Comprehensive assessment covering all 11 lessons: average and instantaneous rates of change, limits, the derivative from first principles, the gradient of a tangent, differentiation rules, the chain rule, product and quotient rules, the second derivative, stationary points, optimisation, and increasing and decreasing functions.
Assessment
Select the best answer for each question. 1 mark each.
The derivative of $f(x)$ from first principles is defined as:
The average rate of change of $f(x) = x^2$ over the interval $[1, 4]$ is:
Evaluate $\lim_{x \to 4} \frac{x^2 - 16}{x - 4}$.
Evaluate $\lim_{x \to \infty} \frac{3x^2 - 2}{2x^2 + 5}$.
Differentiate $y = x^5$.
Find the derivative of $y = \sqrt{x}$.
The gradient of the tangent to $y = x^2 - 3x$ at $x = 2$ is:
Differentiate $y = (2x + 1)^4$ using the chain rule.
Differentiate $y = x^2(x + 3)$ using the product rule.
Find $\frac{dy}{dx}$ for $y = \frac{x}{x + 1}$ using the quotient rule.
If $f(x) = 2x^3 - 3x^2 + x$, then $f''(x)$ equals:
How many stationary points does $f(x) = x^3 - 3x$ have?
For $f(x) = x^3 - 3x^2$, the nature of the stationary point at $x = 2$ is a:
The function $f(x) = x^2 - 4x + 3$ is increasing when:
The equation of the normal to $y = x^2$ at the point where $x = 2$ is:
Short Answer
Use first principles to differentiate $f(x) = x^2 + 3x$. Show all working.
Evaluate $\lim_{x \to 2} \frac{x^2 - x - 2}{x - 2}$. Show all working.
Differentiate $y = (3x^2 + 1)^5$ using the chain rule, and evaluate $\frac{dy}{dx}$ at $x = 0$. Show all working.
Find $\frac{dy}{dx}$ for $y = \frac{2x - 1}{x + 3}$ using the quotient rule. Simplify your answer. Show all working.
Find the equation of the tangent to $y = x^3 - 2x$ at the point where $x = 1$. Show all working.
For the curve $y = x^3 - 3x^2$, find the coordinates of the stationary points and determine the nature of each using the second derivative test. Show all working.
A farmer has $60$ m of fencing to enclose a rectangular paddock against a straight river, so only three sides need fencing. Find the dimensions that give the maximum area, and state that maximum area. Show all working.
A particle moves in a straight line with displacement $x(t) = t^3 - 6t^2 + 9t$ metres after $t$ seconds, $t \geq 0$. Find: (a) the velocity at $t = 0$, and (b) the times at which the particle is momentarily at rest. Show all working.
Q1: A, by definition $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
Q2: C, $\frac{f(4) - f(1)}{4 - 1} = \frac{16 - 1}{3} = 5$.
Q3: D, $\frac{x^2 - 16}{x - 4} = \frac{(x-4)(x+4)}{x-4} = x + 4 \to 8$ as $x \to 4$.
Q4: B, divide numerator and denominator by $x^2$: $\frac{3 - \frac{2}{x^2}}{2 + \frac{5}{x^2}} \to \frac{3}{2}$.
Q5: A, power rule: $\frac{d}{dx}(x^5) = 5x^4$.
Q6: C, write $y = x^{1/2}$, so $\frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.
Q7: B, $y' = 2x - 3$, so at $x = 2$ the gradient is $2(2) - 3 = 1$.
Q8: D, chain rule: $4(2x+1)^3 \cdot 2 = 8(2x+1)^3$.
Q9: A, product rule: $(2x)(x+3) + (x^2)(1) = 2x^2 + 6x + x^2 = 3x^2 + 6x$.
Q10: C, quotient rule: $\frac{(1)(x+1) - (x)(1)}{(x+1)^2} = \frac{1}{(x+1)^2}$.
Q11: D, $f'(x) = 6x^2 - 6x + 1$, so $f''(x) = 12x - 6$.
Q12: B, $f'(x) = 3x^2 - 3 = 3(x-1)(x+1) = 0$ gives $x = \pm 1$, so two stationary points.
Q13: C, $f'(x) = 3x^2 - 6x$, $f''(x) = 6x - 6$. $f''(2) = 6 > 0$, so it is a local minimum.
Q14: D, $f'(x) = 2x - 4 > 0$ when $x > 2$.
Q15: B, at $x = 2$: $y = 4$ and $y' = 2x = 4$, so the normal gradient is $-\frac{1}{4}$. $y - 4 = -\frac{1}{4}(x - 2)$, giving $y = -\frac{1}{4}x + \frac{9}{2}$.
Q16 (3 marks): $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$. $f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h$ [1]. $f(x+h) - f(x) = 2xh + h^2 + 3h$, so $\frac{f(x+h) - f(x)}{h} = 2x + h + 3$ [1]. Taking the limit as $h \to 0$: $f'(x) = 2x + 3$ [1].
Q17 (3 marks): Direct substitution gives $\frac{0}{0}$, so factor the numerator: $x^2 - x - 2 = (x - 2)(x + 1)$ [1]. $\frac{(x-2)(x+1)}{x - 2} = x + 1$ for $x \neq 2$ [1]. Substitute $x = 2$: the limit is $3$ [1].
Q18 (3 marks): Let $u = 3x^2 + 1$, so $y = u^5$ and $\frac{du}{dx} = 6x$ [1]. $\frac{dy}{dx} = 5u^4 \cdot 6x = 30x(3x^2 + 1)^4$ [1]. At $x = 0$: $\frac{dy}{dx} = 30(0)(1)^4 = 0$ [1].
Q19 (3 marks): With $u = 2x - 1$ and $v = x + 3$, $u' = 2$ and $v' = 1$ [1]. Quotient rule: $\frac{dy}{dx} = \frac{2(x + 3) - (2x - 1)(1)}{(x + 3)^2}$ [1]. Simplify the numerator: $2x + 6 - 2x + 1 = 7$, so $\frac{dy}{dx} = \frac{7}{(x + 3)^2}$ [1].
Q20 (4 marks): At $x = 1$: $y = 1^3 - 2(1) = -1$, so the point is $(1, -1)$ [1]. $y' = 3x^2 - 2$, so the gradient at $x = 1$ is $3(1) - 2 = 1$ [1]. Tangent: $y - (-1) = 1(x - 1)$ [1], which simplifies to $y = x - 2$ [1].
Q21 (4 marks): $y' = 3x^2 - 6x = 3x(x - 2) = 0$, so $x = 0$ or $x = 2$ [1]. The $y$-coordinates are $y(0) = 0$ and $y(2) = 8 - 12 = -4$, giving $(0, 0)$ and $(2, -4)$ [1]. $y'' = 6x - 6$: at $x = 0$, $y'' = -6 < 0$, so $(0, 0)$ is a local maximum [1]; at $x = 2$, $y'' = 6 > 0$, so $(2, -4)$ is a local minimum [1].
Q22 (4 marks): Let the two sides perpendicular to the river be $x$ each; the side parallel to the river is $60 - 2x$ [1]. Area $A(x) = x(60 - 2x) = 60x - 2x^2$ [1]. $A'(x) = 60 - 4x = 0$ gives $x = 15$ [1]. Then the parallel side is $60 - 30 = 30$ m, and $A''(x) = -4 < 0$ confirms a maximum. Dimensions are $15$ m by $30$ m, and the maximum area is $15 \times 30 = 450$ m$^2$ [1].
Q23 (4 marks): $v(t) = x'(t) = 3t^2 - 12t + 9$ [1]. (a) $v(0) = 3(0) - 12(0) + 9 = 9$ m/s [1]. (b) At rest, $v(t) = 0$: $3t^2 - 12t + 9 = 0$, so $t^2 - 4t + 3 = 0$, giving $(t - 1)(t - 3) = 0$ [1]. Therefore $t = 1$ s and $t = 3$ s [1].