theory and problems of differential and integral calculus third edition - ayres & mendelson

SCHAUMS OUTLINE OF THEORY AND PROBLEMS OF DIFFERENTIAL AND INTEGRAL CALCULUS Third Edition 0 FRANK AYRES, JR, Ph.D. Formerly Professor and Head Department of Mathematics Dickinson College and ELLIOTT MENDELSON, Ph.D. Professor of Mathematics Queens College 0 SCHAUM’S OUTLINE SERIES McGRAW-HILL New York St. Louis San Francisco Auckland Bogota Caracas Lisbon London Madrid Mexico City Milan Montreal New Delhi San Juan Singapore Sydney Tokyo Toronto FRANK AYRES, Jr., Ph.D., was formerly Professor and Head of the Department of Mathematics at Dickinson College, Carlisle, Pennsyl- vania. He is the author of eight Schaum’s Outlines, including TRI- LEGE MATH, and MATRICES. GONOMETRY, DIFFERENTIAL EQUATIONS, FIRST YEAR COL- ELLIOTT MENDELSON, Ph.D. , is Professor of Mathematics at Queens College. He is the author of Schaum’s Outlines of BEGINNING CAL- CULUS and BOOLEAN ALGEBRA AND SWITCHING CIRCUITS. Schaum’s Outline of Theory and Problems of CALCULUS Copyright 0 1990, 1962 by The McGraw-Hill Companies, Inc. All Rights Reserved. Printed in the United States of America. Except as permitted under the Copyright Act of 1976, no part of this pub- lication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. 9 10 11 12 13 14 15 16 17 18 19 20 BAW BAW 9 8 7 6 ISBN 0-07-002bb2-9 Sponsoring Editor, David Beckwith Production Supervisor, Leroy Young Editing Supervisor, Meg Tobin Library of Congress Catabghg-io-Pubkation Data Ayres, Frank, Schaum’s outline of theory and problems of differential and integral calculus / Frank Ayres, Jr. and Elliott Mendelson. 3rd ed . p. cm. (Schaum’s outline series) ISBN 0-07-002662-9 1. Calculus- -Outlines, syllabi, etc. 2. Calculus- -Problems, exercises, etc. 1. Mendelson, Elliott. 11. Title. QA303.A% 1990 5 15- -dc20 McGraw -Hill 89- 13068 CIP A Division of The McGraw-Hitl Companies - This third edition of the well-known calculus review book by Frank Ayres, Jr., has been thoroughly revised and includes many new features. Here are some of the more significant changes: Analytic geometry, knowledge of which was presupposed in the first two editions, is now treated in detail from the beginning. Chapters 1 through 5 are completely new and introduce the reader to the basic ideas and results. Exponential and logarithmic functions are now treated in two places. They are first discussed briefly in Chapter 14, in the classical manner of earlier editions. Then, in Chapter 40, they are introduced and studied rigorously as is now customary in calculus courses. A thorough treatment of exponential growth and decay also is included in that chapter. Terminology, notation, and standards of rigor have been brought up to date. This is especially true in connection with limits, continuity, the chain rule, and the derivative tests for extreme values. Definitions of the trigonometric functions and information about the important trigonometric identities have been provided. The chapter on curve tracing has been thoroughly revised, with the emphasis shifted from singular points to examples that occur more frequently in current calculus courses. The purpose and method of the original text have nonetheless been pre- served. In particular, the direct and concise exposition typical of the Schaum Outline Series has been retained. The basic aim is to offer to students a collection of carefully solved problems that are representative of those they will encounter in elementary calculus courses (generally, the first two or three semesters of a calculus sequence). Moreover, since all fundamental concepts are defined and the most important theorems are proved, this book may be used as a text for a regular calculus course, in both colleges and secondary schools. Each chapter begins with statements of definitions, principles, and theorems. These are followed by the solved problems that form the core of the book. They give step-by-step practice in applying the principles and provide derivations of some of the theorems. In choosing these problems, we have attempted to anticipate the difficulties that normally beset the beginner. Every chapter ends with a carefully selected group of supplementary problems (with answers) whose solution is essential to the effective use of this book. 1. 2. 3. 4. 5. ELLIO~T MENDELSON Table of Contents Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 Chapter 31 Chapter 32 Chapter 33 Chapter Chapter 35 Chapter 36 Chapter 37 Chapter 38 ABSOLUTE VALUE; LINEAR COORDINATE SYSTEMS; INEQUALITIES THE RECTANGULAR COORDINATE SYSTEM LINES CIRCLES EQUATIONS AND THEIR GRAPHS FUNCTIONS LIMITS CONTINUITY THE DERIVATIVE RULES FOR DIFFERENTIATING FUNCTIONS IMPLICIT DIFFERENTIATION TANGENTS AND NORMALS MAXIMUM AND MINIMUM VALUES APPLIED PROBLEMS INVOLVING MAXIMA AND MINIMA . . RECTILINEAR AND CIRCULAR MOTION RELATED RATES DIFFERENTIATION OF TRIGONOMETRIC FUNCTIONS DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNC- TIONS DIFFERENTIATION OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS DIFFERENTIATION OF HYPERBOLIC FUNCTIONS PARAMETRIC REPRESENTATION OF CURVES CURVATURE PLANE VECTORS CURVILINEAR MOTION POLAR COORDINATES THE LAW OFTHE MEAN INDETERMINATE FORMS DIFFERENTIALS CURVE TRACING FUNDAMENTAL INTEGRATION FORMULAS INTEGRATION BY PARTS TRIGONOMETRIC INTEGRALS TRIGONOMETRIC SUBSTITUTIONS INTEGRATION BY PARTIAL FRACTIONS MISCELLANEOUS SUBSTITUTIONS INTEGRATION OF HYPERBOLIC FUNCTIONS APPLICATIONS OF INDEFINITE INTEGRALS THE DEFINITE INTEGRAL 1 8 17 31 39 52 58 68 73 79 88 91 96 106 112 116 120 129 133 141 145 148 155 165 172 183 190 196 201 206 219 225 230 234 239 244 247 251 CONTENTS Chapter 39 Chapter 40 Chapter 41 Chapter 42 Chapter 43 Chapter 44 Chapter 45 Chapter 46 Chapter 47 Chapter 48 Chapter 49 Chapter 50 Chapter 51 Chapter 52 Chapter 53 Chapter 54 Chapter 55 Chapter 56 Chapter 57 Chapter 58 Chapter 59 Chapter 60 Chapter 61 Chapter 62 Chapter 63 Chapter 64 Chapter 65 Chapter 66 Chapter 67 Chapter 68 Chapter 69 Chapter 70 Chapter 71 Chapter 72 Chapter 73 Chapter 74 Chapter 75 Chapter 76 INDEX PLANE AREAS BY INTEGRATION EXPONENTIAL AND LOGARITHMIC FUNCTIONS; EX- PONENTIAL GROWTH AND DECAY VOLUMES OF SOLIDS OF REVOLUTION VOLUMES OF SOLIDS WITH KNOWN CROSS SECTIONS CENTROIDS OF PLANE AREAS AND SOLIDS OF REVO- LUTION MOMENTS OF INERTIA OF PLANE AREAS AND SOLIDS OF REVOLUTION FLUID PRESSURE WORK LENGTH OF ARC AREAS OF A SURFACE OF REVOLUTION CENTROIDS AND MOMENTS OF INERTIA OF ARCS AND SURFACES OF REVOLUTION PLANE AREA AND CENTROID OF AN AREA IN POLAR COORDINATES LENGTH AND CENTROID OF AN ARC AND AREA OF A SURFACE OF REVOLUTION IN POLAR COORDINATES IMPROPER INTEGRALS INFINITE SEQUENCES AND SERIES TESTS FOR THE CONVERGENCE AND DIVERGENCE OF POSITIVE SERIES SERIES WITH NEGATIVE TERMS COMPUTATIONS WITH SERIES POWER SERIES SERIES EXPANSION OF FUNCTIONS MACLAURIN'S AND TAYLOR'S FORMULAS WITH RE- MAINDERS COMPUTATIONS USING POWER SERIES APPROXIMATE INTEGRATION PARTIAL DERIVATIVES TOTAL DIFFERENTIALS AND TOTAL DERIVATIVES IMPLICIT FUNCTIONS SPACE VECTORS SPACE CURVES AND SURFACE DIRECTIONAL DERIVATIVES; MAXIMUM AND MINIMUM VALUES VECTOR DIFFERENTIATION AND INTEGRATION DOUBLE AND ITERATED INTEGRALS CENTROIDS AND MOMENTS OF INERTIA OF PLANE AREAS VOLUME UNDER A SURFACE BY DOUBLE INTEGRATION AREA OF A CURVED SURFACE BY DOUBLE INTEGRATION TRIPLE INTEGRALS MASSES OF VARIABLE DENSITY DIFFERENTIAL EQUATIONS DIFFERENTIAL EQUATIONS OF ORDER TWO 260 268 272 280 284 292 297 301 305 309 313 316 321 326 332 338 345 349 354 360 367 371 375 380 386 394 398 411 417 423 435 442 448 45 1 456 466 470 476 48 1 Chapter 1 Absolute Value; Linear Coordinate Systems; Inequalities THE SET OF REAL NUMBERS consists of the rational numbers (the fractions alb, where a and b are integers) and the irrational numbers (such as fi = 1.4142 . . . and T = 3.14159 . . .), which are not ratios of integers. Imaginary numbers, of the form x + ym, will not be considered. Since no confusion can result, the word number will always mean real number here. THE ABSOLUTE VALUE 1x1 of a number x is defined as follows: x if x is zero or a positive number Ixl = { x if x is a negative number For example, 131 = 1-31 = 3 and 101 = 0. In general, if x and y are any two numbers, then - 1x1 5 x 5 1x1 I-xl = 1x1 and Ix - yl = ly -XI 1x1 = lyl implies x = *y Ix + yl 5 1x1 + Iyl (Triangle inequality) (1.5) A LINEAR COORDINATE SYSTEM is a graphical representation of the real numbers as the points of a straight line. To each number corresponds one and only one point, and conversely. To set up a linear coordinate system on a given line: (1) select any point of the line as the origin (corresponding to 0); (2) choose a positive direction (indicated by an arrow); and (3) choose a fixed distance as a unit of measure. If x is a positive number, find the point corresponding to x by moving a distance of x units from the origin in the positive direction. If x is negative, find the point corresponding to x by moving a distance of 1x1 units from the origin in the negative direction. (See Fig. 1-1.) 1 11111 I I I1 I I1 1 I 11111 I I II I 11 1 Y -4 -3 -512 -2 -312 -1 0 1/2 1 ~ 2 3r 4 Fig. 1-1 The number assigned to a point on such a line is called the coordinate of that point. We often will make no distinction between a point and its coordinate. Thus, we might refer to “the point 3” rather than to “the point with coordinate 3.” If points P, and P, on the line have coordinates x, and x, (as in Fig. 1-2), then Ix, - x2( = PIP2 = distance between P, and P2 1x1 = distance between P and the origin (1.6) (1.7) As a special case, if x is the coordinate of a point P, then 1 2 ABSOLUTE VALUE; LINEAR COORDINATE SYSTEMS; INEQUALITIES FINITE INTER x2 Fig. 1-2 [CHAP. 1 QLS. Let a and b be two points such that a < b. By the open ',zterval (a, ") we mean the set of all points between a and b, that is, the set of all x such that a < x < b. By the closed interval [a, b] we mean the set of all points between a and b or equal to a or b, that is, the set of all x such that a I x 5 b. (See Fig. 1-3.) The points a and b are called the endpoints of the intervals (a, b) and [a, b]. A 4 - W * U b Open interval (a, b): a < x < b L - - m U b Closed interval [a, b]: a I x I b Fig. 1-3 By a huff-open interval we mean an open interval (a, b) together with one of its endpoints. There are two such intervals: [a, b) is the set of all x such that a 5 x < b, and (a, b] is the set of all x such that a < x 5 b. For any positive number c, 1x1 5 c if and only if -c 5 x I c 1x1 < c if and only if -c < x < c See Fig. 1-4. I n 1 n 1 - * - + W 1 W -C 0 C -C 0 C Fig. 1-4 INFINITE INTERVALS. Let a be any number. The set of all points x such that a < x is denoted by (a, 30); the set of all points x such that a I x is denoted by [a, 00). Similarly, (-00, b) denotes the set of all points x such that x < b, and (-00, b] denotes the set of all x such that x 5 b. INEQUALITIES such as 2x - 3 > 0 and 5 < 3x + 10 I 16 define intervals on a line, with respect to a given coordinate system. EXAMPLE 1 : Solve 2x - 3 > 0. 2~-3>0 2x > 3 (Adding 3) x > (Dividing by 2) Thus, the corresponding interval is ($, 00). CHAP. 11 ABSOLUTE VALUE; LINEAR COORDINATE SYSTEMS; INEQUALITIES 3 EXAMPLE 2: Solve 5 < 3x + 10 5 16. 5<3x+10116 -5< 3x 56 (Subtracting 10) - ;< x 12 (Dividing by 3) Thus, the corresponding interval is (-5/3,2]. EXAMPLE 3: Solve -2x + 3 < 7. -2~+3<7 - 2x < 4 (Subtracting 3) x > -2 (Dividing by - 2) Note, in the last step, that division by a negative number reverses an inequality (as does multiplication by a negative number). Solved Problems 1. Describe and diagram the following intervals, and write their interval notation: (a) - 3 < x<5; (b) 21x56; (c) -4<x50; (d)x>5; (e)xs2; (f) 3x-458; (g) 1<5-3x<11. (a) All numbers greater than -3 and less than 5; the interval notation is (-3,5): (6) All numbers equal to or greater than 2 and less than or equal to 6; [2,6]: (c) All numbers greater than -4 and less than or equal to 0; (-4,0]: (d) All numbers greater than 5; (5,~): (e) All numbers less than or equal to 2; (-W, 21: (f) 3x - 4 I 8 is equivalent to 3x I 12 and, therefore, to x 5 4. Thus, we get (-m, 41: 1 < 5 - 3x < 11 -4< -3x <6 (Subtracting 5) -2 < x < (Dividing by -3; note the reversal of inequalities) Thus, we obtain (-2, $): 4 ABSOLUTE VALUE; LINEAR COORDINATE SYSTEMS; INEQUALITIES [CHAP. 1 2. Describe and diagram the intervals determined by the following inequalities: (a) 1x1 <2; (6) 1x1 > 3; (c) Ix - 31 < 1; (d) Ix - 21 < 6, where 6 > 0; (e) Ix + 21 5 3; (f) 0 < Ix - 41 < 6, where 6 CO. (a) This is equivalent to -2 < x < 2, defining the open interval (-2,2): (6) This is equivalent to x >3 or x < -3, defining the union of the infinite intervals (3, a) and (-m, -3). (c) This is equivalent to saying that the distance between x and 3 is less than 1, or that 2 < x < 4, which defines the open interval (2,4): We can also note that Ix - 31 < 1 is equivalent to - 1 < x - 3 < 1. Adding 3, we obtain 2 < x < 4. (d) This is equivalent to saying that the distance between x and 2 is less than 6, or that 2 - 6 < x < 2 + 6, which defines the open interval (2 - 6,2 + 6). This interval is called the 6-neighborhood of 2: n v 1 0 - 2-6 2 2+6 (e) Ix + 21 < 3 is equivalent to -3 < x + 2 < 3. Subtracting 2, we obtain -5 < x < 1, which defines the open interval (-5, 1): (f) The inequality Ix - 41 < 6 determines the interval 4 - 6 < x < 4 + 6. The additional condition 0 < Ix - 41 tells us that x # 4. Thus, we get the union of the two intervals (4 - 6,4) and (4,4 + 6). The result is called the deleted 6-neighborhood of 4: n n n W - e - 4-6 4 4+6 3. Describe and diagram the intervals determined by the following inequalities: (a) 15 - XI 5 3; (6) 12~ - 31 <5; (c) 11 -4x(< $. (a) Since 15 - XI = Ix - 51, we have Ix - 51 I 3, which is equivalent to -3 5 x - 5 5 3. Adding 5, we get 2 I x 5 8, which defines the open interval (2,s): (6) 12x - 31 < 5 is equivalent to -5 < 2x - 3 < 5. Adding 3, we have -2 < 2x < 8; then dividing by 2 yields - 1 < x < 4, which defines the open interval (- 1,4): v -1 - 4 (c) Since 11 - 4x1 = 14x - 11, we have (4x - 11 < 4, which is equivalent to - 4 < 4x - 1 < 4. Adding 1, we get 5 < 4x < t. Dividing by 4, we obtain Q < x < i, which defines the interval (Q , ): CHAP. 11 ABSOLUTE VALUE; LINEAR COORDINATE SYSTEMS; INEQUALITIES 4. Solve the inequalities (a) 18x - 3x2 > 0, (b) (x + 3)(x - 2)(x - 4) < 0, and 5 (x + l)L(x - 3) > 0, and diagram the solutions. Set 18x - 3x2 = 3x(6 - x) = 0, obtaining x = 0 and x = 6. We need to determine the sign of 18x - 3x‘ on each of the intervals x < 0, 0 < x < 6, and x > 6, to determine where 18x - 3x’ > 0. We note that it is negative when x < 0, and that it changes sign when we pass through 0 and 6. Hence, it is positive when and only when O<x<6: The crucial points are x = -3, x = 2, and x = 4. Note that (x + 3)(x - 2)(x - 4) is negative for x < -3 (since each of the factors is negative) and that it changes sign when we pass through each of the crucial points. Hence, it is negative for x < - 3 and for 2 < x < 4: * -3 2 4 Note that (x + 1)’ is always positive (except at x = - 1, where it is 0). Hence (x + l)*(x - 3) > 0 when and only when x - 3 > 0, that is, for x > 3: 5. Solve 13x - 71 = 8. In general, when c I 0, lul= c if and only if U = c or U = - c. Thus, we need to solve 3x - 7 = 8 and 3x-7=-8, from which wegetx=5orx=-+. 2x + 1 x+3 6. Solve - > 3. Case 2 : x + 3 > 0. Multiply by x + 3 to obtain 2x + 1 > 3x + 9, which reduces to - 8 > x. However, Case 2: x + 3 < 0. Multiply by x + 3 to obtain 2x + 1 < 3x + 9. (Note that the inequality is reversed, Thus, the only solutions are -8 < x < -3. since x + 3 > 0, it must be that x > -3. Thus, this case yields no solutions. since we multiplied by a negative number.) This yields - 8 < x. Since x + 3 < 0, we have x < - 3. 7. solve I f - 31 < 5. 2 The given inequality is equivalent to -5 < - - 3 < 5. Add 3 to obtain -2 < 2/x < 8, and divide by 2 Case I : x > 0. Multiply by x to get -x < 1 < 4x. Then x > j and x > - 1; these two inequalities are Case 2: x < 0. Multiply by x to obtain -x > 1 > 4x. (Note that the inequalities have been reversed, and x < - 1. These two inequalities are Thus, the solutions are x > 4 or x < - 1, the union of the two infinite intervals (4, M) and (-E, - 1). X to get -1 < l/x<4. equivalent to the single inequality x > i. since we multiplied by the negative number x.) Then x < equivalent to x < - 1. 8. Solve 12x - 51 I 3. Let us first solve the negation 12x - 51 < 3. The latter is equivalent to -3 < 2x - 5 < 3. Add 5 to obtain 2 < 2x < 8, and divide by 2 to obtain 1 < x < 4. Since this is the solution of the negation, the original inequality has the solution x 5 1 or x 2 4. 9. Prove the triangle inequality, Ix + yI 5 1x1 + I yl. [...]... - 1, and Q ' A , = 6 - x So 14 THE RECTANGULAR COORDINATE SYSTEM [CHAP 2 Y 6 Fig 2-9 x- - 1 - - and cross-multiplying yields 3x - 3 = 12 - 2 x Hence 5x = 15, whence 6-x 3’y-2 reasoning, - from which it follows that y = 4 7-y 3’ x = 3 By similar Supplementary Problems 8 In Fig 2-1 0, find the coordinates of points A , B, C, D,E, and F Y E e F A e 1 -5 1 -4 D I -3 1 -2 1 -1 -2 -l Ans 9 I A = ( -. .. points (3, - 1) and (2,3) Its slope is m = 3 - ( - 1 ) x-2 Y+l Y-3 2-3 -4 Two point-slope equations of 3 are -= -4 and -= -4 ’ x-3 x-2 ~ 4 = -1 SLOPE-INTERCEPTEQUATION If we multiply (3.1) by x - x,, we obtain the equation y - y, = m(x - x , ) , which can be reduced first to y - y , = mx - mx,, and then to y = mx + ( y , - m x , ) Let 6 stand for the number y, - mx, Then the equation for line Y becomes... 5 - 2 ) 2 = d ~ = m = v D = v( 1- 5 ) * + ( 5 - 6), = d (-4 ),+ (- 1), = = fl BC = I/(4 - 5)’ + (2 - 6), = d (- 1), + (-4 )2 = = fl BC, the triangle is isosceles Is the triangle with vertices A (-5 ,6), B(2,3), and C(5,lO) a right triangle? 13 THE RECTANGULAR COORDINATE SYSTEM CHAP 2 1 - AB = v ( - 5 - 2 ) , x2 + (6 - 3), = v m V% + (-4 )'=VE%FT6=VTE = -= AC=v (-5 -5 )2+( 6-1 0)'=v (-1 0)' BC=~( 2-5 )2+( 3-1 0)2=~ (-3 )2+ (-7 )'=m=V%... parallel Let us calculate the slopes of these lines: li-0 Slope(M,M,) = 2 u+b b 2 U - - 2- - v _ - 2 U Y+u2 Slope(M,M,) = 2 X + U 2 2 u+b - 2 ' 2 - Y X-b - X-b 2 = 2 2 x x+u 2 slope(M,M,) U 2 -U '-y+v 2 - 2 u - u u 2 z -0 slope(M,M,) = 2 x - -b _ ~ 2 Y =- X-b 2 Since slope(M,M,) = slope(M,M,), M , M , and M , M , are parallel Since slope(M,M,) = slope(M,M,), M , M , and M , M , are parallel Thus, M ,... 9 Find a point-slope equation for the line through each of the following pairs of points: ( a ) ( 3 , 6 ) and ( 2 , - 4 ) ; (b) ( 8 , 5 ) and (4,O); (c) ( 1 , 3 ) and the origin; (d) ( 2 , 4 ) and ( - 2 , 4 ) Am 10 = 10; ( b ) y-5 = x-g - * 47@) Y- 3-3 ;(d) x- Y- 4-0 x-2 Find the slope-intercept equation of each line: ( a ) Through the points ( 4 , -2 ) and ( 1 , 7 ) (b) Having slope 3 and y intercept... and the slope-intercept equation of 2 is y = -2 x + 5 ' ~ 3 Determine whether the points A( 1, - l), B ( 3 , 2 ) , and C(7,S) are collinear, that is, lie on the same line A , B , and C are collinear if and only if the line A B is identical with the line A C , which is equivalent 2-( 11) - 3 to the slope of A B being equal to the slope of AC (Why?) The slopes of A B and AC are -8 - (-1 ) - 9 - 3 3-1 2 and. .. corner (See Fig 2-4 .) Quadrunt II consists of all points with negative x coordinate and positive y coordinate Quadrants I U and n/ are also shown in Fig 2-4 10 THE RECTANGULAR COORDINATE SYSTEM [CHAP 2 Y I (+ +I I ( 3-1 ) 1 -3 -2 -1 ( - 2 , -1 ) 1 1 1 0 2 3 X -1 (2 -2 ) -2 I11 (-, - ) IV -1 (+ Fig 2-4 The points on the x axis have coordinates of the form ( a , O ) The y axis consists of the points with... by each of the following conditions: (d) (2x + 1)2> 1 (h) 2x2 > x + 6 - 2 < x < 2 ; (6) x 2 3 or XI -3 ; (c) - 2 1 x 1 6 ; (d) x > O or x < - 1 ; ( e ) x > l o r x < - 4 ; (f) - 4 5 x 5 - 2 ; (g) - 2 < x < 7 ; ( h ) x > 2 o r x < - $ ; (i) - $ < x < f ; ( j ) - 5 < x < O o r x > 2 (a) Solve: ( a ) - 4 < 2 - x < 7 (6) 2x - 1 73 < X x+2 < 1 CHAP 11 Am 15 16 (b)x>Oorx-2 ; (d) - y < x < - z ; (e)x . (x - 4)(2x - 3) < 0 (i) (x - 2)3 > 0 (a) O<x<5; (6) x>6 or x<2; (c) -1 <x<2; (d) x>2 or -3 <x<O; (e) -3 <x< ;-2 orx< ;-4 ; (f)x>2or -l<x<lorx< ;-3 ;. x>O or x< ;-1 ; (e)x>l orx< ;-4 ; (f) -4 5x 5-2 ; (g) -2 <x<7; (h)x>2orx< ;-$ ; (i) -$ <x<f; (j) -5 <x<Oorx>2 2x - 1 X <1 x+2 14. Solve: (a) -4 <2-x<7. -l<x<lorx< ;-3 ; (g)x> ;-4 andx#l; (h) -5 < x < 3; (i) x > 2; (j) x < - 1; (k) - 1 < x < 2; (I) x < 1 and x # - 1; (m)x>forx< ;-; ;