Soalan 6:
$$ \text { (a) Rajah } 5 \text { menunjukkan graf bagi fungsi } f(x)=|x-3|+1 \text { untuk domain }-2 \leqslant x \leqslant 7 \text {. } $$
$$ \text { Cari nilai objek yang satu lagi bagi } h \text {. [2 markah]} $$
$$ \text { (b) Diberi bahawa fungsi } f: x \rightarrow 4 x-1 \text { dan } g: x \rightarrow 2 x+3 \text {, cari } $$
$$ \text { (i) nilai } p \text { jika f }{ }^{-1}(p)=2 \text {, } $$
$$ \text { (ii) nilai } x \text { jika } f g(x)=5 f(2) \text {. [4 markah]} $$
Penyelesaian:
(a)
$$ \begin{aligned} f(7) & =h \\ |7-3|+1 & =h \\ |4|+1 & =h \\ h & =5 \end{aligned} $$
$$ \begin{aligned} f(x) & =5 \\ |x-3|+1 & =5 \\ |x-3| & =4 \end{aligned} $$
$$ \begin{aligned} x-3 & =4 \\ x & =7 \end{aligned} $$
$$ \begin{aligned} x-3 & =-4 \\ x & =-1 \end{aligned} $$
$$ \therefore \text { Object of } h \text { is }-1 \text {. } $$
(b)(i)
$$ \begin{aligned} &f(x)=4 x-1\\ &\begin{aligned} & \text { Katakan } 4 x-1=y \\ & 4 x=y+1 \\ & x=\frac{y+1}{4} \\ & f^1(x)=\frac{x+1}{4} \end{aligned} \end{aligned} $$
$$ \begin{aligned} f^1(p) & =2 \\ \frac{p+1}{4} & =2 \\ p+1 & =8 \\ p & =7 \end{aligned} $$
(b)(ii)
$$ \begin{aligned} f g(x) & =5 f(2) \\ f(2 x+3) & =5[4(2)-1] \\ 4(2 x+3)-1 & =5(8-1) \\ 8 x+12-1 & =5(7) \\ 8 x+11 & =35 \\ 8 x & =24 \\ x & =3 \end{aligned} $$
$$ \text { (a) Rajah } 5 \text { menunjukkan graf bagi fungsi } f(x)=|x-3|+1 \text { untuk domain }-2 \leqslant x \leqslant 7 \text {. } $$
$$ \text { Cari nilai objek yang satu lagi bagi } h \text {. [2 markah]} $$
$$ \text { (b) Diberi bahawa fungsi } f: x \rightarrow 4 x-1 \text { dan } g: x \rightarrow 2 x+3 \text {, cari } $$
$$ \text { (i) nilai } p \text { jika f }{ }^{-1}(p)=2 \text {, } $$
$$ \text { (ii) nilai } x \text { jika } f g(x)=5 f(2) \text {. [4 markah]} $$
Penyelesaian:
(a)
$$ \begin{aligned} f(7) & =h \\ |7-3|+1 & =h \\ |4|+1 & =h \\ h & =5 \end{aligned} $$
$$ \begin{aligned} f(x) & =5 \\ |x-3|+1 & =5 \\ |x-3| & =4 \end{aligned} $$
$$ \begin{aligned} x-3 & =4 \\ x & =7 \end{aligned} $$
$$ \begin{aligned} x-3 & =-4 \\ x & =-1 \end{aligned} $$
$$ \therefore \text { Object of } h \text { is }-1 \text {. } $$
(b)(i)
$$ \begin{aligned} &f(x)=4 x-1\\ &\begin{aligned} & \text { Katakan } 4 x-1=y \\ & 4 x=y+1 \\ & x=\frac{y+1}{4} \\ & f^1(x)=\frac{x+1}{4} \end{aligned} \end{aligned} $$
$$ \begin{aligned} f^1(p) & =2 \\ \frac{p+1}{4} & =2 \\ p+1 & =8 \\ p & =7 \end{aligned} $$
(b)(ii)
$$ \begin{aligned} f g(x) & =5 f(2) \\ f(2 x+3) & =5[4(2)-1] \\ 4(2 x+3)-1 & =5(8-1) \\ 8 x+12-1 & =5(7) \\ 8 x+11 & =35 \\ 8 x & =24 \\ x & =3 \end{aligned} $$