Differential Calculus — NBT

Q1Basic

Topic: Reading the first principles definition

Which expression represents the derivative of $f(x)$ from first principles?

A) $\displaystyle\lim_{h \to \infty} \frac{f(x + h) - f(x)}{h}$

B) $\displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

C) $\displaystyle\lim_{h \to 0} \frac{f(x + h) + f(x)}{h}$

D) $\displaystyle\lim_{h \to 0} \frac{f(x) - f(x + h)}{x}$

Show the worked solution

Answer: B

Explanation

Step 1 — Recall the definition.

The derivative measures the instantaneous rate of change. We find it by computing the gradient of a chord between $x$ and $x + h$ , then shrinking $h$ to zero:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Why the distractors are wrong:

A) $h \to \infty$ gives the gradient of a chord extending infinitely far — this is not a tangent.

C) The numerator uses addition instead of subtraction — this is not the difference in function values.

D) The denominator is $x$ instead of $h$ — the gradient of the chord must be divided by the horizontal gap $h$ , not by $x$ .

Takeaway: The two key features of the definition are: subtraction in the numerator ( $f(x+h) - f(x)$ ) and the limit as $h \to 0$ in the denominator $h$ .

Q2Basic

Topic: Applying first principles to $f(x) = x^2$

Use first principles to find $f'(x)$ if $f(x) = x^2$ .

A) $x$

B) $2x$

C) $2x + h$

D) $x^2$

Show the worked solution

Answer: B

Explanation

Step 1 — Write the definition.

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2 — Substitute $f(x) = x^2$ .

$= \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$

Step 3 — Expand the numerator.

$= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h}$

Step 4 — Cancel $h$ (valid since $h \neq 0$ during the limit process).

$= \lim_{h \to 0} (2x + h)$

Step 5 — Apply the limit.

$= 2x$

Why the distractors are wrong:

A) $x$ — missing the coefficient of 2; power rule gives $2x$ , not $x$ .

C) $2x + h$ — this is the expression before applying the limit. Once $h \to 0$ , the $h$ term vanishes.

D) $x^2$ — this is the original function, not its derivative.

Takeaway: The key step is expanding $(x+h)^2$ fully, then cancelling $h$ before taking the limit.

Q3Intermediate

Topic: First principles with a linear-quadratic function

Use first principles to find $f'(x)$ if $f(x) = 3x^2 - 2x$ .

A) $6x$

B) $3x - 2$

C) $6x + 2$

D) $6x - 2$

Show the worked solution

Answer: D

Explanation

Step 1 — Apply the definition.

$f'(x) = \lim_{h \to 0} \frac{[3(x+h)^2 - 2(x+h)] - [3x^2 - 2x]}{h}$

Step 2 — Expand.

$= \lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 - 2x - 2h - 3x^2 + 2x}{h}$

Step 3 — Simplify (all constant terms cancel).

$= \lim_{h \to 0} \frac{6xh + 3h^2 - 2h}{h}$

Step 4 — Cancel $h$ .

$= \lim_{h \to 0} (6x + 3h - 2) = 6x - 2$

Why the distractors are wrong:

B) $3x - 2$ — halved the coefficient; $\frac{d}{dx}[3x^2] = 6x$ , not $3x$ .

C) $6x + 2$ — sign error on the derivative of $-2x$ ; $\frac{d}{dx}[-2x] = -2$ .

A) $6x$ — forgot to differentiate the $-2x$ term.

Takeaway: Expand fully and collect terms in $h$ before cancelling. Every term in the original function contributes to the derivative.

Q4Intermediate

Topic: Recognising the first principles form

The expression $\displaystyle\lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}$ equals:

A) $x^2$

B) $2x^2$

C) $3x^2$

D) $3x^3$

Show the worked solution

Answer: C

Explanation

Step 1 — Recognise the structure.

This is the first principles definition with $f(x) = x^3$ . So the expression equals $f'(x)$ .

Step 2 — Differentiate $f(x) = x^3$ using the power rule (or verify via expansion).

$f'(x) = 3x^2$

Verification by expansion:

$\frac{(x+h)^3 - x^3}{h} = \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} = 3x^2 + 3xh + h^2 \xrightarrow{h \to 0} 3x^2$

Why the distractors are wrong:

A) $x^2$ — missing the coefficient 3.

B) $2x^2$ — this is the derivative of $x^2$ times an extra factor; doesn't apply here.

D) $3x^3$ — the derivative reduces the power by one; $x^3$ differentiates to $x^2$ , not $x^3$ .

Takeaway: When you see $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ , identify $f(x)$ and differentiate it directly.

Section 10.2 — Rules of Differentiation

First principles works for any function but is slow. The rules of differentiation give the same result in seconds.

Rule	Formula
Power rule	$\dfrac{d}{dx}[x^n] = nx^{n-1}$
Constant	$\dfrac{d}{dx}[c] = 0$
Constant multiple	$\dfrac{d}{dx}[cf(x)] = c \cdot f'(x)$
Sum/difference	$\dfrac{d}{dx}[f \pm g] = f' \pm g'$

Note: Always simplify the expression before differentiating. Fractions, brackets, and roots must be rewritten in $ax^n$ form first.

Q5Basic

Topic: Power rule

If $f(x) = x^5$ , then $f'(x) =$

A) $5x^4$

B) $x^4$

C) $5x^6$

D) $4x^5$

Show the worked solution

Answer: A

Explanation

Apply the power rule: multiply by the exponent, reduce the exponent by 1.

$f'(x) = 5x^{5-1} = 5x^4$

Why the distractors are wrong:

B) Missing the coefficient 5 — you must bring the power down as a multiplier.

C) The exponent increases — differentiation reduces the power by 1, not increases it.

D) The exponent stays the same — $5 - 1 = 4$ , not 5.

Takeaway: Power rule: bring down the exponent as a coefficient, then subtract 1 from the exponent.

Q6Basic

Topic: Differentiating a polynomial

If $g(x) = 4x^3 - 3x^2 + 7x - 5$ , then $g'(x) =$

A) $12x^2 - 6x + 7$

B) $12x^2 - 6x$

C) $12x^3 - 6x^2 + 7$

D) $4x^2 - 3x + 7$

Show the worked solution

Answer: A

Explanation

Differentiate term by term:

$\frac{d}{dx}[4x^3] = 12x^2 \qquad \frac{d}{dx}[-3x^2] = -6x \qquad \frac{d}{dx}[7x] = 7 \qquad \frac{d}{dx}[-5] = 0$

$g'(x) = 12x^2 - 6x + 7$

Why the distractors are wrong:

B) Dropped the $+7$ — the derivative of $7x$ is 7, not 0.

C) Kept the original exponents — each power reduces by 1 after differentiation.

D) Didn't multiply by the exponent — forgot to apply the coefficient from the power rule.

Takeaway: Differentiate every term. Constants vanish; linear terms become constants.

Q7Intermediate

Topic: Simplify before differentiating — fraction

Find $f'(x)$ if $f(x) = \dfrac{3x^3 - 6x}{3x}$ .

A) $x^2 - 2$

B) $x - 1$

C) $2x - 1$

D) $2x$

Show the worked solution

Answer: D

Explanation

Step 1 — Simplify by dividing each term in the numerator by $3x$ .

$f(x) = \frac{3x^3 - 6x}{3x} = \frac{3x^3}{3x} - \frac{6x}{3x} = x^2 - 2$

Step 2 — Differentiate.

$f'(x) = 2x$

Takeaway: Always simplify a fraction before differentiating. Dividing each numerator term by the denominator converts it to standard polynomial form.

Why the distractors are wrong:

B) $x - 1$ — incorrect simplification.

C) $2x - 1$ — brought a phantom constant into the derivative.

A) $x^2 - 2$ — forgot to differentiate after simplifying. This IS the simplified function; the question asks for its derivative.

Q8Intermediate

Topic: Negative exponents

Find $f'(x)$ if $f(x) = \dfrac{4}{x^3}$ .

A) $\dfrac{-12}{x^4}$

B) $\dfrac{12}{x^4}$

C) $\dfrac{-4}{x^2}$

D) $\dfrac{-12}{x^2}$

Show the worked solution

Answer: A

Explanation

Step 1 — Rewrite with negative exponent.

$f(x) = 4x^{-3}$

Step 2 — Apply power rule.

$f'(x) = 4 \cdot (-3) x^{-3-1} = -12x^{-4} = \frac{-12}{x^4}$

Why the distractors are wrong:

B) Sign error — the exponent $-3$ makes the derivative negative.

C) $-4/x^2$ — reduced the power by too little; $-3 - 1 = -4$ , not $-2$ .

D) $-12/x^2$ — correct coefficient but wrong exponent; $x^{-4}$ means $1/x^4$ , not $1/x^2$ .

Takeaway: Rewrite $\frac{a}{x^n}$ as $ax^{-n}$ , then apply the power rule. The derivative of $x^{-n}$ is $-nx^{-n-1}$ .

Q9Intermediate

Topic: Rational exponents (roots)

If $f(x) = \sqrt{x}$ , then $f'(x) =$

A) $\dfrac{1}{\sqrt{x}}$

B) $\dfrac{1}{2\sqrt{x}}$

C) $2\sqrt{x}$

D) $\dfrac{-1}{2\sqrt{x}}$

Show the worked solution

Answer: B

Explanation

Step 1 — Rewrite using exponent notation.

$f(x) = x^{\frac{1}{2}}$

Step 2 — Apply power rule.

$f'(x) = \frac{1}{2} x^{\frac{1}{2} - 1} = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2x^{\frac{1}{2}}} = \frac{1}{2\sqrt{x}}$

Why the distractors are wrong:

A) $1/\sqrt{x}$ — missing the factor of $\frac{1}{2}$ .

C) $2\sqrt{x}$ — this would be the result of integrating $\sqrt{x}$ , not differentiating.

D) $-1/(2\sqrt{x})$ — sign error; the exponent $\frac{1}{2}$ is positive, so the derivative is positive.

Takeaway: Rewrite $\sqrt[n]{x^m}$ as $x^{m/n}$ before differentiating.

Q10Proficient

Topic: Expand before differentiating

Find $h'(x)$ if $h(x) = (2x - 1)^2$ .

A) $2(2x-1)$

B) $8x$

C) $2(2x-1)^2$

D) $8x - 4$

Show the worked solution

Answer: D

Explanation

Step 1 — Expand the bracket first (do not use chain rule — rather expand for NBT level).

$h(x) = (2x-1)^2 = 4x^2 - 4x + 1$

Step 2 — Differentiate term by term.

$h'(x) = 8x - 4$

Why the distractors are wrong:

A) $2(2x-1) = 4x - 2$ — used the coefficient 2 instead of 4 from $(2x)^2 = 4x^2$ .

B) $8x$ — forgot to differentiate the constant $+1$ ; derivative of $-4x$ is $-4$ , giving $8x - 4$ , not $8x$ .

C) $4x - 2$ — differentiated $4x^2 - 4x$ incorrectly; $\frac{d}{dx}[4x^2] = 8x$ , not $4x$ .

Takeaway: Always expand brackets before differentiating polynomial expressions.

Section 10.3 — Tangent and Normal Lines

The derivative $f'(a)$ gives the gradient of the tangent to the curve $y = f(x)$ at $x = a$ . The normal at that point is perpendicular to the tangent, so its gradient is $-\dfrac{1}{f'(a)}$ .

Q11Basic

Topic: Gradient of tangent at a point

Find the gradient of the tangent to $f(x) = x^2 - 4x + 3$ at $x = 3$ .

A) $0$

B) $2$

C) $6$

D) $-2$

Show the worked solution

Answer: B

Explanation

Step 1 — Differentiate.

$f'(x) = 2x - 4$

Step 2 — Substitute $x = 3$ .

$f'(3) = 2(3) - 4 = 2$

Why the distractors are wrong:

A) 0 — this is $f'(2) = 2(2) - 4 = 0$ ; substituted the wrong $x$ -value.

C) 6 — from computing $2(3) = 6$ but forgetting to subtract the $4$ in $f'(x) = 2x - 4$ .

D) $-2$ — forgot to add 4 correctly; sign error.

Takeaway: The gradient of the tangent at $x = a$ is $f'(a)$ . Differentiate first, then substitute.

Q12Basic

Topic: Finding the equation of a tangent

Find the equation of the tangent to $y = x^3$ at the point where $x = 1$ .

A) $y = x + 1$

B) $y = x - 2$

C) $y = 3x + 1$

D) $y = 3x - 2$

Show the worked solution

Answer: D

Explanation

Step 1 — Find the point of tangency.

$y(1) = 1^3 = 1 \quad \Rightarrow \quad \text{point: } (1,\ 1)$

Step 2 — Find the gradient.

$\frac{dy}{dx} = 3x^2 \quad \Rightarrow \quad \frac{dy}{dx}\bigg|_{x=1} = 3$

Step 3 — Use point-gradient form.

$y - 1 = 3(x - 1) \quad \Rightarrow \quad y = 3x - 2$

Why the distractors are wrong:

A) Gradient of 1 — used $f(1)$ as the gradient instead of $f'(1)$ .

C) $y = 3x + 1$ — used $y = mx + c$ with $c = f(1) = 1$ directly; forgot to subtract using the point.

B) Gradient of 1 — same error as A with a different intercept.

Takeaway: Three steps every time: find the point, find the gradient from $f'(a)$ , write $y - y_1 = m(x - x_1)$ .

Q13Intermediate

Topic: Finding $x$ where the gradient equals a given value

At which $x$ -value does $f(x) = x^3 - 6x^2 + 9x$ have a gradient of $9$ ?

A) $x = 0$ or $x = 4$

B) $x = 1$ or $x = 3$

C) $x = 0$ only

D) $x = 4$ only

Show the worked solution

Answer: A

Explanation

Step 1 — Differentiate.

$f'(x) = 3x^2 - 12x + 9$

Step 2 — Set equal to 9.

$3x^2 - 12x + 9 = 9$ $3x^2 - 12x = 0$ $3x(x - 4) = 0$

Step 3 — Solve.

$x = 0 \quad \text{or} \quad x = 4$

Why the distractors are wrong:

B) $x = 1$ or $x = 3$ — these are the stationary points where $f'(x) = 0$ , not where $f'(x) = 9$ .

C) and D) each give only one solution — factoring gives two, and both must be included.

Takeaway: Set $f'(x)$ equal to the required gradient value. Do not set it to zero unless you want stationary points.

Q14Intermediate

Topic: Equation of the normal

The tangent to $y = x^2 + 1$ at $x = 2$ has gradient $4$ . What is the gradient of the normal at this point?

A) $4$

B) $-4$

C) $\dfrac{1}{4}$

D) $-\dfrac{1}{4}$

Show the worked solution

Answer: D

Explanation

Step 1 — Recall the relationship between tangent and normal gradients.

If the tangent gradient is $m_t$ , the normal gradient is:

$m_n = -\frac{1}{m_t}$

Step 2 — Substitute.

$m_n = -\frac{1}{4}$

Why the distractors are wrong:

A) Same as tangent — parallel lines, not perpendicular.

B) $-4$ — negative of the tangent; perpendicular requires the reciprocal AND the negative.

C) $\frac{1}{4}$ — reciprocal but forgot the negative sign.

Takeaway: Tangent gradient $m$ → normal gradient $-1/m$ . Both the sign flip and the reciprocal are required.

Section 10.4 — Stationary Points and Nature

A stationary point occurs where $f'(x) = 0$ — the tangent is horizontal. There are three types:

Local maximum: gradient changes from $+$ to $-$ ; $f''(x) < 0$
Local minimum: gradient changes from $-$ to $+$ ; $f''(x) > 0$
Point of inflection: gradient does not change sign; $f''(x) = 0$

Q15Basic

Topic: Finding stationary points

Find the $x$ -values of the stationary points of $f(x) = x^3 - 3x$ .

A) $x = 0$ only

B) $x = 1$ and $x = -1$

C) $x = 3$ and $x = -3$

D) $x = 1$ only

Show the worked solution

Answer: B

Explanation

Step 1 — Differentiate.

$f'(x) = 3x^2 - 3$

Step 2 — Set $f'(x) = 0$ .

$3x^2 - 3 = 0 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = \pm 1$

Why the distractors are wrong:

A) $x = 0$ — this is the inflection point of $f(x) = x^3$ , not this function.

C) $x = \pm 3$ — forgot to divide by 3 before solving.

D) Only $x = 1$ — missed the negative root; $x^2 = 1$ has two solutions.

Takeaway: Stationary points require $f'(x) = 0$ . Solve the resulting equation completely — don't miss negative roots.

Q16Basic

Topic: Nature using the second derivative

For $f(x) = x^3 - 3x$ , classify the stationary point at $x = 1$ .

A) Local maximum

B) Local minimum

C) Point of inflection

D) Cannot be determined

Show the worked solution

Answer: B

Explanation

Step 1 — Find $f''(x)$ .

$f'(x) = 3x^2 - 3 \quad \Rightarrow \quad f''(x) = 6x$

Step 2 — Substitute $x = 1$ .

$f''(1) = 6(1) = 6 > 0$

Step 3 — Interpret. Since $f''(1) > 0$ , the curve is concave up at $x = 1$ — this is a local minimum.

Why the distractors are wrong:

A) Maximum requires $f'' < 0$ (concave down).

C) Inflection requires $f'' = 0$ and a sign change — $f''(1) = 6 \neq 0$ .

D) The second derivative test gives a clear answer here.

Takeaway: $f'' > 0$ → minimum (cup shape). $f'' < 0$ → maximum (cap shape). $f'' = 0$ → test is inconclusive; use sign of $f'$ .

Q17Intermediate

Topic: Coordinates of turning points

Find the coordinates of the local maximum of $f(x) = 2x^3 - 9x^2 + 12x - 4$ .

A) $(1,\ 1)$

B) $(2,\ 0)$

C) $(1,\ 0)$

D) $(2,\ 1)$

Show the worked solution

Answer: A

Explanation

Step 1 — Differentiate and find stationary points.

$f'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)$

$x = 1 \quad \text{or} \quad x = 2$

Step 2 — Find $f''(x)$ to classify.

$f''(x) = 12x - 18$

$f''(1) = -6 < 0 \quad \Rightarrow \quad \text{local maximum at } x = 1$ $f''(2) = 6 > 0 \quad \Rightarrow \quad \text{local minimum at } x = 2$

Step 3 — Find $y$ -coordinate of maximum.

$f(1) = 2 - 9 + 12 - 4 = 1$

$\text{Local maximum: } (1,\ 1)$

Why the distractors are wrong:

B) $(2,\ 0)$ — this is the local minimum, not the maximum.

C) $(1,\ 0)$ — correct $x$ but wrong $y$ ; $f(1) = 1$ , not $0$ .

D) $(2,\ 1)$ — swapped the coordinates of the two turning points.

Takeaway: Find both stationary points, use $f''$ to classify, then substitute the correct $x$ to find the $y$ -coordinate.

Q18Intermediate

Topic: Point of inflection

At which $x$ -value does $f(x) = x^3 - 6x^2 + 9x + 1$ have its point of inflection?

A) $x = 1$

B) $x = 2$

C) $x = 3$

D) $x = 0$

Show the worked solution

Answer: B

Explanation

Step 1 — Differentiate twice.

$f'(x) = 3x^2 - 12x + 9 \quad \Rightarrow \quad f''(x) = 6x - 12$

Step 2 — Set $f''(x) = 0$ .

$6x - 12 = 0 \quad \Rightarrow \quad x = 2$

Step 3 — Confirm sign change (not required at NBT level but worth knowing).

$f''(1) = -6 < 0$ and $f''(3) = 6 > 0$ — sign changes, confirming inflection.

Why the distractors are wrong:

A) $x = 1$ — this is a stationary point ( $f'(1) = 0$ ), not the inflection point.

C) $x = 3$ — also a stationary point.

D) $x = 0$ — neither a stationary point nor the inflection point for this function.

Takeaway: Point of inflection: set $f''(x) = 0$ . Note: for a cubic, the inflection point is always the midpoint (in $x$ ) of the two turning points.

Q19Proficient

Topic: Using turning point information

A cubic function $f(x) = ax^3 + bx^2 + cx + d$ has a local maximum at $x = -1$ and a local minimum at $x = 3$ . What is the $x$ -coordinate of its point of inflection?

A) $x = 0$

B) $x = -2$

C) $x = 2$

D) $x = 1$

Show the worked solution

Answer: D

Explanation

Step 1 — Use the key property of cubic functions.

The point of inflection of a cubic always lies at the average (midpoint) of the $x$ -coordinates of the two turning points.

Step 2 — Calculate.

$x_{\text{inflection}} = \frac{-1 + 3}{2} = \frac{2}{2} = 1$

Why the distractors are wrong:

A) $x = 0$ — average of $-1$ and $1$ , but 1 is not the minimum; the minimum is at $x = 3$ .

C) $x = 2$ — average of 1 and 3; used the wrong pair.

B) $x = -2$ — not the midpoint of any meaningful pair.

Takeaway: For a cubic $f$ , the $x$ -coordinate of the inflection point = $\dfrac{x_{\text{max}} + x_{\text{min}}}{2}$ .

Section 10.5 — Cubic Functions and Graphs

A cubic function has the form $f(x) = ax^3 + bx^2 + cx + d$ .

If $a > 0$ : falls to the left, rises to the right
If $a < 0$ : rises to the left, falls to the right
Has at most two turning points
Has exactly one point of inflection
$f'(x) > 0$ : function is increasing; $f'(x) < 0$ : function is decreasing

Q20Basic

Topic: Reading a cubic graph — turning points

A cubic graph has a local maximum at $(-1,\ 4)$ and a local minimum at $(2,\ -3)$ . On what interval is the function decreasing?

A) $x < -1$

B) $-1 < x < 2$

C) $x > 2$

D) $x < 2$

Show the worked solution

Answer: B

Explanation

A cubic with positive leading coefficient rises, reaches a local maximum, decreases to the local minimum, then rises again.

The function decreases between the local maximum and local minimum:

$-1 < x < 2$

Why the distractors are wrong:

A) $x < -1$ — the function is increasing here (before the maximum).

C) $x > 2$ — the function is increasing here (after the minimum).

D) $x < 2$ — too broad; the function is increasing for $x < -1$ and only decreasing for $-1 < x < 2$ .

Takeaway: Between the maximum and minimum of a cubic, the function is decreasing. Outside this interval, it is increasing.

Q21Intermediate

Topic: Linking $f$ , $f'$ , and $f''$

The graph of $f'(x)$ has roots at $x = 1$ and $x = 5$ , and opens upward (parabola). Which statement about $f(x)$ is correct?

A) $f$ has a maximum at $x = 1$ and no turning point at $x = 5$

B) $f$ has a minimum at $x = 1$ and a maximum at $x = 5$

C) $f$ has a minimum at $x = 1$ and a minimum at $x = 5$

D) $f$ has a maximum at $x = 1$ and a minimum at $x = 5$

Show the worked solution

Answer: D

Explanation

Read the sign of $f'$ . It is an upward parabola with roots at 1 and 5, so it is negative between them and positive outside.

Translate to $f$ :

At $x=1$ , $f'$ goes $+ \to -$ → local maximum
At $x=5$ , $f'$ goes $- \to +$ → local minimum

Why the others are wrong:

B is reversed — an upward parabola is negative between its roots, so $f$ decreases there.
C claims two minima. An upward $f'$ changes sign $+\to-\to+$ , giving one of each.
A denies a turning point at $x=5$ , but $f'(5)=0$ and the sign changes there.

Takeaway: $f'$ going $+ \to -$ means a maximum; $- \to +$ means a minimum. Sketch the sign of $f'$ first, then read $f$ off it.

Q22Intermediate

Topic: Determining a cubic equation from its graph

A cubic graph cuts the x-axis at $x = -2$ , $x = 1$ , and $x = 3$ , and passes through $(0,\ -6)$ . What is the equation of the cubic?

A) $y = (x+2)(x-1)(x-3)$

B) $y = -(x+2)(x-1)(x-3)$

C) $y = -x(x+2)(x-1)(x-3)$

D) $y = 2(x+2)(x-1)(x-3)$

Show the worked solution

Answer: B

Explanation

Step 1 — Write the general form using the roots.

$y = k(x+2)(x-1)(x-3)$

Step 2 — Use the point $(0,\ -6)$ to find $k$ .

$-6 = k(0+2)(0-1)(0-3) = k(2)(-1)(-3) = 6k \quad \Rightarrow \quad k = -1$

Step 3 — Write the equation.

$y = -(x+2)(x-1)(x-3)$

Why the distractors are wrong:

A) $k = 1$ gives $y = +6$ at $x = 0$ , not $-6$ .

C) Has an extra factor $x$ — this would be degree 4, not cubic.

D) $k = 2$ gives $y = 12$ at $x = 0$ , not $-6$ .

Takeaway: Use known roots to write $y = k(x - r_1)(x - r_2)(x - r_3)$ , then substitute a known point to find $k$ .

Q23Proficient

Topic: Intervals where $f(x) > 0$

The cubic $f(x) = (x+1)(x-2)(x-4)$ has a positive leading coefficient. For which values of $x$ is $f(x) > 0$ ?

A) $x < -1$ or $x > 4$

B) $-1 < x < 2$ or $x > 4$

C) $x < -1$ or $2 < x < 4$

D) $-1 < x < 2$ or $2 < x < 4$

Show the worked solution

Answer: B

Explanation

Step 1 — Identify the roots: $x = -1,\ x = 2,\ x = 4$

Step 2 — Use a sign table (positive leading coefficient means the cubic is negative, positive, negative, positive in consecutive intervals):

Interval	Sign of $f$
$x < -1$	negative
$-1 < x < 2$	positive
$2 < x < 4$	negative
$x > 4$	positive

Step 3 — Select where $f(x) > 0$ :

$-1 < x < 2 \quad \text{or} \quad x > 4$

Why the distractors are wrong:

A) These are where the cubic would be positive if leading coefficient were negative — reversed.

C) Selects the negative intervals.

D) Skips the interval $x > 4$ .

Takeaway: Draw a sign table for each interval. Positive leading coefficient → starts negative for $x <$ smallest root.

Section 10.6 — Optimisation

Optimisation uses calculus to find the maximum or minimum value of a quantity. The method:

Write an expression for the quantity to optimise (area, volume, profit, etc.)
Express it in terms of one variable using any constraint given
Differentiate and set $f'(x) = 0$
Verify using $f''(x)$ (or context)
Answer the question — often requires finding the actual max/min value, not just $x$

Q24Basic

Topic: Finding the maximum of a quadratic function

The profit (in rands) from selling $x$ units is $P(x) = -2x^2 + 40x - 50$ . How many units maximise the profit?

A) $x = 10$

B) $x = 20$

C) $x = 40$

D) $x = 5$

Show the worked solution

Answer: A

Explanation

Step 1 — Differentiate.

$P'(x) = -4x + 40$

Step 2 — Set $P'(x) = 0$ .

$-4x + 40 = 0 \quad \Rightarrow \quad x = 10$

Step 3 — Verify it's a maximum.

$P''(x) = -4 < 0 \quad \Rightarrow \quad \text{maximum confirmed}$

Why the distractors are wrong:

B) $x = 20$ — twice the correct answer; common error from forgetting to divide.

C) $x = 40$ — used the coefficient of $x$ directly without differentiating.

D) $x = 5$ — half the correct value.

Takeaway: Set the derivative of the objective function to zero. Verify with $f''$ or context.

Q25Intermediate

Topic: Optimisation with a constraint — fence problem

A farmer has $120\,\text{m}$ of fencing to enclose a rectangular field. One side uses a river (no fencing needed). Find the length of the side parallel to the river that maximises the area.

A) $30\,\text{m}$

B) $40\,\text{m}$

C) $60\,\text{m}$

D) $80\,\text{m}$

Show the worked solution

Answer: C

Explanation

Step 1 — Set up variables.

Let the side parallel to the river have length $y$ and the two perpendicular sides each have length $x$ .

Step 2 — Write the constraint (only 3 sides need fencing).

$2x + y = 120 \quad \Rightarrow \quad y = 120 - 2x$

Step 3 — Write the area function.

$A = xy = x(120 - 2x) = 120x - 2x^2$

Step 4 — Differentiate and set to zero.

$A'(x) = 120 - 4x = 0 \quad \Rightarrow \quad x = 30$

Step 5 — Find $y$ .

$y = 120 - 2(30) = 60\,\text{m}$

Why the distractors are wrong:

A) 30 m — this is the length of the perpendicular sides, not the side parallel to the river.

B) 40 m — an intermediate calculation error.

D) 80 m — $120 - 2(20)$ ; found $x = 20$ by error.

Takeaway: Label your variables clearly. Write the constraint, use it to eliminate one variable, then differentiate the objective function.

Q26Intermediate

Topic: Box optimisation

An open box is made from a $12\,\text{cm} \times 12\,\text{cm}$ square of cardboard by cutting equal squares of side $x$ from each corner and folding up the sides. Find the value of $x$ that maximises the volume.

A) $x = 1$

B) $x = 2$

C) $x = 3$

D) $x = 4$

Show the worked solution

Answer: B

Explanation

Step 1 — Write the volume formula.

After cutting, the base has dimensions $(12 - 2x) \times (12 - 2x)$ and height $x$ :

$V = x(12 - 2x)^2$

Step 2 — Expand.

$V = x(144 - 48x + 4x^2) = 144x - 48x^2 + 4x^3$

Step 3 — Differentiate.

$V'(x) = 144 - 96x + 12x^2 = 12(x^2 - 8x + 12) = 12(x - 2)(x - 6)$

Step 4 — Set $V'(x) = 0$ .

$x = 2 \quad \text{or} \quad x = 6$

Step 5 — Check validity. Since $x < 6$ (otherwise the box has no base), $x = 2$ .

Why the distractors are wrong:

A) $x = 1$ — $V'(1) = 12(1-2)(1-6) = 60 \neq 0$ .

C) $x = 3$ — $V'(3) = 12(1)(-3) \neq 0$ .

D) $x = 6$ — this makes the base zero; physically invalid.

Takeaway: Always check that your solution is physically valid. Discard solutions outside the feasible domain.

Q27Proficient

Topic: Optimisation — rate of change interpretation

A particle moves along a straight line. Its displacement (in metres) after $t$ seconds is $s(t) = t^3 - 6t^2 + 9t$ . At what time is the particle momentarily at rest?

A) $t = 1$ and $t = 3$

B) $t = 0$ and $t = 2$

C) $t = 2$ only

D) $t = 3$ only

Show the worked solution

Answer: A

Explanation

Step 1 — Recall that velocity $= \dfrac{ds}{dt}$ .

The particle is at rest when velocity $= 0$ .

Step 2 — Differentiate.

$v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t-1)(t-3)$

Step 3 — Set $v(t) = 0$ .

$t = 1 \quad \text{or} \quad t = 3$

Why the distractors are wrong:

B) $t = 0$ and $t = 2$ — these come from setting $s(t) = 0$ , not $s'(t) = 0$ .

C) and D) each give only one value — the quadratic $v(t) = 0$ has two roots.

Takeaway: Velocity is the derivative of displacement. "At rest" means velocity $= 0$ , so set $s'(t) = 0$ .

Section 10.7 — Mixed Practice

M1Basic

Topic: Power rule

If $f(x) = 6x^4 - 2x^3 + x$ , then $f'(x) =$

A) $6x^3 - 2x^2 + 1$

B) $24x^3 - 6x^2$

C) $24x^4 - 6x^3 + 1$

D) $24x^3 - 6x^2 + 1$

Show the worked solution

Answer: D

Explanation

$f'(x) = 24x^3 - 6x^2 + 1$

Derivative of $6x^4 = 24x^3$ ; of $-2x^3 = -6x^2$ ; of $x = 1$ .

Why the others are wrong: A lowers each power but never multiplies by it. C multiplies by the power but never lowers it. B does both correctly and then drops the derivative of the lone $x$ , which is 1.

M2Basic

Topic: Stationary point x-value

At which value does $g(x) = x^2 - 8x + 3$ have its turning point?

A) $x = 3$

B) $x = 4$

C) $x = 8$

D) $x = -4$

Show the worked solution

Answer: B

Explanation

$g'(x) = 2x - 8 = 0 \Rightarrow x = 4$

Why the others are wrong: C uses $b$ where the formula needs $\tfrac{b}{2}$ . D has the sign of $-\tfrac{b}{2a}$ wrong. A reads off the constant term.

M3Basic

Topic: Nature of turning point

If $f''(a) = -7$ , the turning point at $x = a$ is a:

A) Minimum

B) Point of inflection

C) Maximum

D) Cannot be determined

Show the worked solution

Answer: C

Explanation

$f'' < 0$ means concave down — local maximum.

Why the others are wrong: A would need a POSITIVE second derivative. B would need it to be zero. D gives up too early: a non-zero second derivative settles the question on its own.

M4Basic

Topic: Gradient of tangent

The gradient of the tangent to $y = 3x^2 - x$ at $x = 2$ is:

A) $10$

B) $11$

C) $6$

D) $5$

Show the worked solution

Answer: B

Explanation

$y' = 6x - 1$ ; at $x = 2$ : $y'(2) = 12 - 1 = 11$

Why the others are wrong: A drops the derivative of $-x$ , which is $-1$ . C gives the second derivative, 6. D follows from evaluating $6x - 1$ at something other than 2.

M5Basic

Topic: Simplify before differentiating

$f(x) = \dfrac{x^4 - x^2}{x}$ . Then $f'(x) =$

A) $3x^2 - 1$

B) $3x^2$

C) $4x^3 - 2x$

D) $x^3 - x$

Show the worked solution

Answer: A

Explanation

$f(x) = x^3 - x$ ; $f'(x) = 3x^2 - 1$

Why the others are wrong: D is the SIMPLIFIED function, $x^3 - x$ , left undifferentiated. C differentiates the numerator alone and ignores the division. B differentiates correctly but drops the $-1$ .

M6Intermediate

Topic: First principles

Using first principles, $f'(x)$ where $f(x) = 5x$ equals:

A) $5x$

B) $0$

C) $5$

D) $5x + 5h$

Show the worked solution

Answer: C

Explanation

$\lim_{h \to 0} \frac{5(x+h) - 5x}{h} = \lim_{h \to 0} \frac{5h}{h} = 5$

Why the others are wrong: A copies the function back. B treats $5x$ as a constant, which only a number is. D stops before the $h$ is cancelled and the limit taken.

M7Intermediate

Topic: Equation of tangent

The equation of the tangent to $f(x) = x^2 - 3$ at $x = -1$ is:

A) $y = -2x - 4$

B) $y = 2x - 4$

C) $y = -2x - 2$

D) $y = -2x + 4$

Show the worked solution

Answer: A

Explanation

$f(-1) = 1 - 3 = -2$ ; $f'(x) = 2x$ ; $f'(-1) = -2$

Tangent: $y - (-2) = -2(x - (-1)) \Rightarrow y = -2x - 4$

Why the others are wrong: B has the sign of the gradient wrong: $f'(-1) = -2$ , not $+2$ . C takes the point's $y$ -value as the intercept. D has the sign of the intercept wrong. Check by substituting $x = -1$ — the tangent must pass through $(-1;\ -2)$ .

M8Intermediate

Topic: Turning points of cubic

$f(x) = x^3 - 12x + 5$ has turning points at:

A) $x = 2$ and $x = -2$

B) $x = 4$ and $x = -4$

C) $x = 3$ and $x = -3$

D) $x = 6$ only

Show the worked solution

Answer: A

Explanation

$f'(x) = 3x^2 - 12 = 3(x^2 - 4) = 3(x-2)(x+2)$ ; $x = 2$ or $x = -2$

Why the others are wrong: B and C solve $3x^2 = 48$ and $3x^2 = 27$ ; the derivative is $3x^2 - 12$ , so the equation is $3x^2 = 12$ . D halves the 12 without differentiating at all.

M9Intermediate

Topic: Inflection point

The point of inflection of $f(x) = x^3 - 3x^2 + 6x - 1$ occurs at $x =$

A) $1$

B) $2$

C) $3$

D) $0$

Show the worked solution

Answer: A

Explanation

$f''(x) = 6x - 6 = 0 \Rightarrow x = 1$

Why the others are wrong: C reads off the coefficient 3 rather than solving $6x - 6 = 0$ . B and D follow from solving it wrongly. Note that this cubic has no turning points at all — an inflection does not need one.

M10Intermediate

Topic: Increasing/decreasing

For $f(x) = -x^3 + 3x$ , the function is increasing when:

A) $-1 < x < 1$

B) $x < -1$ or $x > 1$

C) $x > 0$

D) All real $x$

Show the worked solution

Answer: A

Explanation

$f'(x) = -3x^2 + 3 = -3(x^2 - 1) = -3(x-1)(x+1)$

$f'(x) > 0$ when $(x-1)(x+1) < 0$ , i.e. $-1 < x < 1$

Why the others are wrong: B names exactly where the function DECREASES — the leading term is $-x^3$ , so the cubic falls at both ends. C ignores that it turns again at $x = 1$ . D would need no turning points at all.

M11Intermediate

Topic: Optimisation — maximum value

Find the maximum value of $f(x) = -x^2 + 6x + 1$ .

A) $3$

B) $9$

C) $10$

D) $7$

Show the worked solution

Answer: C

Explanation

$f'(x) = -2x + 6 = 0 \Rightarrow x = 3$

$f(3) = -9 + 18 + 1 = 10$

Why the others are wrong: A is the $x$ -value of the turning point, not the maximum VALUE — substitute it back. B does substitute, then drops the $+1$ . D adds $6 + 1$ and substitutes nothing.

M12Intermediate

Topic: Normal gradient

The tangent to $f(x) = x^3$ at $x = -1$ has gradient $3$ . The gradient of the normal is:

A) $-3$

B) $3$

C) $\dfrac{1}{3}$

D) $-\dfrac{1}{3}$

Show the worked solution

Answer: D

Explanation

$f'(x) = 3x^2$ ; $f'(-1) = 3$ . Normal gradient $= -\frac{1}{3}$

Why the others are wrong: B gives the tangent's gradient back. C takes the reciprocal but forgets the sign. A changes the sign but forgets the reciprocal. The normal needs both: $m_1 m_2 = -1$ .

M13Intermediate

Topic: Reading $f'$ from a graph

A function $f$ is increasing on $(-\infty,\ 2)$ and decreasing on $(2,\ \infty)$ . Which statement about $f'$ is true?

A) $f'(2) = 0$ and $f'(x) < 0$ for $x < 2$

B) $f'(x) < 0$ for all $x$

C) $f'(2) > 0$

D) $f'(2) = 0$ and $f'(x) > 0$ for $x < 2$

Show the worked solution

Answer: D

Explanation

Increasing means $f' > 0$ . At the maximum, $f' = 0$ .

Why the others are wrong: A has the sign backwards — increasing means $f'$ is POSITIVE. B would make $f$ decrease everywhere, but it rises to the left of 2. C puts a positive gradient at the turning point, where it must be zero.

M14Intermediate

Topic: Second derivative sign

If $f''(x) > 0$ for all $x$ in an interval, then on that interval $f$ is:

A) Decreasing

B) Concave down

C) Concave up

D) At a maximum

Show the worked solution

Answer: C

Explanation

$f'' > 0$ means $f'$ is increasing — the curve is concave up (cup shape).

Why the others are wrong: A is what a negative FIRST derivative means; this is the second. B is the opposite sign — concave down needs $f'' < 0$ . D describes a single point, not a property holding across a whole interval.

M15Intermediate

Topic: Cubic with negative leading coefficient

Which of the following best describes the end behaviour of $f(x) = -2x^3 + x^2 - 5$ ?

A) Rises left, rises right

B) Rises left, falls right

C) Falls left, rises right

D) Falls left, falls right

Show the worked solution

Answer: B

Explanation

Negative leading coefficient ( $a = -2 < 0$ ): for large negative $x$ , $-2x^3 \to +\infty$ (rises left); for large positive $x$ , $-2x^3 \to -\infty$ (falls right).

Why the others are wrong: A and D are even-degree shapes, where both ends go the same way; this is a cubic, so they go opposite ways. C is the right shape for a cubic but the wrong direction — it belongs to a POSITIVE leading coefficient, and this one is $-2$ .

M16Proficient

Topic: Finding $k$ given a stationary point

$f(x) = x^3 - kx$ has a stationary point at $x = 2$ . Find $k$ .

A) $6$

B) $4$

C) $12$

D) $2$

Show the worked solution

Answer: C

Explanation

$f'(x) = 3x^2 - k = 0$ at $x = 2$ :

$3(4) - k = 0 \Rightarrow k = 12$

Why the others are wrong: A differentiates $x^3$ as $3x$ and solves $3x - k = 0$ . B stops at $x^2 = 4$ without multiplying by the 3. D hands the $x$ -value back instead of $k$ .

M17Proficient

Topic: Optimisation — tin can

A cylindrical tin has volume $250\pi\ \text{cm}^3$ . The radius that minimises total surface area satisfies:

A) $r = 25$ cm

B) $r = 10$ cm

C) $r = 2$ cm

D) $r = 5$ cm

Show the worked solution

Answer: D

Explanation

$V = \pi r^2 h = 250\pi \Rightarrow h = \frac{250}{r^2}$

$SA = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \frac{500\pi}{r}$

$\frac{d(SA)}{dr} = 4\pi r - \frac{500\pi}{r^2} = 0$

$4r^3 = 500 \Rightarrow r^3 = 125 \Rightarrow r = 5$

Why the others are wrong: A reports $r^2$ rather than $r$ — the square root was never taken. B and C do not satisfy $4r^3 = 500$ ; substitute each one and check.

M18Proficient

Topic: Discriminant of $f'$ and number of turning points

For $f(x) = x^3 + bx + c$ , how many turning points does $f$ have if $b > 0$ ?

A) Two turning points

B) One turning point

C) No turning points

D) Cannot be determined

Show the worked solution

Answer: C

Explanation

$f'(x) = 3x^2 + b$ . Since $b > 0$ , $f'(x) = 3x^2 + b \geq b > 0$ for all $x$ .

$f'(x)$ never equals zero — no stationary points, therefore no turning points.

Why the others are wrong: A would need $b < 0$ , so that $3x^2 + b = 0$ has two roots. B would need $b = 0$ , and even then $x = 0$ is an inflection rather than a turn. D gives up too early: the sign of $b$ settles it completely.

M19Proficient

Topic: Relating graphs of $f$ and $f'$

At the point where the graph of $f'$ crosses the x-axis from below, $f$ has a:

A) Local maximum

B) Local minimum

C) Point of inflection

D) Vertical asymptote

Show the worked solution

Answer: B

Explanation

$f'$ crosses from below (negative) to above (positive) the x-axis → $f'$ changes from negative to positive → $f$ changes from decreasing to increasing → local minimum.

Why the others are wrong: A is where $f'$ crosses DOWNWARD — from positive to negative. C is where $f''$ changes sign, not $f'$ . D does not follow at all: a derivative passing through zero says nothing about asymptotes.

M20Proficient

Topic: Chain of reasoning

If $f(x) = x^4 - 8x^2$ , at how many points is the tangent to the curve horizontal?

A) 1

B) 2

C) 3

D) 4

Show the worked solution

Answer: C

Explanation

$f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x-2)(x+2) = 0$

$x = 0,\ x = 2,\ x = -2$ — three points with horizontal tangents.

Why the others are wrong: A counts only $x = 0$ . B counts only the outer pair, $x = \pm 2$ . D reads off the degree 4 instead of counting the roots of $4x(x^2 - 4) = 0$ .

M21Proficient

Topic: Using turning point to find coefficients

$f(x) = ax^2 + bx$ has a turning point at $(3,\ -9)$ . Find $a$ and $b$ .

A) $a = 1,\ b = -6$

B) $a = -1,\ b = 6$

C) $a = 1,\ b = 6$

D) $a = 2,\ b = -3$

Show the worked solution

Answer: A

Explanation

Turning point at $x = 3$ : $f'(x) = 2ax + b = 0$ at $x = 3$ :

$6a + b = 0 \Rightarrow b = -6a$

Turning point value $f(3) = -9$ :

$9a + 3b = -9 \Rightarrow 9a + 3(-6a) = -9 \Rightarrow -9a = -9 \Rightarrow a = 1$ , $b = -6$

Why the others are wrong: C has the sign of $b$ wrong, and B has both signs wrong — either way the turning point lands at $x = -3$ , not $3$ . D satisfies neither condition: $f'(3) = 4(3) - 3 = 9$ , which is not zero.

M22Proficient

Topic: Sign of $f''$ and concavity

For $f(x) = x^4 - 6x^2$ , on which interval is $f$ concave down?

A) $-1 < x < 1$

B) $x < -1$ or $x > 1$

C) $x > 0$

D) All real $x$

Show the worked solution

Answer: A

Explanation

$f'(x) = 4x^3 - 12x$ ; $f''(x) = 12x^2 - 12 = 12(x^2 - 1)$

$f''(x) < 0$ when $x^2 < 1$ , i.e. $-1 < x < 1$

Why the others are wrong: B names exactly where the graph is concave UP — the two intervals swap. C includes everything beyond $x = 1$ , which is concave up. D would need $f''$ never to change sign, but a quartic's changes twice.

M23Proficient

Topic: Optimisation — area

A rectangle is inscribed in a right-angled triangle with legs $6$ and $8$ . If one corner is at the right angle, find the width $x$ that maximises the rectangle's area.

A) $x = 3$

B) $x = 6$

C) $x = 2$

D) $x = 4$

Show the worked solution

Answer: D

Explanation

The hypotenuse passes through $(8, 0)$ and $(0, 6)$ , so its equation is $\dfrac{x}{8} + \dfrac{y}{6} = 1$ , giving $y = 6\!\left(1 - \dfrac{x}{8}\right)$ .

$A = xy = x \cdot 6\!\left(1 - \frac{x}{8}\right) = 6x - \frac{3x^2}{4}$

$A' = 6 - \frac{3x}{2} = 0 \quad \Rightarrow \quad x = 4$

$y = 6(1 - 4/8) = 3$ ; maximum area $= 4 \times 3 = 12\,\text{cm}^2$ .

Why the others are wrong: A halves the wrong leg. B takes a whole leg, which leaves the rectangle no height at all. C does give a rectangle, but its area is $9$ , short of the maximum $12$ .

M24Proficient

Topic: Velocity and acceleration

A ball's height is $h(t) = -5t^2 + 20t + 2$ metres after $t$ seconds. What is the maximum height?

A) $20\,\text{m}$

B) $22\,\text{m}$

C) $25\,\text{m}$

D) $2\,\text{m}$

Show the worked solution

Answer: B

Explanation

$h'(t) = -10t + 20 = 0 \Rightarrow t = 2$

$h(2) = -5(4) + 20(2) + 2 = -20 + 40 + 2 = 22\,\text{m}$

Why the others are wrong: A reads off the coefficient of $t$ , which is a speed, not a height. D is the height at $t = 0$ , before the ball is thrown. C is close but not reached by any step here — substituting $t = 2$ gives exactly $22$ .

M25Proficient

Topic: Reading cubic graph features

A cubic $f$ has roots at $x = -3,\ 0,\ 2$ and a negative leading coefficient. Which is the sketch?

A) Falls left, local max between $-3$ and $0$ , local min between $0$ and $2$ , rises right

B) Rises left, local max between $-3$ and $0$ , local min between $0$ and $2$ , falls right

C) Rises left, local min between $-3$ and $0$ , local max between $0$ and $2$ , falls right

D) Falls left, local min between $-3$ and $0$ , local max between $0$ and $2$ , rises right

Show the worked solution

Answer: C

Explanation

Negative leading coefficient: rises left, falls right. The function comes from $+\infty$ on the left, decreases through $x = -3$ , dips to a local minimum between $-3$ and $0$ , rises through $x = 0$ , peaks at a local maximum between $0$ and $2$ , then decreases through $x = 2$ to $-\infty$ .

Checking signs: $f(-1) = -(-1+3)(-1)(-1-2) = -(2)(-1)(-3) = -6 < 0$ and $f(1) = -(1+3)(1)(1-2) = -(4)(1)(-1) = 4 > 0$ , confirming the minimum is in $(-3, 0)$ (function negative) and the maximum is in $(0, 2)$ (function positive).

Why the others are wrong: A falls on the left, which is what a POSITIVE leading coefficient does, and turns the wrong way round as well. B gets the ends right and swaps the two turning points. D gets the turning points right and swaps the ends. Fix the ends first, then the turns.

M26Proficient

Topic: Using the discriminant of $f'$

How many real stationary points does $f(x) = x^3 + 3x^2 + 5x - 1$ have?

A) 0

B) 1

C) 2

D) 3

Show the worked solution

Answer: A

Explanation

$f'(x) = 3x^2 + 6x + 5$

Discriminant of $f'$ : $\Delta = 36 - 4(3)(5) = 36 - 60 = -24 < 0$

No real roots → no stationary points.

Why the others are wrong: C assumes every cubic has two turning points; that needs $f'$ to have two roots, and here its discriminant is $-24$ . D counts the degree. B counts the point of inflection, which is not stationary.

M27Proficient

Topic: Second derivative and concavity change

The graph of $f''(x)$ crosses the x-axis at $x = 3$ (from negative to positive). This means $f(x)$ has:

A) A local maximum at $x = 3$

B) A local minimum at $x = 3$

C) A point of inflection at $x = 3$

D) A stationary point at $x = 3$

Show the worked solution

Answer: C

Explanation

$f''$ changes sign at $x = 3$ → $f$ changes concavity → point of inflection at $x = 3$ .

Note: A point of inflection does not require $f'(x) = 0$ .

Why the others are wrong: A and B are settled by $f'$ , not by $f''$ — the second derivative only tells you the concavity. D needs $f'(3) = 0$ , and nothing here says that.

M28Proficient

Topic: Optimisation — combined constraint

A farmer builds a rectangular pen divided into three equal sections by two interior fences parallel to the width. Total fencing is $480\,\text{m}$ . Find the width $w$ that maximises area.

A) $w = 60$

B) $w = 80$

C) $w = 120$

D) $w = 40$

Show the worked solution

Answer: A

Explanation

Two lengths ( $l$ ) and four widths ( $w$ ): $2l + 4w = 480 \Rightarrow l = 240 - 2w$

$A = lw = (240 - 2w)w = 240w - 2w^2$

$A' = 240 - 4w = 0 \Rightarrow w = 60$

Why the others are wrong: B counts three widths where the pen needs four: two outer sides plus two interior fences. D counts six. C is the LENGTH, not the width the question asks for.

M29Proficient

Topic: Applying calculus to a real-world graph

The velocity of a car is $v(t) = 3t^2 - 18t + 24\ \text{m/s}$ . When is the car decelerating?

A) $t < 2$

B) $t > 4$

C) $2 < t < 4$

D) $t < 3$

Show the worked solution

Answer: D

Explanation

Acceleration $a(t) = v'(t) = 6t - 18$ .

Decelerating means $a(t) < 0$ :

$6t - 18 < 0 \quad \Rightarrow \quad t < 3$

Why the others are wrong: A and B name the intervals where the VELOCITY is positive, which is a direction, not a change. C is where the velocity is negative — the car is reversing there, and for part of it speeding up. Decelerating means the ACCELERATION is negative, so solve $v'(t) < 0$ .

M30Proficient

Topic: Linking all calculus concepts

$f(x) = x^3 - 3x^2 - 9x + 5$ has a local maximum at $x = -1$ and a local minimum at $x = 3$ . The value of $f(-1) - f(3)$ is:

A) $10$

B) $16$

C) $32$

D) $-16$

Show the worked solution

Answer: C

Explanation

$f(-1) = -1 - 3 + 9 + 5 = 10$

$f(3) = 27 - 27 - 27 + 5 = -22$

$f(-1) - f(3) = 10 - (-22) = 32$

Why the others are wrong: A is $f(-1)$ on its own, with the subtraction never done. B and D do not follow from any step here: the two values are $10$ and $-22$ , so the difference is $32$ — and subtracting the other way gives $-32$ .

Chapter 10 — Differential Calculus

Section 10.1 — Limits and First Principles

Section 10.2 — Rules of Differentiation

Section 10.3 — Tangent and Normal Lines

Section 10.4 — Stationary Points and Nature

Section 10.5 — Cubic Functions and Graphs

Section 10.6 — Optimisation

Section 10.7 — Mixed Practice