MTH104 · TU past paper

Basic Mathematics 2081.2 question paper

The complete TU 2081.2 exam paper for Basic Mathematics (MTH104), all 12 questions with solved model answers written to the mark scheme.

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1NumericalContinuity definitionHideAnswer
(question text pending review) # Solution

Main Question [5 marks]

Show that $f(x) = 1 - \sqrt{1 - x^2}$ is continuous on $[-1,1]$

A function is continuous at a point $c$ if $\lim_{x \to
(question text pending review) # Solution

Main Question [5 marks]

Show that $f(x) = 1 - \sqrt{1 - x^2}$ is continuous on $[-1,1]$

A function is continuous at a point $c$ if $\lim_{x \to
Definition: A function $f$ is continuous on a closed interval $[a,b]$ if it is continuous at every interior point and continuous from the right at $a$ and from the left at $b$ . Given: $f(x) = 1 - \sqrt{1 - x^2}$ , interval $[-1, 1]$ . Step...
Definition: A function $f$ is continuous on a closed interval $[a,b]$ if it is continuous at every interior point and continuous from the right at $a$ and from the left at $b$. Given: $f(x) = 1 - \sqrt{1 - x^2}$, interval $[-1, 1]$. Step...
210 marksNumericalLocal extrema and critical pointsHideAnswer
If $f(x) = x^2 + 2x - 1$ and $g(x) = 2x - 3$ , then find $fog(x)$ and $gof(x)$ . Find the local maxima and local minima of the function $f(x) = 3x^4 - 4x^3 - 12x^2 + 5$ . [5+5]
If $f(x) = x^2 + 2x - 1$ and $g(x) = 2x - 3$, then find $fog(x)$ and $gof(x)$. Find the local maxima and local minima of the function $f(x) = 3x^4 - 4x^3 - 12x^2 + 5$. [5+5]
$f(x) = x^2 + 2x - 1$ - $g(x) = 2x - 3$ - $f(x) = 3x^4 - 4x^3 - 12x^2 + 5$ (for Part 2) --- $fog(x) = f(g(x)) = f(2x-3)$ Substitute $(2x-3)$ into $f$ : $= (2x-3)^2 + 2(2x-3) - 1$ $= (4x^2 - 12x + 9) + (4x - 6) - 1$ $$= 4x^2 - 8x +...

$f(x) = x^2 + 2x - 1$ - $g(x) = 2x - 3$ - $f(x) = 3x^4 - 4x^3 - 12x^2 + 5$ (for Part 2) --- $$fog(x) = f(g(x)) = f(2x-3)$$ Substitute $(2x-3)$ into $f$: $$= (2x-3)^2 + 2(2x-3) - 1$$ $$= (4x^2 - 12x + 9) + (4x - 6) - 1$$ $$= 4x^2 - 8x +...
310 marksNumericalArea under curvesHideAnswer
Find the area enclosed by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ .

Evaluate: $\int_0^2 \frac{x}{\sqrt{x^2 + 4}} \, dx$ .

[5+5]
Find the area enclosed by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

Evaluate: $\int_0^2 \frac{x}{\sqrt{x^2 + 4}} , dx$.

[5+5]
Part 1: Ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ Part 2: Integral $\displaystyle\int0^2 \frac{x}{\sqrt{x^2 + 4}}\, dx$ --- (a) Area Enclosed by the Ellipse Compare with standard form $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ : $$a^2 ...
Part 1: Ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ Part 2: Integral $\displaystyle\int0^2 \frac{x}{\sqrt{x^2 + 4}}, dx$ --- (a) Area Enclosed by the Ellipse Compare with standard form $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$: $$a^2 ...
45 marksNumericalSeparable differential equationsHideAnswer
Solve: $xy' = y$ , when $y(1) = 2$ . [5]
Solve: $xy' = y$, when $y(1) = 2$. [5]
Differential equation: $xy' = y$ - Initial condition: $y(1) = 2$ All required data is present. Step 1: Separate variables $x\frac{dy}{dx} = y$ $\frac{dy}{y} = \frac{dx}{x}$ Step 2: Integrate both sides $$\int \frac{dy}{y} = \int \f...

Differential equation: $xy' = y$ - Initial condition: $y(1) = 2$ All required data is present. Step 1: Separate variables $$x\frac{dy}{dx} = y$$ $$\frac{dy}{y} = \frac{dx}{x}$$ Step 2: Integrate both sides $$\int \frac{dy}{y} = \int \f...
55 marksNumericalComparison testsHideAnswer
Determine whether the series $\sum_{n=1}^{\infty} \frac{5}{2n^2 + 4n + 3}$ is convergent or divergent. [5]
Determine whether the series $\sum_{n=1}^{\infty} \frac{5}{2n^2 + 4n + 3}$ is convergent or divergent. [5]
Series: $\displaystyle\sum{n=1}^{\infty} an$ where $an = \dfrac{5}{2n^2 + 4n + 3}$ - All terms positive for $n \geq 1$ (numerator and denominator positive). Choose the Limit Comparison Test. For large $n$ , the dominant term in the deno...

Series: $\displaystyle\sum{n=1}^{\infty} an$ where $an = \dfrac{5}{2n^2 + 4n + 3}$ - All terms positive for $n \geq 1$ (numerator and denominator positive). Choose the Limit Comparison Test. For large $n$, the dominant term in the deno...
65 marksNumericalHideAnswer
Find a unit vector that has the same direction as the given vector $-3\mathbf{i} + 7\mathbf{j}$ . [5]
Find a unit vector that has the same direction as the given vector $-3\mathbf{i} + 7\mathbf{j}$. [5]
Given Data
- Vector: $\mathbf{v} = -3\mathbf{i} + 7\mathbf{j}$
- Components: $v_x = -3$ , $v_y = 7$
Solution

A unit vector in the same direction as $\mathbf{v}$ is given by:

$\mathbf{\hat{u}} = \frac{\mathbf{v}}{|\mathbf{v}|}$

Step 1: Magnitude of the vector

$|\mathbf{v}| = \sqrt{(-3)^2 + (7)^2} = \sqrt{9 + 49} = \sqrt{58}$

Step 2: Divide by the magnitude

$\mathbf{\hat{u}} = \frac{-3\mathbf{i} + 7\mathbf{j}}{\sqrt{58}} = \frac{-3}{\sqrt{58}}\mathbf{i} + \frac{7}{\sqrt{58}}\mathbf{j}$

Step 3: Rationalized form

$\mathbf{\hat{u}} = \frac{-3\sqrt{58}}{58}\mathbf{i} + \frac{7\sqrt{58}}{58}\mathbf{j}$

Numerically: $\sqrt{58} \approx 7.6158$ , so

$\mathbf{\hat{u}} \approx -0.394\,\mathbf{i} + 0.919\,\mathbf{j}$

Verification

$|\mathbf{\hat{u}}| = \sqrt{\frac{9}{58} + \frac{49}{58}} = \sqrt{\frac{58}{58}} = 1 \checkmark$

Final Answer

$\boxed{\mathbf{\hat{u}} = \frac{-3}{\sqrt{58}}\mathbf{i} + \frac{7}{\sqrt{58}}\mathbf{j} = \frac{-3\sqrt{58}}{58}\mathbf{i} + \frac{7\sqrt{58}}{58}\mathbf{j}}$
Given Data
- Vector: $\mathbf{v} = -3\mathbf{i} + 7\mathbf{j}$
- Components: $v_x = -3$, $v_y = 7$
Solution

A unit vector in the same direction as $\mathbf{v}$ is given by:

$$\mathbf{\hat{u}} = \frac{\mathbf{v}}{|\mathbf{v}|}$$

Step 1: Magnitude of the vector

$$|\mathbf{v}| = \sqrt{(-3)^2 + (7)^2} = \sqrt{9 + 49} = \sqrt{58}$$

Step 2: Divide by the magnitude

$$\mathbf{\hat{u}} = \frac{-3\mathbf{i} + 7\mathbf{j}}{\sqrt{58}} = \frac{-3}{\sqrt{58}}\mathbf{i} + \frac{7}{\sqrt{58}}\mathbf{j}$$

Step 3: Rationalized form

$$\mathbf{\hat{u}} = \frac{-3\sqrt{58}}{58}\mathbf{i} + \frac{7\sqrt{58}}{58}\mathbf{j}$$

Numerically: $\sqrt{58} \approx 7.6158$, so

$$\mathbf{\hat{u}} \approx -0.394,\mathbf{i} + 0.919,\mathbf{j}$$

Verification

$$|\mathbf{\hat{u}}| = \sqrt{\frac{9}{58} + \frac{49}{58}} = \sqrt{\frac{58}{58}} = 1 \checkmark$$

Final Answer

$$\boxed{\mathbf{\hat{u}} = \frac{-3}{\sqrt{58}}\mathbf{i} + \frac{7}{\sqrt{58}}\mathbf{j} = \frac{-3\sqrt{58}}{58}\mathbf{i} + \frac{7\sqrt{58}}{58}\mathbf{j}}$$
75 marksNumericalSecond order linear differential equationsHideAnswer
Solve: $y'' - y' - 6y = 0$ . [5]
Solve: $y'' - y' - 6y = 0$. [5]
Given data: - Differential equation: $y'' - y' - 6y = 0$ - Constant coefficients: $1, -1, -6$ - Homogeneous, second-order, linear ODE. All data present. Characteristic equation: Assume $y = e^{rx}$ , so $y' = re^{rx}$ , $y'' = r^2 e^{rx}$ ....
Given data: - Differential equation: $y'' - y' - 6y = 0$ - Constant coefficients: $1, -1, -6$ - Homogeneous, second-order, linear ODE. All data present. Characteristic equation: Assume $y = e^{rx}$, so $y' = re^{rx}$, $y'' = r^2 e^{rx}$....
85 marksNumericalChain ruleHideAnswer
Use the chain rule to find $\frac{dz}{dt}$ when $z = \cos(x + 4y)$ , $x = 5t^4$ , $y = \frac{1}{t}$ . [5]
Use the chain rule to find $\frac{dz}{dt}$ when $z = \cos(x + 4y)$, $x = 5t^4$, $y = \frac{1}{t}$. [5]
$z = \cos(x + 4y)$ - $x = 5t^4$ - $y = \dfrac{1}{t}$ Required: $\dfrac{dz}{dt}$ All data present. --- $\frac{dz}{dt} = \frac{\partial z}{\partial x}\cdot\frac{dx}{dt} + \frac{\partial z}{\partial y}\cdot\frac{dy}{dt}$ $$\frac{\partia...

$z = \cos(x + 4y)$ - $x = 5t^4$ - $y = \dfrac{1}{t}$ Required: $\dfrac{dz}{dt}$ All data present. --- $$\frac{dz}{dt} = \frac{\partial z}{\partial x}\cdot\frac{dx}{dt} + \frac{\partial z}{\partial y}\cdot\frac{dy}{dt}$$ $$\frac{\partia...
95 marksNumericalImplicit differentiation in multiple variaHideAnswer
Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ if $x^3 + y^3 + z^3 + 6xyz = 1$ . [5]
Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ if $x^3 + y^3 + z^3 + 6xyz = 1$. [5]
Implicit relation: $F(x,y,z) = x^3 + y^3 + z^3 + 6xyz - 1 = 0$ Required: $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$ , where $z$ is an implicit function of $x$ and $y$ . No numeric values are missing; this is a ...
Implicit relation: $$F(x,y,z) = x^3 + y^3 + z^3 + 6xyz - 1 = 0$$ Required: $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$, where $z$ is an implicit function of $x$ and $y$. No numeric values are missing; this is a ...
105 marksNumericalMaclaurin seriesHideAnswer
Find the Maclaurin’s series expansion of $f(x) = \ln x$ . [5]
Find the Maclaurin’s series expansion of $f(x) = \ln x$. [5]
Function: $f(x) = \ln x$ - Required: Maclaurin's series expansion (expansion about $x = 0$ ) The Maclaurin series is a Taylor expansion about $x = 0$ : $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$ For ...

Function: $f(x) = \ln x$ - Required: Maclaurin's series expansion (expansion about $x = 0$) The Maclaurin series is a Taylor expansion about $x = 0$: $$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$ For ...
115 marksNumericalContinuity definitionHideAnswer
Test whether the function $f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & \text{if } x \ne 2 \\ 4 & \text{if } x = 2 \end{cases}$ is continuous or discontinuous at $x = 2$ . Explain. [5]
Test whether the function $$f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & \text{if } x \ne 2 \ 4 & \text{if } x = 2 \end{cases}$$ is continuous or discontinuous at $x = 2$. Explain. [5]
$f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2} & \text{if } x \ne 2 \\[2mm] 4 & \text{if } x = 2 \end{cases}$ Point to test: $x = 2$ . A function $f$ is continuous at $x = a$ if: 1. $f(a)$ is defined, 2. $\lim{x \to a} f(x)$ exists, 3.
$$f(x) = \begin{cases} \dfrac{x^2 - 4}{x - 2} & \text{if } x \ne 2 \[2mm] 4 & \text{if } x = 2 \end{cases}$$ Point to test: $x = 2$. A function $f$ is continuous at $x = a$ if: 1. $f(a)$ is defined, 2. $\lim{x \to a} f(x)$ exists, 3.
125 marksDomain and range of functionsHideAnswer
Sketch the graph of $f(x) = \sqrt{x}$ . Also, find the domain and range. [5]
Sketch the graph of $f(x) = \sqrt{x}$. Also, find the domain and range. [5]
The function f(x) = √x is defined only for non-negative real numbers (since we cannot take the square root of negative numbers in the real number system). Domain: [0, ∞) or {x ∈ ℝ x ≥ 0} Since √x produces only non-negative outputs: - Whe...

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Basic Mathematics 2081.2 question paper

Main Question [5 marks]

Main Question [5 marks]

Given Data

Solution

Step 1: Magnitude of the vector

Step 2: Divide by the magnitude

Step 3: Rationalized form

Verification

Final Answer

Given Data

Solution

Step 1: Magnitude of the vector

Step 2: Divide by the magnitude

Step 3: Rationalized form

Verification

Final Answer

Main Question [5 marks]

Given Data

Solution

Step 1: Magnitude of the vector

Step 2: Divide by the magnitude

Step 3: Rationalized form

Verification

Final Answer