. Overview.. Limits of a fuctio Let f be a fuctio defied i a domai wic we take to be a iterval, say, I. We sall study te cocept of it of f at a poit a i I. We say f ( ) is te epected value of f at a give te values of f ear to te a left of a. Tis value is called te left ad it of f at a. We say f( ) + is te epected value of f at a give te values of f ear to te a rigt of a. Tis value is called te rigt ad it of f at a. If te rigt ad left ad its coicide, we call te commo value as te it of f at a ad deote it by f ( ). a Some properties of its Let f ad g be two fuctios suc tat bot f ( ) ad g ( ) eist. Te a a (i) [ f( ) + g( )] f ( ) + g( ) a a a (ii) [ f( ) g( )] f ( ) g( ) (iii) a a a For every real umber α ( α f)( ) α f( ) a a (iv) [ f ( ) g( )] [ f ( ) g( )] a a a f ( ) f ( ) a a g () g (), provided g () 0 a Capter LIMITS AND DERIVATIVES ot to be republised
6 EXEMPLAR PROBLEMS MATHEMATICS Limits of polyomials ad ratioal fuctios If f is a polyomial fuctio, te f ( ) eists ad is give by a f ( ) f ( a ) a A Importat it A importat it wic is very useful ad used i te sequel is give below: a a a a Remark Te above epressio remais valid for ay ratioal umber provided a is positive. Limits of trigoometric fuctios To evaluate te its of trigoometric fuctios, we sall make use of te followig its wic are give below: (i) si (ii) cos (iii) si 0 0 0 0.. Derivatives Suppose f is a real valued fuctio, te f () f + f 0 ( ) ( ) is called te derivative of f at, provided te it o te R.H.S. of () eists.... () Algebra of derivative of fuctios Sice te very defiitio of derivatives ivolve its i a rater direct fasio, we epect te rules of derivatives to follow closely tat of its as give below: Let f ad g be two fuctios suc tat teir derivatives are defied i a commo domai. Te : (i) (ii) Derivative of te sum of two fuctio is te sum of te derivatives of te fuctios. ot to be republised d d d d + f( ) + g( ) d d [ f () g() ] Derivative of te differece of two fuctios is te differece of te derivatives of te fuctios. d d d d f( ) g( ) d d [ f () g() ]
LIMITS AND DERIVATIVES 7 (iii) Derivative of te product of two fuctios is give by te followig product rule. (iv) d d d d f ( ) g( ) + f( ) g( ) d d [ f () g() ] Tis is referred to as Leibitz Rule for te product of two fuctios. Derivative of quotiet of two fuctios is give by te followig quotiet rule (werever te deomiator is o-zero). d f () d g( ) d d f ( ) g( ) f( ) g( ) d d ( ). Solved Eamples Sort Aswer Type g ( ) ( ) Eample Evaluate + Solutio We ave ( ) + ( ) ( )( ( ) ( ) ( )( ) 5 + 6 ( )( ) ot to be republised ( )( ) ( )( ) [ 0] ( )
8 EXEMPLAR PROBLEMS MATHEMATICS + Eample Evaluate 0 Solutio Put y + so tat we 0, y. Te 0 + y y y () Eample Fid te positive iteger so tat 08. Solutio We ave () Terefore, () 08 (7) () Comparig, we get Eample Evaluate (sec ta ) Solutio Put y. Te y 0 as. Terefore (sec ta ) [sec( y) ta ( y)] y 0 ot to be republised (cosec y cot y) y 0 y 0 si y si cosy y cosy y 0 si y
LIMITS AND DERIVATIVES 9 y 0 y si y y si cos Eample 5 Evaluate Solutio (i) We ave 0 y cos y sice, si y y si y si cos y ta 0 y 0 si( + ) si( ) si ( + ) si( ) 0 (+ + ) (+ + ) cos si 0 cos si 0 si cos cos 0 si as 0 Eample 6 Fid te derivative of f () a + b, were a ad b are o-zero costats, by first priciple. Solutio By defiitio, f () f + f 0 ( ) ( ) ( ) ( ) b b 0 a + + b a + b 0 ot to be republised Eample 7 Fid te derivative of f () a + b + c, were a, b ad c are oe-zero costat, by first priciple. Solutio By defiitio, f () f + f 0 ( ) ( )
0 EXEMPLAR PROBLEMS MATHEMATICS b + a + a 0 0 a( + ) + b( + ) + c a b c 0 a + a + b b + a Eample 8 Fid te derivative of f (), by first priciple. Solutio By defiitio, f () f + f 0 0 ( ) ( ) ( + ) 0 + + ( + ) ( + ( + )) 0 Eample 9 Fid te derivative of f () Solutio By defiitio, f () by first priciple. f + f 0 ( ) ( ) + 0 0 ( + ). ot to be republised Eample 0 Fid te derivative of f () si, by first priciple. Solutio By defiitio, f () f + f 0 ( ) ( )
LIMITS AND DERIVATIVES si ( + ) si 0 + cos si 0 ( + ) si cos 0 0 cos. cos Eample Fid te derivative of f (), were is positive iteger, by first priciple. Solutio By defiitio, f () f( + ) f ( ) ( + ) Usig Biomial teorem, we ave ( + ) C 0 + C +... + C ( + ) Tus, f () 0 0 Eample Fid te derivative of +. ( +... + ] Solutio Let y + Differetiatig bot sides wit respect to, we get dy d d d ( ) + ( ) d d + 0. ot to be republised
EXEMPLAR PROBLEMS MATHEMATICS 8 + Terefore, d ( + ) d 8 +. Eample Fid te derivative of cos. Solutio Let y cos Differetiatig bot sides wit respect to, we get dy d d ( cos ) d Log Aswer Type Eample Evaluate Solutio Note tat Terefore, d d (cos ) + cos ( ) d d ( si) + cos () cos si si + si si si + 6 si + si ( si ) (si + ) si si + ( si ) (si ) si + si + (si ) (si + ) (si ) (si ) 6 si si 6 si + si 6 + si 6 si 6 (as si 0) ot to be republised
LIMITS AND DERIVATIVES Eample 5 Evaluate Solutio We ave 0 Eample 6 Evaluate Solutio We ave a ta si si 0 ta si si si cos si 0 cos cos si 0 a+ a a+ a+ a a+ si cos si cos 0. a+ a + + a a+ a + + a+ a ( a + )( a+ + ) ( a ) ( a+ + ) ( a + + )( a+ ) ( a + + ) ot to be republised a ( a ) a+ + ( a + + )( a + )
EXEMPLAR PROBLEMS MATHEMATICS Eample 7 Evaluate cos a cosb 0 cos c a a ( a+ b) ( a b) si si Solutio We ave 0 si c Eample 8 Evaluate Solutio We ave 0. 9 ( a+ b) ( a b) si si 0 c si ( a+ b) ( a b) c si si c 0 ( a + b) ( a b) c si a+ b a b 0 0 a+ b a b c a b c ( a+ ) si( a+ ) a sia ( a+ ) si( a+ ) a sia ot to be republised ( a + + a)[si acos + cosasi ] a si a a si a(cos ) a cos asi [ + + ( + a)(siacos+ cosasi )] 0
LIMITS AND DERIVATIVES 5 a si a( si ) a cos asi + + ( + a) si ( a+ ) 0 0 0 a si a 0 + a cos a () + a si a a cos a + a si a. Eample 9 Fid te derivative of f () ta (a + b), by first priciple. Solutio We ave f () ( ) f + f 0 ( ) ( ) ta a ( + ) + b ta ( a + b) 0 si ( a + a + b) si ( a + b) cos ( a + a + b) cos ( a + b) 0 si ( a + a + b) cos ( a + b) si ( a + b) cos ( a + a + b) 0 cos ( a + b) cos ( a + a+ b) asi ( a) 0 a cos ( a + b) cos ( a + a+ b) a sia 0 cos ( a + b) cos ( a + a + b) a 0 a a a sec cos ( a + b) (a + b). [as 0 a 0] Eample 0 Fid te derivative of f () si, by first priciple. Solutio By defiitio, f () ot to be republised f + f 0 ( ) ( )
6 EXEMPLAR PROBLEMS MATHEMATICS si ( + ) si 0 0 ( si ( + ) si ) ( si ( + ) + si ) ( si ( + ) + si ) si ( + ) si 0 ( si ( + ) + si ) + cos si si( + ) + si 0 cos si ( ) cot si Eample Fid te derivative of Solutio Let y cos + si cos. + si Differetiatig bot sides wit respects to, we get dy d d cos d + si d d ( + si ) (cos ) cos ( + si ) d d (+ si ) ot to be republised ( + si ) ( si ) cos (cos ) ( + si )
LIMITS AND DERIVATIVES 7 si si cos ( + si ) ( + si ) ( + si ) + si Objective Type Questios Coose te correct aswer out of te four optios give agaist eac Eample to 8 (M.C.Q.). Eample si 0 ( + cos ) (A) 0 (B) is equal to Solutio (B) is te correct aswer, we ave Eample si 0 ( + cos ) si is equal to cos (C) (D) si cos cos 0 ta 0 ot to be republised (A) 0 (B) (C) (D) does ot eit Solutio (A) is te correct aswer, sice si cos si y y cos y takig y 0
8 EXEMPLAR PROBLEMS MATHEMATICS Eample is equal to 0 cos y si y y 0 y ta y 0 0 y si y 0 y y si cos (A) (B) (C) 0 (D) does ot eists Solutio (D) is te correct aswer, sice R.H.S + 0 ad L.H.S 0 Eample 5 [ ], were [.] is greatest iteger fuctio, is equal to (A) (B) (C) 0 (D) does ot eists Solutio (D) is te correct aswer, sice R.H.S [ ] 0 ad L.H.S [ ] Eample 6 si is equals to 0 (A) 0 (B) (C) Solutio (A) is te correct aswer, sice 0 ad 0 + ot to be republised si, by Sadwitc Teorem, we ave (D) does ot eist
LIMITS AND DERIVATIVES 9 si 0 0 + + +... + Eample 7, N, is equal to (A) 0 (B) (C) + + +... + Solutio (C) is te correct aswer. As ( + ) Eample 8 If f() si, te + f is equal to (D) (A) 0 (B) (C) (D) Solutio (B) is te correct aswer. As f () cos + si So,. EXERCISE Sort Aswer Type Evaluate :.. 9 0 ( + ) f cos + si. 5.. ot to be republised 6 ( + ) ( ) + 6. 0 + 5 5 ( + ) ( a + ) a a
0 EXEMPLAR PROBLEMS MATHEMATICS 7. 9.. 5. Fid, if 6. 0 si + 8 + 7 + si cosm 9. 0 cos. 5. 7. 6 6 8. 0.. + 7 5 + +. 8 + 80, N 5. si cos 6 cot cosec cos 7. 0 0. si si + si5 0 k 8. If k k cos6 si +. 0 + ta + cos 6. 0 si + 0 si a si 7 si si 8. 0.., te fid te value of k. Differetiate eac of te fuctios w. r. to i Eercises 9 to. 9. + + + 0. + si cos si si a a a ot to be republised. ( + 5) ( + ta)
LIMITS AND DERIVATIVES +. (sec ) (sec + ).. 5 7+ 9 5. 7. cos si a+ bsi c+ dcos 6. (a + cot) (p + q cos) 5 cos si 8. (si + cos) 9. ( 7) ( + 5) 0. si + cos. si cos. a + b + c Log Aswer Type Differetiate eac of te fuctios wit respect to i Eercises to 6 usig first priciple.. cos ( + ). 6. cos a + b c + d 5. Evaluate eac of te followig its i Eercises 7 to 5. 7. ( + y) sec( + y) sec y 0 y (si( α+β ) + si( α β ) + si α) 8. 0 cos β cos α 9. ta ta cos + 5. Sow tat 50. does ot eists si cos cos si ot to be republised
EXEMPLAR PROBLEMS MATHEMATICS kcos we 5. Let f() ad if f ( ) f ( ), fid te value of k. + 5. Let f (), fid c if li m f ( ) eists. c > Objective Type Questios Coose te correct aswer out of optios give agaist eac Eercise 5 to 76 (M.C.Q). 5. 55. 56. 57. 58. si is (A) (B) (C) (D) 0 cos is cos (A) (B) ( + ) is 0 (C) (D) (A) (B) (C) (D) 0 m is ot to be republised (A) (B) 0 cosθ cos6 θ is m (C) m (D) m
LIMITS AND DERIVATIVES 59. (A) (B) (C) (D) 9 cosec cot is 0 (A) (B) si 60. is 0 + 6. 6. (C) (D) (A) (B) 0 (C) (D) sec is ta (A) (B) (C) 0 (D) ( )( ) + (A) 0 6. If f () 6. is (B) 0 (C) (D) Noe of tese si[ ],[ ] 0 [ ], were [.] deotes te greatest iteger fuctio, 0,[ ] 0 te f ( ) is equal to 0 (A) (B) 0 (C) (D) Noe of tese ot to be republised si is 0 (A) (B) (C) does ot eist(d) Noe of tese, 0 < < 65. Let f (), te quadratic equatio wose roots are f( ) ad +, < f ( ) + is
EXEMPLAR PROBLEMS MATHEMATICS 66. (A) 6 + 9 0 (B) 7 + 8 0 (C) + 9 0 (D) 0 + 0 0 ta is si (A) (B) 67. Let f () []; R, te (A) 68. If y f is (C) (D) (B) (C) 0 (D) +, te dy at is d (A) (B) 69. If f () (A) 5, te f () is (B) + 70. If y, te dy d is (A) ( ) 5 (B) (C) (D) 0 (C) (D) 0 ot to be republised si + cos 7. If y, te dy at 0 is si cos d (C) (D)
LIMITS AND DERIVATIVES 5 (A) (B) 0 (C) si( + 9) 7. If y, te dy at 0 is cos d (A) cos 9 (B) si 9 (C) 0 (D) 00 7. If f () + + +... +, te f () is equal to 00 (A) 00 a 7. If f () a (B) 00 (C) does ot eist (D) 0 for some costat a, te f (a) is (D) does ot eist (A) (B) 0 (C) does ot eist (D) 75. If f () 00 + 99 +... + +, te f () is equal to (A) 5050 (B) 509 (C) 505 (D) 5005 76. If f () +... 99 + 00, te f () is euqal to (A) 50 (B) 50 (C) 50 (D) 50 Fill i te blaks i Eercises 77 to 80. 77. If f () ta, te f( ) 78. 79. if 80. si m cot, te m 0 ot to be republised y + + + +..., te dy + []!!! d