Home / IB Mathematics SL 5.3 The derivative of functions AI SL Paper 1- Exam Style Questions

IB Mathematics SL 5.3 The derivative of functions AI SL Paper 1- Exam Style Questions- New Syllabus

IB DP Maths AI SL Paper 1 - All Topics

Question 

Amara investigates the rate at which a cubical block of sugar dissolves in hot coffee.
Initially, the cube has side lengths of 10 mm. This information is illustrated in the following diagrams.
Amara predicts that, as the block of sugar dissolves, each side length will decrease at a constant rate of 0.2 mm per second.
(a) According to this model, obtain
(i) the length of one side of a block of sugar, 20 seconds after it is placed in hot coffee.
(ii) the volume of a block of sugar, 20 seconds after it is placed in hot coffee. [3]
Let the function \(V(t)\) represent the volume of the block of sugar, mm\(^3\), \(t\) seconds after it is placed in hot coffee. \(V(t)\) is given by
\(V(t)=1000-60t+1.2t^2-0.008t^3,\quad\text{for }0\le t\le 50.\)
(b) Obtain \(V'(t)\). [2]
(c) Obtain the rate of change of the volume of the block of sugar at \(t=20\). [2]
(d) State one reason why the side length of the cube may not always decrease at a constant rate. [1]
▶️ Answer/Explanation
Markscheme (with detailed working)

(a) Linear side-length model \((\text{rate}=0.2\ \text{mm s}^{-1})\)

(i) Side length after \(t\) s: \(s(t)=10-0.2t\) (mm). For \(t=20\):
\(\displaystyle s(20)=10-0.2(20)=10-4=\boxed{6\ \text{mm}}.\) M1 A1

(ii) Volume \(=s^3\). With \(s=6\ \text{mm}\): \(V=6^3=\boxed{216\ \text{mm}^3}.\) A1

(b) Differentiate \(V(t)=1000-60t+1.2t^2-0.008t^3\)

\(\displaystyle V'(t) = -60 + 2.4t – 0.024t^2.\) M1 A1

(c) Instantaneous rate of change at \(t=20\)

\(\displaystyle V'(20) = -60 + 2.4(20) – 0.024(20)^2 = -60 + 48 – 9.6 = \boxed{-21.6\ \text{mm}^3\text{/s}}.\) M1 A1
(Negative value indicates the volume is decreasing.)

(d) Reason for non-constant decrease of side length

For example: the surface area changes as the cube dissolves (edges/corners round off), so the dissolution rate varies over time; hence the side length need not decrease at a constant rate. R1
Total Marks: 8

Question

Consider the function \( g(x) = x^2 – \frac{3}{x}, x \neq 0 \).
(a) Calculate \( g'(x) \). [2]
(b) Use your answer to part (a) to obtain the gradient of line L, which is a tangent to \( g(x) \) at the point \((1, -2)\). [2]
(c) Calculate the number of lines parallel to L that are tangent to \( g(x) \). Justify your answer. [3]
▶️ Answer/Explanation
Markscheme
(a) Given: \( g(x) = x^2 – \frac{3}{x} \). \[ g(x) = x^2 – 3x^{-1} \] \[ \begin{aligned} g'(x) &= \frac{d}{dx}(x^2) – \frac{d}{dx}(3x^{-1}) \\ &= 2x – 3 \times (-1)x^{-2} \\ &= 2x + \frac{3}{x^2} \end{aligned} \] Answer: \( g'(x) = 2x + \frac{3}{x^2} \) A1 A1 [2 marks]
(b) From part (a): \( g'(x) = 2x + \frac{3}{x^2} \). \[ g'(1) = 2 \times 1 + \frac{3}{1^2} = 2 + 3 = 5 \] Answer: Gradient of L = 5 M1 A1 [2 marks]

(c) EITHER

Solve for points where the tangent is parallel to \( L \): \( g'(x) = 5 \). \[ 2x + \frac{3}{x^2} = 5 \] M1
\[ 2x \times x^2 + 3 = 5x^2 \] \[ 2x^3 – 5x^2 + 3 = 0 \] Solutions: \( x \approx -0.686, 1, 2.19 \). A1
OR
Sketch of \( y = g'(x) \) with \( y = 5 \).
M1
Three points of intersection at \( x \approx -0.686, 1, 2.19 \). A1
THEN there are two other tangent lines to \( g(x) \) parallel to \( L \). A1

[3 marks]

Total Marks: 7
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