functions 02

Additional Additional MathematicsMathematicsFunctions

QuestionsQuestions1.Given function f : x mx + 4 , x=n. x – nIf f(2) = 10 and f(8) = 4, finda) the values of m and nb) the image of 3c) the object of 11 Answer 4.The graph shows the absolute values function f(x) = I hx – k Ia) Find the possible values of h and kb) If the value of f(m) = 6, find the value of m Answer

2.Given that f(x) = 6 x = -b ,where ax + b aa and b are constants, f(1) = 2 and f(5) = 6/7, find the value of a and b Answer

3. Given that f:x 1 , x= k x – kand f(3) = 1, Finda) the value of kb) the values of a such that f(a ) = f(3a)² Answer

5.Sketch the graph of the function g:x I3x – 5I for the domain 0≤x≤4 and state the range of g(x) corresponding to the domain. Answer

QuestionsQuestions6.Given the functions f:x ax + b and g : x 5 – x. if fg(x) = 19 – 3x, find a) the values of a and bb) the value of x such that fg(-x) = f(7) Answer

9.Given that f(x) = 3x – 8 and fg(x) = 3x - 2, find the ²function g. Answer7.Given that f:x 2x + 1 and g:x 3 xx = 0, find the composite functiona) fgb) gfc) f²d) g ²Answer

10.Given that g(x) = 2 – x, fg(x) = 2x + x + 8, find²a) the function fb) the value of g(x) if fg(x) = 9 Answer

8.Given that h:x2x + 1 and k:x 2x +3 3Finda) the value of x if hk(x) = xb) the value of x if kh(x) = -3c) the value of x when hk(x) = 3 kh(x) Answer

11. f(x) = 3x + 2 and gf(x)= 9x+12x ,²find a) g(x)b)fg(2) Answer

QuestionsQuestions12.Given that h(x) = 3 + x and hk(x) = 1 , x ≠3,find 3 – xa) k(x)b) the value of k(x) such that x = 4 Answer

15. Given that f:x 3x – 1 and gf:x 2x - 6x + 4, find²a) the function gb) the values of x such that f(x) = 9g(x) Answer

13) Given that f:x 2 +x, g(x) ax + b, where a and b are constants, and gf(x) = 6 + 10x, finda) the value of a and bb) the value of x if g(x) = gf(x) Answer

14) Given that f(x) = 2x + x² and g(x) = 4 – 3x,Find a) the values of x such that f(x) = 8b) the composite function fgc) the values of x if f(x) = 3g(x) Answer

Solution for Question 1(a)Solution for Question 1(a) Subtracting (1) and (2)Where 2m + 10n = 16- 2m + n = 7 9n = 9 9n = 9 n = 1Substitute n = 1 to (2)2m + 1 = 7 2m = 6 m = 3

Given f(x) = mx + 4 x – nf(2) = 10 and f(8) = 4f(2) = 10 (substitute x = 2, f(x) = 10) 2m + 4 = 102 – n 2m + 4 = 20 – 10n2m + 10 = 16 ……. (1)f(8) = 48m + 4 = 4(Substitute x = 8, f(x)=4) 8 – n (bring 8 – n to the right)8m + 4 = 32 – 4n8m + 4n = 28(Simplify by dividing with 4)2m + n = 7 ……(2)Therefore, f(x) = 3x + 4 x – 1 Menu

Solutions for Question 1(b)Solutions for Question 1(b) To find the object of 11f(x) = 11 3x + 4 = 11x – 13x + 4 = 11x – 118x = 15x = 15 8 (answer)

Solution for Question 1(c) Solution for Question 1(c) to find the image of 3, x = 3f(3) = 3(3) + 4 3 – 1 = 13 2 (answer)

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Solution for Question 2Solution for Question 2Given f(x) = 6 . ax + bf(1) = 2 and f(5) = 6/7f(1) = 2 ( substitute x = 1, f(x) = 2) 6 = 2ax + b (bring a + b to the right)6 = 2a + 2ba + b = 3 …… (1)Then f(5) = 6/7( substitute x = 5, f(x) = 6/7) 6 = 6ax + b 7 (eliminate 6 on both sides) 1 =15a + b 7

after we eliminate 6, we cross multiplytherefore, 7 = 5a + b 5a + b = 7 …… (2)Subtracting (1) from (2) 5a + b = 7- a + b = 3 4a = 4a = 1Substitute a=1 into (1)1 + b = 3b = 2Therefore, a = 1, b = 2 (answer) Menu

Solution for Question 3(a)Solution for Question 3(a)b) find the values of a such that

f(a²) = f(3a) 1 = 1

a² - 2 3a – 2 (cross multiply for both sides)

3a – 2 = a² - 2a² = 3a(both sides cancel out one a)

a = 3 (answer)

Solution for question 3(b)Solution for question 3(b)a) find the value of kf(x) = 1 x – kf(3) = 1 (substitute x = 3, f(x) = 1)1 = 1 3 – k ( bring 3 – k over the right)3 – k = 1k = 2Therefore f(x) = 1 x – 2 (answer)Menu

Solution for Question 4(a)Solution for Question 4(a) If k = 8, 0 = I 4h – 8 I 4h = 8 h = 2Therefore, h = -2 or 2Then f(x) = I -2x + 8 I or f(x) = I2x – 8I (answer)

a) Find the possible values of h and kf(x) = I hx – k IGiven f(0) = 8 , f(4) = 08 = I h(0) – k II-kI = 8-k = 8 or -k = -8 when modulus sign, I I is removed, the resulting value can be negative or positive.therefore, k = -8 or 8When f(4) = 0 x = 4 and f(x) = 0If k = -8, 0 = I 4h + 8 I -4h = 8 h = -2Menu

Solution for Question 4(b)Solution for Question 4(b) or 6 = 2x – 82x = 14 x = 7From the graph, m>4so, x = 7 (answer)

b) If the value of f(m) 6, findthe value of mTake either one of thefunction,Let f(x) = I 2x – 8 IGiven f(m) = 66 = I 2x – 8 I-6 = 2x – 8 or 6 = 2x – 8 when modulus sign, I I isremoved, the resultingvalue can be negative or positiveObtain answers separately-6 = 2x – 82x = 2 x = 1 Menu

Solution for Question 5Solution for Question 5 When g(x) = 0 0 = I 3x – 5 I3x – 5 = 03x = 5x = 5/3Steps to sketch the graphGiven g(x) = I 3x – 5 I-First, find the value of g(x),when x = 0-Second, find the value of g(x),when x = 4. This is to checkthe maximum value of g(x)-Finally, find the value of x,when g(x) = 0. This is to checkthe point where the turning point exists.When x = 0,g(0) = I 3(0) – 5 I = I -5 I = 5When x = 4, g(4) = I 3(4) – 5 I = 7

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Solution for Question Solution for Question 6(a)6(a) Substitute a = 35(3) + b = 1915 + b = 19b = 4Therefore, a = 3, b = 4f(x) = 3x + 4 (answer)b) Find the value of xsuch that fg(-x) = f(4)19 + 3x = 3(4) + 419 + 3x = 163x = 3x = -1( answer)

a)the values of a and bGiven f(x) = ax + bg(x) = 5 – xfg(x) = 19 – 3x ……(1)fg(x) = f[g(x)] = a(5 – x) + b = 5a – ax + b (factorize a) = (5a + b) – ax……(2)Compare the x-coefficient of (1) and (2)-a = -3a = 3Compare the constant of (1) and (2)5a + b = 19

Solution for Question 6(b)Solution for Question 6(b)

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Solution for Question Solution for Question 7(a)7(a)a)fg fg(x) = f[g(x)] substitute g(x) = 3 x = f(3/x) = 2(3/x) + 1 =6 + 1 (answer) xb) gf gf(x) = g[f(x)](substitute f(x) = 2x + 1) = g(2x + 1) = 3 (answer) 2x + 1

c) f ²f (x) = ff(x)² = f[f(x)](substitute f(x) = 2x + 1) = f(2x + 1) = 2(2x + 1) + 1 = 4x + 3 (answer)d) g² g (x) = g[g(x)]² (substitute g(x) = 3/x) = g(3/x) = 3 3/x = x (answer)

Solution for Question Solution for Question 7(b)7(b)

Solution for Question Solution for Question 7(c)7(c)

Solution for Question Solution for Question 7(d)7(d)

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Solution for Question Solution for Question 8(a)8(a)a)Find the value of x, if hk(x) = xFirst, find hk(x)hk(x)= h[k(x)]= h(2x + 3)= 2(2x + 3) + 1 3= 4x + 6 + 1 3= 4x + 7 3From hk(x) = x, so4x + 7 = x 34x + 7 = 3xx = -7 (answer)

b) Find the value of x if kh(x) = -3First, find kh(x),kh(x) = k[h(x)] = k ( 2x + 1) 3 = 2 ( 2x + 1 ) + 3 3 = 4x + 2 + 3 3kh(x) = -34x + 2 + 3 = -3 34x + 2 + 9 = -94x = -20x = -5 (answer)


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Solution for Question Solution for Question 8(c)8(c)c) find the value of x when hk(x) = 3kh(x)From (a) From (b)hk(x) = 3kh(x) 4x + 7 = (4x + 2 + 3) 3 3 34x + 7 = 3(4x + 2 + 9)4x + 7 = 12x + 6 + 27-26 = 8xx = - 13/4 (answer)

fg(x) = 3x - 2²f[g(x)] = 3x - 2²Substitute f(x) = 3x – 83g(x) – 8 = 3x - 2²3g(x) = 3x + 6²g(x) = x + 2 (answer)²

Solution for Question Solution for Question 99

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Solution for Question Solution for Question 10(a)10(a)a)fg(x) = 2x + x + 8²f(2-x) = 2x + x + 8²Let y = 2 – xHence x = 2 – y f(y) = 2(2 – y) + (2 – y) + ²8 = 2(4–4y+y )+ 2– y ²+ 8 = 8 – 8y+2y + 2 – y ²+ 8 = 2y - 9y + 18² f(x) = 2x - 9x + ²18(answer)

b) fg(x) = 92x + x + 8 = 9²2x + x – 1 = 0²(2x – 1)(x + 1) = 02x – 1 = 0 or x + 1 = 0x = x = -½1When x = ½g( ) = 2 – ½ ½ = 3/2When x = -1g(-1) = 2 – (-1) = 3So, g(x) = 3/2 or 3 (answer)


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Solution for Question Solution for Question 11(a)11(a)a)Find g(x)gf(x) = 9x + 12x²g[f(x)] = 9x + 12x²(substitute f(x) = 3x + 2)g(3x + 2) = 9x + 12x²Let y = 3x + 2 3x = y – 2Hence x = y – 2 3Let u = y – 2 3g(u) = 9 (y – 2) + 12 (y – 2)² 3 3 =3y – 6 + 12 (y - 4y +4² ) 9 = 4y - 7y -2² 3so, g(x) = 4x - 7x – 2² 3 (answer)

b) fg(x) = f[g(x)] = 3(4x - 7x – ²2 ) + 2 3 = 4x - 7x – 2 ²+ 2 = 4x - 7x ²(answer)


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Solution for Question 12 Solution for Question 12 (a)(a)a) Find k(x),hk(x) = 1 (substitute h(x) = 3 + x) 3 – xh[k(x)] = 1 3 – x3 + k(x) = 1 3 – x[3 + k(x)][3 – x] = 19 – 3x + 3k(x) – k(x)x = 1So, 8 – 3x = k(x)x – 3k(x)8 – 3x = k(x) (x – 3)k(x) = 8 – 3x x – 3 (answer)

b) Find the value of k(4)k(4) = 8 – 3(4) 4 – 3 = 8 – 12 = -4

Solution for Question Solution for Question 12 (b)12 (b)

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Solution for Question Solution for Question 13(a)13(a)a)Find the values of a and b gf(x) = 6 + 10xg[f(x)] = 6 + 10x (substitute f(x) = 2 + 5x)9(2 + 5x) + b = 6 + 10x2a + 5ax + b = 6 + 10x5ax + 2a + b = 6 + 10xCompare the x-coefficient5a = 10a = 2Compare the constant2a + b = 6 (replace a = 2)2(2) + b = 64 + b = 6b = 2 ( answer)

b) the value of x if g(x) = gf(x)g(x) = gf(x)2x + 2 = 6 + 10x8x = -4 (Bring the 8 to the right)x = - 4 = - 1 8 2


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Solution for Question Solution for Question 14(a)14(a)a)Find the values of x suchthat f(x) = 8f(x) = 82x + x = 8²x + 2x – 8 = 0²(x + 4)(x – 2) = 0So, x + 4 = 0x = -4ORx – 2 = 0x = 2

b) Find the composite fgfg(x) = f[g(x)] ( substitute g(x) = 4–x) = f[ 4 – 3x] = 2(4 – 3x) + (4 – 3x)² = 8 – 6x + 16 + 9x - ²24x = 9x - 30x + 24 ²(answer)c)Find the values of x if f(x) = 3g(x)f(x) = 3g(x)2x + x = 3(4 – 3x)²x + 2x = 12 – 9x²x + 11x – 12 = 0²So, (x + 12)(x – 1) = 0(x + 12) = 0 OR (x – 1) = 0x = -12 x = 1


Solution for Question Solution for Question 14(c)14(c)

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Solution for Question Solution for Question 15(a)15(a)g[3x – 1] = 2x - 6x + 4²Let y = 3x – 1 3x = y + 1Hence x = y + 1 3Let u = y + 1 3g(u) = 2(y + 1) -6(² y + 1)+4 3 3 =2(y +2y+1² )-2y–2+ 4 9 = 2y +4y+2² –2y–2+ 4 9 = 2y +4y+2–18y + 18² 9 = 2y - 14y + 20² 9g(x) = 2x - 14x + 20² 9

b) f(x) = 9g(x)3x – 1 = 9(2x -14x+20² ) 93x – 1 = 2x -14x+20²2x -17x+21=0²(2x-3)(x-7) =02x – 3 = 0 or x – 7 = 0x=3 x = 7 2


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