AIEEE Mathematics - Limits And Derivatives

AIEEE LIMITS AND DERIVATIVES MATHEMAT I C S S T U D Y M A T E R I A L NARAYANA INSTITUTE OF CORRESPONDENCE COURSES F N S H O U S E , 6 3 K A L U S A R A I M A R K E T S A R V A P R I Y A V I H A R , N E W D E L H I -1 1 0 0 1 6 PH.: (011) 32001131/32/50 • FAX : (011) 41828320 Website : w w w . n a r a y a n a i c c . c o m E-mail : i n f o @n a r a y a n a i c c . c o m 2004 NARAYANA GROUP This study material is a part of NARAYANA INSTITUTE OF CORRESPONDENCE COURSES for AIEEE, 2007-08. This is meant for the personal use of those students who are enrolled with NARAYANA INSTITUTE OF CORRESPONDENCE COURSES, FNS House, 63, Kalu Sarai Market, New Delhi-110016, Ph.: 32001131/32/50. All rights to the contents of the Package rest with NARAYANA INSTITUTE. No other Institute or individual is authorized to reproduce, translate or distribute this material in any form, without prior information and written permission of the institute.PREFACE Dear Student, Heartiest congratulations on making up your mind and deciding to be an engineer to serve the society. As you are planning to take various Engineering Entrance Examinations, we are sure that this STUDY PACKAGE is going to be of immense help to you. At NARAYANA we have taken special care to design this package according to the Latest Pattern of AIEEE, which will not only help but also guide you to compete for AIEEE & other State Level Engineering Entrance Examinations. The salient features of this package include : ! Power packed division of units and chapters in a scientific way, with a correlation being there. ! Sufficient number of solved examples in Physics, Chemistry & Mathematics in all the chapters to motivate the students attempt all the questions. ! All the chapters are followed by various types of exercises (Level-I, Level-II, Level-III and Questions asked in AIEEE and other Engineering Exams). These exercises are followed by answers in the last section of the chapter. This package will help you to know what to study, how to study, time management, your weaknesses and improve your performance. We, at NARAYANA, strongly believe that quality of our package is such that the students who are not fortunate enough to attend to our Regular Classroom Programs, can still get the best of our quality through these packages. We feel that there is always a scope for improvement. We would welcome your suggestions & feedback. Wish you success in your future endeavours. THE NARAYANA TEAM ACKNOWLEDGEMENT While preparing the study package, it has become a wonderful feeling for the NARAYANA TEAM to get the wholehearted support of our Staff Members including our Designers. They have made our job really easy through their untiring efforts and constant help at every stage. We are thankful to all of them. THE NARAYANA TEAMCONTENTS LIMITS AND DERIVATIVES Theory Solved Examples Exercises Level – I Level – II Level – III Questions asked in AIEEE and other Engineering Exams Answers C O N T E N T S FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 1 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES LIMITS AND DERIV DERIVATIVES ATIVES AIEEE Syllabus Limits, Differentiation of the sum, difference, product and quotient of two functions, differentiation of trigonometric, inverse trigonometric, logarithmic, exponential, composite and implicit functions, derivatives of order up to two. CONTENTS ♦ Definition of a limit ♦ Trigonometric limits ♦ Exponential and logarithmic limits ♦ Approximations ♦ Some useful expansions ♦ Indeterminate forms ♦ Limit of greatest integer function ♦ Sandwich Theorem ♦ Derivative of a function ♦ Some differentiation formulae ♦ Algebra of differentiation ♦ Differentiation of implicit functions ♦ Derivative of parametric functions ♦ Derivative of a function w.r.t. another function ♦ Use of log in finding derivatives of the function of type (f(x))g(x) ♦ Differentiation using trigonometrical substitutions ♦ Higher order differentiation ♦ Derivative of infinite series ♦ Differentiation of a determinant function INTRODUCTION This chapter is an introduction to calculus. Calculus is that branch of mathematics which mainly deals with the study of change in the value of a function as the points in the domain change. In this chapter we define limit and some algebra of limits. Also we study derivative and algebra of derivatives and derivatives of certain standard functions.Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 2 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 1. DEFINITION If y = f(x) is any function which is defined in a neighbourhood of a then for some ‘∈’ greater than zero there exists a δ > 0 such that |f(x) – l| < ∈ |x a| ⇒ − <δ then l is said to be limit of the function when x-approches a. It is symbolically written as x a Lt f (x) → = l 2. STANDARD FORMULA n n n 1 x ax a Lt na ; x a x a − → − = ≠ − ; n is a rational number or integer. Remark : m m m n n n x ax a m Lt a x a n − → − = − 3. TRIGONOMETRIC LIMITS (i) x 0 sin x lim 1 x → = (ii) x 0 lim cos x 1 → = (iii) x 0 tan x lim 1 x → = (iv) 1 x 0 sin x lim 1 x− → = (v) 1 x 0 tan x lim 1 x− → = (vi) o x 0 sin x lim x 180 → π = 4. EXPONENTIAL AND LOGARITHMIC LIMITS (i) x x 0e 1 Lt 1 x → − = (ii) x e x 0a 1 Lt log a x → − = (a > 0) (iii) x x e x 0a b a lim log , a, b 0 x b → −   = >     (iv) n x 0 (1 x) 1 lim n x → + − = (v) x 1/x x 0 x 1 Lt (1 x) e Lt 1 x → →∞  + = = +     (vi) 1/x a x 0 lim (1 ax) e → + = (vii) x a x a lim 1 e x →∞  + =     (viii) f (x) x 1 lim 1 e f (x) →∞  + =     , where f (x)→∞ as x→∞ (ix) ( )1/f (x) x a lim 1 f (x) e → + = (x) m x log x lim 0 x →∞ = (m > 0) (xi) a a x 0 log (1 x) lim log e x → + = (a 0, a 1) > ±FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 3 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 5. APPROXIMATIONS (i) sin ax ax ! (ii) 2 2 a x cos ax 1 2 − ! (iii) tan ax ax ! (iv) ax e 1 ax + ! (v) ax e 1 ax − − ! (vi) log(1 + ax) ! ax (vii) ax ! 1 + (logea)x (viii) sinhax ! ax (ix) tanhax ! ax (x) coshax ! 2 2 a x 1 2 + (xi) n x 1 x 1 n ± ± ! , |x| < 1 6. SOME USEFUL EXPANSIONS If x → 0 and there is at least one function in the given expansion which can be expanded, then we express numerator and denominator in the ascending powers of x and remove the common factor there, the following expansions of some standard functions should be remembered. (a) 2 x x x e 1 ...... 1 2 = + + + (b) 2 3 x x x x e 1 ..... 1 2 3 − = − + − + (c) 2 2 x (loga)x (loga) x a 1 ...... 1 2 = + + + (d) 2 3 x x log(1 x) x ..... 2 3 + = − + − (e) 2 3 x x log(1 x) x ..... 2 3   − =− + + +     (f) 2 4 x x cos x 1 ....... 2 4 = − + − (g) 3 5 x x sin x x ....... 3 5 = − + − (h) 3 5 x x sin h x x ....... 3 5 = + + + (i) 3 5 x 2x tan x x ....... 3 15 = − + − (j) 3 5 x 2x tan h x x ....... 3 15 = + + + (k) 2 4 x x cosh x 1 ....... 2 4 = + + + (l) 3 5 1 x 9x cos x x ...... 2 3 5 −   π = − + + +     (m) 3 5 1 x 9x sin x x ...... 3 5 − = + + + (n) 3 5 7 1 x x x tan x x ...... 3 5 7 − = − + − + (o) n 2 n(n 1) (1 x) 1 nx x ....... 2− + = + + + where n z+ ∈ (p) x 2 1 x 11 1 e 1 x ...... x 2 24     + = − + +        Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 4 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 7. INDETERMINATE FORMS The forms which cannot be defined exactly are called indeterminate forms, they are 00 , ∞∞ , 0×∞ , ∞ −∞ , 00, 0 ∞ and 1∞ L’ HOSPITAL’S RULE If x a f (x) Lt g(x) → takes the form of 00 or ∞∞ then the limit of the function is x a f (x) Lt g (x) → ′′ , if x a f (x) Lt g (x) → ′′ itself takes the form again 00 , ∞∞ then the limit of the function is x a f (x) Lt g (x) → ′′′′ and the process is continued till 0 , 0 ∞∞ is eliminated then limit is obtained. 1. If 0×∞ form is given, convert it in the form of 00 , ∞∞ by taking one term to the denominater then apply L’Hospital’s Rule. 2. If ( ) ∞−∞ form is given, take L.C.M convert it in the form of 00 or ∞∞ form, then take the help of L’Hospital’s Rule. 3. 00 and 0 ∞ form is given take the help of logarithms convert the problem again in the form of 00 or ∞∞ form and then use L’Hospital’s Rule. 4. If [ ]g(x) x a Lt f (x) → takes the form of 1∞ then write it as ( ) x a Lt g(x)[f (x) 1] g(x) x a Lt f (x) e → − → = 8. LIMIT OF GREATEST INTEGER FUNCTION Greatest integer function is denoted by [.] Let a R ∈ then two cases arise. Case (1) if a ∈ integer then we have 1. x a Lt [x] a + → = 2. x a Lt [x] a 1 − → = − 3. x a Lt [x] → does not exist Case (2) If a∉ integer then x c Lt [x] c → =FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 5 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Example : If sin[x] f (x) , [x] 0 [x] = ≠ = 0 [x] = 0 where [x] denotes the greatest integer less than or equal to x, then find x 0 Lt f (x) → Solution : h 0 h 0 Lt f (0 h) Lt f (0 h) → → − = + ⇒ h 0 h 0 sin[ h] sin[h] Lt Lt [ h] [h] → → − = − ⇒ sin1 1 ≠ 9. SANDWICH THEOREM Suppose that g(x) ≤ f(x) ≤ h(x) for all x in some open interval containing c, except possibly at ‘c’ itself. Suppose also that x c x c Lt g(x) Lt h(x) → → = =! then x c Lt f (x) → = ! h(x) f(x) g(x) c y O! This is called sandwich theorem. 10. SPECIAL TYPES OF LIMITS 1. Use of Leibnitz’s formula for evaluating the limit Consider the integral (x) (x) g(x) f (t)dt ψφ = ∫ then g (x) f[ (x)] (x) f ( (x)) (x) ′ ′ ′ = ψ ψ − φ φ 2. Summation of series using definite integral as the limit of a sum. It is used in the expression of the form b n n r 1 a 1 r Lt f f (x)dx n n →∞ =  =     ∑ ∫ To Evaluate such limits we note the following (a) ∑ is replaced by sign of integration (b) r x n → (r = x , n = 1) (c) 1 dx n → (d) Lower limit is always zero. (e) Upper limit is Coefficient of n in the upper limit of ΣMathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 6 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES DERIVATIVES 11. DERIVATIVE OF A FUNCTION Let y = f(x) be a function defined on an interval [a, b]. Let for a small increment δx in x, the corresponding increment in the value of y be δy. Then y = f(x) and y + δy = f(x + δx) On subtraction, we get δy = f(x + δx) – f(x) or y f (x x) f (x) x x δ +δ − = δ δ Taking limit on both sides when x 0 δ → we have, x 0 x 0 y f (x x) f (x) lim lim x x δ → δ → δ +δ − = δ δ if this limit exists, is called the derivative or differential coefficient of y with respect to x and is written as dy dx or f (x) ′ . Thus ∴ x 0 x 0 dy y f (x x) f (x) lim lim dx x x δ → δ → δ +δ − = = δ δ . This is called Differentiation from first principle. Derivative at a point: The value of f ′(x) obtained by putting x = a, is called the derivative of f(x) at x = a and it is denoted by f ′(a) or x a dy dx =       Note : dy dx is ( ) d y dx in which d dx is simply a symbol of operation and not 'd' divided by dx. 12. SOME DIFFERENTIATION FORMULAE (i) dx d (constant) = 0 (ii) dx d (xn) = nxn–1 (iii) dx d (ex) = ex (iv) dx d (ax) = ax loge a (v) dx d (logex) = 1x (vi) dx d (loga x) = a log x 1e (vii) dx d (sin x) = cos x (viii) dx d (cos x) = – sinxFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 7 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES (ix) dx d (tan x) = sec2 x (x) dx d (cot x) = – cosec2x (xi) dx d (sec x) = sec x tan x (xii) dx d (cosec x) = – cosecx cotx (xiii) dx d (sin–1x) = 2 x 1 1− (xiv) dx d (cos–1x) = – 2 x 1 1− (xv) dx d (tan–1x) = 2 x 1 1 + (xvi) dx d (cot–1x) = – 2 x 1 1 + (xvii) dx d (sec–1x) = 2 1 | x | x 1 − (xviii) dx d (cosec–1x) = 2 1 | x | x 1 − − (xix) dx d (eax sin bx) = eax (a sin bx + b cos bx) = 2 2 b a + eaxsin (bx + tan–1 ab ) (xx) dx d (eax cos bx) = eax (a cos bx – b sin bx) = 2 2 b a + eaxcos (bx + tan–1 ab ) (xxi) dx d |x| = | x | x or | x | x : ≠ x 0 (xxii) dx d log |x| = x1 13. ALGEBRA OF DIFFERENTIATION (i) Sum and difference rule 1 2 1 2 d d d [f (x) f (x)] [f (x)] [f (x)] dx dx dx ± = ± (ii) Scalar multiple rule d d [k f(x)] k [f(x)] dx dx = , where k is any constant (iii) Product rule 1 2 1 2 2 1 d d d [f (x).f (x)] [f (x)] [f (x)] [f (x)] [f (x)] dx dx dx = + (iv) Quotient rule −  =     2 1 1 2 1 2 2 2 d d f (x) [f (x)] f (x) [f (x)] f (x) d dx dx dx f (x) [f (x)] (v) Chain rule if y = f1(u), u = f2(v) and v = f3(x) then x d v d . v d u d . u d y d x d y d =Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 8 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 14. DERIVATIVE OF PARAMETRIC FUNCTIONS Let x and y are two functions of variable 't' (parameter) such that x = f(t) and y = g(t). then dy dy g (t) dt dx dx f (t) dt     ′ = =   ′       Example -1 If x = a(cosθ + θsinθ), y = a(sinθ – θcosθ) then find dy dx . Solution : dx a[ sin sin cos ] a cos d = − θ+ θ+θ θ = θ θ θ ; dy a(cos cos sin ) a sin d = θ− θ+θ θ=θ θ θ ∴ dy tan dx= θ 15. DIFFERENTIATION OF IMPLICIT FUNCTIONS If in an equation, x and y both occur together. i.e. f(x, y) = 0 or f(x, y) = c and this function can not be solved either for 'y' or 'x' then f(x, y) is called the implicit function of x (or y). If xy + yx = ab, then y 1 x y x1 f dy (yx y logy) x f dx (x logx xy ) y − − ∂ − − + ∂ = = ∂ + ∂ 15.1. WORKING RULE FOR FINDING THE DERIVATIVE Method – 1 (i) Differentiate every term of f(x, y) = 0 with respect to 'x'. (ii) Collect the coefficients of dy dx and obtain the value of dy dx . Method – 2 If f(x, y) = constant, then xy f f dy x f dx f y ∂ − ∂ = =− ∂∂ where fx ∂∂ and fy ∂∂ are partial differential coefficients of f(x, y) with respect to x and y respectively. Note : An implicit function can be differentiated either with respect to 'x' or with respect to 'y' 16. DERIVATIVE OF A FUNCTION WITH RESPECT TO ANOTHER FUNCTION Let y = f(x) and z = g(x) be two functions of 'x' then the derivative of f(x) w.r.t g(x) or derivative of y is denoted by dy dz i.e. dy dy f (x) dx dz dz g (x) dx     ′   = = ′      FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 9 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Example -2 Find the derivative of 1 2 2x sin 1 x −    +   with respect to 2 1 2 1 x cos 1 x −  −   +   Solution : Let f(x) = 1 1 2 2x sin 2tan x 1 x − −  =   +   ∴ 2 2 f (x) 1 x ′ = + Let 2 1 1 2 1 x g(x) cos 2tan x 1 x − −   − = =   +   ∴ 2 2 g (x) 1 x ′ = + Hence the derivative of f(x) with respect to g(x) is 22 f (x) 2/1 x 1 g (x) 2/1 x ′ + = = ′ + 17. USE OF LOG IN FINDING DERIVATIVES OF THE FUNCTION OF TYPE ( )g(x) f(x) Let [ ]g(x) y f(x) = Taking log on both sides we get log y = g(x) . log f(x) Differentiating we get [ ]g(x) dy f (x) f(x) g (x) logf(x) g(x) dx f(x) ′   ′ = ⋅ +     Example -3 { } tanx tanx 2 d tanx x x secxlogx dx x   = +     18. DIFFERENTIATION USING TRIGONOMETRICAL SUBSTITUTIONS (i) sin–1 x ± sin–1 y = sin–1 ] x 1 y y 1 x [ 2 2 − ± − (ii) cos–1 x ± cos–1 y = cos–1 2 2 [xy (1 x )(1 y )] − − ∓ (iii) tan–1 x ± tan–1 y = tan–1     ± y x 1 y x∓ (iv) 2 sin–1 x = sin–1 ) x 1 x 2 ( 2 − (v) 2 cos–1 x = cos–1 (2x2 – 1) (vi) 2 tan–1 x = tan–1     − 2 x 1 x 2 (vii) 2 tan–1 x = sin–1       + 2 x 1 x 2 (viii) 2 tan–1 x = cos–1     +− 22 x 1 x 1 (ix) 4π – tan–1 x = tan–1       +− x 1 x 1 (x) 3 sin–1 x = sin–1 (3x – 4x3)Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 10 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES (xi) 3 cos–1 x = cos–1 (4x3 – 3x) (xii) 3 tan–1 x = tan–1     −− 23 x 3 1 x x 3 (xiii) sin–1x + cos–1x = 2π (xiv) tan–1x + cot–1x = 2π (xv) sec–1x + cosec–1x = 2π 19. SUITABLE SUBSTITUTION (i) If the function involve the term 2 2 x a − , then put x = a sin θ, or x = a cos θ (ii) If the function involve the term 2 2 a x + , then put x = a tan θ (iii) If the function involve the term 2 2 a x − , then put x = a sec θ (iv) If the function involve the term     +− x a x a , then put x = a cos θ 20. nth DIFFERENTIATION OF SUITABLE FUNCTION (1) Dn(ax + b)m = m(m – 1) (m – 2) ........... (m – n + 1) an(ax + b)m–n (2) If m ∈ N and m > n, then Dn(ax + b)m = ! ) n m ( ! m− an(ax + b)m–n n m m n x ! ) n m ( ! m ) x ( D − − = (3) Dn(ax + b)n = n ! an Dn(xn) = n ! (4) + −  =   + +   n n n n 1 1 (1)n!a D ax b (ax b) + −  =     n n n 1 1 ( 1)n! D x x (5) Dn{log (ax +b)} = − − − + n 1 n ( 1) (n 1)! (ax b) an Dn(log x) = − − − n 1 n ( 1) (n 1)! x (6) Dn(eax) = an eax (7) Dn (amx) = (log a)n amx . mnFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 11 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES (8) Dn{sin (ax + b)} = an sin (ax + b + n π2 ) Dn(sin x) = sin (x + n π2 ) (9) Dn{cos (ax + b)} = an cos (ax + b + n π2 ) Dn(cos x) = cos (x + n π2 ) (10) Dn{eax sin (bx + c)} = (a2 + b2)n/2 eax sin (bx + c + n tan–1 ba ) (11) Dn{eax cos (bx + c)} = (a2 + b2)n/2 eax cos (bx + c + n tan–1 ba ) (12) Dn(tan–1 xa ) = n 1 n n ( 1) (n 1) ! sin sin n a − − − θ θ Where θ = tan–1 ( ax ) (13) Dn(tan–1 x) = (–1)n–1 (n–1) ! sinn θ sin nθ Where θ = tan–1 ( 1x ) Example -4 If x = a cos3 θ, y = a sin3 θ, then find d y d x . Solution : Since x = a cos3 θ ∴ θ θ = θ θ θ 3 d x d (cos ) d(cos ) a . d (cos) d (Using chain rule) = 3a cos2 θ (–sin θ) = –3a cos2 θ sin θ and y = a sin3 θ ∴ 3 d y d (sin ) d(sin ) a . d d(sin) d θ θ = θ θ θ = 3a sin2 θ . cos θ Now,     θ θ θ   = = =− θ − θ θ     θ   2 2 d y d d y 3a sin cos tan d x 3a cos sin d x d 21. HIGHER ORDER DIFFERENTIATION If y = f(x) be a differentiable function of x, such that whose second, third............, nth derivatives exist. The first, second, third ........., nth derivatives of y = f(x) are denoted respectively by 2 2 dxy d , x d y d , ......., n n dxy dMathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 12 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Also denoted by y', y"........y(n) y1, y2, y3..............,yn f', f".............fn dy, d2y, d3y,.........dny Example -5 If y = xx ln then find 2 2 d y dx Solution : We have y = xx ln xy = ln x .....(1) Differentiating both sides w.r.t. x, we get x x1 1 . y x d y d = + ⇒ x2 x d y d + xy = 1 ⇒ x2 x d y d + ln x = 1 [From (1)] .....(2) Again differentiating both sides w.r.t. x, we get x2 0 x1 x 2 . x d y d dxy d 2 2 = + + ⇒ 0 1 x d y d x 2 dxy d x 2 2 2 3 = + + ⇒ x3 2 2 dxy d + 2 (1 – ln x) + 1 = 0 [from (2)] Hence ) 3 x ln 2 ( dxy d x 2 2 3 − = or 3 2 2 x 3 x ln 2 dxy d − = 22. DERIVATIVE OF INFINITE SERIES If taking out one or more than one terms form an infinite series. (A) If y = ∞ + + + ......... ) x ( f ) x ( f ) x ( f then y = y ) x ( f + ⇒ (y2 – y) = f(x) Differentiating both sides w.r.t. x, we get (2y – 1) x d y d = f '(x) (B) If y = { }{ }....... f (x) f (x) {f(x)} ∞ then y = {f(x)}y ⇒ y = ey !n f(x)FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 13 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Differentiating both sides w.r.t. x, we get x d y d = ey ln f(x) 1 dy y f'(x) lnf(x) f(x) dx   ⋅ ⋅ + ⋅     ⇒ x d y d = {f(x)}y y dy f '(x) ln f(x) f(x) dx   ⋅ + ⋅     ⇒ {1 – {f(x)}y ln f(x)} x d y d = y {f(x)}y–1 . f '(x) (C) y = f(x)f(x) then ′ = ⋅ + f (x) dy f(x) f (x)[1 logf(x)] dx = + x x d (x ) x (1 logx) dx = + sin x sin x d(sinx ) (sin x) .cos x[1 logsinx] (D) y = f(x)g(x) then ′   ′ = + ⋅     g(x) dy f (x) f(x) g (x)logf(x) g(x) dx f(x) { }   = +     logx logx d 1 (sinx) (sinx) log(sinx) logx.cot x dx x (E) 1 y f(x) 1 f(x) ..... f(x) = + + ∞ + then ′ = + 22 dy y f (x) dx y 1 (F)   + =   −   1 f(x) y log 1 f(x) then ′ = − 2 dy 2f (x) dx 1 f (x) +   =   −   1 tanx y log 1 tanx then = − 22 dy 2sec x dx 1 tan x Example -6 If y x y x y ....... = + + + + ∞ then find x d y d Solution : We have y = ∞ + + + + ...... y x y x ⇒ y = y y x + + ⇒ y2 – x = y 2 ⇒ (y2 –x)2 = 2yMathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 14 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Differentiating both sides w.r.t. x, we get 2 (y2 – x) x d y d 2 1 x d y d y 2 =       − ⇒ (2y (y2 – x) –1) x d y d = (y2 – x) Hence ) 1 xy 2 y 2 ( ) x y ( x d y d 3 2 − − − = 23. DIFFERENTIATION OF A DETERMINANT FUNCTION If F(x) = w v u n m h g f! Where f, g, h, !, m, n, u, v, w are functions of x and differentiable then F'(x) = ' w ' v ' u n m h g f w v u ' n ' m ' h g f w v u n m ' h ' g ' f ! ! ! + + or F'(x) = f ' g h f g' h f g h' ' m n m' n m n' u' v w u v' w u v w' + + ! ! !FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 15 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Example -1 x x x 0a b Lt tan x → − = (1) log(a/b) (2) log(b/a) (3) logab (4) a/b Solution : Ans. (1)x x x 0 (a 1) (b 1) x Lt x tanx → − − −       x x x 0 a 1 b 1 Lt x x → − − − x 0 tan x ( Lt 1) x → = ∵ = loga – logb = log(a/b) Example -2 1/x x 0 Lt (cos x a sin bx) → + = (1) eab (2) ea/b (3) log(a/b) (4) logab Solution : Ans. (1) Given limit is in the form 1∞ x 0 1 Lt [cos x a sin bx 1] x e → + − 2 x cos x 1 2     −       ∵ " 2 x 0 x 1 abx1 2 Lt ab x e e →  − + −       = = (sin bx bx " ) Example -3 n 1 1 1 Lt .... 3.5 5.7 (2n 1)(2n 3) →∞  + + + =   + +   (1) 12 (2) 13 (3) 16 (4) 14 Solution : Ans. (3) n 1 1 1 1 1 1 1 Lt .... 2 3 5 5 7 2n 1 2n 3 →∞  − + − + + −   + +   1 1 1 2 3 6 = × = SOLVED EXAMPLESMathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 16 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Example -4 10 10 10 10 10 10 x (x 1) (x 4) (x 9) ... (x 100) Lt x 100 →∞ + + + + + + + + = + (1) 100 (2) 1000 (3) 10 (4) 1 Solution : Ans. (3) 10 10 10 10 10 x 10 10 1 4 100 x 1 1 .... 1 x x x Lt 100 x 1 x →∞          + + + + + +                     +    = 1 + 1 + ..... + 10 times = 10 Note : If the degree of numerater and denomenater are equal, then the ratio of constant terms is the limit when x 0 → and the ratio of coefficients of highest degree terms is the limit when x→∞. Example -5 2 x0 3 x 0 sin t dt Lt x → = ∫ (1) 32 (2) 13 (3) 12 (4) 23 Solution : Ans. (4) 2 x0 x 0 3 d sin t dt dx Lt d (x ) dx →           ∫ = 2 x 0sin x (2x) Lt 3x → (Use Leibnitz’s rule and sinx " x) 2 x 02x x 2 Lt 3x 3 → ⋅ = = Example -6 2 x [x] [2x] .... [nx] Lt n →∞ + + + is, where [.] denotes the greatest integer function (1) x3 (2) x6 (3) does not exist (4) x2 Solution : Ans. (4) Using the fact nx 1 [nx] nx − < ≤FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 17 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES ∴ (nx 1) [nx] nx − < ≤ ∑ ∑ ⇒ n r 1 n(n 1) n(n 1) x n [rn] x 2 2 = + + − < ≤ ⋅ ∑ ∴ n 2 2 2 n n n r 1 x n(n 1) n x.n(n 1) Lt Lt [rx] Lt 2 n n 2n →∞ →∞ →∞ = + +   ⋅ − < ≤     ∑ ⇒ n n r 1 x x Lt [rx] 2 2 →∞ = ≤ ≤ ∑ ∴ n n r 1 x Lt [rx] 2 →∞ = = ∑ (Using sandwich theorem) Example -7 2 2 x 0 1 1 Lt sin x sinh x →  − =     (1) 23 (2) 0 (3) 13 (4) 23 − Solution : Ans. (1) 2 2 2 2 2 2 4 x 0 x 0 sinh x sin x sinh x sin x Lt Lt sinh xsin x x → → − − =3 x 0 2sinhxcosh x 2sinxcosx Lt (L.H.R) 4x → − = 3 x 0 sinh 2x sin 2x Lt 4x → − = 2x 2x 3 x 0e e sin 2x 2 Lt 4x − → − − = 3 2x 3 2x 3 x 0 2 e ( 2) e 3 2 sin 2x 2 2 Lt 4.3! − → ⋅ − − π   − +     = (∵ the degree of denominator is 3, we take the 3rd order derivative) 8 8 8 16 2 2 4.3.2 8.3 3 + + = == = Example -8 Let a, b be the distinct roots of ax2 + bx + c = 0, then 2 2 x 1 cos(ax bx c) Lt (x ) →α− + + −α is equal to (1) 2 ( ) 2 α−β (2) 2 2 a ( ) 2 − α−β (3) 0 (4) 2 2 a ( ) 2 α−βMathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 18 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Solution : Ans. (4) 2 x 1 cosa(x )(x ) Lt (x ) →α− −α −β −α (∵ 2 ax bx c a(x )(x ) + + ≡ −α −β ) 2 2 a x cosax 1 2   −     " 2 2 2 2 2 2 x a (x ) (x ) a ( ) Lt 2(x ) 2 →α −α −β α−β = = −α Example -9 3 x 2 (1 tan x /2)(1 sin x) Lt (1 tan x /2)( 2x) π → − − = + π− (1) 18 (2) 0 (3) 1 32 (4) ∞ Solution : Ans. (3) 3 x 2 (1 tan x /2)(1 sin x) lim (1 tan x /2)( 2x) π → − − + π− If x 2π → then h 0 → put x h 2π = − 3 h 0 h 1 tan 1 cosh 4 2 lim h (2h) 1 tan 4 2 → π  − −   −   = ⋅ π  + −     3 h 0 h 1 cosh lim tan 4 4 2 8h → π π −   = + + ⋅     2 h cosh 1 2     −       " 2 2 3 3 h 0 h 0 h h h h tan 1 2 2 2 2 lim lim 8h 8h 32 → → ⋅ ⋅ = = = Example -10 2 4 2 4 4 x 0 tan[e ]x tan[ e ]x Lt sin x → − − = (1) 0 (2) 15 (3) 8 (4) 7 Solution : Ans. (2) e2 = (2.718)2 = 7.3875 [e2] = 7, [–e2] = –8 ( tanax ax ∵ " , sin x x " ) ∴ 4 4 4 4 4 x 0 tan 7x tan8x 15x Lt 15 sin x x → + = = Example -11          −+ − x 1 x 1 cot sin x dd 1 2 = (1) 21 − (2) 0 (3) 21 (4) –1FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 19 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Solution : Ans. (1) Let y = sin2     −+ − x 1 x 1 cot 1 . Put x = cos 2θ. ∴ y = sin2 −     + θ         − θ       1 1 cos 2 cot 1 cos 2 = sin2 cot–1 (cot θ) ∴ y = sin2 θ = 2 2 cos 1 θ − = 2x 1− = 2x 21 − ∴ 21 x d y d − = . Example -12 If ) b x )( x a ( y − − = -(a – b) tan–1     −− b x x a then find x d y d . (1)     −− b x a x (2)     +− b x x a (3)     −− b x x a (4) x a x b ++ Solution : Ans. (3) Let x = a cos2θ + b sin2θ ∴ a – x = a – a cos2θ – b sin2θ = (a – b) sin2θ .....(1) and x – b = a cos2θ + b sin2θ – b = (a – b) cos2θ .....(2) ∴ y = (a – b) sin θ cos θ – (a – b) tan–1 tanθ = 2 ) b a ( − sin2θ – (a – b) θ Then θ = θ θ − = θ − − − θ − =       θ       θ = tan 2 sin 2 cos 1 2 sin ) a b ( ) b a ( 2 cos ) b a ( d x dd y d x d y d =     −− b x x a [From (1) and (2)] Example -13 Derivative of sec–1     −   21 2x 1 w.r.t x 3 1+ at x = 31 − is-(1) 0 (2) 21 (3) 31 (4) 61Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 20 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Solution : Ans. (1) Let y = sec–1     −   21 2x 1 and z = x 3 1+ ∴ 2 2 2 2 2 2 2 4x dy 2 4x 1 3x dy (2x 1) 1 (2x 1) dx 3 dz 3 dx 1 1 1 1 dx 2 1 3x 2x 1 2x 1 − − × + − ⋅ − = = ⋅ =     − −     + − −     ∴ 1 x 3 d y 0 d x − =     =       . Example -14 If x = θ – θ1 and y = θ + θ1 , then x d y d = (1) yx (2) xy (3) yx − (4) yx − Solution : Ans. (1) x = θ – θ1 ⇒ θ d x d = 1 + 2 1 θ , y = θ + θ1 ⇒ θ d y d = 1 – 2 1 θ ∴ yx 11 1 1 1 1 d x dd y d x d y d 22 = θ + θ θ − θ = θ + θ − = θθ = Example -15 If y =              +− − − 2x tan b a b a tan . ) b a ( 2 1 2 2 then find 2 2 d y dx . (1) 2 ) x cos b a ( x cos b + (2) 2 ) x cos b a ( x sin b + (3) 2 ) x cos a b ( x sin b + (4) ( )2 bcosx a bcosx − + Solution : Ans. (2) We have y = ) b a ( 2 2 2 − . tan–1              +− 2x tan b a b aFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 21 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Let         +− = 2x tan b a b a u .............(1) ∴ u tan . ) b a ( 2 y 1 2 2 − − = ∴ 2 2 2 u 1 1 . ) b a ( 2 u d y d + − = = 2 2 2 u 1 1 . ) b a ( 2 + − =     +− +       − 2x tan b a b a 1 1 . ) b a ( 2 2 2 2 [From (1)] =            +−     +− + − x cos 1 x cos 1 b a b a 1 1 . ) b a ( 2 2 2 = 2 2 2 (a b) (1 cos x) .{(a b) (1 cos x) (a b) (1 cos x)} (a b ) + + + + + − − − = ) x cos b 2 a 2 ( ) x cos 1 ( ) b a ( . ) b a ( 2 2 2 + + + − .....(2) and ) x cos 1 ( 1 . b a b a 2x sec 21 . b a b a x d u d 2 +     +− =         +− = ........(3) ∴ x d u d . u d y d x d y d = =     +− ++     −+ b a b a . ) x cos b a ( ) x cos 1 ( . b a b a [from (2) and (3)] = x cos b a 1 += + 22 2 d y b sin x dx (a b cos x) Example -16 If 1 2 tan 2 y x x e x −  −    =      , then x d y d equals-(1) x [1 + tan (log x) + sec2 x] (2) 2x [1 + tan (log x)] + sec2 x (3) 2x [1 + tan (log x)] + sec x (4) none of these Solution : Ans. (4)Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 22 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Taking log on both sides, we get log x = tan–1     −2 2 x x y ⇒ tan (log x) = (y – x2) /x2 ⇒ y = x2 + x2 tan (log x) ∴ x d y d = 2x + 2x tan (log x) + x sec2 (log x) = 2x [1 + tan (log x)] + x sec2 (log x) Example -17 If x2 ey + 2xyex + 13 = 0, then x d y d equals (1) ) 2 xe ( x ) 1 x ( y 2 xe 2 x y x y + + + − − − (2) ) 2 xe ( x ) 1 x ( y 2 xe 2 x y y x + + +− − (3) ) 2 xe ( x ) 1 x ( y 2 xe 2 x y y x + + + − − − (4) none of these Solution : Ans. (1) Using partial derivatives, we have   + + = −  +   y x x 2 y x dy 2xe 2ye 2xye dx x e 2xe = – y x 2 y x 2xe 2y 2xy x e 2x − −   + +   +   = y x y x 2xe 2y (x 1) x (xe 2) − −   + + −  +   Example -18 If ........ x sin x sin y + + = , then x d y d equals-(1) 1 y 2 x sin+ (2) 1 y 2 x cos− (3) 1 y 2 x cos+ (4) none of these Solution : Ans. (2) y x sin y + = ⇒ y2 = sin x + y ⇒ y2 – y – sin x = 0 ∴ 1 y 2 x cos 1 y 2 x cos x d y d − = − − − =FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 23 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Example -19 If x2 + y2 = t – t1 , x4 + y4 = t2 + 2 t1 , then x d y d equals-(1) y x12 (2) 3 y x 1 (3) 31 x y (4) 3 y x 1 − Solution : Ans. (3) Squaring the first equation, we have x4 + y4 + 2x2y2 = t2 + 2 t1 – 2 ⇒ t2 + 2 t1 + 2x2y2 = t2 + 2 t1 – 2 (from second equation) ⇒ x2y2 = –1 ⇒ y2 = 2 x1 − ∴ 2y x d y d = 3 x2 ⇒ x d y d = y x13 Example -20 If y2 = p(x) is a polynomial of degree 3, then     2 2 3 dxy d y x dd 2 is equal to-(1) p′′′ (x) p′(x) (2) p′′(x) p′′′(x) (3) p(x) p′′′(x) (4) none of these Solution : Ans. (3) Giveny2 = p(x) ...(i) p′(x) = 2yy′ p′′(x) = 2yy′′ + 2y′2 p′′′(x) = 2yy′′′ + 6y′y′′ ...(ii) Also ) ' ' y y ( x dd 2 x d y d y x dd 2 3 2 2 3 =     = 2 [y3y’’’ + 3y2y’y’’] = y2 [2yy’’’ + 6y’y’’] = p(x) p’’’(x) from (i) and (ii) Example -21 If + +  =     x y f(x) f(y) f 2 2 ∀ ∈ x,y R and ′ = − f (0) 1, f(0) = 1, then f(2) = (1) 12 (2) 1 (3) –1 (4) − 12Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 24 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Solution : Ans. (3) + +   = =     2x 0 f(2x) f(0) f(x) f 2 2 ∴ h 0 f(x h) f(x) f (x) Lt h → + − = ′ h 0 2x 2h f f(x) 2 Lt h → +  −     = h 0 f(2x) f(2h) f(2x) f(0) Lt /h 2 2 → + +   = −     → − ′ = = 2h 0 f(2h) f(0) Lt f (0) 2h ∴ ′ ′ = =− ⇒ =− + f (x) f (0) 1 f(x) x c ⇒ f(0) = c ⇒ c = 1 ∴ f(2) = –2 + 1 = –1 Example -22 If f(x + y) = f(x) f(y) and f(x) = 1 + xg(x) H(x) where → = x 0 Lt g(x) 2 and → = x 0 Lt H(x) 3 then ′ = f (x) (1) f(x) (2) 2f(x) (3) 3f(x) (4) 6f(x) Solution : Ans. (4)h 0 h 0 h 0 f(x h) f(x) f(x)f(h) f(x) f(h) 1 f (x) Lt Lt f(x) Lt h h h → → → + − − − = = = ′ → → + − = = h 0 h 0 1 hg(h)H(h) 1 f(x) Lt f(x) Lt g(h)H(h) h = f(x) (2 × 3) = 6f(x) Example -23 If f(a) = 2, ′ = f (a) 1, g(a) = – 1, ′ = g (a) 2 then the value of → −− x a g(x)f(a) g(a)f(x) Lt x a is (1) – 5 (2) 15 (3) 5 (4) 4 Solution : Ans. (3) Use L.H.R x a g (x)f(a) g(a)f (x) Lt g (a)f(a) g(a)f (a) 1 → ′ ′ − ′ ′ = − = 2(2) – (–1) (1) = 4 + 1 = 5 Example -24 If ′ = − ∫ x 2 1 g(x) {3t 2g (t)} dt x then g (2) ′ = (1) –2/3 (2) –3/2 (3) 2/3 (4) 3/2FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 25 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES Solution : Ans. (4) Given ′ = − ∫ x 2 1 g(x) {3t 2g (t)} dt x ⇒ x2 x g(x) {3t 2g(t)} dt ′ = − ∫ ⇒ g(x) xg (x) {3x 2g (x)} (1) 0 ′ ′ + = − − ⇒ ′ ′ + = − g(2) 2 g (2) 6 2g (2) ′ = − = − = 4g (2) 6 g(2) 6 0 6 ∴ ′ = 3 g (2) 2 Example -25 Let → f :R R be such that f(1) = 3 and ′ = f (1) 6 then →  + =    1/x x 0 f(1 x) Lt f(1) (1) 1 (2) e1/2 (3) e2 (4) e3 Solution : Ans. (3) Let 1/x f (1 x) y f (1)   + =    ⇒ { } 1 log y log f (1 x) log f (1) x = + − ⇒ ( ) x 0 x 0f (1 x) f (1) 6 log Lt (y) Lt 2 f (1 x) f (1) 3 → → ′ ′ + = = = = + ∴ 2 x 0 Lt (y) e → =Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 26 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES LEVEL -I 1. − →  =    sin x x sinx x 0 sinx Lt x (1) 2 1 e (2) 1e (3) e2 (4) e 2. →∞ + + + + = + + + + 2 n n 2 n 1 1 1 1 .... 2 2 2 Lt 1 1 1 1 ..... 3 3 3 (1) 43 (2) 23 (3) 13 (4) 12 3. → + + − = 2 2 h 0 (a h) sin(a h) a sina Lt h (1) a2 cosa + 2a sina (2) a (cosa + 2 sina) (3) a2 (cosa + 2 sina) (4) 0 4. If α, β be the roots of ax2 + bx + c = 0, then −α →α + + + 12 x xLt (1 ax bx c) is (1) a(α – β) (2) log | a(α – β)| (3) ea(α – β) (4) ea(α + β) 5. → + + = + −3 1/x 3 1/x x 0 (1 a ) 8e Lt 2 1 (1 b )e Then (1) a = 1, b = 2 (2) a = 1, b = –31/3 (3) a = 1, b = –1/2 (4) a = 2, b R ∈ 6. →∞ θ = n sinn Lt n (1) 0 (2) ∞ (3) 1 (4) n EXERCISESFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 27 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 7. →∞          ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =                    n n x x x x Lt cos cos cos cos 2 4 8 2 (1) 1 (2) sin x x (3) x sin x (4) –1 8. →∞  + + + =   − − −   2 2 2 n 1 2 n Lt ..... 1 n 1 n 1 n (1) 0 (2) − 12 (3) 12 (4) 13 9. If → +3 x 0 sin2x asinx Lt x be finite, then the value of ‘a’ and the limit are given by (1) –2, 1 (2) – 2, –1 (3) 2, 1 (4) 2, –1 10. → − = 3 2 x 0 cos x cos x Lt sin x (1) 16 (2) 13 (3) − 1 12 (4) − 1 14 11. − → − = 1 2 x 1 1 x Lt (cos x) (1) 14 (2) 12 (3) 14 − (4) 13 12. If [x] denotes the greatest integer x ≤ , then { } →∞ + + 2 2 2 3 n [1 x] [2 x] ....[n x] Lt n (1) x2 (2) x3 (3) x6 (4) 0Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 28 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 13. − − −   − =   +  1 1 1 d x x cos d x x x (1) 2 x 1 1 + (2) 2 x 1 1 +− (3) 2 x 1 2 + (4) 2 x 1 2 +− 14. If ∞ + + + = ..... x e x e x e y , then x d y d = (1) y 1 y− (2) y 1 1− (3) y 1 y+ (4) 1 y y− 15. If y = 1 + x + ! 3x ! 2x 3 2 + + ....... ∞, then x d y d = (1) y (2) y – 1 (3) y + 1 (4) none of these 16. Differential coefficient of sec–1 1 x 2 12 − w.r.t. 2 x 1− at x = 21 is-(1) 2 (2) 4 (3) 6 (4) 1 17. If y = x x1 1     + , then x d y d = (1)     + −     +     + x 1 1 x1 1 log x1 1 x (2)         +     + x1 1 log x1 1 x (3) ( )     + − −     + 1 x x 1 x log x1 x x (4)     + +     +     + x 1 1 x1 1 log x1 x x 18. If f ′(x) = sin (log x) and y =     −+ x 2 3 3 x 2 f , then x d y d = (1) 2 ) x 2 3 ( x ) x (log cos 9 − (2) 2 ) x 2 3 ( x x 2 3 3 x 2 log cos 9 −     −+ (3) 2 2 ) x 2 3 ( x 2 3 3 x 2 log sin 9 −     − + (4) 2 12 2x 3 sin log 3 2x (3 2x)   +       − −    FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 29 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 19. If y = cot–1 (cos 2x)1/2, then the value of x d y d at x = 6π will be-(1) 2 /1 32     (2) 2 /1 31     (3) (3)1/2 (4) (6)1/2 20. If y = logcos x sin x, then x d y d is equal to-(1) 2 ) x cos (log x sin log x tan x cos log x cot + (2) 2 ) x cos (log x sin log x cot x cos log x tan + (3) 2 ) x sin (log x sin log x tan x cos log x cot + (4) 2 cot x (logsin x) 21. If n y sin xcosnx = , then dy dx equals (1) ( ) n 1 nsin xcos n 1 x − + (2) ( ) n 1 nsin x sin n 1 x − + (3) ( ) n 1 nsin xcos n 1 x − − (4) n 1 nsin xcosnx − 22. If ( ) f x cos x cos2x cos4x cos8x cos16x = then f ' 4π       is (1) 2 (2) 12 (3) 1 (4) None of these 23. The value of the derivative of |x – 2| + |x – 3| at x = 3 is (1) 2 (2) –2 (3) 0 (4) 1 24. Let 3 2 3 x sinx cosx f(x) 6 1 0 p p p = − where p is constant, then ( ) 33 d f(x) dx at x = 0 is (1) p (2) p + p2 (3) p + p3 (4) Independent of p 25. If f (x) g(x) ′ = and g (x) f(x) ′ = − for all x and f(2) = 4 = f (2) ′ then f2(19) + g2(19) is (1) 16 (2) 32 (3) 64 (4) 8Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 30 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES LEVEL -II 1. x x a a x Lt f (x)dx x a → = − ∫ (1) 2 f(a) (2) f(a) (3) a f(a) (4) 0 2. 2 3 3 3 3 n 1 4 9 n Lt .... n 1 n 1 n 1 n 1 →∞  + + + + =   + + + +   (1) 1 (2) 2/3 (3) 1/3 (4) 0 3. x 0 log(1 {x}) Lt {x} → + = (where {x} denotes the fractional part of x) (1) 1 (2) 0 (3) 2 (4) does not exist 4. x 1/x x 1/x x 0a a Lta a → −+ , a > 1 is (1) 4 (2) 2 (3) –1 (4) 0 5. n n n n n a b Lt a b →∞ +− , where a > b > 1, is equal to (1) –1 (2) 1 (3) 0 (4) none of these 6. { }2 2 2 2 sin x 1/sin x 1/sin x 1/sin x x 0 Lt 1 2 .... n → + + + is (1) ∞ (2) 0 (3) n(n 1) 2+ (4) n 7. If 2 G(x) 25 x = − − , then x 1 G(x) G(1) Lt x 1 → −− is (1) 1 24 (2) 15 (3) 24 − (4) 124FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 31 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 8. If f(9) = 9, f (9) 4 ′ = , then x 9 f (x) 3 Lt x 3 → − = − (1) 3 (2) 4 (3) 6 (4) 8 9. 2 x 0 x tan2x 2xtan x Lt (1 cos 2x) → − − is (1) 2 (2) –2 (3) 1/2 (4) –1/2 10. sin x x 2 sin x (sin x) Lt 1 sin x log(sin x) π → − = − + (1) 1 (2) 2 (3) 3 (4) 4 11. Given f (2) 6 ′ = , and f (1) 4 ′ = , 2 2 h 0 f (2h 2 h ) f (2) Lt f (h h 1) f (1) → + + − − + − is (1) 3/2 (2) 3 (3) 5/2 (4) –3 12. Let g(x) = sin2[π3]x2 – sin2[–π3]x2 , [.] represents greatest integer function then 4 x 0 g(x) Lt sin x → = (1) –63 (2) 63 (3) 1 (4) –1 13. If y = x |x| then = dy dx (1) |x| (2) | x | x (3) x|x| (4) 2 |x| 14. If 2x + 2y = 2x + y then = dy dx (1) 2x – y (2) 2y – x (3) –2x – y (4) –2y – x 15. If x3y2 = (x + y)5 then = dy dx (1) xy (2) yx (3) − xy (4) logxMathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 32 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 16. { } + + + + = 2 2 4 4 8 8 d (x a)(x a )(x a )(x a ) dx (1) − + − 16 15 16 2 15x 16x a a (x a) (2) − + − 16 15 16 2 x x a a (x a) (3) −− 16 16 x a x a (4) 16x15 – a16 17. If = 2 32 x x x f(x) 1 2x 3x 0 2 6x then ′ = f (x) (1) 0 (2) 1 (3) x2 (4) 6x2 18. If f(x) = cos2x + 3 cos x sinx sin x 3 3 π π     + − +         and γ (5/4) = 3 then (gof)(x) = (1) 0 (2) 1 (3) cosx + π  +     cos x 3 (4) 3 19. Let f(x) be a polynomial function of second degree if f(1) = f(–1) and a, b, c are in A.P then ′ ′ ′ f (a), f (b), f (c) are in (1) A.P (2) G.P (3) H.P (4) A.G.P 20. If       ⋅ ⋅ ∞ =             2 3 x x x sinx cos cos cos .........to 2 x 2 2 then (1)       + + + ∞ = − +             2 2 3 3 1 x 1 x 1 x 1 tan tan tan ...... cot x 2 2 x 2 2 2 2 (2)       + + + ∞ = −             2 2 3 3 1 x 1 x 1 x 1 tan tan tan ...... cot x 2 2 x 2 2 2 2 (3)     + + + ∞ = −         2 2 2 2 4 2 2 1 x 1 x 1 sec sec ...... cosec x 2 2 2 2 x (4)     + + ∞ = + +         2 2 2 2 4 2 2 3 1 x 1 x 1 1 sec sec ...... cosec x 2 2 2 2 x x 21. A triangle has two of its vertices at P(a, 0), Q(0, b) and the third vertex R(x, y) is moving along the st.line y = x, if A be the area of the ∆, then = dA dx (1) − a b 2 (2) − a b 4 (3) + a b 2 (4) + a b 4FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 33 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 22. If ′ = φ f (x) (x) and ′φ = (x) f(x) ∀x , also f(3) = 5 and ′ = f (3) 4 , then the value of [f(10)]2 – [φ(10)]2 = (1) 0 (2) 9 (3) 41 (4) 25 23. If π  = −    5 f(x) sin [x] x 2 , 1 < x < 2 and [x] denotes the greatest integer less than or equal to x, then   π ′ =       5 f 2 (1) π      4/5 5 2 (2) π   −    4/5 5 2 (3) 0 (4) π      4/5 3 2 24 If − +   =   −   1 asinx bcosx y tan acosx bsinx then = dy dx (1) 1 (2) –1 (3) 0 (4) − a acosx bsinx 25. If = + + + + x y x a x b x a b ........ then = dy dx (1) +b a(b 2y) (2) +b b 2y (3) +a b(b 2y) (4) +ab a b2yMathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 34 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES LEVEL -III 1. xLt x cos sin 4x 4x →∞  π π     =             (1) 1π (2) 4π (3) 1 (4) π 2. x [x] Lt x →∞ = (1) 1 (2) 3 (3) –1 (4) does not exist 3. x 2 x 2 Lt secx tan x π → π  −     = − (1) 1 (2) –1 (3) 2 (4) 3 4. 3x 2 x 0 xe x Lt 1 x 1 → − = + − (1) 3 (2) 6 (3) 4 (4) 1 5. |x| x 0e 1 Lt x → − = (1) 1 (2) –1 (3) does not exist (4) 0 6. nx 1/x 1/x 1/x x 1 2 .... n Lt n →∞  + + +     is (1) n (2) n (3) n 1 − (4) 0 7. xLt x x x x →∞  + + − =     (1) 12 (2) 1 (3) 0 (4) does not existFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 35 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 8. x 3x x /2 1 x x 22 2 6 Lt 2 2− − − → + − = − (1) 8 (2) 4 (3) 2 (4) 6 9. If n n 1 n 4 3a a 3 2a + + = + then n nLt a →∞ = (1) 0 (2) –2 (3) 2 (4) 2 − 10. If 1 1 1 f (x) x .... .....nterms 1 x (1 x)(1 2x) (1 2x)(1 3x)   = + + + +   + + + + +   x > 0, then nLt f (x) →∞ = (1) 1 1 x − (2) 1 1 x + (3) 1 (4) 0 11. The value of 4 2 3 x x sin(1/x) x lim 1 |x| →−∞  +   +   is (1) 1 (2) –1 (3) 0 (4) ∞ 12. The value of 2 2 2 2 cos x 1/cos x 1/cos x 1/cos x x 2 lim 1 2 .... n π →  + + +   is (1) 0 (2) n (3) ∞ (4) n(n 1) 2+ 13. If siny = xcos(a + y) then dy dx = (1) 2 sin (a y) sina+ (2) 2 cos (a y) cosa+ (3) 2 cos (a y) cosa− (4) 2 cos (a y) sina+ 14. If sin2mx + cos2ny = a2 then = dy dx (1) msin2mx nsin2ny (2) nsin2ny msin2mx (3) nsin2mx msin2ny (4) −msin2mx nsin2nyMathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 36 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 15. If x x 1 x x f(x) cot 2 − −  − =     then f (1) ′ = (1) –1 (2) 1 (3) log2 (4) –log2 16. If 1 cot x tan x 2 4x y log tanx log cot x tan 4 x −  = ⋅ +   −   then dy dx = (1) 2 1 4 x + (2) 2 4 4 x + (3) 2 1 4 x − (4) 2 4 4 x − 17. If 3f(cosx) + 2f(sinx) = 5x then f (cosx) ′ = (1) 5 cos x − (2) 5 cos x (3) 5 sin x − (4) 5 sin x 18. If f(x) = (cosx + sinx) (cos3x + isin3x) .... (cos[(2n – 1)x] + isin(2n – 1)x] then f (x) ′′ = (1) n2f(x) (2) –n4f(x) (3) –n2f(x) (4) n3f(x) 19. Let (x) φ be the inverse of the function f(x) and f (x) ′ = 5 1 1 x + then d [ (x)] dx φ = (1) 5 1 1 ( (x)) + φ (2) ( )5 1 1 f(x) + (3) ( )5 1 (x) + φ (4) ( )5 1 f(x) + 20. If 1 2 2 tan (y/x) x y ae ,a 0 − + = > then y (0) ′′ is (1) /2 a e 2 −π (2) /2 aeπ (3) /2 2 e a −π − (4) /2 2 e a π 21. If y = a cos(logx) + b sin(logx) then 2 2 2 d y dy x xdx dx + = (1) 0 (2) y (3) –y (4) 1FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 37 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 22. If siny = x sin(a + y) and 2 dy A dx 1 x 2xcosa = + − then the value of A is (1) 2 (2) cosa (3) sina (4) –2 23. If 2 ax b y x c + = + , where a, b, c are constants then (2xy y)y ′ ′′′ + is equal to (1) 3(xy y )y ′′ ′ ′′ + (2) 3(xy y )y ′ ′′ ′′ + (3) 3(xy y )y ′′ ′ ′ + (4) none of these 24. If 6 6 3 3 3 1 x 1 y a (x y ) − + − = − , then dy dx is equal to (1) 2 6 2 6 x 1 y y 1 x −− (2) 2 6 2 6 y 1 y x 1 x −− (3) 2 6 2 6 x 1 x y 1 y −− (4) none of these 25. Let u(x) f(x) log , u (2) 4, (2) 2 (x)   ′ ′ = = ν =   ν  , u(2) = 2, v(2) = 1, then f (2) ′ is (1) 0 (2) 1 (3) -1 (4) 2Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 38 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES QUESTIONS ASKED IN AIEEE & OTHER ENGINEERING EXAMS 1. If m n mn x y (x y) + ⋅ = + , then dy dx is (1) yx (2) x y xy + (3) xy (4) xy [AIEEE -2006] 2. 2 2 2 2 2 2 2 n 1 1 2 4 1 lim sec sec .... sec 1 n n n n n →∞  + + +     is (1) 1 sec1 2 (2) 1cosec1 2 (3) tan1 (4) 1 tan1 2 [AIEEE -2005] 3. Let α and β be the distinct roots of ax2 + bx + c = 0 then 2 2 x 1 cos(ax bx c) lim (x a) →α − + + − is equal to (1) 2 2 a ( ) 2 α −β (2) 0 (3) 2 2 a ( ) 2 − α −β (4) 2 1( ) 2 α −β [AIEEE -2005] 4. If 2x 2 2 x a b lim 1 e x x →∞  + + =     , then the values of a and b, are (1) a R,b 2 ∈ = (2) a 1,b R = ∈ (3) a R,b R ∈ ∈ (4) a = 1 and b = 2. [AIEEE -2004] 5. x /2 3 x 1 tan [1 sinx] 2 lim x 1 tan [ 2x] 2 →π     − −             + π−         is (1) 18 (2) 0 (3) 1 32 (4) ∞ [AIEEE -2003]FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 39 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 6. If x 0 log(3 x) log(3 x) lim k x → + − − = , the value of k is (1) 0 (2) 13 − (3) 23 (4) 23 − [AIEEE -2003] 7. Let f(a) = g(a) = k and their nth derivatives fn(a), gn(a) exist and are not equal for some n. Further if x a f(a)g(x) f(a) g(a)f(x) g(a) lim 4 g(x) f(x) → − − + = − , then the value of k is (1) 4 (2) 2 (3) 1 (4) 0 [AIEEE -2003] 8. If = n f(x) x , then the value of − ′ ′′ ′′′ − + − + + n n f (1) f (1) f (1) ( 1) f (1) f(1) .... 1! 2! 3! n! is (1) 2n (2) 2n–1 (3) 0 (4) 1 [AIEEE -2003] 9. If + ∞ + = y ....to y e x e , x > 0 then dy dx is (1) − 1 x x (2) 1x (3) +x 1 x (4) + 1 x x [AIEEE -2003] 10. x 0 1 cos2x lim 2 x → − is (1) λ (2) –1 (3) zero (4) does not exist [AIEEE -2002] 11. x 22 x x 5x 3 lim x x 2 →∞  + +   + +   is (1) 4 e (2) 2 e (3) e3 (4) e [AIEEE -2002] 12. For x R ∈ , x x x 3 lim x 2 →∞ −       + is (1) e (2) e–1 (3) e–5 (4) e5 [AIEEE -2002]Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 40 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 13. Let f(2) = 4 and f (2) 4 = ′ . Then x 2 xf(2) 2f(x) lim x 2 → −− is given by (1) 2 (2) –2 (3) –4 (4) 3 [AIEEE -2002] 14. If = + + 2 n y (x 1 x ) , then + + 2 2 2 d y dy (1 x ) x dx dx is (1) 2 n y (2) –n2y (3) –y (4) 2x2y [AIEEE -2002] 15. If = + siny x sin(a y) , then dy dx is (1) + 2 sina sinasin (a y) (2) + 2 sin (a y) sina (3) + 2 sinasin (a y) (4) − 2 sin (a y) sina [AIEEE -2002] 16. If − = y xy x e , then dy dx is (1) + +1 x 1 logx (2) −+ 1 logx 1 logx (3) not defined (4) + 2 logx (1 logx) [AIEEE -2002] 17. x sinx x 0 e e lim x sinx →  −   −   is equal to (1) – 1 (2) 0 (3) 1 (4) none of these [UPSEAT -2004] 18. 3 6 x 0 cos2x 1 lim sin 2x → − is equal to (1) 1 16 (2) 1 16 − [CEET (Haryana) -2004] (3) 1 32 (4) 1 32 − 19. 3x 1 x 4 lim 1 x 1 − →∞  −   −   is equal to: (1) 12 e (2) 12 e− (3) 4 e (4) 3 e [CET (Karnataka) -2004]FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 41 Mathematics : Limits and Derivatives NARAYANA INSTITUTE OF CORRESPONDENCE COURSES 20. n n a lim 1 sin n →∞  +     equals to: (1) ea (2) e [CEE (Delhi) -2004] (3) e2a (4) 0 21. The differential coefficient of f(sinx) with respect to x where f(x) = logx is (1) tanx (2) cotx (3) f(cosx) (4) 1x [CET (Karnataka) -2004] 22. If x = A cos 4t + B sin4t, then 2 2 d x dt is equal to (1) –16x (2) 16x (3) x (4) –x [CET (Karnataka) -2004] 23. If 1 x b b a x b a a x ∆ = and 2 x b a x ∆ = are given, then (1) 2 1 2 3( ) ∆ = ∆ (2) 1 2 d 3 dx∆ = ∆ (3) 2 1 2 d 3( ) dx∆ = ∆ (4) 3/2 1 2 3( ) ∆ = ∆ [UPSEAT -2000] 24. If e y xlog x = then dy dx at x = e is (1) 1e (2) 1e (3) e (4) none of these [BIT (Mesra) -2000] 25. If 2 1 1 2 log(e /x ) 3 2log x y tan tan , log(ex ) 1 6log x − −     + = +     −     then 2 2 d y dx = (1) 2 (2) 1 (3) 0 (4) –1 [CEE (Delhi) -2004]Mathematics : Limits and Derivatives FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016 ! Ph.: (011) 32001131/32 Fax : (011) 41828320 42 NARAYANA INSTITUTE OF CORRESPONDENCE COURSES ANSWERS EXERCISES LEVEL -I 1. (2) 2. (1) 3. (1) 4. (3) 5. (2) 6. (1) 7. (2) 8. (2) 9. (2) 10. (3) 11. (1) 12. (2) 13. (4) 14. (1) 15. (1) 16. (1) 17. (1) 18. (4) 19. (1) 20. (1) 21. (1) 22. (1) 23. (3) 24. (4) 25. (2) LEVEL -II 1. (3) 2. (3) 3. (4) 4. (3) 5. (2) 6. (3) 7. (4) 8. (2) 9. (3) 10. (2) 11. (2) 12. (2) 13. (4) 14. (4) 15. (2) 16. (1) 17. (4) 18. (4) 19. (1) 20. (1) 21. (3) 22. (2) 23. (2) 24. (1) 25. (1) LEVEL -III 1. (2) 2. (1) 3. (3) 4. (2) 5. (3) 6. (1) 7. (1) 8. (1) 9. (3) 10. (3) 11. (2) 12. (2) 13. (2) 14. (1) 15. (1) 16. (2) 17. (3) 18. (2) 19. (3) 20. (3) 21. (3) 22. (3) 23. (1) 24. (1) 25. (2) QUESTIONS ASKED IN AIEEE & OTHER ENGINEERING EXAMS 1. (2) 2. (4) 3. (1) 4. (2) 5. (3) 6. (3) 7. (1) 8. (3) 9. (1) 10. (4) 11. (1) 12. (3) 13. (3) 14. (1) 15. (2) 16. (4) 17. (3) 18. (4) 19. (2) 20. (1) 21. (2) 22. (1) 23. (2) 24. (2) 25. (3)