Matematik Tambahan SPM 2025, Kertas 1 (Soalan 4 – 6)


Soalan 4:
(a) Rajah 4 menunjukkan dua buah segi empat selari dengan tinggi p cm. 
$$ \text { Diberi bahawa } G F=(3+\sqrt{2}) \mathrm{cm}, E D=\sqrt{8} \mathrm{~cm} \text { dan jumlah luas segi empat selari itu ialah }(9 r+3 r \sqrt{2}) \mathrm{cm}^2 \text {. } $$
$$ \text { Ungkapkan } p \text { dalam bentuk } a+b \sqrt{2} \text { dalam sebutan } r \text {, dengan keadaan } a \text { dan } b \text { 1alah pemalar. } $$
[4 markah]

(b) $$ \text { Diberi bahawa } \log _p 3-\log _p x=\frac{1}{3} \log _p(2-y) \text {, ungkapkan } x \text { dalam sebutan } y \text {. } $$
[4 markah]

(c) $$ \text { Diberi bahawa } \frac{\left(x^m\right)^n}{(x+y)^m}-\frac{x^{m n+1}}{(x+y)^{m+1}}=y^2\left(x^m\right)^n+y x^{m n+1} \text {, cari nilai } m \text {. } $$
[4 markah]


Jawapan:
(a) $$ \begin{aligned} \text { ABFG }+ \text { BCDE } & =9 r+3 r \sqrt{2} \\ (3+\sqrt{2})(p)+(\sqrt{8})(p) & =9 r+3 r \sqrt{2} \\ p(3+\sqrt{2}+\sqrt{8}) & =9 r+3 r \sqrt{2} \\ p[3+\sqrt{2}+(\sqrt{4})(\sqrt{2}] & =9 r+3 r \sqrt{2} \\ p(3+\sqrt{2}+2 \sqrt{2}) & =9 r+3 r \sqrt{2} \\ p(3+3 \sqrt{2}) & =9 r+3 r \sqrt{2} \\ p & =\frac{9 r+3 r \sqrt{2}}{3+3 \sqrt{2}} \\ p & =\frac{3(3 r+r \sqrt{2})}{3(1+\sqrt{2})} \\ p & =\frac{3 r+r \sqrt{2}}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}} \\ p & =\frac{3 r-3 r \sqrt{2}+r \sqrt{2}-2 r}{1-2} \\ p & =\frac{r-2 r \sqrt{2}}{-1} \\ p & =-r+2 r \sqrt{2} \text { atau }  2 r \sqrt{2}-r \end{aligned} $$


(b) $$ \begin{aligned} \log _p 3-\log _p x & =\frac{1}{3} \log _p(2-y) \\ \log _p\left(\frac{3}{x}\right) & =\log _p(2-y)^{\frac{1}{3}} \\ \frac{3}{x} & =(2-y)^{\frac{1}{3}} \\ \frac{x}{3} & =\frac{1}{(2-y)^{\frac{1}{3}}} \\ x & =\frac{3}{(2-y)^{\frac{1}{3}}} \end{aligned} $$


(c) $$ \begin{aligned} \frac{\left(x^m\right)^n}{(x+y)^m}-\frac{x^{m n+1}}{(x+y)^{m+1}} & =y^2\left(x^m\right)^n+y x^{m n+1} \\ \frac{\left(x^{m n}\right) \times(x+y)}{(x+y)^m \times(x+y)}-\frac{x^{m n+1}}{(x+y)^{m+1}} & =y^2\left(x^m\right)^n+y x^{m n+1} \\ \frac{\left(x^{m n}\right)(x)+\left(x^{m n}\right)(y)}{(x+y)^{m+1}}-\frac{x^{m n+1}}{(x+y)^{m+1}} & =y^2\left(x^{m n}\right)+y\left(x^{m n}\right)(x) \\ \frac{x^{m n+1}+y x^{m n}-x^{m n+1}}{(x+y)^{m+1}} & =y x^{m n}(y+x) \\ \frac{y x^{m n}}{(x+y)^{m+1}} & =\frac{y x^{m n}(y+x)}{1} \\ \frac{y x^{m n}}{y x^{m n}} & =\frac{(x+y) \times(x+y)^{m+1}}{1} \\ 1 & =(x+y)^{m+2} \\ (x+y)^0 & =(x+y)^{m+2} \\ 0 & =m+2 \\ m & =-2 \end{aligned} $$


Soalan 5:
Fungsi kuadratik f ditakrifkan oleh f(x) = k(x p)2 + q, dengan keadaan k, p dan q ialah pemalar. 

(a) Jika k = 2, nyatakan persamaan bagi paksi simetri graf f(x). [1 markah]
(b) Jika k ditambah daripada 2 kepada 3, bagaimanakah kedudukan titik minimum graf itu berubah? [1 markah]


Jawapan:
(a) $$ \begin{aligned} &\begin{aligned} & f(x)=k(x-p)^2+q \\ & f(x)=2(x-p)^2+q \end{aligned}\\ &\text { Paksi simetri: }\\ &\begin{aligned} x-p & =0 \\ x & =p \end{aligned} \end{aligned} $$

(b) Kedudukan titik minimum tidak berubah. 


Soalan 6:
Rajah 6 menunjukkan x dan y dilukis pada grid segi empat sama dengan sisi 1 unit. 

(a) $$ \text { Pada Rajah 6, lukis dan label } 2 \underline{x}-\underline{y} \text {. } $$
[2 markah]
(b) $$ \text { Seterusnya, nyatakan }|2 \underline{x}-\underline{y}| \text {. } $$
[1 markah]

Jawapan:
(a)


(b)

$$ \begin{aligned} |2 \underline{x}-\underline{y}| & =\sqrt{x^2+y^2} \\ & =\sqrt{0^2+3^2} \\ & =3 \end{aligned} $$

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