Soalan 16:
Diberi fungsi h : x → 3x + 1, dan gh : x → 9x² + 6x – 4, cari
(a) h^-1 (x),
(b) g(x).

Penyelesaian:
(a)
$\begin{array}{l}\text{Katakan }{h}^{-1}\left(x\right)=y,\\ \text{oleh itu }h\left(y\right)=x\\ 3y+1=x\\ \text{ 3}y=x-1\\ y=\frac{x-1}{3}\\ \therefore h^{-1}\left(x\right)=\frac{x-1}{3}\\ h^{-1}:x\mapsto \frac{x-1}{3}\end{array}$

(b)
$\begin{array}{l}g\left[h\left(x\right)\right]=9x^2+6x-4\\ g\left(3x+1\right)=9x^2+6x-4\\ \text{Katakan }y=3x+1\\ \text{oleh itu }x=\frac{y-1}{3}\\ \text{ g}\left(y\right)=9{\left(\frac{y-1}{3}\right)}^2+6\left(\frac{y-1}{3}\right)-4\\ =\frac{\cancel{9}{\left(y-1\right)}^2}{\cancel{9}}+2\left(y-1\right)-4\\ =y^2-2y+1+2y-2-4\\ =y^2-5\\ \therefore g\left(x\right)=x^2-5\end{array}$

Soalan 18 (4 markah):
Rajah menunjukkan hubungan antara set A, set B dan set C.

Rajah

Diberi bahawa set A dipetakan kepada set B oleh fungsi $\frac{x+1}{2}$dan dipetakan kepada set C oleh fg : x → x² + 2x + 4.
(a) Tulis fungsi yang memetakan set A kepada set B dengan menggunakan tatatanda fungsi.
(b) Cari fungsi yang memetakan set B kepada set C.

Penyelesaian:

(a)
$g:x\to \frac{x+1}{2}$

(b)

$\begin{array}{l}g\left(x\right)=\frac{x+1}{2}\\ fg\left(x\right)=x^2+2x+4\\ f\left[g\left(x\right)\right]=x^2+2x+4\\ f\left(\frac{x+1}{2}\right)=x^2+2x+4\\ \\ \text{Katakan }\frac{x+1}{2}=y\\ x+1=2y\\ x=2y-1\\ \\ \therefore f\left(y\right)={\left(2y-1\right)}^2+2\left(2y-1\right)+4\\ f\left(y\right)=4y^2-4y+1+4y-2+4\\ f\left(y\right)=4y^2+3\\ f\left(x\right)=4x^2+3\\ \\ \text{Maka, fungsi yang memetakan set }B\\ \text{kepada set }C\text{ ialah }f\left(x\right)=4x^2+3.\end{array}$

Soalan 13:
Diberi fungsi g(x) = 3x dan h(x) = m – nx, dengan keadaan m dan n ialah pemalar.
Ungkapkan m dalam sebutan n dengan keadaan hg(1) = 4.

Penyelesaian:
$\begin{array}{l}hg\left(x\right)=h\left(3x\right)\\ =m-n\left(3x\right)\\ =m-3nx\\ \\ hg\left(1\right)=4\\ m-3n\left(1\right)=4\\ m-3n=4\\ m=4+3n\end{array}$

Soalan 14:
Diberi fungsi g : x → 3x – 2, cari  
(a) nilai x apabila g(x) memeta kepada diri sendiri,
(b) nilai k dengan keadaan g(2 – k) = 4k.

Penyelesaian:
(a)
$\begin{array}{l}g\left(x\right)=x\\ 3x-2=x\\ 3x-x=2\\ 2x=2\\ x=1\end{array}$

(b)
$\begin{array}{l}g\left(x\right)=3x-2\\ g\left(2-k\right)=4k\\ 3\left(2-k\right)-2=4k\\ 6-3k-2=4k\\ -7k=-4\\ k=\frac{4}{7}\end{array}$

Soalan 15:
Diberi fungsi f : x → px + 1, g : x → 3x – 5 dan fg(x) = 3px + q.  
Ungkapkan p dalam sebutan q.

Penyelesaian:
$\begin{array}{l}f\left(x\right)=px+1,g\left(x\right)=3x-5\\ fg\left(x\right)=p\left(3x-5\right)+1\\ =3px-5p+1\\ \\ \text{Diberi }fg\left(x\right)=3px+q\\ 3px-5p+1=3px+q\\ -5p+1=q\\ -5p=q-1\\ 5p=1-q\\ p=\frac{1-q}{5}\end{array}$

Soalan 5:
$\begin{array}{l}\text{Fungsi }f\text{ ditakrifkan oleh }f:x\to \frac{1+x}{1-x},x\ne 1.\\ \text{Cari }{f}^2,f^3,f^4\text{ dan seterusnya tulis fungsi bagi }{f}^{51}\text{ dan }{f}^{52}\text{.}\end{array}$

Penyelesaian:
$\begin{array}{l}f\left(x\right)=\frac{1+x}{1-x},x\ne 1\\ f^2\left(x\right)=f\left[f\left(x\right)\right]=f\left(\frac{1+x}{1-x}\right)\\ =\frac{1+\left(\frac{1+x}{1-x}\right)}{1-\left(\frac{1+x}{1-x}\right)}=\frac{\frac{1-x+1+x}{\cancel{1-x}}}{\frac{1-x-1-x}{\cancel{1-x}}}\\ =\frac{2}{-2x}=-\frac{1}{x}\\ \\ f^3\left(x\right)=f\left[f^2\left(x\right)\right]=f\left(-\frac{1}{x}\right)\\ =\frac{1+\left(-\frac{1}{x}\right)}{1-\left(-\frac{1}{x}\right)}=\frac{\frac{x-1}{\cancel{x}}}{\frac{x+1}{\cancel{x}}}\\ =\frac{x-1}{x+1}\\ \\ f^4\left(x\right)=f\left[f^3\left(x\right)\right]=f\left(\frac{x-1}{x+1}\right)\\ =\frac{1+\left(\frac{x-1}{x+1}\right)}{1-\left(\frac{x-1}{x+1}\right)}=\frac{\frac{x+1+x-1}{\cancel{x+1}}}{\frac{x+1-x+1}{\cancel{x+1}}}\\ =\frac{2x}{2}=x\\ \\ f^5\left(x\right)=f\left[f^4\left(x\right)\right]=f\left(x\right)=\frac{1+x}{1-x}\left(\text{berulang}\right)\\ \\ \therefore f^{51}\left(x\right)=f^3\left[f^{48}\left(x\right)\right]=f^3\left(x\right)\\ =\frac{x-1}{x+1}\\ f^{52}\left(x\right)=f^4\left[f^{48}\left(x\right)\right]=f^4\left(x\right)=x\end{array}$

Soalan 6:
Dalam rajah di bawah, fungsi g memetakan set P kepada set Q dan fungsi h memetakan set Q kepada set R.



Cari
(a) dalam sebutan x, fungsi
(i) yang memetakan set Q kepada set P,
(ii) h(x).

(b) nilai x dengan keadaan gh(x) = 8x + 1.


Penyelesaian:
(a)(i)
$\begin{array}{l}g\left(x\right)=3x+2\\ \text{Katakan }{g}^{-1}\left(x\right)=y\\ g\left(y\right)=x\\ 3y+2=x\\ y=\frac{x-2}{3}\\ g^{-1}\left(x\right)=\frac{x-2}{3}\end{array}$

(a)(ii)
$\begin{array}{l}hg\left(x\right)=12x+5\\ h\left(3x+2\right)=12x+5\to \boxed{g\left(x\right)=3x+2}\\ \text{Katakan }u=3x+2\\ x=\frac{u-2}{3}\\ h\left(u\right)=12\left(\frac{u-2}{3}\right)+5\\ =4u-8+5\\ =4u-3\\ h\left(x\right)=4x-3\end{array}$

(b)
$\begin{array}{l}gh\left(x\right)=g\left(4x-3\right)\\ =3\left(4x-3\right)+2\\ =12x-9+2\\ =12x-7\\ 12x-7=8x+1\\ 4x=8\\ x=2\end{array}$

Soalan 3:
Fungsi f dan g ditakrifkan oleh
$\begin{array}{l}f:x\mapsto 2x-3\\ g:x\mapsto \frac{2}{x};x\ne 0\end{array}$
Ungkapkan dalam bentuk yang serupa
(a) ff,
(b) gf,
(c) f² , Hitungkan nilai x supaya ff(x) = gf(x).

Penyelesaian:
(a)
$\begin{array}{l}ff\left(x\right)=f\left[f\left(x\right)\right]\\ =f\left(2x-3\right)\\ =2\left(2x-3\right)-3\\ =4x-9\\ \text{Jadi, }ff:x\mapsto 4x-9\end{array}$


(b)
$\begin{array}{l}gf\left(x\right)=g\left[f\left(x\right)\right]\\ =g\left(2x-3\right)\\ =\frac{2}{2x-3}\\ \text{Jadi, }gf:x\mapsto \frac{2}{2x-3}\end{array}$

(c)
$\begin{array}{l}\text{Katakan }{f}^{-1}\left(x\right)=y,\\ \text{maka}f\left(y\right)=x\\ 2y-3=x\\ y=\frac{x+3}{2}\\ \text{maka }{f}^{-1}\left(x\right)=\frac{x+3}{2}\\ f^{-1}:x\mapsto \frac{x+3}{2}\\ \\ \text{Apabila }ff\left(x\right)=gf\left(x\right),\\ 4x-9=\frac{2}{2x-3}\\ \left(4x-9\right)\left(2x-3\right)=2\\ 8x^2-30x+27=2\\ 8x^2-30x+25=0\\ \left(4x-5\right)\left(2x-5\right)=0\\ 4x-5=0\text{ atau }2x-5=0\\ x=\frac{5}{4}\text{ atau }x=\frac{5}{2}\end{array}$

Soalan 4:
Fungsi f dan g ditakrifkan oleh
$\begin{array}{l}f\left(x\right)=3x-2\\ g\left(x\right)=\frac{3}{x},x\ne 0\\ \text{Cari}\\ \text{(a) }{f}^{-1}\left(2\right),\\ \text{(b) g}f\left(-3\right),\\ \text{(c) fungsi }h\text{ jika diberi }hf\left(x\right)=3x+2,\\ \text{(d) fungsi }k\text{ jika diberi }fk\left(x\right)=4x-7.\end{array}$

Penyelesaian:
(a)
$\begin{array}{l}\text{Katakan }{f}^{-1}\left(2\right)=x,\\ \text{maka}f\left(x\right)=2\\ \text{ 3}x-2=2\\ 3x=4\\ x=\frac{4}{3}\\ f^{-1}\left(2\right)=\frac{4}{3}\end{array}$

(b)
$\begin{array}{l}gf\left(-3\right)=g\left[3\left(-3\right)-2\right]\\ =g\left(-11\right)\\ =-\frac{3}{11}\end{array}$

(c)
$\begin{array}{l}h\left[f\left(x\right)\right]=3x+2\\ h\left(3x-2\right)=3x+2\\ \text{Katakan }y=3x-2\\ \text{Maka }x=\frac{y+2}{3}\\ h\left(y\right)=3\left(\frac{y+2}{3}\right)+2\\ =y+2+2\\ =y+4\\ \text{Maka }h\left(x\right)=x+4\end{array}$

(d)
$\begin{array}{l}f\left[k\left(x\right)\right]=4x-7\\ 3k\left(x\right)-2=4x-7\\ 3k\left(x\right)=4x-5\\ k\left(x\right)=\frac{4x-5}{3}\end{array}$