MTH104 · TU past paper
Basic Mathematics 2077 question paper
The complete TU 2077 exam paper for Basic Mathematics (MTH104), all 12 questions with solved model answers written to the mark scheme.
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- 1NumericalEven and odd functionsHideAnswer
Even and Odd Functions
Definition: A function $f(x)$ is even if $f(-x) = f(x)$ for every $x$ in its domain. Symmetry: The graph is symmetric about the y-axis. Example: $f(x) = x^2$ $$f(-x) = (-x)^2 = x^2 = f(x) \checkmark$$ Definition: A function $f(x)$ is odd...
- 210 marksNumericalTaylor seriesHideAnswer
Find the Taylor's series generated by $f(x) = \frac{1}{x}$ at $a = 2$. Where, if anywhere, does the series converge to $\frac{1}{x}$? [10]
Taylor's Series for $f(x) = \frac{1}{x}$ at $a = 2$
Step 1 - Given Data
- Function: $f(x) = \dfrac{1}{x}$
- Center of expansion: $a = 2$
Step 2 - Solve
Derivatives of $f(x) = x^{-1}$
$$f'(x) = -x^{-2}, \quad f''(x) = 2x^{-3}, \quad f'''(x) = -6x^{-4}$$
General form:
$$f^{(n)}(x) = (-1)^n , n! , x^{-(n+1)}$$
Evaluate at $a = 2$
$$f(2) = \frac{1}{2}, \quad f'(2) = -\frac{1}{4}, \quad f''(2) = \frac{2}{8} = \frac{1}{4}, \quad f'''(2) = -\frac{6}{16} = -\frac{3}{8}$$
General coefficient:
$$f^{(n)}(2) = (-1)^n , n! , 2^{-(n+1)}$$
Form the Taylor series
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n = \sum_{n=0}^{\infty} \frac{(-1)^n , n! , 2^{-(n+1)}}{n!}(x-2)^n$$
$$= \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(x-2)^n$$
Writing out the first terms:
$$f(x) = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \frac{1}{16}(x-2)^3 + \cdots$$
This can be written as a geometric series:
$$f(x) = \frac{1}{2}\sum_{n=0}^{\infty}\left(-\frac{x-2}{2}\right)^n$$
Convergence Analysis
Geometric ratio:
$$r = -\frac{x-2}{2}$$
Convergence requires $|r| < 1$:
$$\left|\frac{x-2}{2}\right| < 1 ;\Rightarrow; |x-2| < 2 ;\Rightarrow; 0 < x < 4$$
Sum verification (inside the interval):
$$f(x) = \frac{1}{2}\cdot\frac{1}{1-\left(-\frac{x-2}{2}\right)} = \frac{1}{2}\cdot\frac{1}{1+\frac{x-2}{2}} = \frac{1}{2}\cdot\frac{2}{2+(x-2)} = \frac{1}{x}$$
Endpoints:
- At $x = 0$: $r = 1$, terms do not tend to $0$, series diverges (also $f$ is undefined here).
- At $x = 4$: $r = -1$, series $\frac{1}{2}(1 - 1 + 1 - \cdots)$ diverges.
Final Answer
$$\boxed{\dfrac{1}{x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(x-2)^n = \frac{1}{2} - \frac{1}{4}(x-2) + \frac{1}{8}(x-2)^2 - \cdots}$$
The series converges to $\dfrac{1}{x}$ precisely for $x \in (0, 4)$, i.e. $|x - 2| < 2$, and diverges elsewhere including both endpoints.
- 310 marksNumericalEpsilon-delta definitionHideAnswer
Sketch the graph of the function $f(x) = x^2$
Shifted vertically up to 1 and -2 units and horizontally up to 3 and -2 units.
Find the $\delta$ algebraically for the following functions.
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$\lim_{x \to 5} \sqrt{x-1}$, and $L = 2$, $\epsilon = 1$
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$\lim_{x \to 2} (2x - 2)$, and $L = 6$, $\epsilon = 0.02$
[5+5]
Part 1 (Sketching): Base function $f(x) = x^2$. - Vertical shifts: up 1 unit and down 2 units - Horizontal shifts: right 3 units and left 2 units Part 2 (Finding $\delta$): - Problem 1: $\lim{x \to 5} \sqrt{x-1}$, $L = 2$, $\epsilon = 1$...
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- 45 marksNumericalInitial value problemsHideAnswer
Find the initial value problem in $\frac{dy}{dx} + 2y = 3$, $y(0) = 1$. [5]
- Differential equation: $\dfrac{dy}{dx} + 2y = 3$ - Initial condition: $y(0) = 1$ All required data is present. Identify form: First-order linear ODE $$\frac{dy}{dx} + P(x)y = Q(x), \quad P(x) = 2, \quad Q(x) = 3$$ Integrating factor:
- 55 marksNumericalIntegral testHideAnswer
Determine the convergence or divergence of the series $\sum_{n=1}^{\infty} n^2 e^{-n}$. [5]
Series: $\displaystyle\sum{n=1}^{\infty} n^2 e^{-n}$ General term: $an = n^2 e^{-n} = \dfrac{n^2}{e^n}$ All terms are positive, so tests for positive-term series (Ratio Test, Root Test) apply. For a positive series, compute $$L = \lim{n ...
- 65 marksNumericalContinuous extension of functionsHideAnswer
Show that $f(x)=\frac{x^2+x-6}{x^2-4}$, $x \neq 2$ has a continuous extension to $x=2$ [5]
- Function: $f(x) = \dfrac{x^2 + x - 6}{x^2 - 4}$, defined for $x \neq 2$. - Target point for extension: $x = 2$. Direct substitution: $$f(2) = \frac{2^2 + 2 - 6}{2^2 - 4} = \frac{0}{0}$$ This is indeterminate, so $f$ is not defined at
- 75 marksNumericalLimit evaluation techniquesHideAnswer
Evaluate the following:
$$\lim_{x \to \infty} (x-\sqrt{x^2 + 16})$$
$$\lim_{x \to 1} \left(\frac{\sqrt{6x+10}-5}{x^2}\right)$$
[2.5+2.5]
- Limit 1: $\lim{x \to \infty} (x - \sqrt{x^2 + 16})$ - Limit 2: $\lim{x \to 1} \dfrac{\sqrt{6x+10} - 5}{x^2}$ --- Step 1: Identify the form As $x \to \infty$, this is of the form $\infty - \infty$ (indeterminate). Step 2: Multiply by co...
- 85 marksNumericalDirectional derivativesHideAnswer
Find the derivatives of the function $f(x,y) = x^3 - xy^2 + x^2y - y^3$ at the point $p_0(5,5)$ in the direction of $\vec{u} = 4\vec{i} + 3\vec{j}$. [5]
- Function: $f(x,y) = x^3 - xy^2 + x^2y - y^3$ - Point: $p0(5,5)$ - Direction: $\vec{u} = 4\vec{i} + 3\vec{j}$ $$D{\vec{u}}f = \nabla f \cdot \hat{u}$$ $$\frac{\partial f}{\partial x} = 3x^2 - y^2 + 2xy$$ $$\frac{\partial f}{\partial y} ...
- 95 marksNumericalTrigonometric integralsHideAnswer
Integrate the following: $$\int_{0}^{\frac{\pi}{4}} \frac{dx}{1-\sin x}$$ [5]
- Integrand: $\dfrac{1}{1-\sin x}$ - Limits: from $0$ to $\dfrac{\pi}{4}$ (the LaTeX "x/4" is a rendering artifact for the standard upper limit $\pi/4$) $$I = \int{0}^{\pi/4} \frac{dx}{1-\sin x}$$ Multiply numerator and denominator by th...
- 105 marksNumericalSurface area of revolutionHideAnswer
Find the area of the surface generated by revolving the curve $y = 2\sqrt{x}$, $1 \leq x \leq 2$, about the x-axis. [5]
- Curve: $y = 2\sqrt{x}$ - Interval: $1 \le x \le 2$ - Axis of revolution: x-axis $$S = 2\pi \inta^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} , dx$$ $$y = 2x^{1/2} \implies \frac{dy}{dx} = x^{-1/2} = \frac{1}{\sqrt{x}}$$ $$\left(\frac{...
- 115 marksNumericalConcavity and inflection pointsHideAnswer
Determine the concavity and find the inflection point of the function $f(x) = x^3 - 3x^2 + 2$. [5]
- Function: $f(x) = x^3 - 3x^2 + 2$ $$f'(x) = 3x^2 - 6x$$ $$f''(x) = 6x - 6$$ Set $f''(x) = 0$: $$6x - 6 = 0 \implies x = 1$$ - For $x < 1$ (test $x = 0$): $f''(0) = 6(0) - 6 = -6 < 0$ → concave down on $(-\infty, 1)$ - For $x 1$ (test
- 125 marksNumericalChain ruleHideAnswer
Find $\frac{dy}{dx}$ for:
- $Y = 2u^3$, $u = 8x - 1$
- $Y = \sin u$, $u = x - x\cos x$
[5]
Problem 1: $Y = 2u^3$, $u = 8x - 1$ Problem 2: $Y = \sin u$, $u = x - x\cos x$ Chain rule: $\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}$ --- Step 1: Differentiate $Y$ with respect to $u$ $$\frac{dy}{du} = 2 \cdot 3u^2 = 6u^2$$ S...