[ 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 CauchySchwarz 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 CauchySchwarz 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 CauchySchwarz 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 CauchySchwarz 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 CauchySchwarz 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 oegative, so that the sum is oegative. 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
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