DHS Equations & Inequalities (9758) (Revision Solutions)

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 1 2. Equations and Inequalities (solutions) (I) Equations 1 140dfg++= …(1) 20 20gdf dfg=++ ⇒+−= − …(2) 21 42 10 2900d fg++= …(3) Solving the 3 equations, 20, 40, 80dfg= = = 2 Let x, y, and z be the number of trays of blueberry, strawberry and chocolate cupcakes respectively. Time: 8 7 6 17 60 1020xyz++=×= Amt: 0.6 0.6 0.8 96xyz++= Price: 12 (1) 12 (0.9) 12 (0.8) 1572xy z++= Using GC, 50, 50, 45xyz= = = 3 4 Let x be no. of chickens. Let y be no. of horses. Let z be no. of sheep. 2zx= 20 0x yz⇒ + −= --------------(1) 2x + 4y + 4z = 1250 2 4 4 1250xyz⇒++= ---------(2) Case 1: If 250xyz++= -------------(3) By GC, x = 125− , y = 625, z = 250− (rejected) (Alternative): Reject 250xyz++= , because any combination of 250 animals will never have 1250 legs. (Maximum no of legs = 250 x 4 = 1000)

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 2 Case 2: If 350xyz++= -------------(3) By GC, x = 75, y = 125, z = 150 ∴ Correct number of chickens =75, horses =125, sheep = 150 5 Finding table values in SGD, Cheese/kg Chocolate/kg Candy/kg Price/SGD 4 6 6 Price/SGD 8 10 4 Price/SGD 8 5 7 Let x, y, z be the number of three kg packs bought from Denmark, England and Russia respectively. 4 8 8 84 6 10 5 85 6 4 7 77 xyz x yz xyz ++= + += ++= 5,x= 3,y= 5z= She should buy 13 packs in total. 6 Sub and into . 1 ----- (1) 8 4 2 2 -----(2) abcd abc d +++= + + += Since (2,2) is also the stationary point, h'(2) 0= . i.e. 12 4 0 ----- (3)a bc+ += Using the GC, 11 24 35 24 2 ad bd cd = −− = + =− 0 11 35 24 24 02 ab c dd d ≤   −− +     ≤− 6{ : 2 or 0} 5dd d∈ ≤− − ≤ < (1, 1) (2, 2) h( )yx= 0

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 3 7 22 22 22 : 0 02 2 44 M x y Ax By C A B ABxy C + + + +=   ⇒ + + + −−+ =     Centre of M : ,22 AB−− 2( 1)yx= −+ passes through the centre: 2122 2 4 ----- (1 ) BA AB −= −−+  += At intersection between | |yx= and M, we have 22 || || 0x x Ax B x C+ + + += At x = –2, 22 ( 2) (| 2 |) ( 2) (| 2 | ) 0 2 2 8 ----- (2) AB C A BC − +− + −+ − += − + += − At x = –8, 22 ( 8) (| 8 |) ( 8) (| 8 | ) 0 8 8 128 ----- (3) AB C A BC − +− + −+ − += − + += − Solving (2), (3) and (4) using GC: A = 8, B = –12, C = 32 22: 8 12 32 0Mx y x y++− += 8 Let x : y : z be the ratio for the servings of fish fillet, salad and fries. 150x + 15y + 5z = 4k 60x + 30y + 250z = 8k 25x + 5y + 110z = 3k where k is a constant. 150x + 15y + 5z = 4 60x + 30y + 250z = 8 25x + 5y + 110z = 3 Solving matrix or simultaneous equations x = 0.02, y = 0.06, z =0.02 Ratio is 1:3:1 (ans) 9 (i) Let 32 nu an bn cn d= + ++ 1 32.1u abcd=+++=

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 4 2 8 4 2 17u abc d= + + += 3 27 9 3 0.7u a b cd= + ++= 4 64 16 4 7.8u a b cd= + + += − Using GC, 1.5a= , 9.6b=− , 3.2c= , 37d = 321.5 9.6 3.2 37nunnn∴= − + + (ii) 32555 1.5 9.6 3.2 518 0nu nnn>⇒ − +−> Method 1 From GC graphing, 9.7871n> ∴ least value of n 10= Method 2 From GC Table, ∴ least value of n 10= (II) Inequalities 10 ( ) 2 31 12 31 and 12 31 01 32 01 x xx x xxx x xx x xx x + <<− + <<− +− − <− +− <− ( )( ) 2 23 01 13 0 1 xx x xx x −− >− +− >− 1 1 or 3 xx∴− < < > 1 3

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 5 11Since , 1 22xx< −< < 11 (i) ( ) 2 2 21 042 x xx + ≥−− ( ) 2 2 21 042 x xx + ≤−+ ( ) ( ) ( ) ( ) ( ) 2 2 2 21 0 22 21 0 22 22 x x x xx + ≤ −− + ≤ −+ −− 2 2 < 2 2 x− <+ or 1 2x = − (ii) Replace x with x 12 2 < 2 2 or (rej) 2 0.343 < 11.7 xx x − <+ = − <

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 6 12 2 2 2 2 2 2 2 61 (*) 6 ( 1) 0 6 0 6 0 ( 3)( 2) 0 3 or 2 x xx xx x xx x xx x xx x xx +≤ −−− −+ ≤ − −+ ≤ +− ≥ +− ≥ ≤− ≥ Replace x with 2x− in (*), obtain 2 2 6 ( 2) 1 ( 2) 2 61 (2 ) 2 x xx x xx −+≤−− −≤−− 2 3 or 2 2 1 or 4 xx xx − ≤− − ≥ ≤− ≥ 13 5 22 xx ≤+− 5 ( 2) 02 xx −+≤− 5 ( 2)( 2) 02 xx x −− + ≤− (3 )(3 ) 02 xx x −+ ≤− Solution set = { : 3 or 3 2}xx x∈ ≥ −≤< “Hence method” Replace x by x2 The solutions for 2 2 5 2 2 x x ≥+ − are x2 ≤ −3 or 2 < x2 ≤ 3 2 3 or 3 2xx∴ <≤ − ≤< − “Otherwise method” Solve algebraically but method is longer.

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 7 14 2 ( 4)( 1) 4 ( 2) ( 3 4 4 8) 0 xx x xx xx x x + −≥ + + −− − ≥ 2( 12) 0xx x −− ≥ ( 3)( 4) 0xx x+ −≥ 3 0 or 4xx−≤ ≤ ≥ (ii) Replace x by x , 3 0 or 4 Since 0, xx x ∴− ≤ ≤ ≥ ≥ 0 or 4 or 4.x xx⇒ = ≥ ≤− 15 2 2 2 15 066 xx xx −+ ≥−+ Since the discriminant of 2 2 15xx−+ = 4 – 4(1)(15) = -56 < 0 and the coefficient of x2 is positive, we know 2 2 15 0xx−+> for all real values of x. Since 2 2 15 0xx−+> for all real values of x, ( ) ( ) 2 2 2 2 15 0 6 6066 33 330 3 3 or 33 xx xxxx xx xx −+ ≥ ⇒ − +>−+ ⇒ −− −+ > <− >+ Replace x by |x|, 3 3 or 3 3 3 3 3 3 or 3 3 or 3 3 xx xx x <− >+ −+ < <− < −− >+ 16 From GC, 0 < x < 4.42806 or x > 13.706 0 < x < 4.43 or x > 13.7 Replacing x by x2: x2 2 > ln x2 x > 2 2 ln x From abov e, 0 < x2 < 4.42806 or x2 > 13.706 0 < x < 2.1043 or x > 3.702 0 < x < 2.10 or x > 3.70 (ii) Replace x by 2 x , 3 0 or 422 xx∴− ≤ ≤ ≥ 6 0 or 8xx⇒∴− ≤ ≤ ≥ 0 -3 4

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 8 17 To find intersection point, 2 ( 2)xa x a+= −− 3 ax= From the graph, for 22x a xa−<+ , 3 ax> Replace x by –x and let a = 2 in the above inequality, ( ) 2(2) 2( ) 2xx−− <−+ becomes 4 22xx+<− Thus 22 33xx− > ⇒ <−

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 9 18 53 5 15 5 3yx y x+ = ⇒=− ( )15 533 , 2 2 xxx x −−≤ ≠ − ⇒ 15 532 3 xxx −− −≤ ⇒ ( )( ) 53 25 3xx x− − ≤− ⇒ 2 5565 3xx x− +≤− From GC, the x-coordinates of the points of the intersections of the 2 graphs are 1 3x= or x = 3. From the graph, 2 55 65 3xx x− +≤− for 1 33 x≤≤ Since 2,x≠ therefore the solution for 15 533 2 xx x −−≤ − is 1 2 or 2 33 xx≤< <≤ (OR 1 3, 23 xx≤≤ ≠ ) y 1 3 o x 6 5 3 2 2 56yx x=−+ 3 5 15yx+=

Y5 Topical Revision Package 1 2. Equations and Inequalities (solutions) 10 19 (i) ( ) ( ) 2222 4 5 2 52 2 10xx x x+ + = ++ − = ++ > , x∀∈  . (shown) (ii) ( ) ( ) ( ) ( ) ( ) 2 2 25 0, 1 45 1 254 50 0 1 x x xx x xxx x − ≤≠ ++ − −+ +>⇒ ≤ − 51 2x⇒<≤ . (b) Sketch the graphs of ( )( )23yx x= +− and 21yx= − . From the graph, solution to the inequality ( )( )2 321xx x+ −> − is 3.19x<− or 4.19x> . 21yx= − ( )( )23yx x= +− (4.193 , 7.385)Q ( 3.193 , 7.385)P −