�

1 Functions

1. Diagram(i): Arrowdiagram Diagram(ii): Orderedpairs Diagram(iii): Graph

2. (a) 3(b) 2(c) (i) {2,�,–�} (ii) {–3,3,6} (iii) {–3,3,6}(d) one-to-one

3. (a) many-to-one(b) (i) {8,9,�6} (ii) {9,�6}

4. Sincex-coordinateisonemorethany,then(a) x=�2,(b) y=�4.

5. (a) one-to-many(b) SinceimageinsetBisthesquarerootofobject

insetA,

theny=±AB9 =±3

6. (a) many-to-many(b) Range={d,e,f}

7. Diagram(i): one-to-oneDiagram(ii): one-to-oneDiagram(iii): isnotafunctionbecauseobjectchas noimage.Diagram(iv): is not a function because it is one-to-manyrelation.Diagram(v): many-to-one

8. (a) 5(b) 4(c) f :x→x+3orf (x)=x+3(d) (i) {2,4,5} (ii) {5,7,8}

9. f (x)=3x2

Imageforobject–�=f (–�) =3(–�)2

=3

10. Givenf (x)=2x–5andf (x)=�0.

2x–5=�0

x=�5—–2

Therefore,theobjectis�5—–2

.

11. h(x)=sinx h(90°)=sin90° =�

12. g(x)= x–7——–5

(a) g(2)= 2–7——–5

=|–�| =�

(b) g(x)=4

x–7——–5 =4

x–7——–

5=4 or

x–7——–5

=–4

x–7=20 x–7=–20 x=27 x=–�3

13. f (x)=|x–5|

Whenx=–2,f (x)=|–2–5| =|–7| =7

Whenx=7,f (x) =|7–5| =2

Whenx–5=0 x=5

Whenx=0,f (x)=0–5 =–5 =5

2

x –2 0 5 7

f (x) 7 5 0 2

f (x)

x–2 5 7

2

0

5

7

Range=0<f (x)<7

14. (a)f (x)

x

1

090° 180° 270° 360°

f(x) = cos x

(b) Whenx=�20° f (�20°) =cos�20°

=– �—2

=�—2

Hence,therangeoff (x)is0<f (x)<�.

15. (a) (i) f (x)=x2–4 Imageforobject3=f (3) =32–4 =5 (ii) Imageforobject–4=f (–4) =(–4)2–4 =�2(b) (i)

f (x)

x–4 –2 0 2 3

5

12

4

Hence,therangeoff (x)is0<f (x)<�2. (ii) f (x) =5 x2–4 =5 x2 =9 x =±AB9 =±3

16. f (x)=–2x–�Whenx=0,f (0)=––� =–�

Whenx=3,f (3)=–6–�

=–5When2x–�=0,

x=�—2

x 0�—2

3

f (x) –� 0 –5

f (x)

x–1

–5

1

2–

3

0

f (x) = – 2x – 1

Therefore,therangeoff (x)is–5<f (x)<0.

17. (a) f (7) =3(7)–� =20

(b) f (3) =5+3—3

=6

18. Sincethegraphisastraightline,

thegradient= 3–0———–0–(–2)

=3—2

,f (x)-intercept=3

Therefore,theequationisf (x)= 3—2

x+3.

19. (a) fg(x)=f (�–6x) =3(�–6x) =3–�8x

(b) gf (x)=g(3x) =�–6(3x) =�–�8x

(c) f 2(x)=ff (x) =f (3x) =3(3x) =9x

(d) g2(x)=gg(x) =g(�–6x) =�–6(�–6x) =�–6+36x =36x–5

20. (a) hp(x)=h(x2–2x) =2(x2–2x)+3 =2x2–4x+3

(b) ph(x)=p(2x+3) =(2x+3)2–2(2x+3) =(2x+3)(2x+3–2) =(2x+3)(2x+�)

3

21. (a) fg(–2)=f [�–4(–2)2] =f [�–�6] =f (–�5)

=–�5—–2

+3

=–9—2

(b) gf (–2)=g1–2—2

+32 =g(2) =�–4(2)2

=–�5

22. (a) fg(x)=4 f (�–3x)=4 5(�–3x)=4 5–�5x=4 –�5x=–�

x=�—–

�5

(b) gf (x)=–3+x g(5x)=–3+x �–3(5x)=–3+x �–�5x=–3+x �6x=4

x=4—–

�6

=�—4

(c) f 2(x)=8x+� ff (x)=8x+� f (5x)=8x+� 5(5x)=8x+� 25x=8x+� �7x=�

x=�—–

�7

23. hg(x)=4x2–2x+53g(x)–�=4x2–2x+5 3g(x)=4x2–2x+6

g(x)= 2—3

(2x2–x+3)

24. gh(x)=x—2

+�

g(2+5x)=x—2

+�........................1

Let 2+5x=y

x=y–2——–

5

From1,

g(y)=1 y–2——–

5 2———–

2+�

=y–2——–

�0+�

=y+8——–

�0

Therefore,g(x)=x+8——–

�0

25. (a) f (x)=2x+� f (2)=t 2(2)+�= t t=5(b) f –�(t)=2

26. (a) g–�(5)=0

(b) g(r)=8 5+2r=8 2r=3

r=3—2

(c) Letg–�(7)=x, then 7=g(x) =5+2x 2x=2 x=�

Hence,g–�(7)=�

27. (a) Letf –�(x) =y, then x =f (y) =2y

y =x—2

Hence,f –�(x)=x—2

(b) Letg–�(x)=y, then x=g(y)

=y—5

y=5x Hence,g–�(x)=5x

(c) Leth–�(x)=y, then x=h(y) =3y+� 3y=x–�

y=x–�——–

3

Hence,h–�(x)=x–�——–

3

(d) Letp–�(x)=y, then x=p(y)

=y—2

+�

y—2

=x–�

y=2(x–�)

Hence,p–�(x) =2(x–�) =2x–2

4

28. (a) t–3=0 t=3

(b) Letf –�(x)=y, then x=f (y)

=2——–

y–3

y–3=2—x

y=2—x +3

Therefore,f –�(x)=2—x +3,x ≠ 0

(c) Let g –�(x)=y, then x=g(y)

=y——–

y+� x(y+�)=y xy+x=y y–xy=x y(�–x)=x

y=x——–

�–x

Therefore,g–�(x)=x——–

�–x ,x ≠ 1

29. (a) The inverse function of f, f –�, does not existfor the domain –5 < x < 5 because f –� is amany-to-onetypeofrelation.

(b) The inverse functionof f, f –�,existsbecause f –�isaone-to-onetypeofrelation.

1. (a) many-to-many(b) Range={d,f,g}

2. (a) f (2)=–�(b) g(–�)=3(c) gf (2)=g(–�) =3

3. (a) Theobjectsof5are0and�.(b) Theimagesof2are9and�3.

4. (a) one-to-one(b) f:x→x2orf (x)=x2

5. f (x)=x–n——–x

f (4)=–2—3

4–n——–

4=–

2—3

4–n=–8—3

n=4+8—3

=20—–3

6. (a) fg(x) =5x 4g(x)–� =5x 4g(x) =5x+�

g(x) =�—4

(5x+�)

(b) gf (x) =9 g(4x–�) =9

�—4

[5(4x–�)+�] =9

5(4x–�)+� =36 5(4x–�) =35 4x–� =7

x =8—4

=2

7. (a) hf (3)=2(b) h–�(2)=6

8. Letf –�(x)=y,then x=f (y) =3–4y 4y=3–x

y=3–x——–

4

Therefore,f –�(x)=3–x——–

4

Hence,f –�g(x) =f –�(�+2x)

=3–(�+2x)—————–

4

=3–�–2x————–

4

=2–2x———

4

=2(�–x)———–

4

=�–x–—––

2

9. (a) Letf –�(4)=k, then 4=f (k) =�+2k

k=4–�——–

2

=3—2

f –�(4)=3—2

5

(b) gf (x)=g(�+2x) =(�+2x)2+4(�+2x)–3 =�+4x+4x2+4+8x–3 =4x2+�2x+2

10. (a) Letp–�(x)=y, then x=p(y)

=4——–

y+� x(y+�)=4

y+�=4—x

y=4—x –�

Hence,p–�(x)= 4—x

–�,x ≠ 0

(b) p–�(5)= 4—5

–�

=– �—5

11. Letf –�(x)=y,then x=f (y) =p–3y 3y=–x+p

y=– x—3

+p—3

\f –�(x)=– x—3

+p—3

.....................1

Compare1tothef –�(x)=q—2

x+ 2—3

Therefore,q—2

=– �—3

q=– 2—3

and p=2

12. (a) Let f –�(x)=y, then x=f (y) =4+5y

y=x–4——–

5

Hence,f –�(x)=x–4——–

5

(b) gf –�(x)=g1 x–4——–5 2

=1 x–4——–

5 2———–

2–�

= x–4——–�0

–�

=x–4–�0————–

�0

=x–�4———

�0

(c) hg(x)=6–5x

h1 x—2

–�2=6–5x................1

Let x—2

–�=y,

then x—2

=�+y

x=2+2y...............2

Therefore,1becomesh(y) =6–5(2+2y) =6–�0–�0y =–4–�0y Hence,h(x)=–4–�0x

1. (a) �—2

(b) x=�2sinceelementinsetB ishalfofsetA.

2. (a) f –�(�5)=3(b) codomain={�0,�5,20,30}

3. Let f (x)=y,then x=f –�(y)

x=y+�——–

4 y+�=4x y=4x–�

Hence,f (x)=4x–�

4. fg(x)=f 1 x—2

+32 =�+51 x—

2+32

=�+ 5—2

x+�5

= 5—2

x+�6

5. gf (3)=g(5) =6

6. (a) many-to-one(b) f:x→x2orf (x)=x2

7. f (x)=2x–2 f (–2)=2(–2)–2 =–6 =6Therefore,therangeis2<f (x)<6.

6

8. h(x)=x+5——–

2 h(q)=8

q+5——–

2 =8

q+5=�6 q=��

9. fg(x)= x—4

+7

3g(x)–�= x—4

+7

3g(x)= x—4

+8

g(x)= x—–�2

+ 8—3

10. h2(x)=hh(x) =h(ax+b) =a(ax+b)+b =a2x+ab+b.................................. 1

Compare1withh2(x)=4x+9Therefore, a2=4 a=±2and ab+b=9...................................... 2Sincea.0,hencea=2

Substitutea=2into2,2b+b=9 3b=9 b=3

Hence,a=2andb=3.

11. hg(x)=h1 x—2 2

=31 x—2 2–�

= 3—2

x–�

Hence,p(x)=hg(x)

12. (a) (i) Let h–�(x)=y, then x=h(y) =4y–3

y= x+3——–4

h–�(x)=x+3——–

4

h–�g(2)=h–�3�– 2—2 4

=h–�(0)

=0+3——–

4

= 3—4

(ii) Let g–�(x) =y, then x =g(y)

=�–y

—2

y

—2

=�–x

y =2–2x g–�(x) =2–2x

hg–�(x)=h(2–2x) =4(2–2x)–3 =8–8x–3 =5–8x

(b) gh(x)=g(4x–3)

=�– 4x–3———2

=�–12x– 3—2 2

=�–2x+ 3—2

= 5—2

–2x

=5–4x———

2........................... 1

Comparegh(x)=m–nx———

2 with1,

Hence,m=5andn=4.

13.x –� 0 2—

34

f (x) 5 2 0 �0

f (x)=2–3x

Whenx=–�,f (–�)=2–3(–�) =2+3 =5

Whenx=4,f (4)=2–�2 =–�0 =�0

Whenf (x)=0,2–3x=0

x= 2—3

Whenx=0,f (0)=2–0 =2

f (x)

x–1 42

3–

0

5

10

2

7

(a) Therangefor0<x<4is0<f (x)<�0.(b) Whenf (x)=5 2–3x=5 or 2–3x=–5

x= 7—3

3x=–3 x=–�

Therefore,thedomainis–�<x< 7—3

.

14. (a) f (2)=7

2—a +b=7................................... 1

f (–2)=2

–2—–a +b=2.................................. 2

1+2, 2b=9

b= 9—2

Substituteb= 9—2

into1,

2—a + 9—2

=7

2—a =7– 9—2

=�4–9———

2

= 5—2

5a=4

a= 4—5

(b) f (x) = x—–4—5

+ 9—2

= 5x—–4

+ 9—2

Letf –�(4)=p, then 4=f (p)

= 5—4

p+ 9—2

5—4

p=4– 9—2

=– �—2

p=1– �—2 2×1 4—

5 2 =– 2—

5

Hence,f –�(4)=– 2—5

15. (a) fg(x)=4x–7 f (x–3)=4x–7.......................... 1

Let x–3=y, then x=y +3

Therefore,1becomesf (y) =4(y+3)–7 =4y+�2–7 =4y+5

Hence,f (x)=4x+5

(b) Let f –�(5)=p, then 5=f (p) =4p+5 p=0

Therefore,gf –�(5)=g(0) =0–3 =–3

(c) Let g–�(x)=y, and Let f –�(x)=y,then x=f (y) =4y+5

y= x–5——–4

f –�(x)= x–5——–4

then x=g(y) =y–3 y=x+3 g –�(x)=x+3

f –�g–�(x)=2x–9 f –�(x+3)=2x–9

(x+3)–5————–

4 =2x–9

x–2=8x–36 7x=34

x=34—–7

16. (a) f 2(x)=ff (x)

=f 1 x——–x+2 2

=1 x——–

x+2 2—————–

1 x——–x+2 2+2

=

x——–x+2——————–

x+2(x+2)—————–x+2

=

x———x+2—–——3x+4———x+2

=x———

3x+4 ,x ≠ – 4—3

(b) Let f –�(x)=y, then x=f (y)

=y——–

y+2

x(y+2)=y xy+2x=y y–xy=2x y(�–x)=2x

y=2x——–

�–x

Hence,f –�(x)=2x——–

�–x ,x ≠ 1

8

17. (a) f (x)=x x2–3x=x x2–4x=0 x(x–4)=0 x=0,4

(b) f (x)=2g(x) x2–3x=2x2+2 x2+3x+2=0 (x+�)(x+2)=0 x=–�,–2

18. (a) f (x)=x+�——–

2 f (r)=4

r+�——–

2 =4

r+�=8 r=7

(b) g(4)=p 4=g–�(p)

4=p—4 +5

p—4 =–�

p=–4

(c) Let g(x)=y, then x=g–�(y)

=y

—4 +5

y

—4 =x–5

y=4x–20 g(x)=4x–20

gf (x)=g1 x+�——–2 2

=41 x+�——–2 2–20

=2(x+�)–20 =2x+2–20 =2x–�8

19. (a) f (x) ,8 x–5+�,8 x–5 ,7 –7 ,x–5,7 –7+5 , x ,7+5 –2 , x ,�2

(b) (i) f 2(x)=ff (x) =f (x–2) =(x–2)–2 =x–4

(ii) f 3(x)=f 2f (x) =f 2(x–2) =(x–2)–4 =x–6

Hence,f 30(x) =x–2×30 =x–60