of x 5x x 50x 1 0 ad hece solve the equatios. 7 p q 7 p q pq ( p) x x 0x 16x 4 0 You ca see that 4( x 5x 6) is H.C.F. of the two equatios ad hece, the commo roots are the roots of x 5x 6 0 i. e., x or x Now, ad 4 x 5x x 50x 1 0 (1) 4 x x 0x 16x 4 0 () have ad as their commo roots. If the other roots of Eq. (1) are ad, the 5 5, 10 from eq. (1) 6 1 So, ad are also roots of the quadratic equatio

14 [ 4 ] x 10x x 5 So the roots of Eq. (1) are,, 5, 5. For Eq. (), if 1 ad 1 be the roots of Eq. 9), the we have So 1 ad 1 are the roots of or 11 4 x 6x x 5 So the roots of Eq. () are,, 5, 5. Questio 4. Solve the system : Solutio : i terms of L. Addig the three equatios, we get ( x y) ( x y z) 18 ( y z) ( x y z) 0 ( z x) ( x y z) L ( x y z) 48 L or x y z 4 L Dividig the three equatios by ( x y z) 4 L, we get ad solvig we get, ad x y, y z, z x 4 L 4 L 4 L x y z L 4 0 L 6, 4 L 4 L (4 L) L 4 L, 4 L 4 L 4 L 18 L 6. 4 L 4 L 4

15 [ 5 ] Questio 5. If x 1 ad x are o zero roots of the equatio ax bx c 0 ad ax bx c 0 respectively, prove that Solutio : x 1 ad x are roots of a x bx c 0 has a root betwee x 1 ad x. ad respectively. We have ad Let Thus, Addig Subtractig ax bx c 0 (1) ax bx c 0 () ax bx c 0 ax bx c a f ( x) x bx c. a f ( x1 ) x1 bx1 c () a f ( x ) x bx c (4) 1 1 ax i Eq. (), we get 1 f ( x1 ) ax1 ax1 bx1 c 0 ax from Eq. (4), we get 1 f ( x1 ) ax1 (5) f ( x) ax ax bx c 0 f ( x ) ax. Thus f ( x 1) ad f ( x ) have opposite sigs ad, hece, f ( x ) must have a root betwee x 1 ad x. Questio 6. Fid all real values of m such that both roots of the equatio greater tha but less tha +4. Solutio : The roots are m 1 i.e., ( m 1), ( m 1) ( m 1) ( m 1) 4 gives 1 m. x mx ( m 1) 0 are 5

16 [ 6 ] Questio 7. Solutio : 5 4 The roots of the equatio x 40x px qx rx s 0 are i G.P. The sum of their reciprocal is 10. Compute the umerical value of s. Let the roots be Sum of be reciprocals Dividig (1) by (), a, a, a, ar, ar r r 1 1 Sum of the root a 1 r r 40 r r r r 1 10 a r r a 4 a Sice s is the ve of the product of the roots 5 (1) () () s a (4) s or s (5) Questio 8. Let 4 P( x) x ax bx cx d where a, b, c, d are costats. If compute P(1) P( 8). 10 P(1) 10, P() 0, P() 0 Solutio : We use a trick Q( x) p( x) 10x (1) The Q(1) Q() Q() 0 () Q( x ) i.e., divisible by ( x 1) ( x ) ( x ) () Sice Q( x ) is a 4 th degree polyomial Q( x) ( x 1) ( x ) ( x ) ( x r ) ad P( x) ( x 1) ( x ) ( x ) ( x r) 10x (4) P(1) P( 8) Questio 9. Let P( x) 0 be the polyomial equatio of least possible degree with ratioal coefficiets, havig 7 49 as a root, Compute the product of all the roots of P( x) 0. Solutio : Let x 7 49 i.e., Thus, x x x P( x) x ad the product of the root is 56. Questio 10. The equatios x 5x px q 0 ad x 7x px r 0 have two roots i commo. If the third root of each equatio is represeted by x 1 ad x respectively, compute the ordered pair ( x1, x ). 6

17 [ 7 ] Solutio : Commo roots must be the roots of Their sum is 0. x ( r q ) 0 (Differece of equatio) The the third root of the first equatio must be 5 ad of the secod equatio is 7. x1 x (, ) ( 5, 7). Questio 11. If a, b, c, x, y, z are all real ad a b c a b c ax by cz 0, fid the value of. x y z 5, x y z 6 ad Solutio : a b c ax by cz x z z Thus a x b y c z a x 5 6 a kx where 5 k ; b ky ad c kz. 6 a b c ( ) k x y z k x y z x y z k Questio 1. If the iteger A its reduced by the sum of its digits, the result is B. If B is icreased by the sum of its digits, the result is A. Compute the largest -digt umber A with this property. Solutio : 5 6 A (sum of the digits) must be divisible by 9. The B + (sum of the digits) does ot satisfy must be divisible by 9. Now cosider 999 : = 97 (so defied sum of 7) Aswer is 990. Questio 1. The roots of 990 : = 97 (so defied sum of 18) 4 x kx kx lx m 0) are a, b, c, d. If k, l, m are real umbers, compute the miimum value of the sum a b c d. Solutio : Sum of the roots = k; Sum of the roots take two at a lie = k The k ( a b c d) ( a b c d ) ( ab ac ad bc bd cd ) ( a b c d ) k Thus a b c d k k (1) 7

18 [ 8 ] Thus miimum value of k k 1. x x Questio 14. If 0, 6 6 the it must be true that a x b for some itegers a ad b. Compute (a, b) where (b a) as small as possible. Note : [x] represets the greatest iteger fuctio. Solutio : Replacig x 6 by y ad solvig, y y 0 0 x 4 6 which meas 4 x 18 As. ( 4, 18) Questio 15. The roots of 5 y or 4 x px qx 19 0 are each oe more tha the roots of x Ax Bx C 0. If A, B, C, P, Q are costats, compute A B C. Solutio : Now ( a 1) ( b 1) ( c 1) 19. The A B C ( a b c) ( ab bc ca) ( abc ) ( a 1) ( b 1) ( c 1) Questio 16. Fid all ordered pairs of positive itegers (x, z) that Solutio : x z 10 x z 10. ( x z) ( x z) x 1; z 9; x 17, z 1; x 1, z 7; x 11, z 1 Required ordered pairs are : (1, 9), (17, 1), (1, 7), (11, 1). 8

Objective Mathematics

Objective Mathematics . If sum of '' terms of a sequece is give by S Tr ( )( ), the 4 5 67 r (d) 4 9 r is equal to : T. Let a, b, c be distict o-zero real umbers such that a, b, c are i harmoic progressio ad a, b, c are i arithmetic