Secondary 1 Mathematics Calculus Quiz

Secondary 1 Mathematics Quiz - Calculus: Answer Key

Total Marks: 40

---

Section A: Direct Calculation and Concept Recall

---

1. Find the value of $3^4$.

Answer: $81$

Working: $3^4 = 3 \times 3 \times 3 \times 3 = 81$

Teaching note: This tests basic index evaluation. $3^4$ means 3 multiplied by itself 4 times. Common error: thinking $3^4 = 3 \times 4 = 12$.

Marks: [2]

---

2. Given $y = x^5$, find $\frac{dy}{dx}$.

Answer: $5x^4$

Working: Using the power rule: $\frac{d}{dx}(x^n) = nx^{n-1}$

Here $n = 5$, so $\frac{dy}{dx} = 5 \times x^{5-1} = 5x^4$

Teaching note: The power rule is fundamental: bring the power down as a coefficient, then subtract 1 from the power. The gradient function $5x^4$ tells us the slope at any point on $y = x^5$.

Marks: [2]

---

3. Evaluate $\frac{d}{dx}(7x)$.

Answer: $7$

Working: $\frac{d}{dx}(7x) = \frac{d}{dx}(7x^1) = 7 \times 1 \times x^{1-1} = 7 \times x^0 = 7 \times 1 = 7$

Teaching note: A linear term $7x$ has constant gradient 7. Using the power rule formally: $7x = 7x^1$, so the derivative is $7 \times 1 \times x^0 = 7$. This makes sense as $y = 7x$ is a straight line with slope 7.

Marks: [2]

---

4. Differentiate $y = 12$ with respect to $x$.

Answer: $0$

Working: $y = 12 = 12x^0$

$\frac{dy}{dx} = 0 \times 12 \times x^{0-1} = 0$

Teaching note: A constant has zero gradient — its graph is a horizontal line. The power rule gives 0 when applied to any constant (since the "power" is effectively 0, and $0 \times \text{(anything)} = 0$). Common error: thinking the answer is 12.

Marks: [2]

---

5. Find the gradient function of $y = 2x^3$.

Answer: $6x^2$

Working: $\frac{dy}{dx} = 2 \times 3 \times x^{3-1} = 6x^2$

Teaching note: The coefficient 2 multiplies the result of differentiating $x^3$. "Gradient function" is another name for the derivative — it gives the slope at any value of $x$.

Marks: [2]

---

6. Given $f(x) = x^2 + 3x$, find $f'(x)$.

Answer: $2x + 3$

Working: Differentiate term by term:

• $\frac{d}{dx}(x^2) = 2x^{2-1} = 2x$
• $\frac{d}{dx}(3x) = 3 \times 1 \times x^{1-1} = 3$

So $f'(x) = 2x + 3$

Teaching note: The prime notation $f'(x)$ means the same as $\frac{df}{dx}$. Differentiate each term separately (sum rule). Common error: combining terms before differentiating, e.g., writing $x^2 + 3x = x^3$ then differentiating.

Marks: [2]

---

7. Find $\frac{d}{dx}(x^7)$.

Answer: $7x^6$

Working: $\frac{d}{dx}(x^7) = 7 \times x^{7-1} = 7x^6$

Teaching note: Straightforward power rule application. The pattern: the new power is always one less than the original power.

Marks: [2]

---

8. Differentiate $y = \frac{1}{2}x^4$ with respect to $x$.

Answer: $2x^3$

Working: $\frac{dy}{dx} = \frac{1}{2} \times 4 \times x^{4-1} = \frac{4}{2}x^3 = 2x^3$

Teaching note: The fraction coefficient $\frac{1}{2}$ stays as a multiplier. Multiply it by the power 4 to get $\frac{4}{2} = 2$. Watch for arithmetic errors in multiplying fractions by integers.

Marks: [2]

---

Section B: Application and Problem Solving

---

9. Given $y = 5x^2 - 3x + 7$, find $\frac{dy}{dx}$.

Answer: $10x - 3$

Working: Differentiate term by term:

• $\frac{d}{dx}(5x^2) = 5 \times 2 \times x^{2-1} = 10x$
• $\frac{d}{dx}(-3x) = -3 \times 1 \times x^{1-1} = -3$
• $\frac{d}{dx}(7) = 0$

So $\frac{dy}{dx} = 10x - 3$

Teaching note: Apply the power rule to each term separately. The constant term 7 disappears (derivative of constant is 0). Be careful with negative coefficients: $-3x$ differentiates to $-3$, not $+3$ or $-3x^0$ left unsimplified.

Marks: [2]

---

10. Differentiate $f(x) = 2x^3 - 4x^2 + x - 9$ with respect to $x$.

Answer: $6x^2 - 8x + 1$

Working: Differentiate term by term:

• $\frac{d}{dx}(2x^3) = 2 \times 3 \times x^{3-1} = 6x^2$
• $\frac{d}{dx}(-4x^2) = -4 \times 2 \times x^{2-1} = -8x$
• $\frac{d}{dx}(x) = 1 \times x^{1-1} = 1$
• $\frac{d}{dx}(-9) = 0$

So $f'(x) = 6x^2 - 8x + 1$

Teaching note: Note that $x = x^1$, so its derivative is $1 \times x^0 = 1$. The $-9$ vanishes. Watch signs: the $-4x^2$ gives $-8x$, not $+8x$.

Marks: [2]

---

11. Find the gradient of the curve $y = x^3 - 2x$ at the point where $x = 2$.

Answer: $10$

Working:

Step 1: Find the gradient function. $\frac{dy}{dx} = 3x^{3-1} - 2 \times 1 \times x^{1-1} = 3x^2 - 2$

Step 2: Substitute $x = 2$. $\frac{dy}{dx}\bigg|_{x=2} = 3(2)^2 - 2 = 3 \times 4 - 2 = 12 - 2 = 10$

Teaching note: "Gradient of the curve at a point" means evaluate the derivative at that specific $x$-value. First differentiate, then substitute — never substitute before differentiating. Common error: finding $y$ when $x = 2$ (which gives 4) instead of finding the gradient.

Marks: [2]

---

12. Given $y = 4x^2 + \frac{3}{x}$, find $\frac{dy}{dx}$. (Express your answer using positive index notation.)

Answer: $8x - \frac{3}{x^2}$

Working:

Step 1: Rewrite $\frac{3}{x}$ using negative index: $\frac{3}{x} = 3x^{-1}$

Step 2: Differentiate term by term. $\frac{dy}{dx} = 4 \times 2 \times x^{2-1} + 3 \times (-1) \times x^{-1-1} = 8x - 3x^{-2}$

Step 3: Convert back to positive index: $-3x^{-2} = -\frac{3}{x^2}$

$\frac{dy}{dx} = 8x - \frac{3}{x^2}$

Teaching note: Terms like $\frac{3}{x}$ must be rewritten as $3x^{-1}$ before differentiating. The power rule applies to $x^{-1}$: bring down $-1$, then subtract 1 to get $-2$. Remember to convert back to the requested positive index form. Common error: thinking $\frac{d}{dx}(x^{-1}) = x^{-2}$ (forgetting the coefficient $-1$).

Marks: [2]

---

13. A curve has equation $y = x^2 - 6x + 5$.

(a) Find $\frac{dy}{dx}$.

Answer: $2x - 6$

Working: $\frac{dy}{dx} = 2x^{2-1} - 6 \times 1 \times x^{1-1} + 0 = 2x - 6$

Teaching note: Straightforward differentiation. The $+5$ disappears.

Marks: [1]

(b) Find the $x$-coordinate of the stationary point on the curve.

Answer: $3$

Working:

Step 1: At a stationary point, $\frac{dy}{dx} = 0$.

Step 2: Set gradient equal to zero. $2x - 6 = 0$

Step 3: Solve. $2x = 6$ $x = 3$

Teaching note: A stationary point is where the curve temporarily stops rising or falling — the gradient is zero. Set the derivative equal to 0 and solve. The corresponding $y$-coordinate would be $3^2 - 6(3) + 5 = 9 - 18 + 5 = -4$, so the stationary point is $(3, -4)$, but the question only asks for $x$.

Marks: [2]

---

14. Differentiate $y = \frac{2}{x^3}$ with respect to $x$. (Express your answer using negative index notation.)

Answer: $-6x^{-4}$ or $\frac{-6}{x^4}$

Working:

Step 1: Rewrite: $y = \frac{2}{x^3} = 2x^{-3}$

Step 2: Differentiate. $\frac{dy}{dx} = 2 \times (-3) \times x^{-3-1} = -6x^{-4}$

Teaching note: The term $\frac{2}{x^3}$ becomes $2x^{-3}$. Applying the power rule: multiply by $-3$, subtract 1 from power. The question asks for negative index notation, so $-6x^{-4}$ is acceptable; $\frac{-6}{x^4}$ is also correct. Common error: getting the sign wrong ($+6$ instead of $-6$) or adding 1 to the power instead of subtracting.

Marks: [2]

---

15. Given $y = (2x + 1)(x - 3)$, find $\frac{dy}{dx}$.

Answer: $4x - 5$

Working:

Step 1: Expand the brackets first. $y = (2x + 1)(x - 3) = 2x \times x + 2x \times (-3) + 1 \times x + 1 \times (-3)$ $y = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3$

Step 2: Differentiate term by term. $\frac{dy}{dx} = 4x - 5$

Teaching note: At Secondary 1 level, we do not yet have the product rule, so we must expand brackets first, then differentiate. This is a key exam technique: when you see a product of linear factors, expand before differentiating. Common error: trying to "differentiate each bracket separately" — this is not valid at this level.

Marks: [2]

---

16. Find the value of $x$ for which the gradient of the curve $y = x^2 + 4x - 7$ is equal to 6.

Answer: $x = 1$

Working:

Step 1: Find the gradient function. $\frac{dy}{dx} = 2x + 4$

Step 2: Set gradient equal to 6. $2x + 4 = 6$

Step 3: Solve. $2x = 2$ $x = 1$

Teaching note: This combines differentiation with solving equations. The "gradient" is $\frac{dy}{dx}$, so we set our derivative equal to the given value 6. Check: when $x = 1$, gradient $= 2(1) + 4 = 6$ ✓

Marks: [3]

---

Section C: Reasoning and Extension

---

17. The curve $y = x^3 + ax^2 + 5x - 1$ passes through the point $(1, 8)$.

(a) Find the value of $a$.

Answer: $a = 3$

Working:

Step 1: Substitute $x = 1$ and $y = 8$ into the equation. $8 = (1)^3 + a(1)^2 + 5(1) - 1$

Step 2: Simplify. $8 = 1 + a + 5 - 1$ $8 = a + 5$

Step 3: Solve. $a = 3$

Teaching note: "Passes through" means the coordinates satisfy the equation. This is a substitution problem before calculus is applied — a common exam structure where part (a) sets up part (b).

Marks: [2]

(b) Find the gradient of the curve at the point where $x = 1$.

Answer: $14$

Working:

Step 1: With $a = 3$, the equation is $y = x^3 + 3x^2 + 5x - 1$.

Step 2: Find $\frac{dy}{dx}$. $\frac{dy}{dx} = 3x^2 + 6x + 5$

Step 3: Substitute $x = 1$. $\frac{dy}{dx}\bigg|_{x=1} = 3(1)^2 + 6(1) + 5 = 3 + 6 + 5 = 14$

Teaching note: Part (b) depends on part (a). Must substitute $a = 3$ before differentiating. The gradient at $(1, 8)$ is 14. Notice we didn't need to verify $y = 8$ again — the point was given to find $a$.

Marks: [2]

---

18. The diagram shows the curve $y = x^2 - 4x + 3$.

Expected visual from Q18-fig1: Parabola opening upwards with x-intercepts at (1,0), (3,0), y-intercept at (0,3), minimum point at (2,-1).

(a) Find $\frac{dy}{dx}$.

Answer: $2x - 4$

Working: $\frac{dy}{dx} = 2x - 4$

Teaching note: Straightforward differentiation of a quadratic.

Marks: [1]

(b) Find the coordinates of the minimum point of the curve.

Answer: $(2, -1)$

Working:

Step 1: At minimum point, $\frac{dy}{dx} = 0$. $2x - 4 = 0$ $x = 2$

Step 2: Find $y$-coordinate by substituting $x = 2$ into original equation. $y = (2)^2 - 4(2) + 3 = 4 - 8 + 3 = -1$

Minimum point is $(2, -1)$.

Teaching note: The minimum point is a stationary point where the curve changes from decreasing to increasing. For a parabola $y = ax^2 + bx + c$ with $a > 0$, the single stationary point is the minimum. The expected diagram shows this visually: the lowest point is at $(2, -1)$, confirming our calculation.

Marks: [2]

---

19. Given that $y = \frac{x^4 + 2x^3}{x}$ for $x \neq 0$, find $\frac{dy}{dx}$ in its simplest form.

Answer: $3x^2 + 4x$

Working:

Step 1: Simplify by dividing each term by $x$ (valid since $x \neq 0$). $y = \frac{x^4}{x} + \frac{2x^3}{x} = x^{4-1} + 2x^{3-1} = x^3 + 2x^2$

Step 2: Differentiate. $\frac{dy}{dx} = 3x^2 + 4x$

Teaching note: Always simplify algebraic fractions before differentiating — it makes the differentiation much easier. Dividing by $x$ subtracts 1 from each power. Common error: trying to use quotient rule (not taught at this level) or incorrectly dividing only the first term.

Marks: [2]

---

20. A particle moves in a straight line so that its displacement $s$ metres from a fixed point $O$ after $t$ seconds is given by $s = t^3 - 6t^2 + 9t$.

(a) Find an expression for the velocity, $v$, of the particle in terms of $t$.

Answer: $v = 3t^2 - 12t + 9$

Working: Velocity is rate of change of displacement with respect to time. $v = \frac{ds}{dt} = 3t^2 - 12t + 9$

Teaching note: In kinematics, velocity $v = \frac{ds}{dt}$. This is an important application of differentiation — the derivative of displacement gives velocity. The coefficients 3, -12, 9 are typical of exam questions that factorise nicely in part (c).

Marks: [1]

(b) Find the velocity of the particle when $t = 4$.

Answer: $9$ m/s

Working: $v = 3(4)^2 - 12(4) + 9 = 3 \times 16 - 48 + 9 = 48 - 48 + 9 = 9$

Teaching note: Substitute into the velocity expression from part (a). Units are m/s (metres per second) since $s$ is in metres and $t$ in seconds.

Marks: [1]

(c) Find the value(s) of $t$ when the particle is instantaneously at rest.

Answer: $t = 1$ and $t = 3$

Working:

Step 1: Instantaneously at rest means $v = 0$. $3t^2 - 12t + 9 = 0$

Step 2: Divide by 3 to simplify. $t^2 - 4t + 3 = 0$

Step 3: Factorise. $(t - 1)(t - 3) = 0$

Step 4: Solve. $t = 1 \quad \text{or} \quad t = 3$

Teaching note: "Instantaneously at rest" means velocity is zero at that instant — the particle stops momentarily before changing direction. The quadratic factorises neatly because the coefficients were chosen to work well. Common error: setting $s = 0$ instead of $v = 0$ (this finds when the particle is at $O$, not when it is at rest).

Marks: [2]

---

END OF ANSWER KEY