Question
A company’s profit per year was found to be changing at a rate of
\(\frac{dP}{dt}=3t^2-8t\)
where P is the company’s profit in thousands of dollars and t is the time since the company
was founded, measured in years.
(a) Determine whether the profit is increasing or decreasing when t = 2.
One year after the company was founded, the profit was 4 thousand dollars.
(b) Find an expression for P(t), when t ≥ 0.
▶️Answer/Explanation
Ans:
(a) METHOD 1
(when t = 2)
\(\frac{dP}{dt} = -4 \) OR \(\frac{dP}{dt}<0\) (equivalent in words) OR \(3(2)^2-8(2)=-4\)
therefore P is decreasing
METHOD 2
sketch with t = 2 indicated in 4th quadrant OR t-intercepts identified
therefore P is decreasing
(b) \((P(t)=)t^3 – 4t^2 (+c)\)
\(4=1^3 – 4(1)^2 + c\)
c =7
\(P(t) = t^3 – 4t^2 + 7\)
Question
The figure below shows the graphs of functions \(f_1 (x) = x\) and \(f_2 (x) = 5 – x^2\).
a.(i) Differentiate \(f_1 (x) \) with respect to x.
(ii) Differentiate \(f_2 (x) \) with respect to x.[3]
▶️Answer/Explanation
Markscheme
(i) \(f_1 ‘ (x) = 1\) (A1)
(ii) \(f_2 ‘ (x) = – 2x\) (A1)(A1)
(A1) for correct differentiation of each term. (C3)[3 marks]
\(1 = – 2x\) (M1)
\(x = – \frac{1}{2}\) (A1)(ft) (C2)[2 marks]
(A1) is for the tangent drawn at \(x = \frac{1}{2}\) and reasonably parallel to the line \(f_1\) as shown.
(A1) (C1)[1 mark]
Question
The table given below describes the behaviour of f ′(x), the derivative function of f (x), in the domain −4 < x < 2.
a.State whether f (0) is greater than, less than or equal to f (−2). Give a reason for your answer.[2]
b.The point P(−2, 3) lies on the graph of f (x).
Write down the equation of the tangent to the graph of f (x) at the point P.[2]
c.The point P(−2, 3) lies on the graph of f (x).
From the information given about f ′(x), state whether the point (−2, 3) is a maximum, a minimum or neither. Give a reason for your answer.[2]
▶️Answer/Explanation
Markscheme
greater than (A1)
Gradient between x = −2 and x = 0 is positive. (R1)
OR
The function is increased between these points or equivalent. (R1) (C2)
Note: Accept a sketch. Do not award (A1)(R0).[2 marks]
y = 3 (A1)(A1) (C2)
Note: Award (A1) for y = a constant, (A1) for 3.[2 marks]
minimum (A1)
Gradient is negative to the left and positive to the right or equivalent. (R1) (C2)
Note: Accept a sketch. Do not award (A1)(R0).[2 marks]
Question
Consider the curve \(y = {x^2} + \frac{a}{x} – 1,{\text{ }}x \ne 0\).
a.Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).[3]
b.The gradient of the tangent to the curve is \( – 14\) when \(x = 1\).
Find the value of \(a\).[3]
▶️Answer/Explanation
Markscheme
\(2x – \frac{a}{{{x^2}}}\) (A1)(A1)(A1) (C3)
Notes: Award (A1) for \(2x\), (A1) for \( – a\) and (A1) for \({x^{ – 2}}\).
Award at most (A1)(A1)(A0) if extra terms are present.
\(2(1) – \frac{a}{{{1^2}}} = – 14\) (M1)(M1)
Note: Award (M1) for substituting \(1\) into their gradient function, (M1) for equating their gradient function to \( – 14\).
Award (M0)(M0)(A0) if the original function is used instead of the gradient function.
\(a = 16\) (A1)(ft) (C3)
Note: Follow through from their gradient function from part (a).
Question
A function \(f\) is given by \(f(x) = 4{x^3} + \frac{3}{{{x^2}}} – 3,{\text{ }}x \ne 0\).
a.Write down the derivative of \(f\).[3]
b.Find the point on the graph of \(f\) at which the gradient of the tangent is equal to 6.[3]
▶️Answer/Explanation
Markscheme
\(12{x^2} – \frac{6}{{{x^3}}}\) or equivalent (A1)(A1)(A1) (C3)
Note: Award (A1) for \(12{x^2}\), (A1) for \( – 6\) and (A1) for \(\frac{1}{{{x^3}}}\) or \({x^{ – 3}}\). Award at most (A1)(A1)(A0) if additional terms seen.[3 marks]
\(12{x^2} – \frac{6}{{{x^3}}} = 6\) (M1)
Note: Award (M1) for equating their derivative to 6.
\((1,{\text{ }}4)\)\(\,\,\,\)OR\(\,\,\,\)\(x = 1,{\text{ }}y = 4\) (A1)(ft)(A1)(ft) (C3)
Note: A frequent wrong answer seen in scripts is \((1,{\text{ }}6)\) for this answer with correct working award (M1)(A0)(A1) and if there is no working award (C1).[3 marks]