A Level H1 Mathematics Graphs Coordinate Geometry Quiz

A-Level Maths H1 Quiz - Graphs Coordinate Geometry (Answer Key)

Total Marks: 40

Section A: Basic Skills and Graph Sketching

1.
(i) As $x \to -\infty$, $e^x \to 0$. Thus, $y \to -5$.
Answer: $y = -5$ [1]
(ii) At $x$-intercept, $y = 0$.
$0 = 2e^x - 5 \implies 2e^x = 5 \implies e^x = 2.5 \implies x = \ln(2.5)$.
Answer: $(\ln 2.5, 0)$ [1]

2.
(i) Vertical asymptote occurs where argument of ln is zero: $x - 2 = 0 \implies x = 2$.
Answer: $x = 2$ [0.5]
(ii) $x$-intercept: $0 = \ln(x-2) + 1 \implies \ln(x-2) = -1 \implies x-2 = e^{-1} \implies x = 2 + e^{-1} \approx 2.37$.
Answer: $(2 + e^{-1}, 0)$ or approx $(2.37, 0)$ [0.5]
(iii) e.g., if $x=3$, $y = \ln(1)+1 = 1$. Point $(3,1)$.
Answer: Sketch showing correct shape (increasing, concave down), asymptote at $x=2$, and intercepts. [1]

3.
Discriminant $\Delta = b^2 - 4ac = (-4)^2 - 4(3)(5) = 16 - 60 = -44$.
Since $\Delta < 0$, there are no real roots.
Since coefficient of $x^2$ ($a=3$) is positive, the parabola opens upwards.
Answer: Always positive. [2] (1 for discriminant calc/conclusion, 1 for justification via $a>0$)

4.
$x^2 - 5x + 6 < 0$
$(x-2)(x-3) < 0$
Critical values: $x=2, x=3$.
Since inequality is $<0$, solution is between roots.
Answer: $2 < x < 3$ [2]

5.
(i) $y = 0^3 - 3(0)^2 = 0$.
Answer: $(0,0)$ [1]
(ii) $y' = 3x^2 - 6x$. $y'' = 6x - 6$.
At $x=0$, $y'' = -6 < 0$.
Answer: Maximum point. [1]

Section B: Calculus and Coordinate Geometry Applications

6.
$y = e^{2x} + 1$.
$\frac{dy}{dx} = 2e^{2x}$.
At $x=0$, $y = e^0 + 1 = 2$. Point $(0,2)$.
Gradient $m = 2e^0 = 2$.
Equation: $y - 2 = 2(x - 0) \implies y = 2x + 2$.
Answer: $y = 2x + 2$ [3] (1 for derivative, 1 for point/gradient, 1 for equation)

7.
(i) $y = 4x^{-1} + x$.
$\frac{dy}{dx} = -4x^{-2} + 1 = 1 - \frac{4}{x^2}$. [1]
(ii) Stationary points when $\frac{dy}{dx} = 0$.
$1 - \frac{4}{x^2} = 0 \implies x^2 = 4 \implies x = \pm 2$.
If $x=2, y = 4/2 + 2 = 4$. Point $(2,4)$.
If $x=-2, y = 4/(-2) - 2 = -4$. Point $(-2,-4)$.
Answer: $(2,4)$ and $(-2,-4)$ [2]

8.
(i) Intersection: $\sqrt{x} = x \implies x = x^2 \implies x^2 - x = 0 \implies x(x-1)=0$.
$x=0 \implies y=0$.
$x=1 \implies y=1$.
Answer: $(0,0)$ and $(1,1)$ [2]
(ii) Area $= \int_{0}^{1} (\sqrt{x} - x) \, dx$.
$= [\frac{2}{3}x^{3/2} - \frac{1}{2}x^2]_0^1$
$= (\frac{2}{3}(1) - \frac{1}{2}(1)) - 0 = \frac{4-3}{6} = \frac{1}{6}$.
Answer: $\frac{1}{6}$ units$^2$ [2]

9.
$\int_{1}^{e} (\frac{1}{x} + 2x) \, dx = [\ln|x| + x^2]_1^e$
$= (\ln e + e^2) - (\ln 1 + 1^2)$
$= (1 + e^2) - (0 + 1) = e^2$.
Answer: $e^2$ [3]

10.
$y = \ln(2x)$. $\frac{dy}{dx} = \frac{1}{2x} \cdot 2 = \frac{1}{x}$.
At $x=1$, $y = \ln 2$. Gradient of tangent $m_t = 1/1 = 1$.
Gradient of normal $m_n = -1$.
Equation of normal: $y - \ln 2 = -1(x - 1) \implies y = -x + 1 + \ln 2$.
At $x$-axis, $y=0$: $0 = -x + 1 + \ln 2 \implies x = 1 + \ln 2$.
Answer: $1 + \ln 2$ [3]

11.
$x + 2 = x^2 - 4 \implies x^2 - x - 6 = 0$.
$(x-3)(x+2) = 0$.
$x=3 \implies y=5$.
$x=-2 \implies y=0$.
Answer: $(3,5)$ and $(-2,0)$ [2]

12.
$f'(x) = e^x - 3$.
$e^x - 3 > 0 \implies e^x > 3 \implies x > \ln 3$.
Answer: $x > \ln 3$ [2]

Section C: Data Analysis and Regression

13.
Answer: Scatter plot with $x$-axis labeled "Ad Spend ($000s)" and $y$-axis labeled "Sales Revenue ($000s)". Points plotted correctly according to table. [2]

14.
Using GC:
Answer: $r \approx 0.994$ (accept 0.993 - 0.995) [1]

15.
Answer: There is a strong positive linear correlation between advertising spend and sales revenue. [1]

16.
Using GC for linear regression $y = ax+b$:
$a \approx 5.37$, $b \approx 4.66$.
Answer: $y = 5.37x + 4.66$ [2]

17.
Answer: For every additional $1,000 spent on advertising, sales revenue increases by approximately $5,370 on average. [1]

18.
$x = 3.2$.
$y = 5.37(3.2) + 4.66 = 17.184 + 4.66 = 21.844$.
Answer: $21,800 (or 21.8 in $000s) [1]

19.
Answer: $15,000 ($x=15$) is outside the range of the observed data ($x$ ranges from 1.5 to 5.0). This is extrapolation, which is unreliable as the linear relationship may not hold. [1]

20.
(i) Answer: The value of $r$ will decrease (become weaker). The point $(6,10)$ deviates significantly from the existing strong positive linear trend (high $x$ but low $y$). [2]
(ii) Answer: Yes, it is an outlier because it does not follow the general pattern of the rest of the data. [1]