Matematik Tambahan SPM 2017, Kertas 2 (Soalan 1 & 2)

Soalan 1 (5 markah):
Selesaikan persamaan serentak berikut:
x – 3y = 1,
x² + 3xy + 9y² = 7

 

Penyelesaian:

$\begin{array}{l}x-3y=\mathrm{1\dots \dots \dots \dots \dots \dots .}\left(1\right)\\ x^2+3xy+9y^2=\mathrm{7\dots \dots \dots \dots \dots \dots .}\left(2\right)\\ \text{Daripada }\left(1\right):x=3y+\mathrm{1\dots \dots \dots \dots \dots \dots .}\left(3\right)\\ \text{Gantikan }\left(3\right)\text{ ke dalam }\left(2\right):\\ {\left(3y+1\right)}^2+3\left(3y+1\right)y+9y^2=7\\ 9y^2+6y+1+9y^2+3y+9y^2-7=0\\ 27y^2+9y-6=0\\ 9y^2+3y-2=0\\ \left(3y-1\right)\left(3y+2\right)=0\\ y=\frac{1}{3}\text{ atau }-\frac{2}{3}\\ \end{array}$

$\begin{array}{l}\text{Gantikan }y\text{ ke dalam }\left(3\right):\\ \text{Apabila }y=\frac{1}{3}\\ x=3\left(\frac{1}{3}\right)+1=2\\ \text{Apabila }y=-\frac{2}{3}\\ x=3\left(-\frac{2}{3}\right)+1=-1\\ \\ \text{Maka, penyelesaian ialah }x=2,y=\frac{1}{3}\text{ atau }x=-1,y=-\frac{2}{3}.\end{array}$

Soalan 2 (7 markah):
Diberi bahawa persamaan suatu lengkung ialah  $y=\frac{5}{x^2}.$
(a) Cari nilai  $\frac{dy}{dx}$  apabila x = 3.
(b) Seterusnya, anggarkan nilai bagi $\frac{5}{{\left(2.98\right)}^2}.$

Penyelesaian:
(a)

$\begin{array}{l}y=\frac{5}{x^2}=5x^{-2}\\ \frac{dy}{dx}=-10x^{-3}\\ =-\frac{10}{x^3}\\ \text{Apabila }x=3\\ \frac{dy}{dx}=-\frac{10}{3^3}=-\frac{10}{27}\end{array}$

(b)

$\begin{array}{l}\delta x=2.98-3=-0.02\\ \delta y=\frac{dy}{dx}.\delta x\\ =-\frac{10}{27}\times \left(-0.02\right)\\ =0.007407\\ \\ \text{Nilai bagi }\frac{5}{{\left(2.98\right)}^2}\\ =y+\delta y\\ =\frac{5}{x^2}+\left(0.007407\right)\\ =\frac{5}{3^2}+\left(0.007407\right)\\ =0.56296\end{array}$