A Level H2 Mathematics Graphs Coordinate Geometry Quiz

A-Level Maths H2 Quiz - Graphs Coordinate Geometry (Answer Key)

1. [2 marks]
Vertical Asymptote: $x = -3$ [1]
Horizontal Asymptote: $y = 2$ [1]
(Note: VA from denominator zero; HA from ratio of coefficients of highest powers)

2. [2 marks]
Image Point: $(3, 7)$ [1]
(Translation vector $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ applied to $(2,5)$)
New Asymptote: $y = 3$ [1]
(Old HA $y=1$ shifted up by 2)

3. [2 marks]
Sketch: "W" shape touching x-axis at $x=\pm 2$, y-intercept at $(0,4)$. [1]
Coordinates: Intercepts $(\pm 2, 0)$ and $(0, 4)$. Minima at $(\pm 2, 0)$, Local Max at $(0, 4)$. [1]

4. [2 marks]
From $y = 2t - 1 \Rightarrow t = \frac{y+1}{2}$. [1]
Substitute into $x$: $x = (\frac{y+1}{2})^2 + 1 \Rightarrow x - 1 = \frac{(y+1)^2}{4} \Rightarrow (y+1)^2 = 4(x-1)$. [1]
(Accept $y^2 + 2y + 1 = 4x - 4$ or similar equivalent forms)

5. [2 marks]
$\frac{1}{x} < 2 \Rightarrow \frac{1-2x}{x} < 0$.
Critical values: $x = 1/2, x = 0$.
Since $x > 0$, we look at interval $(0, \infty)$. Test $x=1 \Rightarrow -1 < 0$ (True). Test $x=0.1 \Rightarrow 8 > 0$ (False).
Solution: $x > \frac{1}{2}$. [2]
(Set notation: $\{ x \in \mathbb{R} : x > 0.5 \}$)

6. [4 marks]
(i) VA: $x = 1$. [1]
Oblique Asymptote: By division, $\frac{x^2}{x-1} = x + 1 + \frac{1}{x-1}$. So $y = x + 1$. [1]
(ii) $f'(x) = \frac{(x-1)(2x) - x^2(1)}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2}$. [1]
Set $f'(x)=0 \Rightarrow x(x-2)=0 \Rightarrow x=0, x=2$.
Points: $(0, 0)$ and $(2, 4)$. [1]

7. [3 marks]
Sketch showing:

• VA at $x=1$, OA at $y=x+1$. [1]
• Stationary points at $(0,0)$ (Max) and $(2,4)$ (Min). [1]
• Correct shape in 3 regions (left of VA, between VA and min, right of min). [1]

8. [3 marks]

• VA of $f$ ($x=2$) becomes VA of $1/f$ ($x=2$). [1]
• HA of $f$ ($y=0$) becomes HA of $1/f$ ($y \to \infty$? No, $1/0 \to \infty$, so VA remains. Wait. If $y \to 0$, $1/y \to \infty$. So $x=2$ is VA. As $x \to \infty, f(x) \to 0 \Rightarrow 1/f(x) \to \infty$? No. If $f(x) \to 0$, $1/f(x)$ grows large. The HA of $f$ is $y=0$. This means as $x \to \infty$, $f(x) \to 0$. Thus $1/f(x) \to \infty$. There is no HA for $1/f$ at infinity unless $f$ had a zero.
  Correction: The question implies standard transformation.
  Points: $(0, -1) \to (0, -1)$. $(3, 2) \to (3, 0.5)$. [1]
  Shape: Inverted relative to x-axis signs. [1]

9. [3 marks]
$\frac{2x-1}{x+2} - 1 \ge 0 \Rightarrow \frac{2x-1 - (x+2)}{x+2} \ge 0 \Rightarrow \frac{x-3}{x+2} \ge 0$. [1]
Critical values: $x = 3, x = -2$. [1]
Positive regions: $x < -2$ or $x \ge 3$. [1]
Answer: $x < -2$ or $x \ge 3$.

10. [3 marks]
(i) $y = 2\sin\theta\cos\theta$. Since $x=\cos\theta$, $\sin\theta = \pm\sqrt{1-x^2}$.
$y = 2x(\pm\sqrt{1-x^2}) \Rightarrow y^2 = 4x^2(1-x^2)$. [2]
(ii) Since $x = \cos\theta$, range is $[-1, 1]$. [1]

11. [2 marks]

1. Reflection in the y-axis ($x \to -x$). [1]
2. Translation by vector $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ ($x \to x-2$ in argument? No. $f(-(x-2)) = f(2-x)$).
   Alternative valid sequence:
3. Translation by vector $\begin{pmatrix} -2 \\ 0 \end{pmatrix}$ ($x \to x+2$).
4. Reflection in the y-axis ($x \to -x$). $f(-x+2)$. [1]

12. [4 marks]
(i) Complete square: $(x-3)^2 - 9 + (y+2)^2 - 4 - 12 = 0 \Rightarrow (x-3)^2 + (y+2)^2 = 25$.
Centre $(3, -2)$, Radius $5$. [2]
(ii) Distance from Centre $(3, -2)$ to line $x - y + 10 = 0$.
$d = \frac{|3 - (-2) + 10|}{\sqrt{1^2 + (-1)^2}} = \frac{15}{\sqrt{2}} \approx 10.6$. [1]
Since $d > r$ ($10.6 > 5$), the line does not intersect the circle. [1]

13. [3 marks]
(i) Circle with centre $(3, 4)$ and radius $5$. [1]
(ii) Max $|z|$ is distance from origin to centre + radius.
Distance $OC = \sqrt{3^2+4^2} = 5$.
Max $|z| = 5 + 5 = 10$. [2]

14. [3 marks]
Perpendicular bisector of segment joining $(0, 2)$ and $(4, 0)$. [1]
Midpoint $(2, 1)$. Gradient of segment $\frac{0-2}{4-0} = -\frac{1}{2}$.
Gradient of perp bisector $= 2$.
Eq: $y - 1 = 2(x - 2) \Rightarrow y = 2x - 3$. [2]
(Or algebraic: $x^2 + (y-2)^2 = (x-4)^2 + y^2 \dots$)

15. [2 marks]
(i) Plot $y$ against $x^2$. [1]
(ii) Gradient $A = -2$. Intercept $B = 5$. [1]

16. [4 marks]
(i) $x^3 - 3x^2 + 2 = 0$. Try $x=1 \Rightarrow 1-3+2=0$. Factor $(x-1)$.
$(x-1)(x^2 - 2x - 2) = 0$.
Roots: $x=1$, $x = \frac{2 \pm \sqrt{4+8}}{2} = 1 \pm \sqrt{3}$.
Points: $(1,0), (1+\sqrt{3}, 0), (1-\sqrt{3}, 0)$. [2]
(ii) Find stationary points: $y' = 3x^2 - 6x = 3x(x-2)$.
Stat points at $x=0 (y=2)$ and $x=2 (y=-2)$.
For 3 roots, line $y=k$ must be between local min and max.
$-2 < k < 2$. [2]

17. [2 marks]
$f'(x)$ is quadratic (since $f$ is cubic-like).
Zeros of $f'$ correspond to stationary points of $f$.
Max at $x=1 \Rightarrow f'(1)=0$ and changes from + to -.
Graph is a downward parabola crossing x-axis at $x=1$?
Wait, $f$ has max at $A(1,4)$. So $f'(1)=0$.
Does $f$ have other stat points? Not stated, but cubic usually has 2.
Assuming standard cubic shape with roots -1, 3 and max at 1, there must be a min between 1 and 3? No, root at 3.
Actually, just sketch derivative: Positive before 1, Zero at 1, Negative after 1.
X-intercept at 1. [2]

18. [3 marks]
Square both sides: $(2x-3)^2 < (x+1)^2$.
$4x^2 - 12x + 9 < x^2 + 2x + 1$.
$3x^2 - 14x + 8 < 0$.
$(3x - 2)(x - 4) < 0$. [1]
Critical values $x = 2/3, x = 4$. [1]
Solution: $\frac{2}{3} < x < 4$. [1]

19. [3 marks]
Intersection: $x^2 - 4x + 5 = mx + 1 \Rightarrow x^2 - (4+m)x + 4 = 0$.
Tangent $\Rightarrow$ Discriminant $\Delta = 0$.
$(4+m)^2 - 4(1)(4) = 0$.
$(4+m)^2 = 16 \Rightarrow 4+m = \pm 4$.
$m = 0$ or $m = -8$. [3]

20. [3 marks]
(i) $V = \pi \int_0^1 \left(\frac{1}{x+1}\right)^2 dx$. [1]
(ii) $\int (x+1)^{-2} dx = [-(x+1)^{-1}]_0^1$. [1]
$= \pi [ -\frac{1}{2} - (-\frac{1}{1}) ] = \pi [ \frac{1}{2} ] = \frac{\pi}{2}$. [1]