Secondary 1 Mathematics Calculus Quiz

Secondary 1 Mathematics Quiz - Calculus (Answer Key)

Total Marks: 50

Section A: Gradient of a Straight Line (Questions 1 – 5)

Question 1

(a) Gradient $= \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2$ [2 marks]

(b) The gradient represents the vertical change (rise) divided by the horizontal change (run). For every 1 unit moved horizontally, the line rises 2 units vertically. [1 mark]

Teaching Note: The gradient measures how steep a line is. It is calculated as the ratio of the change in $y$ to the change in $x$ between any two points on the line.

Question 2

(a) Using $y = mx + c$ with $m = -3$ and point $(1, 8)$: $8 = -3(1) + c$ $8 = -3 + c$ $c = 11$ Equation: $y = -3x + 11$ [2 marks]

(b) The $y$-intercept is 11. [1 mark]

Teaching Note: The $y$-intercept is the value of $c$ in the equation $y = mx + c$. It is the point where the line crosses the $y$-axis (when $x = 0$).

Question 3

(a) Using the points $(0, 1)$ and $(4, 3)$: Gradient $= \frac{3 - 1}{4 - 0} = \frac{2}{4} = \frac{1}{2}$ [2 marks]

(b) The $y$-intercept is 1 (from point $(0, 1)$). Equation: $y = \frac{1}{2}x + 1$ [1 mark]

Teaching Note: When a point has $x = 0$, the $y$-value is the $y$-intercept directly. This makes finding the equation straightforward.

Question 4

(a) Vertical change $= 2$ units. [1 mark]

(b) Since the line passes through the origin, $c = 0$. Equation: $y = \frac{2}{5}x$ [1 mark]

Teaching Note: The gradient $\frac{2}{5}$ means for every 5 units moved horizontally, the line rises 2 units vertically. A line through the origin has no $y$-intercept term.

Question 5

(a) Using $y = mx + c$ with $m = -\frac{4}{3}$ and point $(3, -2)$: $-2 = -\frac{4}{3}(3) + c$ $-2 = -4 + c$ $c = 2$ Equation: $y = -\frac{4}{3}x + 2$ [2 marks]

(b) Substitute $x = 6$ into the equation: $y = -\frac{4}{3}(6) + 2 = -8 + 2 = -6$ Since $y = -6$, the point $(6, -6)$ does lie on the line. [2 marks]

Teaching Note: To check if a point lies on a line, substitute the $x$-coordinate into the equation and verify the resulting $y$-value matches the point's $y$-coordinate.

Section B: Linear Functions and Graphs (Questions 6 – 10)

Question 6

[2 marks: 1 mark for 2-3 correct values, 2 marks for all correct]

Working:

• $x = 0$: $y = 2(0) - 3 = -3$
• $x = 1$: $y = 2(1) - 3 = -1$
• $x = 2$: $y = 2(2) - 3 = 1$
• $x = 3$: $y = 2(3) - 3 = 3$

Teaching Note: Substitute each $x$-value into the formula to find the corresponding $y$-value. This creates ordered pairs that can be plotted on a graph.

Question 7

(a) Gradient $= -\frac{1}{2}$ [1 mark]

(b) $y$-intercept $= 4$ [1 mark]

(c) The graph should be a straight line passing through $(0, 4)$ and $(4, 2)$, sloping downwards from left to right. [2 marks: 1 mark for correct points plotted, 1 mark for correct straight line drawn]

Teaching Note: To draw the graph, plot the $y$-intercept $(0, 4)$ first. Then use the gradient $-\frac{1}{2}$ to find another point: move 2 units right and 1 unit down to reach $(2, 3)$, or move 4 units right and 2 units down to reach $(4, 2)$.

Question 8

(a) Gradient $a = \frac{7 - 1}{4 - (-2)} = \frac{6}{6} = 1$

Using point $(4, 7)$: $7 = 1(4) + b$, so $b = 3$

$a = 1$, $b = 3$ [3 marks: 1 mark for gradient, 1 mark for substitution, 1 mark for $b$]

(b) Equation: $y = x + 3$ [1 mark]

Teaching Note: First find the gradient using the two points, then substitute one point into $y = mx + c$ to find the $y$-intercept.

Question 9

(a) $C = 3 + 2d$ [1 mark]

(b) C = 3 + 2(12) = 3 + 24 = \27$ [1 mark]

(c) $27 = 3 + 2d$ $24 = 2d$ $d = 12$ km [2 marks: 1 mark for forming equation, 1 mark for solving]

Teaching Note: This is a real-world application of a linear function. The fixed fee is the $y$-intercept and the rate per km is the gradient.

Question 10

(a) Using the points $(-3, 4)$ and $(3, -2)$: Gradient $= \frac{-2 - 4}{3 - (-3)} = \frac{-6}{6} = -1$ [2 marks]

(b) Using $y = mx + c$ with $m = -1$ and point $(3, -2)$: $-2 = -1(3) + c$ $-2 = -3 + c$ $c = 1$ Equation: $y = -x + 1$ [2 marks]

Teaching Note: A negative gradient means the line slopes downwards from left to right. The $y$-intercept can also be read from the graph where the line crosses the $y$-axis.

Section C: Rate of Change and Applications (Questions 11 – 15)

Question 11

(a) Rate $= \frac{25 - 10}{7 - 2} = \frac{15}{5} = 3$ litres per minute [2 marks]

(b) Using the point $(2, 10)$ and rate $= 3$: $10 = 3(2) + c$ $10 = 6 + c$ $c = 4$ litres [2 marks]

Teaching Note: The rate of change is the gradient of the linear relationship. To find the initial amount, work backwards using the equation of the line.

Question 12

(a) Speed $= \frac{20 - 0}{40 - 0} = \frac{20}{40} = 0.5$ km/min [2 marks]

(b) $0.5 \text{ km/min} \times 60 \text{ min/h} = 30$ km/h [2 marks]

Teaching Note: To convert from km/min to km/h, multiply by 60 since there are 60 minutes in an hour. The gradient of a distance-time graph gives the speed.

Question 13

(a) Rate $= \frac{32 - 20}{3 - 1} = \frac{12}{2} = 6^\circ\text{C}$ per hour [2 marks]

(b) From 3:00 pm to 5:00 pm is 2 hours. Temperature increase $= 6 \times 2 = 12^\circ\text{C}$ Temperature at 5:00 pm $= 32 + 12 = 44^\circ\text{C}$ [2 marks]

Teaching Note: The rate of change is constant, so we can extend the linear pattern. The time difference between 1:00 pm and 3:00 pm is 2 hours, and between 3:00 pm and 5:00 pm is also 2 hours.

Question 14

(a) Initial height $= 20$ cm [1 mark]

(b) $h = 20 - 0.4(15) = 20 - 6 = 14$ cm [1 mark]

(c) When the candle burns completely, $h = 0$: $0 = 20 - 0.4t$ $0.4t = 20$ $t = 50$ minutes [2 marks]

Teaching Note: The initial height is the value of $h$ when $t = 0$. The gradient $-0.4$ represents the rate at which the candle burns (0.4 cm per minute). To find when it burns completely, set $h = 0$.

Question 15

(a) Speed $= 65$ km/h [1 mark]

(b) $d = 65(3.5) = 227.5$ km [1 mark]

(c) $260 = 65t$ $t = \frac{260}{65} = 4$ hours [2 marks]

Teaching Note: In the formula $d = 65t$, the coefficient of $t$ (65) represents the speed. This is a direct proportion relationship where distance is proportional to time.

Section D: Gradient and Equation of a Line (Questions 16 – 20)

Question 16

(a) Gradient $= \frac{9 - 2}{4 - 4} = \frac{7}{0}$, which is undefined. [1 mark]

(b) The line is vertical. [1 mark]

Teaching Note: When the $x$-coordinates are the same, the line is vertical and has an undefined gradient. Division by zero is not possible in mathematics.

Question 17

(a) Gradient $= \frac{5 - 5}{7 - 2} = \frac{0}{5} = 0$ [1 mark]

(b) The line is horizontal at $y = 5$. Equation: $y = 5$ [1 mark]

Teaching Note: When the $y$-coordinates are the same, the line is horizontal with a gradient of 0. The equation is simply $y =$ (the common $y$-value).

Question 18

(a) Substitute $(2, 11)$ into $y = 3x + c$: $11 = 3(2) + c$ $11 = 6 + c$ $c = 5$ [2 marks]

(b) Equation: $y = 3x + 5$ [1 mark]

Teaching Note: When the gradient is given and a point is known, substitute the coordinates into the equation to find the unknown $y$-intercept.

Question 19

(a) Using $y = mx + c$ with $m = -\frac{1}{4}$ and point $(-4, 6)$: $6 = -\frac{1}{4}(-4) + c$ $6 = 1 + c$ $c = 5$ Equation: $y = -\frac{1}{4}x + 5$ [2 marks]

(b) At the $x$-intercept, $y = 0$: $0 = -\frac{1}{4}x + 5$ $\frac{1}{4}x = 5$ $x = 20$ The $x$-intercept is 20. [2 marks]

Teaching Note: The $x$-intercept is where the line crosses the $x$-axis, which occurs when $y = 0$. Substitute $y = 0$ into the equation and solve for $x$.

Question 20

(a) Using points $A(1, 3)$ and $B(4, 9)$: Gradient $= \frac{9 - 3}{4 - 1} = \frac{6}{3} = 2$ [2 marks]

(b) Using $y = mx + c$ with $m = 2$ and point $A(1, 3)$: $3 = 2(1) + c$ $c = 1$ Equation: $y = 2x + 1$ [2 marks]

(c) When $x = 10$: $y = 2(10) + 1 = 21$ [1 mark]

Teaching Note: Since all three points lie on the same straight line, the gradient between any two points will be the same. The equation can be verified by checking that point $C(7, 15)$ satisfies $y = 2x + 1$: $2(7) + 1 = 15$ ✓

END OF ANSWER KEY