HAND BOOK OF MATHEMATICS (Definitions and Formulae) CLASS 12 SUBJECT: MATHEMATICS

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1 HAND BOOK OF MATHEMATICS (Definitions nd Formule) CLASS 12 SUBJECT: MATHEMATICS D.SREENIVASULU PGT(Mthemtics) KENDRIYA VIDYALAYA D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 1

2 CLASS 12 : CBSE MATHEMATICS RELATIONS AND FUNCTIONS TYPES OF RELATIONS: EMPTY RELATION: A reltion R in set A is clled empty reltion, if no element of A is relted to ny element of A, i.e., R = A A. UNIVERSAL RELATION: A reltion R in set A is clled universl reltion, if ech element of A is relted to every element of A, i.e., R = A A. TRIVIAL RELATIONS: Both the empty reltion nd the universl reltion re sometimes clled trivil reltions. A reltion R in set A is clled ) Reflexive, if (x, x) R for every x A b) Symmetric, if (x, y) R implies tht (y, x) R for ll x, y A c) Trnsitive, if (x, y) R nd (y, z) Rimplies tht (x, z) R for ll x, y, z A EQUIVALENCE RELATION: A reltion R in set A is sid to be n equivlence reltion if R is reflexive, symmetric nd trnsitive. EQUIVALENCE CLASS: Let R be n equivlence reltion on non-empty set A nd A. Then the set of ll those elements of A which re relted to, is clled the equivlence clss determined by nd is denoted by []. i.e [] = {x A (x, ) R} TYPES OF FUNCTIONS: ONE-ONE (INJECTIVE) FUNCTION: A function f X Y is defined to be one-one (or injective), if the imges of distinct elements of X under f re distinct, i.e., for every x 1, x 2 X, f(x 1 ) = f( x 2 ) implies x 1 = x 2. Otherwise, f is clled mny-one. ONTO (SURJECTIVE) FUNCTION : A function f X Y is sid to be onto (or surjective), if every element of Y is the imge of some element of X under f. i.e., for every y Y, there exists n element x in X such tht f(x) = y. NOTE: f : X Y is onto if nd only if Rnge of f = Codomin. BIJECTIVE FUNCTION: A function f X Y is sid to be bijective, if f is both oneone nd onto. COMPOSITION OF FUNCTIONS: Let f A B nd g B C be two functions. Then the composition of f nd g, denoted by gof, is defined s the function gof A C given by gof (x) = g(f (x)), x A. INVERTIBLE FUNCTION : A function f X Y is defined to be invertible, if there exists function g Y X such tht gof = I x nd fog = I y (i.e gof(x) = x nd fog(y) = y) The function g is clled the inverse of f nd is denoted by f 1 D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 2

3 NOTE: If f is invertible, then f must be one-one nd onto nd conversely, if f is one-one nd onto, then f must be invertible. BINARY OPERATIONS: BINARY OPERATION : A binry opertion on set A is function A A A. We denote (, b) by b Note: In generl, is sid to be binry opertion on A if, b A b A COMMUTATIVE: A binry opertion on the set X is clled commuttive, if b = b, for every, b X. ASSOCIATIVE: A binry opertion A A A is sid to be ssocitive if ( b) c = (b c),, b, c, A. IDENTITY: Given binry opertion A A A, n element e A, if it exists, is clled identity for the opertion, if e = = e A. INVERSE: Given binry opertion A A A with the identity element e in A, n element A is sid to be invertible with respect to the opertion, if there exists n element b in A such tht b = e = b nd b is clled the inverse of nd is denoted by 1. INVERSE TRIGONOMETRIC FUNCTIONS PRINCIPAL VALUE BRANCHES: FUNCTION DOMAIN RANGE (Principl Vlue Brnch) sin 1 x [ 1, 1] [ π 2, π 2 ] cos 1 x [ 1, 1] [0, π] tn 1 x R ( π 2, π 2 ) cosec 1 x R ( 1, 1) [ π 2, π 2 ] {0} sec 1 x R ( 1, 1) [0, π] { π 2 } cot 1 x R (0, π) PROPERTIES OF INVERSE TRIGONOMETRIC FUNCTIONS: sin(sin 1 x) = x, x [ 1, 1] D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 3

4 sin 1 (sin x) = x, x [ π 2, π 2 ] sin 1 1 x = cosoc 1 x, x 1 or x 1 cos 1 1 x = sec 1 x, x 1 or x 1 tn 1 1 = x cot 1 x, x > 0 sin 1 ( x) = sin 1 x, x [ 1, 1] cosec 1 ( x) = cosec 1 x, x 1 tn 1 ( x) = tn 1 x, x R cos 1 ( x) = π cos 1 x, x [ 1, 1], sec 1 ( x) = π sec 1 x, x 1 cot 1 ( x) = π cot 1 x, x R sin 1 x + cos 1 x = π 2, x [ 1, 1] tn 1 x + cot 1 x = π, x R 2 cosec 1 x + sec 1 x = π 2, x 1 tn 1 x + tn 1 y = tn 1 ( x+y 1 xy ), xy < 1 tn 1 x + tn 1 y = π + tn 1 ( x+y 1 xy ), xy > 1 tn 1 x tn 1 y = tn 1 ( x y ), xy > 1 1+xy 2 tn 1 x = tn 1 ( 2x 1 x2), x 1 2 tn 1 x = cos 1 ( 1 x2 1+x2), x 0 2 tn 1 x = sin 1 ( 2x 1+x2), x ( 1, 1) D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 4

5 MATRICES ORDER OF A MATRIX : A generl mtrix of order m n cn be written s = [ ij ] m n, where i = 1,2, m nd j = 1,2, n Number of rows = m nd number of columns = n TYPES OF MATRICES: COLUMN MATRIX: A mtrix is sid to be column mtrix if it hs only one column. Exmples: A = [ 2 ] order of mtrix A is nd B = [ 5 0 ] order of mtrix B is ROW MATRIX: A mtrix is sid to be row mtrix if it hs only one row Exmples: A = [14 26] order of mtrix A is 1 2 B = [0 7 12] order of mtrix B is 1 3 SQUARE MATRIX: A mtrix in which the number of rows is equl to the number of columns, is sid to be squre mtrix. Exmples: A = [ 2 4 ] order of mtrix A is X = [ 2 14] order of mtrix X is DIAGONAL MATRIX: A squre mtrix A = [ ij ] m n is sid to be digonl mtrix if ll its non-digonl elements re zero Exmple: A = [ 0 6 0] D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 5

6 SCALAR MATRIX : A digonl mtrix is sid to be sclr mtrix if its digonl elements re equl Exmple : A = [ 0 5 0] IDENTITY MATRIX: A squre mtrix in which elements in the digonl re ll 1 nd rest re ll zero is clled n identity mtrix Exmple : A = [ 0 1 0], generlly it is denoted by I ZERO MATRIX: A mtrix is sid to be zero mtrix or null mtrix if ll its elements re zero. Exmples : [0], [ ], [ 0 0 0] re l Zero mtrices, generlly denoted by O EQUALITY OF MATRICES: Two mtrices A = [ ij ] m n nd B = [b ij ] m n re sid to be equl if (i) they re of the sme order (ii) ech element of A is equl to the corresponding element of B, tht is ij = b ij for ll i nd j Exmple: Let A = [ 8 6 ] nd B = [ 8 6 ], we sy tht A = B OPERATION OF MATRICES: ADDITION OF MATRICES: Let A = [ ij ] m n nd B = [b ij ] m n be two mtrices of the sme order. Then A + B is defined to be the mtrix of order of m n obtined by dding corresponding elements of A nd B i.e A + B = [ ij + b ij ] m n DIFFERENCE OF MATRICES: Let A = [ ij ] m n nd B = [b ij ] be two mtrices of m n the sme order. Then A B is defined to be the mtrix of order of m n obtined by subtrcting corresponding elements of A nd B i.e A B = [ ij b ij ] m n MULTIPLICATION OF MATRICES: The product of two mtrices A nd B is defined if the number of columns of A is equl to the number of rows of B. D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 6

7 Let A = [ ij ] m n nd B = [b jk ] n p. Then the product of the mtrices A nd B is the mtrix C of order m p. To get the (i, k) th element c ik mtrix C, we tke the i th row of A nd k th column of B, multiply them elementwise nd tke the sum of n ll these products. i.e C ik = j=1 ij. b jk Exmple: Let A = [ ] nd B = [ 6 9] AB = [ ] = [ ] = [ ] MULTIPLICATION OF A MATRIX BY A SCALAR: Let A = [ ij ] m n nd k is sclr, then ka = k[ ij ] m n = [k. ij ] m n (4) 3( 5) Exmple: A = [ ] 3A = [3(2) ] = [ y z x 3y 3z 3x 3y 3z 3x ] TRANSPOSE OF A MATRIX: If A = [ ij ] be n m n mtrix, then the mtrix m n obtined by interchnging the rows nd columns of A is clled the trnspose of A. Trnspose of the mtrix A is denoted by A or A T. If A = [ ij ], then m n A = [ ji ] n m Exmple: A = [ 4 7 9] A T = [ 2 7 1] SYMMETRIC MATRIX: A squre mtrix If A = [ ij ] is sid to be symmetric if A = A, tht is, [ ij ] =[ ji ] for ll possible vlues of i nd j Exmple: A = [ ], clerly A = A SKEW-SYMMETRIC MATRIX: A squre mtrix A = [ ij ] is sid to be skew symmetric mtrix if A = A, tht is ij = ji for ll possible vlues of i nd j nd ii = 0 for ll i.(ll the digonl elements re zero) Exmple: A = [ ], clerly A = A D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 7

8 TRANSFORMATION OF A MATRIX: The interchnge of ny two rows or two columns. The interchnge of i th nd j th rows is denoted by R i R j nd interchnge of i th nd j th column is denoted by C i C j. The multipliction of the elements of ny row or column by non-zero number. The multipliction of ech element of the i th row by k, where k 0 is denoted by Ri k R i. The corresponding column opertion is denoted by C i kc i The ddition to the elements of ny row or column, the corresponding elements of ny other row or column multiplied by ny non-zero number. The ddition to the elements of i th row, the corresponding elements of j th row multiplied by k is denoted by R i R i + kr j. The corresponding column opertion is denoted by C i C i + kc j. INVERTIBLE MATRICES: If A is squre mtrix of order m, nd if there exists nother squre mtrix B of the sme order m, such tht AB = BA = I, then B is clled the inverse mtrix of A nd it is denoted by A 1. In tht cse A is sid to be invertible. PROPERTIES OF MATRICES: A + B = B + A A B B A AB BA (AB)C = A(BC) (A ) = A AI = IA = A AB = BA = I, then A 1 = B nd B 1 = A AB = 0 it is not necessry tht one the mtrix is zero. A(B + C) = AB + AC Every squre mtrix cn possible to express s the sum of symmetric nd skew-symmetric mtrices. D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 8

9 A = 1 (A + A ) + 1 (A A ), where (A + A ) is symmetric mtrix nd (A A ) is 2 2 skew-symmetric mtrices. Apply sequence of row opertion on A = IA till we get, I = BA. The mtrix B will be the inverse of A. Similrly, if we wish to find A 1 using column opertions, then, write A = AI nd pply sequence of column opertions on A = AI till we get, I = AB. The mtrix B will be the inverse of A. After pplying one or more elementry row (column) opertions on A = IA (A = AI), if we obtin ll zeros in one or more rows of the mtrix A on L.H.S., then A 1 does not exist. DETERMINANTS DETERMINANT: b Let A = [ ], then det(a) = A = d bc c d b c Let A = [ d e f e f f e ], then A = b d + c d g h k h k g k g h PROPERTIES OF DETERMINANTS: If rows nd columns re interchnged, then the vlue of the determinnt remins sme. If ny two rows (or columns) of determinnt re interchnged, then sign of determinnt chnges If ny two rows (or columns) of determinnt re identicl (ll corresponding elements re sme), then vlue of determinnt is zero. If ech element of row (or column) of determinnt is multiplied by constnt k, then its vlue gets multiplied by k. If, to ech element of ny row or column of determinnt, the equimultiples of corresponding elements of other row (or column) re dded, then vlue of D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 9

10 determinnt remins the sme, i.e., the vlue of determinnt remin sme if we pply the opertion R i R i + kr j or C i C i + k C j. If some or ll elements of row or column of determinnt re expressed s sum of two (or more) terms, then the determinnt cn be expressed s sum of two (or more) determinnts MINORS: Minor of n element ij of determinnt is the determinnt obtined by deleting its i th row nd jth column in which element ij lies. Minor of n element ij is denoted by M ij. CO-FACTORS: Cofctor of n element ij, denoted by A ij is defined by A ij = ( 1) i+j. M ij, where M ij is minor of ij ADJOINT OF A MATRIX: The djoint of squre mtrix A = [ ij ] is defined s the trnspose of the mtrix [A ij ], where A ij is the cofctor of the element ij. Adjoint of the mtrix A is denoted by dj A. INVERSE OF A MATRIX: Let A be squre mtrix. A 1 = 1 A dja SOLUTION OF SYSTEM OF LINEAR EQUATIONS BY USING MATRIX METHOD: Let the system of liner equtions be 1 x + b 1 y + c 1 z = d 1 2 x + b 2 y + c 2 z = d 2 3 x + b 3 y + c 3 z = d 3 These equtions cn be written s 1 b 1 c 1 x d 1 [ 2 b 2 c 2 ] [ y] = [ d 2 ] 3 b 3 c 3 z d 3 AX = B X = A 1 B A 1 exists, if A 0 i. e the solution existsnd it is unique. The system of equtions is sid to be consistent if the solution exists. if A = 0, then we clculte (dja)b. D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 10

11 If A = 0 nd (dja)b O, (O being zero mtrix), then solution does not exist nd the system of equtions is clled inconsistent. If A = 0 nd (dja)b = O, then system my be either consistent or inconsistent ccording s the system hve either infinitely mny solutions or no solution. IMPORTANT NOTES: The mtrix A is singulr if A = 0 λa = λ n A, where n = order of mtrix A A(djA) = (dja)a = A I dja = A n 1, where n = order of mtrix A A(djA) = A n, where n = order of mtrix A AB = A B (AB) 1 = B 1 A 1 A 1 = A 1 A T = A CONTINUITY AND DIFFERENTIABLITY CONTINUITY: Suppose f is rel function on subset of the rel numbers nd let be point in the domin of f. Then f is continuous t lim x f(x) = f() i.e LHL = RHL = f() lim x f(x) = lim f(x) = f() + x DIFFERENTIATION: FIRST PRINCIPLE: Let y = f(x), then dy = lim h 0 y = constnt dy = 0 y = x n dy = nxn 1 y = sinx dy = cosx y = cosx dy = sinx f(x+h) f(x) h D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 11

12 y = tnx dy = sec2 x y = cosecx dy = cosecx. cotx y = secx dy = secx. tnx y = cotx dy = cosec2 x y = sin 1 x dy = 1 1 x 2 y = cos 1 x dy = 1 1 x 2 y = tn 1 x dy = 1 1+x 2 y = cosec 1 x dy = 1 x x 2 1 y = sec 1 x dy = 1 x x 2 1 y = cot 1 x dy = 1 1+x 2 y = e x dy = ex y = x dy = x. log y = logx dy = 1 x Product Rule: y = u. v dy = u. v + v. u, where u = du, v = dv Quotient Rule: y = u v dy = v.u u.v v 2, where u = du Chin Rule: Let y = f(t)nd x = g(t) then dy = dy. dt dt y = f(x + b) dy =. f (x + b), Ex: y = sin(4x + 9) dy = 4. cos (4x + 9) y = [f(x)] n dy = n. [f(x)]n 1. f (x) Logrithmic Differentition: Let y = [u(x)] v(x) logy = v. logu 1 y. dy = v u. u + v. logu dy = y [v u.. u + v. logu], v = dv D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 12

13 Men Vlue Theorem: Let f : [, b] R be continuous function on [, b] nd differentible on (, b). Then there exists some c in (, b) such tht some c (, b) sucht tht f (c) = f(b) f() b Rolle s Theorem: Let f : [, b] R be continuous on [, b] nd differentible on (, b), such tht f() = f(b), where nd b re some rel numbers. Then there exists some c in (, b) such tht f (c) = 0. Note: In bove ll formuls, logx = log e x APPLICATION OF DERIVATIVES RATE OF CHANGE OF QUANTITIES 1) Are of circle (A) = πr 2 Rte of chnge of re = da dt = 2πr dr dt 2) Circumference of circle (C) = 2πr Rte of chnge of Circumference = dc dt = 2π dr dt 3) Perimeter of rectngle (P) = 2(x + y), where x = length, y = width Rte of chnge of Perimeter = dp dt = 2( xt + dy dt ) 4) Are of rectngle (A) = x. y, where x = length, y = width Rte of chnge of re = da dt dy = x. + y. dt dt 5) Volume of cube (V) =x 3, where x = edge of cube Rte of chnge of Volume = dv 6) Surfce re of cube (S) = 6x 2 dt = 3x2 dt Rte of chnge of Surfce re = ds 7) Volume of sphere (V) = 4 3 πr3 dt = 6x dt Rte of chnge of Volume = dv = 4 dt 3 (3πr2 ) dr dt D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 13

14 8) Surfce re of Sphere (S) = 4πr 2 Rte of chnge of Surfce re = ds dt = 8πr dr dt 9) Totl cost = C(x), where C(x) isi Rupees of the production of x units Mrginl cost = dc 10) Totl Revenue = R(x) Mrginl Revenue = dr INCREASING AND DECREASING FUNCTION Let I be n intervl contined in the domin of rel vlued function f. Then f is sid to be (i) incresing on I if x < y in I then f(x) f(y), for ll x, y I. (ii) strictly incresing on I if x < y in I then f(x) < f(y), for ll x, y I (iii) decresing on I if x < y in I then f(x) f(y), for ll x, y I. (iv) strictly decresing on I if x < y in I then f(x) > f(y), for ll x, y I () f is strictly incresing in (, b) if f (x) > 0 for ech x (, b) (b) f is strictly decresing in (, b) if f (x) < 0 for ech x (, b) A function will be incresing or decresing in R if it is so in every intervl of R f is constnt function in [, b] if f (x) = 0 for ech x (, b) TANGENTS AND NORMALS Let given curve be y=f(x) Slope of tngent to the curve t (x 1, y 1 ) is m = [ dy Slope of norml to the curve t (x 1, y 1 ) = 1 m ] x=x 1 Eqution of tngent t (x 1, y 1 ) is y y 1 = m(x x 1 ) Eqution of norml t (x 1, y 1 ) is y y 1 = 1 m (x x 1) D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 14

15 APPROXIMATION Let y = f(x) y = dy. x y + y = y + dy. x i.e f(x + x) = f(x) + f (x). x MAXIMA AND MINIMA First Derivtive Test: Let f be function defined on n open intervl I. Let f be continuous t criticl point c in I. Then (i) If f (x) chnges sign from positive to negtive s x increses through c, then c is point of locl mxim nd mximum vlue of f(x) = f( c). (ii) If f (x) chnges sign from negtive to positive s x increses through c, then c is point of locl minim nd minimum vlue of f(x) = f( c). (iii) If f (x) does not chnge sign s x increses through c, then c is neither point of locl mxim nor point of locl minim. Infct, such point is clled point of inflexion. Second Derivtive Test Let f be function defined on n intervl I nd c I. Let f be twice differentible t c. Then (i) x = c is point of locl mxim if f (c) = 0 nd f (c) < 0 The vlues f (c) is locl mximum vlue of f. (ii) (ii) x = c is point of locl minim if f (c) = 0 nd f (c) > 0 In this cse, f (c) is locl minimum vlue of f. (iii) The test fils if f (c) = 0 nd f (c) = 0. In this cse, we go bck to the first derivtive test nd find whether c is point of mxim, minim or point of inflexion. Absolute mxim nd bsolute minim (mxim nd minim in closed intervl) Given f(x) nd intervl [, b] Find f (x) Let f (x) = 0 D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 15

16 Find criticl vlues. (i.e find the vlues of x if f (x) = 0 ). sy x = x 1 nd x = x 2 Clculte f(), f(x 1 ), f(x 2 ) nd f(b). Identify mxim nd minim vlues of f(x). INTEGRALS INDEFINITE INTEGRALS 1) 1 = x + c 2) x = x2 2 + c 3) x n = xn+1 + c, n 1 n+1 4) sinx = cosx + c 5) cosx = sinx + c 6) tnx = log secx + c 7) cosecx = log cosecx cotx + c 8) secx = log secx + tnx + c 9) cotx = log sinx + c 10) sec 2 x = tnx + c 11) cosec 2 x = cotx + c 12) secx. tnx = secx + c 13) cosecx. cotx = cosecx + c 14) 1 1 x 2 = sin 1 x + c or cos 1 x + c 15) 1 1+x 2 = tn 1 x + c or cot 1 x + c 1 16) = x x 2 1 sec 1 x + c or cosec 1 x + c 17) e x = e x + c 18) 1 = log x + c x 19) x = x log + c D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 16

17 20) 1 = 1 log x 2 2 x + c 2 x+ 21) 1 = 1 log 2 x 2 +x + c 2 x 22) 1 x = 1 tn 1 x + c 1 23) = log x + x 2 2 x2 2 + c 1 24) = log x + x x2 + 2 = +c 1 25) = 2 x 2 sin 1 x + c 26) x 2 2 = x 2 x ) x = x 2 x log x + x2 2 + c 2 log x + x c 28) 2 x 2 = x 2 2 x sin 1 x + c 29) e x [f(x) + f (x)] = e x f(x) + c 30) u. v = u v [u v] + c, where u = u(x) nd v = v(x) 31) [f(x) ± g(x)] = f(x) ± g(x) 32) k. f(x) = k f(x), where k is constnt. Note: Let ntiderivtive of f(x) = F(x) i.e. f(x) = F(x) + c, then f(x + b) = 1 F(x + b) + c Prtil frctions The rtionl function P(x) is sid to be proper if the degree of Q(x) is Q(x) less thn the degree of P(x) Prtil frctions cn be used only if the integrnd is proper rtionl function S.No Form of rtionl function Form of Prtil frction 1 1 (x )(x b) A x + b x b D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 17

18 2 px + q (x )(x b) 3 1 (x )(x b)(x c) 4 px + q (x )(x b)(x c) 5 px 2 + qx + r (x )(x b)(x c) 6 1 (x ) 2 (x b) 7 px + q (x ) 2 (x b) 8 px 2 + qx + r (x ) 2 (x b) 9 px 2 + qx + r (x )(x 2 + bx + c) where x 2 + bx + c cnnot A x + b x b A x + b x b + 1 x c A x + b x b + 1 x c A x + b x b + 1 x c A x + A x + A x + B (x ) 2 + B (x ) 2 + B (x ) 2 + C x b C x b C x b A x + Bx + C x 2 + bx + c be fctorized further Integrl of the type Liner Qudrtic, Liner Qudrtic,Liner Qudrtic Liner = A d (Qudrtic) + B DEFINITE INTEGRALS Definite integrls the limit of sum f(x) b where h = = lim h 0 h. [f() + f( + h) + f( + 2h) + + f( + (n 1)h] b n nh = b Properties Of Definite Integrls 1) f(x) = 0 D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 18

19 b 2) f(x) b 3) f(x) b 4) f(x) b 5) f(x) 6) f(x) 0 2 7) f(x) 0 2 8) f(x) 0 9) f(x) CURVE LINE = b f(t) dt = f(x) b = c f(x) = b f( + b x) = f( x) 0 = f(x) 0 b + f(x), where < c < b c + f(2 x) 0 = { f(x), if f(2 x) = f(x) 0 0, if f(2 x) = f(x) = { 2 0 f(x), if f(x)is even. i. e f( x) = f(x) 0, if f(x)is odd. i. e f( x) = f(x) APPLICATION OF INTEGRALS CIRCLE LINE PARABOLA LINE ELLIPSE - LINE CURVE CURVE PARABOLA PARABOLA PARABOLA CIRCLE CIRCLE -CIRCLE AREA OF TRIANGLE CO-ORDINATES OF VERTICES ARE GIVEN EQUATIONS OF SIDES ARE GIVEN D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 19

20 STEPS DRAW THE DIAGRAM MAKE A SHADED REGION FIND INTERSECTION POINTS IDENTIFY THE LIMITS WRITE THE INTEGRAL(S) FOR THE REGION EVALUATE THE INTEGRAL THE VALUE SHOULD BE POSITIVE 1) Are of shded region = f(x) b D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 20

21 DIFFERENTIAL EQUATIONS Methods of solving First Order nd First Degree Differentil Equtions Differentil Equtions with Vribles seperbles Homogeneous differentil equtions Liner differentil equtions. Differentil Equtions with Vribles seprbles Let the differentil eqution be dy = f(x) g(y) then g(y)dy = f(x) then integrte on both sides g(y)dy = f(x) Let the differentil eqution be dy = g(y) f(x) then dy = g(y) f(x) then integrte on both sides dy = g(y) f(x) Let the differentil eqution be dy = f(x). g(y) then dy g(y) = f(x) then integrte on both sides dy g(y) = f(x) Homogeneous differentil equtions A function F(x, y) is sid to be homogeneous function of degree n if F(λx, λy) = λ n F(x, y) A differentil eqution of the form dy = F(x, y) is sidto be homogeneous if F(x,y) is homogeneous function of degree zero i.e. if F(λx, λy) = λ 0 F(x, y) Steps to solve the homogeneous differentil eqution of the type: dy = f(y x ) Let y = vx dy dv = v + x Substitute y = vx nd dy dv dy = v + x in = f(y) x Then use vribles nd seprbles in terms of y nd v only Steps to solve the homogeneous differentil eqution of the type: dy = f(x y ) Let x = vy D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 21

22 dy = v + y dv dy Substitute x = vy nd dv = v + y in = dy dy dy f(x) y Then use vribles nd seprbles in terms of x nd v only Liner differentil eqution Steps to solve the Liner differentil eqution of the type: dy + P(x)y = Q(x) dy + P(x)y = Q(x) Integrtin Fctor (IF) = e p(x) Solution is y. (IF) = (IF). Q(x) Steps to solve the Liner differentil eqution of the type: + P(y)x = Q(y) dy + P(y)x = Q(y) dy Integrtin Fctor (IF) = e p(y)dy Solution is x. (IF) = (IF). Q(y)dy VECTORS Position vector of the point A(, b, c ) is OA = xi + yj + zk AB = OB OA Let = xi + yj + zk then = x 2 + y 2 + z 2 Unit vector of =, is denoted by Let = i + bj + ck is sid to be unit vector if = 1 Projection of on b =. b b THREE DIMENSIONAL GEOMETRY Direction cosines of line re the cosines of the ngles mde by the line with the positive directions of the coordinte xes. Let line mking the ngles with x, y, z xis re α, β, γ repectively. Direction cosines re l = cosα, m = cosβ, n = cosγ If l, m, n re the direction cosines of line, then l 2 + m 2 +n 2 = 1. Direction rtios of line joining two points P(x 1, y 1, z 1 ) nd Q(x 2, y 2, z 2 ) re = x 2 x 1, b = y 2 y 1, = y 2 y 1 D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 22

23 If l, m, n re the direction cosines nd, b, c re the direction rtios of line then l = ±, m = ±, n = ± 2 +b 2 +c 2 2 +b 2 +c 2 Direction cosines of line joining two points b D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 23 c 2 +b 2 +c 2 P(x 1, y 1, z 1 ) nd Q(x 2, y 2, z 2 ) re x 2 x 1, y 2 y 1, z 2 z 1 PQ PQ PQ where PQ= (x 2 x 1 ) 2 + (y 2 y 1 ) 2 + (z 2 z 1 ) 2 Direction rtios of line re the numbers which re proportionl to the direction cosines of line. Skew lines re lines in spce which re neither prllel nor intersecting. They lie in different plnes. Angle between skew lines is the ngle between two intersecting lines drwn from ny point (preferbly through the origin) prllel to ech of the skew lines. If l 1, m 1, n 1 nd l 2, m 2, n 2 re the direction cosines of two lines; nd θ is the cute ngle between the two lines; then cosθ l 1 l 2 + m 1 m 2 + n 1 n 2 If 1, b 1, c 1 nd 2, b 2, c 2 re the direction rtios of two lines nd θ is the cute ngle between the two lines; then cosθ = 1 2 +b 1 b 2 +c 1 c b 1 2 +c b 2 2 +c 2 2 Vector eqution of line tht psses through the given point whose position vector is nd prllel to given vectorb is r = + λb. Eqution of line through point (x 1, y 1, z 1 ) nd hving direction cosines l, m, n is x x 1 l = y y 1 = z z 1 m n The vector eqution of line which psses through two points whose position vectors re nd b is r = + λ(b ) Crtesin eqution of line tht psses through two points (x 1, y 1, z 1 ) nd (x 2, y 2, z 2 ) is x x 1 = y y 1 = z z 1. x 2 x 1 y 2 y 1 z 2 z 1 If θ is the cute ngle between r = 1 + λb 1 nd r = 2 + λb 2, then cosθ = b 1.b 2 b 1 b 2 If x x 1 = y y 1 = z z 1 nd x x 2 = y y 2 = z z 2 re the equtions of two lines, then the l 1 m 1 n 1 l 2 m 2 n 2 cute ngle between the two lines is given by cosθ = l 1 l 2 + m 1 m 2 + n 1 n 2. Shortest distnce between two skew lines is the line segment perpendiculr to both the lines. Shortest distnce between r = 1 + λb 1 nd r = 2 + μb 2 is (b 1 b 2 ).( 2 b 1 b 2 1 ) Shortest distnce between the lines: x x 1 1 = y y 1 b 1 = z z 1 c 1 nd x x 2 2 = y y 2 b 2 = z z 2 x 2 x 1 y 2 y 1 z 2 z 1 1 b 1 c 1 2 b 2 c 2 (b 1 c 2 b 2 c 1 ) 2 +(c 1 2 c 2 1 ) 2 +( 1 b 2 2 b 1 ) 2 Distnce between prllel lines r = 1 + λb nd r = 2 + μb is b ( 2 1 ) b c 2 is

24 In the vector form, eqution of plne which is t distnce p from the origin, nd n is the unit vector norml to the plne through the origin is r. n = p. Eqution of plne which is t distnce of d from the origin nd the direction cosines of the norml to the plne s l, m, n is lx + my + nz = d. The eqution of plne through point whose position vector is nd perpendiculr to the vector N is (r ). N = 0. Eqution of plne perpendiculr to given line with direction rtios A,B,C nd pssing through given point (x 1, y 1, z 1 ) is A(x x 1 ) + B(y y 1 ) + C(z z 1 ) = 0 Eqution of plne pssing through three non colliner points x x 1 y y 1 z z 1 (x 1, y 1, z 1 ), (x 2, y 2, z 2 )nd (x 3, y 3, z 3 ) is x 2 x 1 y 2 y 1 z 2 z 1 = 0 x 3 x 1 y 3 y 1 z 3 z 1 Vector eqution of plne tht contins three non colliner points hving position vectors, b nd c is (r ). [(b ) (c )] = 0 Eqution of plne tht cuts the coordintes xes t (, 0,0), (0, b, 0) nd (0,0, c) is x + y b + z c = 1 Vector eqution of plne tht psses through the intersection of plnes r. n 1 = d 1 nd r. n 2 = d 2 is r. (n 1 + λn 2 ) = d 1 + λd 2, where λis ny nonzero constnt. Vector eqution of plne tht psses through the intersection of two given plnes A 1 x + B 1 y + C 1 z + D 1 = 0 nd A 2 x + B 2 y + C 2 z + D 2 = 0 is (A 1 x + B 1 y + C 1 z + D 1 ) + λ(a 2 x + B 2 y + C 2 z + D 2 ) = 0 Two plnes r = 1 + λb 1 nd r = 2 + μb 2 re coplnr if ( 2 1 ). (b 1 b 2 ) = 0 Two plnes 1 x + b 1 y + c 1 z + d 1 = 0 nd 2 x + b 2 y + c 2 z + d 2 = 0 re coplnr if x 2 x 1 y 2 y 1 z 2 z 1 1 b 1 c 1 = 0. 2 b 2 c 2 In the vector form, if θ is the ngle between the two plnes,. n 1 n 2 r. n 1 = d 1 nd r. n 2 = d 2, then θ = cos 1 n 1.n 2 The ngle φ between the line r = + λb nd the plne r. n = d is sin φ = b.n b n The ngle θ between the plnes A 1 x + B 1 y + C 1 z + D 1 = 0 nd A 2 x + B 2 y + C 2 z + D 2 = 0 is given by cos θ = A 1 A 2 +B 1 B 2 +C 1 C 2 A 1 2 +B 1 2 +C 1 2 A 2 2 +B 2 2 +C 2 2 The distnce of point whose position vector is from the plne r. n = d is d. n The distnce from point (x 1, y 1, z 1 ) to the plne Ax + By + Cz + D = 0 is Ax 1+By 1 +Cz 1 +D A 2 +B 2 +C 2 D.SREENIVASULU, M.Sc.,M.Phil.,B.Ed. PGT(MATHEMATICS), KENDRIYA VIDYALAYA. Pge 24