MT1710 - 2023/24: GROUP WORK - Week 1: Hints to Solutions 1. (a) y = 2- x2 is the parabola with the y-axis as axis and vertex the point (0, 2) and passing through the point (v2, 0). y = 1 is a line segment parallel to the x-axis (slope zero) defined between x = 1 and x = 2. y 1 Vx y = (2 - x)2 1 2 3 x (b) By definition, |3x] = ? 3x; -3x; x<0, x?0, with 3 - x= 3-x; x ?3, x-3; x>3. Hence, we are interested in the three intervals -1 <<< 0, 03x?3 and 3<x ?4 for which we have 0?x ?3, 3<x ?4. 8 -3x-(3-x) =- (2x+3); -1?x < 0, 3x-(3-x)=4x - 3; y = : 3x-(x-3)=2x+ 3; 2. The function y = f(x) is given by (-(x+2); x< - 2 y = { (x + 2); 2: -2?x<0 x ? 0. The first two may be combined to give y= 12; x+2|; x< 0 x?0. Either strict inequality or weak inequality is acceptable in this example. (b) 3. Either working from the diagram, or letting f1(-x) = f1(x) since f1 is even gives: 1 f1(-x) = {2 - x2 = 2 - (-2)2; 0 <x ? 1 1?x ?2' so that - 1; -2<x?-1 f?(x) ={2-x2; - 1<x<1 1 1; . 1<x?2 1
y 2 y = 2 - 22 x= 2 - x2 1 1 -2 -1 1 2 x -2 -1 1 2 x -] y = x2-2 -2 f1(x) Similarly, letting g1(-x) = - g1(x) since g1 is odd gives: g?(x) g?(-x) = 1-1 [-(2-x2) = (-x)2 - 2; 0< x ? 1 1?x ?2' so that 8 x2 - 2; -1; -2?x?-1 -1 5 x < 0 g1 (x) = { 2 - x2; 0 = x ? 1 - . 1; 1<x?2 Note that the inequality ? has to be modified to the strict inequality < when discussing the interval [-1, 0), to avoid the f2 taking two distinct values at the origin, thereby violating the definition of a function. 4. For all of these use the properties of periodic functions, () =y (2a+ +a) =y (za) and y (-) = y ( a -a) = y (a). For each of the particular cases this gives: (a) y()= 4 and y(2) = 4 . y a -a N/A a a 2 3a 2 2a 5a 2 3a y()= 2 and yl 7 (b) 1 = 12 . 2 x
y a -a a 30 2 2a 5a 2 3a x 5. Setting u = 0 and v = x gives 2f(0)cosx= f(x)+f(-x). Then f(0) = a gives (i). Setting v = " and u = "2 - x gives (ii), i.e. 0= f(I-x)+ f(-x) . Setting u = 2 and v = 1 - x gives 2f (2) cos (2 - x) = f (1 - x) + f(x). Setting f(5) = b and using cos(5 - x) = sin x gives (iii).