1. **Determine the center and radius of circle L given by** $$x^2 + y^2 - 4x + 6y - 3 = 0$$ Complete the square for $x$ and $y$: $$x^2 - 4x + y^2 + 6y = 3$$ $$x^2 - 4x + 4 + y^2 + 6y + 9 = 3 + 4 + 9$$ $$(x - 2)^2 + (y + 3)^2 = 16$$ Center is $(2, -3)$ and radius is $4$. - Pusat lingkaran L adalah (2, -3): **Benar** - Jari-jari lingkaran L adalah 4: **Benar** - Check if point (0, -6) lies on circle: $$(0 - 2)^2 + (-6 + 3)^2 = 4 + 9 = 13 \neq 16$$ So, (0, -6) is **not** on the circle: **Salah** 2. **Dilations and reflections on quadrilateral ABCD with vertices A(1,1), B(5,1), C(5,4), D(1,4)** - Area of ABCD is width × height = $(5-1) imes (4-1) = 4 imes 3 = 12$ - After dilation with scale factor $k=2$, area scales by $k^2 = 4$, so new area = $12 imes 4 = 48$ - Reflection about x-axis does not change area **Answer: 48 satuan luas (A)** 3. **Equation of circle centered at P(-2,1) passing through Q(1,5)** Radius $r = \sqrt{(1+2)^2 + (5-1)^2} = \sqrt{3^2 + 4^2} = 5$ General form: $$(x + 2)^2 + (y - 1)^2 = 25$$ Expand: $$x^2 + 4x + 4 + y^2 - 2y + 1 = 25$$ $$x^2 + y^2 + 4x - 2y + (4 + 1 - 25) = 0$$ $$x^2 + y^2 + 4x - 2y - 20 = 0$$ **Answer: D** 4. **Range of function** $$f(x) = \sqrt{x - 3} + 2$$ - Domain: $x \geq 3$ - Minimum value at $x=3$ is $f(3) = 0 + 2 = 2$ - Since square root increases, range is $y \geq 2$ **Answer: D** 5. **Tangent line to circle** $$(x + 2)^2 + (y - 1)^2 = 16$$ passing through point $(2,1)$ - Center $C(-2,1)$, radius $4$ - Point $(2,1)$ lies outside circle since distance $= |2+2|=4$ along x-axis, so point is on circle edge? Distance from center to point: $$\sqrt{(2+2)^2 + (1-1)^2} = 4$$ Point lies on circle, so tangent line at $(2,1)$ is perpendicular to radius Radius vector from center to point is horizontal, so tangent line is vertical: $$x = 2$$ **Answer: D** 6. **Matrix multiplication problem:** Given $$\begin{pmatrix}4 & 1 \\ 3 & a\end{pmatrix} \begin{pmatrix} -1 & a \\ 2b + a & 7 \end{pmatrix} = \begin{pmatrix} 1 & 15 \\ 7 & 20 \end{pmatrix}$$ Calculate element (1,1): $$4(-1) + 1(2b + a) = 1 \Rightarrow -4 + 2b + a = 1 \Rightarrow 2b + a = 5$$ Element (1,2): $$4a + 1 \times 7 = 15 \Rightarrow 4a + 7 = 15 \Rightarrow 4a = 8 \Rightarrow a = 2$$ Substitute $a=2$ into first equation: $$2b + 2 = 5 \Rightarrow 2b = 3 \Rightarrow b = \frac{3}{2}$$ Check element (2,1): $$3(-1) + a(2b + a) = 7$$ $$-3 + 2(2 imes \frac{3}{2} + 2) = 7$$ Calculate inside: $$2b + a = 2 \times \frac{3}{2} + 2 = 3 + 2 = 5$$ So: $$-3 + 2 imes 5 = -3 + 10 = 7$$ correct. Element (2,2): $$3a + a imes 7 = 20$$ $$3 imes 2 + 2 imes 7 = 6 + 14 = 20$$ correct. Since $b=1.5$ is not an option, check if $b=1$ (A) or $b=2$ (B) fits better. Recalculate with $b=1$: $$2b + a = 2(1) + 2 = 4 \neq 5$$ With $b=2$: $$2(2) + 2 = 6 \neq 5$$ Closest is $b=1.5$, but since options are integers, answer is **not listed exactly**. Assuming typo, closest is **(A) 1**. 7. **Reflection of point P(a,b) about line $x=2$ gives $P'(0,4)$** Reflection formula about vertical line $x=2$: $$x' = 2 - (x - 2) = 4 - x$$ Given $x' = 0$, so: $$0 = 4 - a \Rightarrow a = 4$$ $y$ coordinate does not change in reflection about vertical line, so: $$b = 4$$ **Answer: (4,4) (A)** 8. **Matrix A is singular:** $$A = \begin{pmatrix} 0 & -1 & -1 \\ 2x & 1 & x-3 \\ 1 & 5 & 6 \end{pmatrix}$$ Singular means determinant = 0. Calculate determinant: $$\det(A) = 0 \times \det \begin{pmatrix}1 & x-3 \\ 5 & 6 \end{pmatrix} - (-1) \times \det \begin{pmatrix}2x & x-3 \\ 1 & 6 \end{pmatrix} + (-1) \times \det \begin{pmatrix}2x & 1 \\ 1 & 5 \end{pmatrix}$$ Calculate minors: $$M_1 = 1 \times 6 - 5(x-3) = 6 - 5x + 15 = 21 - 5x$$ $$M_2 = 2x \times 6 - 1 \times (x-3) = 12x - x + 3 = 11x + 3$$ $$M_3 = 2x \times 5 - 1 \times 1 = 10x - 1$$ So determinant: $$0 - (-1)(11x + 3) + (-1)(10x - 1) = (11x + 3) - (10x - 1) = 11x + 3 - 10x + 1 = x + 4$$ Set to zero: $$x + 4 = 0 \Rightarrow x = -4$$ **Answer: E** 9. **Limit:** $$\lim_{x \to -\infty} \frac{3x - 12}{\sqrt{9x^2 + 3x - 7 + 8}}$$ Simplify denominator: $$\sqrt{9x^2 + 3x + 1}$$ For large negative $x$, dominant term is $9x^2$, so: $$\sqrt{9x^2 + 3x + 1} \approx 3|x| = -3x$$ (since $x \to -\infty$, $|x| = -x$) So limit: $$\lim_{x \to -\infty} \frac{3x - 12}{-3x} = \lim_{x \to -\infty} \frac{3x}{-3x} - \frac{12}{-3x} = -1 + 0 = -1$$ **Answer: A** 10. **Number of full rotations of motorcycle wheel with diameter 70 cm to cover 110 m** - Circumference $C = \pi d = \frac{22}{7} \times 70 = 220$ cm = 2.2 m - Number of rotations = total distance / circumference = $110 / 2.2 = 50$ **Answer: D** 11. **Given matrices:** $$P = \begin{pmatrix}1 & 1 \\ 2 & 4\end{pmatrix}, Q = \begin{pmatrix}a & 1 \\ 2 & b\end{pmatrix}$$ and $$PQ^{-1} = I$$ This implies: $$P = Q$$ Check if $PQ^{-1} = I$ means $P = Q$ or $P = Q$ times $Q^{-1}$? Actually, $PQ^{-1} = I$ means $P = Q$. So $P = Q$: $$a = 1, b = 4$$ Product $a imes b = 1 imes 4 = 4$ **Answer: A** 12. **Area of circular park with circumference 132 m** - Circumference $C = 2 \pi r = 132$ - Radius $r = \frac{132}{2 \pi} = \frac{132}{2 \times \frac{22}{7}} = \frac{132 \times 7}{44} = 21$ - Area $= \pi r^2 = \frac{22}{7} \times 21^2 = \frac{22}{7} \times 441 = 22 \times 63 = 1386$ **Answer: B** 13. **Limit:** $$\lim_{x \to 0} \frac{(1 - \cos 5x) 2 \cos x}{4 x^2 \sqrt{9 - x}}$$ Use approximations near 0: $$1 - \cos 5x \approx \frac{(5x)^2}{2} = \frac{25 x^2}{2}$$ $$\cos x \approx 1$$ $$\sqrt{9 - x} \approx 3$$ Substitute: $$\lim_{x \to 0} \frac{\frac{25 x^2}{2} \times 2 \times 1}{4 x^2 \times 3} = \lim_{x \to 0} \frac{25 x^2}{4 x^2 \times 3} = \frac{25}{12}$$ **Answer: E** **Total questions answered: 13