Dr. Emily Chan Page 1 GE1359 Problem Set 1 Solution Summer 2022 1. π΄ , π΅ , πΆ and π· are equal, since they only contain the elements 1, 2, 3. πΈ is not equal to π΄ , π΅ , πΆ or π· , since 4 β πΈ but 4 is not an element of π΄ , π΅ , πΆ , π· 2. π β π΄ & π β π΅ β π β π΄ β© π΅ = {π, π, π} π β πΆ , so there is at least one element in π which does not belong to πΆ Only the element π in the set {π, π, π} is not in πΆ , so π must contain the element π β΄ π could be {π} , {π, π} , {π, π} or {π, π, π} β΄ Among the sets π΄ , π΅ , πΆ , π· , πΈ , πΉ , only π· and πΈ can equal π 3. (a) π΄ β© π΅ = {π, π} , π΄ β© πΆ = {π, π} , π΅ β© πΆ = {π, π} (b) π΄ βͺ π΅ = {π, π, π, π, π, β} , π΄ βͺ πΆ = {π, π, π, π, π, π} , π΅ βͺ πΆ = {π, π, π, π, π, β} (c) π΄\π΅ = {π, π} , π΅\π΄ = {π, β} , π΅\πΆ = {π, β} , πΆ\π΅ = {π, π} , π΄\πΆ = {π, π} , πΆ\π΄ = {π, π} 4. (a) π΄ βͺ π΅ = [1, 4] (b) π΄ β© π΅ = (2, 3] (c) (β\π΄) β© π΅ = (3, 4] (d) (β\π΅) β© π΄ = [1, 2] (e) (β\π΄) β© (β\π΅) = (ββ, 1) βͺ (4, β) (f) (β\π΄) βͺ (β\π΅) = (ββ, 2] βͺ (3, β) (g) π΅ βͺ [π΄ β© (β\π΅)] = [1, 4] (h) [(β\π΄) β© π΅] βͺ [(β\π΅) β© π΄] = (3, 4] βͺ [1, 2] 5. π(π₯) = π₯ 3 + 2 , π·ππ(π) = β π(π₯) = 2 π₯β1 , π·ππ(π) = β\{1} β(π₯) = βπ₯ , π·ππ(β) = [0, β) (a) (π + π)(π₯) = π(π₯) + π(π₯) = π₯ 3 + 2 + 2 π₯β1 π·ππ(π + π) = β\{1} (b) ( π π ) (π₯) = π(π₯) π(π₯) = 2 π₯β1 π₯ 3 +2 = 2 (π₯β1)(π₯ 3 +2) Dr. Emily Chan Page 2 π π is defined when (π₯ β 1)(π₯ 3 + 2) β 0 β π₯ β 1 β 0 and π₯ 3 + 2 β 0 β π₯ β 1 and π₯ β ββ2 3 = β β2 3 π·ππ ( π π ) = β\{β β2 3 , 1} (c) (π β π)(π₯) = π(π(π₯)) = π(π₯ 3 + 2) = 2 (π₯ 3 +2)β1 = 2 π₯ 3 +1 which is only defined when π₯ 3 + 1 β 0 , i.e. when π₯ β ββ1 3 = β1 π·ππ(π β π) = β\{β1} (d) (π β π)(π₯) = π(π(π₯)) = π ( 2 π₯β1 ) = ( 2 π₯β1 ) 3 + 2 π·ππ(π β π) = β\{1} (e) (π β π β β)(π₯) = π(π(β(π₯))) = π (π(βπ₯ )) = π ( 2 βπ₯ β1 ) = ( 2 βπ₯ β1 ) 3 + 2 which is only defined when βπ₯ β 1 β 0 and π₯ β₯ 0 , i.e. when βπ₯ β 1 and π₯ β₯ 0 β π₯ β 1 and π₯ β₯ 0 π·ππ(π β π β β) = [0, β)\{1} or written as [0, 1) βͺ (1, β) 6. πΉ(π₯) = 3 2+π₯ , πΊ(π₯) = 1 1+ 1 π₯ (a) πΉ(π₯) is defined when 2 + π₯ β 0 β π₯ β β2 π·ππ(πΉ) = β\{β2} πΊ(π₯) is defined when π₯ β 0 and 1 + 1 π₯ β 0 β π₯ β 0 and 1 π₯ β β1 β π₯ β 0 and π₯ β β1 π·ππ(πΊ) = β\{β1, 0} (b) (πΉ β πΊ)(π₯) = πΉ(πΊ(π₯)) = πΉ ( 1 1+ 1 π₯ ) = 3 2+ 1 1+1 π₯ which is defined when π₯ β π·ππ(πΊ) and 2 + 1 1+ 1 π₯ β 0 β π₯ β β\{β1, 0} and 1 1+ 1 π₯ β β2 β π₯ β β\{β1, 0} and 1 + 1 π₯ β 1 β2 β π₯ β β\{β1, 0} and 1 π₯ β β 3 2 β π₯ β β\{β1, 0} and π₯ β β 2 3 Dr. Emily Chan Page 3 π·ππ(πΉ β πΊ) = β\ {β1, β 2 3 , 0} 7. (a) π(π₯) = π₯ 2 β 2π₯ β 3 = (π₯ β 1) 2 β 4 π·ππ(π) = β For every π₯ β π·ππ(π) = β , (π₯ β 1) 2 β₯ 0 β (π₯ β 1) 2 β 4 β₯ β4 π
ππ(π) = [β4, β) (b) π(π₯) = π₯β3 π₯+2 = (π₯+2)β5 π₯+2 = 1 β 5 π₯+2 π is defined when π₯ + 2 β 0 β π₯ β β2 π·ππ(π) = β\{β2} For every π₯ β π·ππ(π) = β\{β2} , β 5 π₯β2 can be any real number except 0, thus 1 β 5 π₯+2 β 1 + 0 = 1 π
ππ(π) = β\{1} (c) β(π₯) = βπ₯ 2 + 4 β is defined when π₯ 2 + 4 β₯ 0 , which is always true for all real values of π₯ π·ππ(β) = β For every π₯ β π·ππ(β) = β , π₯ 2 β₯ 0 β π₯ 2 + 4 β₯ 0 + 4 = 4 β βπ₯ 2 + 4 β₯ β4 = 2 π
ππ(β) = [2, β) 8. (a) π(π₯) = 5 β 4π₯ β π₯ 2 π·ππ(π) = β (b) π(π₯) = 1 5β4π₯βπ₯ 2 π is defined when 5 β 4π₯ β π₯ 2 β 0 β (1 β π₯)(5 + π₯) β 0 β π₯ β 1 and π₯ β β5 π·ππ(π) = β\{β5, 1} (c) β(π₯) = β5 β 4π₯ β π₯ 2 β is defined when 5 β 4π₯ β π₯ 2 β₯ 0 β (1 β π₯)(5 + π₯) β₯ 0 π₯ < β5 π₯ = β5 β5 < π₯ < 1 π₯ = 1 π₯ > 1 Sign of (1 β π₯) + + + 0 β Sign of (5 + π₯) β 0 + + + Sign of (1 β π₯)(5 + π₯) β 0 + 0 β π·ππ(β) = [β5, 1] Dr. Emily Chan Page 4 9. (a) If π(π₯) is an odd function and π(π₯) is an even function, then π(βπ₯) = βπ(π₯) and π(βπ₯) = π(π₯) for all π₯ β π·ππ(π) β© π·ππ(π) (ππ)(βπ₯) = π(βπ₯)π(βπ₯) = βπ(π₯)π(π₯) = β(ππ)(π₯) β΄ (ππ)(π₯) is an odd function. (b) If π(π₯) and π(π₯) are both odd functions, then π(βπ₯) = βπ(π₯) and π(βπ₯) = βπ(π₯) for all π₯ β π·ππ(π) β© π·ππ(π) (ππ)(βπ₯) = π(βπ₯)π(βπ₯) = [βπ(π₯)][βπ(π₯)] = π(π₯)π(π₯) = (ππ)(π₯) β΄ (ππ)(π₯) is an even function. (c) If π(π₯) and π(π₯) are both even functions, then π(βπ₯) = π(π₯) and π(βπ₯) = π(π₯) for all π₯ β π·ππ(π) β© π·ππ(π) (ππ)(βπ₯) = π(βπ₯)π(βπ₯) = π(π₯)π(π₯) = (ππ)(π₯) β΄ (ππ)(π₯) is an even function. 10. (a) π 1 (π₯) = sin(π₯ 3 +π₯) π₯ 4 +3 π 1 (βπ₯) = sin[(βπ₯) 3 + (βπ₯)] (βπ₯) 4 + 3 = sin(βπ₯ 3 β π₯) π₯ 4 + 3 = β sin(π₯ 3 + π₯) π₯ 4 + 3 = βπ 1 (π₯) β΄ π 1 (π₯) is an odd function. (b) π 2 (π₯) = |π₯ 5 + 1| π 2 (βπ₯) = |(βπ₯) 5 + 1| = |βπ₯ 5 + 1| β π 2 (π₯) nor β π 2 (π₯) β΄ π 2 (π₯) is neither even nor odd. (c) π 3 (π₯) = cos 3 (2π₯) π 3 (βπ₯) = cos 3 [2(βπ₯)] = cos 3 (β2π₯) = [cos(β2π₯)] 3 = [cos(2π₯)] 3 = cos 3 (2π₯) = π 3 (π₯) β΄ π 3 (π₯) is an even function. 11. πΉ(π₯) = π₯ 2 + 2π₯ β 3 , π₯ β₯ 0 (a) πΉ(π₯) = π₯ 2 + 2π₯ β 3 = (π₯ + 1) 2 β 4 = (π₯ + 3)(π₯ β 1) 0 1 β3 π₯ π¦ Dr. Emily Chan Page 5 By the Horizontal Line Test, no horizontal intersects the graph more than once, therefore πΉ(π₯) is one-to-one. π
ππ(πΉ) = [β3, β) (b) Since πΉ(π₯) is one-to-one, its inverse exists. Let π¦ = π₯ 2 + 2π₯ β 3 , π₯ β₯ 0 Then π¦ = (π₯ + 1) 2 β 4 β π¦ + 4 = (π₯ + 1) 2 β π₯ + 1 = βπ¦ + 4 or π₯ + 1 = ββπ¦ + 4 (rejected since π₯ β₯ 0 so that π₯ + 1 β₯ 1 > 0 ) β π₯ = βπ¦ + 4 β 1 β΄ πΉ β1 (π₯) = βπ₯ + 4 β 1 π·ππ(πΉ β1 ) = π
ππ(πΉ) = [β3, β) π
ππ(πΉ β1 ) = π·ππ(πΉ) = [0, β) 12. (a) π(π₯) = 3π₯ β 2 , π₯ β β Let π(π₯ 1 ) = π(π₯ 2 ) . Then 3π₯ 1 β 2 = 3π₯ 2 β 2 β 3π₯ 1 = 3π₯ 2 β π₯ 1 = π₯ 2 is the only solution. Hence, π(π₯) is one-to-one and so its inverse exists. Let π¦ = 3π₯ β 2 . Then π¦ + 2 = 3π₯ β π₯ = π¦+2 3 β΄ π β1 (π₯) = π₯+2 3 For every π₯ β π·ππ(π) = β , 3π₯ β 2 can be any real number, i.e. π
ππ(π) = β β΄ π·ππ(π β1 ) = π
ππ(π) = β (b) π(π₯) = π₯ 2 β 1 , π₯ β₯ 0 Let π(π₯ 1 ) = π(π₯ 2 ) Then π₯ 1 2 β 1 = π₯ 2 2 β 1 β π₯ 1 2 = π₯ 2 2 β π₯ 1 = π₯ 2 or π₯ 1 = βπ₯ 2 (rejected since π₯ β₯ 0 ) β π₯ 1 = π₯ 2 is the only solution. Hence, π(π₯) is one-to-one and so its inverse exists. Let π¦ = π₯ 2 β 1 , π₯ β₯ 0 Then π¦ + 1 = π₯ 2 β π₯ = βπ¦ + 1 or π₯ = ββπ¦ + 1 (rejected since π₯ β₯ 0 ) β΄ π β1 (π₯) = βπ₯ + 1 Dr. Emily Chan Page 6 For every π₯ β π·ππ(π) = [0, β) , π₯ 2 β₯ 0 β π₯ 2 β 1 β₯ β1 , i.e. π
ππ(π) = [β1, β) β΄ π·ππ(π β1 ) = π
ππ(π) = [β1, β) - End - GE1359