LIMITS AND DERIVATIVES

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Capter LIMITS AND DERIVATIVES. 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 (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 a

6 EXEMPLAR PROBLEMS MATHEMATICS Limits of polyomials ad ratioal fuctios If f is a polyomial fuctio, te f ( ) eists ad is give by f ( ) f( a) 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 0 (ii) cos (iii) si 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) Derivative of te sum of two fuctio is te sum of te derivatives of te fuctios. (ii) 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) (iv) Derivative of te product of two fuctios is give by te followig product rule. 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 ( ) g ( ). Solved Eamples Sort Aswer Type ( ) Eample Evaluate + Solutio We ave ( ) + ( ) ( )( ( ) ( ) ( )( ) 5+ 6 ( )( ) ( )( ) ( )( ) [ 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 Solutio We ave 08. () 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 (cosec y cot y) y 0 cosy y y 0 si y si cosy y 0 si y

LIMITS AND DERIVATIVES 9 y si y 0 y y si cos y cosy sice, si y y si y si cos y ta y 0 0 Eample 5 Evaluate Solutio (i) We ave si ( + ) si( ) 0 si ( + ) si( ) 0 ( + + ) ( + + ) cos si 0 cos si 0 si cos si cos as 0 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 () 0 f ( + ) f( ) a( + ) + b ( a+ b) 0 0 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, b b f () 0 f ( + ) f( )

0 EXEMPLAR PROBLEMS MATHEMATICS a( + ) + b( + ) + c a b c 0 b + a + a 0 0 a + a + b b + a Eample 8 Fid te derivative of f (), by first priciple. Solutio By defiitio, f () 0 f ( + ) f( ) ( + ) 0 + + ( + ) 0 0 ( + ( + )) Eample 9 Fid te derivative of f() by first priciple. Solutio By defiitio, f () 0 f ( + ) f( ) 0 + + 0 ( ). Eample 0 Fid te derivative of f () si, by first priciple. Solutio By defiitio, f () 0 f ( + ) f( )

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 Eample Fid te derivative of +. Solutio Let y + Differetiatig bot sides wit respect to, we get ( +... + ] 0. dy d d d ( ) + ( ) d d + 0

EXEMPLAR PROBLEMS MATHEMATICS Terefore, 8 + d ( + ) 8 d +. 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 d d (cos ) + cos ( ) d d ( si) + cos () cos si Eample Evaluate Solutio Note tat si + si si si + 6 si + si ( si ) (si + ) si si + ( si ) (si ) Terefore, si + si + (si )(si+ ) (si )(si ) 6 si si 6 si + si 6 + si 6 si 6 (as si 0)

LIMITS AND DERIVATIVES Eample 5 Evaluate Solutio We ave ta si 0 si ta si 0 si si cos si 0 cos cos si 0 si 0 cos si cos. Eample 6 Evaluate a+ a a+ Solutio We ave a+ a a+ a+ a+ + a a+ a+ + a a+ a ( a+ )( a+ + ) ( a ) ( a+ + ) ( a+ + )( a+ )( a+ + ) a ( a ) a+ + ( a+ + )( a+ )

EXEMPLAR PROBLEMS MATHEMATICS a a. 9 Eample 7 Evaluate cos a cosb 0 cos c ( a+ b) ( a b) si si Solutio We ave 0 si c ( 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 a+ b a b c a b c Eample 8 Evaluate ( a+ ) si( a+ ) a sia 0 Solutio We ave ( a + ) si( a+ ) a sia 0 ( a + + a)[siacos+ cosasi ] a sia 0 a si a(cos ) a cosasi [ + + ( + 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. f ( + ) f( ) Solutio We ave f () 0 ( ) ta a( + ) + b ta ( a+ b) 0 0 si ( a + a + b) si ( a + b) cos ( a + a + b) cos ( a + b) 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 si a 0 cos ( a + b) cos ( a + a + b) a 0 a [as 0 a 0] a cos ( a + b) a sec (a + b). Eample 0 Fid te derivative of f ( ) si, by first priciple. Solutio By defiitio, f () 0 f ( + ) f( )

6 EXEMPLAR PROBLEMS MATHEMATICS 0 0 si ( + ) si ( si ( + ) si ) ( si ( + ) + si ) ( si ( + ) + si ) si ( + ) si 0 ( si ( + ) + si ) + cos si 0 si( + ) + si ( ) cos si cot si Eample Fid te derivative of cos. + si Solutio Let y 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 ) ( + 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 ) is equal to (A) 0 (B) Solutio (B) is te correct aswer, we ave (C) (D) si 0 ( + cos ) si cos 0 cos ta 0 Eample si is equal to cos (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 cosy y 0 si y y si y 0 y y si cos y ta y 0 0 Eample 0 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 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 0 is equals to (A) 0 (B) (C) Solutio (A) is te correct aswer, sice (D) does ot eist 0 ad 0 si, by Sadwitc Teorem, we ave

LIMITS AND DERIVATIVES 9 si 0 0 + + +... + Eample 7, N, is equal to (A) 0 (B) (C) (D) + + +... + Solutio (C) is te correct aswer. As ( + ) + Eample 8 If f() si, te f is equal to (A) 0 (B) (C) (D) Solutio (B) is te correct aswer. As f () cos + si So, f cos + si. EXERCISE Sort Aswer Type Evaluate :. 9.. 0 +. ( + ) 0 5. 6 ( + ) ( ) + 6. 5 5 ( + ) ( a+ ) a a

0 EXEMPLAR PROBLEMS MATHEMATICS 7. 8. + 9. + 8 0. 7 5 + +. + 0. 5 + 7 +. 8 +. Fid, if 80, N 5. si a si 7 6. si 0 si cos 7. 0 si si 8. 0 cosm 9. 0 cos 0. cos6. si cos. 6 si cos 6 si +. 0 + ta. si si a a a 5. 6 cot cosec + cos 6. 0 si 7. si si + si 5 0 k 8. If k k, te fid te value of k. Differetiate eac of te fuctios w. r. to i Eercises 9 to. 9. + + + 0. +. ( + 5) ( + ta)

LIMITS AND DERIVATIVES +. (sec ) (sec + ). 5 7+ 9. 5 cos si 5. 7. cos si a+ bsi c+ dcos 6. (a + cot) (p + q cos) 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 ( + ). a + b c + d 5. 6. cos 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

EXEMPLAR PROBLEMS MATHEMATICS k cos we 5. Let f () fid te value of k. ad if f( ) f( ), 5. Let f () +, fid c if f ( ) eists. c > Objective Type Questios Coose te correct aswer out of optios give agaist eac Eercise 5 to 76 (M.C.Q). 5. si is (A) (B) (C) (D) 55. 0 cos cos is (A) (B) (C) (D) 56. ( + ) is 0 (A) (B) (C) (D) 0 57. m is (A) (B) m (C) m (D) m 58. 0 cosθ cos6 θ is

LIMITS AND DERIVATIVES 59. 60. (A) 9 (B) cosec cot is 0 (A) si 0 + (B) is (C) (C) (D) (D) (A) (B) 0 (C) (D) 6. sec ta is 6. (A) (B) (C) 0 (D) ( ) ( ) + is (A) 0 (B) 0 (C) (D) Noe of tese si[ ],[ ] 0 6. If f () [ ], were [.] deotes te greatest iteger fuctio, 0,[ ] 0 te f ( ) is equal to 0 (A) (B) 0 (C) (D) Noe of tese 6. si 0 is (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 (A) 6 + 9 0 (B) 7 + 8 0 (C) + 9 0 (D) 0 + 0 66. 0 ta si is (A) (B) (C) (D) 67. Let f () []; R, te f is (A) (B) (C) 0 (D) 68. If y +, te dy d at is (A) (B) (C) (D) 0 69. If f (), te f () is (A) 5 (B) 5 (C) (D) 0 70. If + y, te dy d is (A) ( ) (B) (C) (D) 7. If si + cos y si cos, te dy d at 0 is

LIMITS AND DERIVATIVES 5 (A) (B) 0 (C) (D) does ot eist 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 (B) 00 (C) does ot eist (D) 0 7. If f( ) a a for some costat a, te f (a) is (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. si m cot, te m 0 79. if y + + + +..., te dy!!! d 80. + [ ]

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