110 marksNumericalPrecise definition of LimitAnswer

A function is defined by $f(x) = |x|$

Calculate $f(-3)$ , $f(4)$ , and sketch the graph.

Prove that the limit does not exist. $\lim_{x \to 2} \frac{|x-2|}{x-2}$ [5+0+5]

A function is defined by $f(x) = |x|$

Calculate $f(-3)$, $f(4)$, and sketch the graph.

Prove that the limit does not exist. $$\lim_{x \to 2} \frac{|x-2|}{x-2}$$ [5+0+5]

Function: $f(x) = x$ - Required: $f(-3)$ , $f(4)$ , sketch of $y = x$ - Limit to analyze: $\displaystyle\lim{x \to 2} \frac{x-2}{x-2}$ - Marks split: [5 + 0 + 5] --- The absolute value function is defined piecewise as: $$f(x) = x = \begi...

Function: $f(x) = x$ - Required: $f(-3)$, $f(4)$, sketch of $y = x$ - Limit to analyze: $\displaystyle\lim{x \to 2} \frac{x-2}{x-2}$ - Marks split: [5 + 0 + 5] --- The absolute value function is defined piecewise as: $$f(x) = x = \begi...

25 marksNumericalRange and domain of functions and their GrAnswer

Find the domain and sketch the graph of the function $f(x) = x^2 - 6x$

Estimate the area between the curve $y = x^2$ and the line y = 1 and y = 2. [3+2]

Find the domain and sketch the graph of the function $f(x) = x^2 - 6x$

Estimate the area between the curve $y = x^2$ and the line y = 1 and y = 2. [3+2]

(a) Domain and Sketch of $f(x) = x^2 - 6x$ - Function: $f(x) = x^2 - 6x$ $f(x) = x^2 - 6x$ is a polynomial, defined for all real $x$ . $\text{Domain} = (-\infty, +\infty) = \mathbb{R}$ Intercepts: - $x$ -intercepts:

(a) Domain and Sketch of $f(x) = x^2 - 6x$ - Function: $f(x) = x^2 - 6x$ $f(x) = x^2 - 6x$ is a polynomial, defined for all real $x$. $$\text{Domain} = (-\infty, +\infty) = \mathbb{R}$$ Intercepts: - $x$-intercepts:

310 marksNumericalMultiple integralsAnswer

Question

If $f(x,y) = \frac{y}{x}$ , show that $\lim_{(x,y)\to(0,0)} \frac{f(x,y)}{x}$ does not exist, justify.

Calculate $\iint_R f(x,y) dA$ , for $f(x,y) = 100 - 6x^2y$ , and $R: 0 \leq x \leq 2, -1 \leq y \leq 1$ . [5+5]

Question

If $f(x,y) = \frac{y}{x}$, show that $\lim_{(x,y)\to(0,0)} \frac{f(x,y)}{x}$ does not exist, justify.

Calculate $\iint_R f(x,y) dA$, for $f(x,y) = 100 - 6x^2y$, and $R: 0 \leq x \leq 2, -1 \leq y \leq 1$. [5+5]

$f(x,y) = \dfrac{y}{x}$ - Required: examine $\displaystyle\lim{(x,y)\to(0,0)} \frac{f(x,y)}{x}$ $\frac{f(x,y)}{x} = \frac{y/x}{x} = \frac{y}{x^2}$ For a limit to exist, it must yield the same finite value along every path to $(0,0)$ ....

$f(x,y) = \dfrac{y}{x}$ - Required: examine $\displaystyle\lim{(x,y)\to(0,0)} \frac{f(x,y)}{x}$ $$\frac{f(x,y)}{x} = \frac{y/x}{x} = \frac{y}{x^2}$$ For a limit to exist, it must yield the same finite value along every path to $(0,0)$....

410 marksNumericalTaylor's and Maclaurin's seriesAnswer

Find the Maclaurin series for $\cos x$ and prove that it represents $\cos x$ for all x. Define initial value problem. Solve that initial value problem of $y' + 2y = 3$ , $y(0) = 1$ . Find the volume of a sphere of radius $r$ . [4+4+2]

Find the Maclaurin series for $\cos x$ and prove that it represents $\cos x$ for all x. Define initial value problem. Solve that initial value problem of $y' + 2y = 3$, $y(0) = 1$. Find the volume of a sphere of radius $r$. [4+4+2]

Part 1: Function $f(x) = \cos x$ ; find Maclaurin series and prove it represents $\cos x$ for all $x$ . - Part 2: Define IVP; solve $y' + 2y = 3$ , with initial condition $y(0) = 1$ . - Part 3: Find the volume of a sphere of radius $r$ . Al...

Part 1: Function $f(x) = \cos x$; find Maclaurin series and prove it represents $\cos x$ for all $x$. - Part 2: Define IVP; solve $y' + 2y = 3$, with initial condition $y(0) = 1$. - Part 3: Find the volume of a sphere of radius $r$. Al...

55 marksNumericalCombination of functionsAnswer

If $f(x) = \sqrt{2-x}$ and $g(x) = \sqrt{x}$ , find $f \circ f$ and $f \circ g$ . [5]

If $f(x) = \sqrt{2-x}$ and $g(x) = \sqrt{x}$, find $f \circ f$ and $f \circ g$. [5]

$f(x) = \sqrt{2-x}, \qquad g(x) = \sqrt{x}$ Required: $f \circ f$ and $f \circ g$ . --- $(f \circ f)(x) = f(f(x)) = \sqrt{2 - f(x)} = \sqrt{2 - \sqrt{2-x}}$ Domain: Inner root requires: $2 - x \geq 0 \implies x \leq 2$ Outer root re...

$$f(x) = \sqrt{2-x}, \qquad g(x) = \sqrt{x}$$ Required: $f \circ f$ and $f \circ g$. --- $$(f \circ f)(x) = f(f(x)) = \sqrt{2 - f(x)} = \sqrt{2 - \sqrt{2-x}}$$ Domain: Inner root requires: $$2 - x \geq 0 \implies x \leq 2$$ Outer root re...

65 marksNumericalContinuityAnswer

Define continuity on an interval. Show that the function is continuous on the interval $[-1,1]$ . $f(x) = 1 - \sqrt{1 - x^2}$ [5]

Define continuity on an interval. Show that the function is continuous on the interval $[-1,1]$. $f(x) = 1 - \sqrt{1 - x^2}$ [5]

Function: $f(x) = 1 - \sqrt{1 - x^2}$ - Interval: $[-1, 1]$ (the problem writes $[1,-1]$ , which is the closed interval between $-1$ and $1$ ) A function $f$ is continuous on a closed interval $[a, b]$ if: 1. $f$ is continuous at every i...

Function: $f(x) = 1 - \sqrt{1 - x^2}$ - Interval: $[-1, 1]$ (the problem writes $[1,-1]$, which is the closed interval between $-1$ and $1$) A function $f$ is continuous on a closed interval $[a, b]$ if: 1. $f$ is continuous at every i...

75 marksNumericalMean value theoremAnswer

Verify Mean value theorem of $f(x) = x^3 - 3x + 2$ for [-1,2]. [5]

Mean Value Theorem Verification for $f(x) = x^3 - 3x + 2$ on $[-1, 2]$

Given Data

Function: $f(x) = x^3 - 3x + 2$
Interval: $[a, b] = [-1, 2]$ , so $a = -1$ , $b = 2$

Statement of MVT

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then there exists at least one $c \in (a,b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}$

Step 1: Verify Conditions

$f(x) = x^3 - 3x + 2$ is a polynomial.

Polynomials are continuous everywhere, so $f$ is continuous on $[-1, 2]$ . ✓
Polynomials are differentiable everywhere, so $f$ is differentiable on $(-1, 2)$ . ✓

Both conditions hold, so MVT applies.

Step 2: Compute $f(a)$ and $f(b)$

$f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4$

$f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4$

Step 3: Compute the Slope of the Chord

$\frac{f(b) - f(a)}{b - a} = \frac{4 - 4}{2 - (-1)} = \frac{0}{3} = 0$

Step 4: Differentiate and Solve $f'(c) = 0$

$f'(x) = 3x^2 - 3$

Set $f'(c) = 0$ :

$3c^2 - 3 = 0 \implies c^2 = 1 \implies c = \pm 1$

Step 5: Select $c$ in $(-1, 2)$

$c = -1$ is an endpoint, not in the open interval $(-1, 2)$ .
$c = 1$ lies in $(-1, 2)$ . ✓

Conclusion

There exists $c = 1 \in (-1, 2)$ such that:

$f'(1) = 3(1)^2 - 3 = 0 = \frac{f(2) - f(-1)}{2 - (-1)}$

Hence, the Mean Value Theorem is verified for $f(x) = x^3 - 3x + 2$ on $[-1, 2]$ , with $c = 1$ .

Mean Value Theorem Verification for $f(x) = x^3 - 3x + 2$ on $[-1, 2]$

Given Data

Function: $f(x) = x^3 - 3x + 2$
Interval: $[a, b] = [-1, 2]$, so $a = -1$, $b = 2$

Statement of MVT

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c \in (a,b)$ such that:

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

Step 1: Verify Conditions

$f(x) = x^3 - 3x + 2$ is a polynomial.

Polynomials are continuous everywhere, so $f$ is continuous on $[-1, 2]$. ✓
Polynomials are differentiable everywhere, so $f$ is differentiable on $(-1, 2)$. ✓

Both conditions hold, so MVT applies.

Step 2: Compute $f(a)$ and $f(b)$

$$f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4$$

$$f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4$$

Step 3: Compute the Slope of the Chord

$$\frac{f(b) - f(a)}{b - a} = \frac{4 - 4}{2 - (-1)} = \frac{0}{3} = 0$$

Step 4: Differentiate and Solve $f'(c) = 0$

$$f'(x) = 3x^2 - 3$$

Set $f'(c) = 0$:

$$3c^2 - 3 = 0 \implies c^2 = 1 \implies c = \pm 1$$

Step 5: Select $c$ in $(-1, 2)$

$c = -1$ is an endpoint, not in the open interval $(-1, 2)$.
$c = 1$ lies in $(-1, 2)$. ✓

Conclusion

There exists $c = 1 \in (-1, 2)$ such that:

$$f'(1) = 3(1)^2 - 3 = 0 = \frac{f(2) - f(-1)}{2 - (-1)}$$

Hence, the Mean Value Theorem is verified for $f(x) = x^3 - 3x + 2$ on $[-1, 2]$, with $c = 1$.

85 marksNumericalNewton's methodAnswer

Starting with $x_1 = 2$ , find the third approximation $x_3$ to the root of the equation $x^3 - 2x - 5 = 0$ . [5]

Starting with $x_1 = 2$, find the third approximation $x_3$ to the root of the equation $x^3 - 2x - 5 = 0$. [5]

Equation: $f(x) = x^3 - 2x - 5$ - Initial approximation: $x1 = 2$ - Required: third approximation $x3$ Iteration formula: $x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 3x^2 - 2$ $f(2) = 8 - 4 - 5 = -1$ $$f'(2) = 3(4) - 2 = 10...

Equation: $f(x) = x^3 - 2x - 5$ - Initial approximation: $x1 = 2$ - Required: third approximation $x3$ Iteration formula: $$x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 3x^2 - 2$$ $$f(2) = 8 - 4 - 5 = -1$$ $$f'(2) = 3(4) - 2 = 10...

95 marksNumericalImproper integralsAnswer

Evaluate $\int_0^{\infty} x^3 \sqrt{1-x^4} dx$ [5]

Integrand: $f(x) = x^3\sqrt{1-x^4}$ - Limits: from $0$ to $\infty$ For the square root $\sqrt{1-x^4}$ to be real, we need: $1 - x^4 \geq 0 \implies x^4 \leq 1 \implies x \leq 1$ On $[0,\infty)$ this means the integrand is real only f...

Integrand: $f(x) = x^3\sqrt{1-x^4}$ - Limits: from $0$ to $\infty$ For the square root $\sqrt{1-x^4}$ to be real, we need: $$1 - x^4 \geq 0 \implies x^4 \leq 1 \implies x \leq 1$$ On $[0,\infty)$ this means the integrand is real only f...

105 marksNumericalVolumes of cylindrical cellsAnswer

Find the volume of the resulting solid which is enclosed by the curve $y = x$ and $y = x^2$ , is rotated about the x-axis. [5]

Find the volume of the resulting solid which is enclosed by the curve $y = x$ and $y = x^2$, is rotated about the x-axis. [5]

Curves: $y = x$ and $y = x^2$ - Axis of rotation: the x-axis $x = x^2 \implies x^2 - x = 0 \implies x(x-1) = 0$ $\therefore x = 0, \quad x = 1$ On $[0,1]$ , since $x \geq x^2$ , the line $y = x$ is the upper (outer) curve and

Curves: $y = x$ and $y = x^2$ - Axis of rotation: the x-axis $$x = x^2 \implies x^2 - x = 0 \implies x(x-1) = 0$$ $$\therefore x = 0, \quad x = 1$$ On $[0,1]$, since $x \geq x^2$, the line $y = x$ is the upper (outer) curve and

115 marksNumericalConvergence tests and power seriesAnswer

Determine whether the series converges or diverges $\sum_{n=1}^{\infty} \frac{n^2}{5n^2+4}$ [5]

Convergence/Divergence of $\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}$

Step 1 - Given Data

Series: $\displaystyle \sum_{n=1}^{\infty} a_n$ where $a_n = \dfrac{n^2}{5n^2 + 4}$

Step 2 - Solve

Test to apply: Divergence Test (nth Term Test)

Theorem: If $\lim_{n \to \infty} a_n \neq 0$ (or does not exist), then $\sum a_n$ diverges.

Compute the limit of the general term:

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n^2}{5n^2 + 4}$

Divide numerator and denominator by $n^2$ :

$= \lim_{n \to \infty} \frac{1}{5 + \dfrac{4}{n^2}}$

As $n \to \infty$ , $\dfrac{4}{n^2} \to 0$ :

$= \frac{1}{5 + 0} = \frac{1}{5}$

Apply the Divergence Test:

$\lim_{n \to \infty} a_n = \frac{1}{5} \neq 0$

Conclusion

Since $\lim_{n \to \infty} \dfrac{n^2}{5n^2+4} = \dfrac{1}{5} \neq 0$ , by the Divergence Test the series

$\sum_{n=1}^{\infty} \frac{n^2}{5n^2+4} \quad \textbf{diverges.}$

Convergence/Divergence of $\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}$

Step 1 - Given Data

Series: $\displaystyle \sum_{n=1}^{\infty} a_n$ where $a_n = \dfrac{n^2}{5n^2 + 4}$

Step 2 - Solve

Test to apply: Divergence Test (nth Term Test)

Theorem: If $\lim_{n \to \infty} a_n \neq 0$ (or does not exist), then $\sum a_n$ diverges.

Compute the limit of the general term:

$$\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n^2}{5n^2 + 4}$$

Divide numerator and denominator by $n^2$:

$$= \lim_{n \to \infty} \frac{1}{5 + \dfrac{4}{n^2}}$$

As $n \to \infty$, $\dfrac{4}{n^2} \to 0$:

$$= \frac{1}{5 + 0} = \frac{1}{5}$$

Apply the Divergence Test:

$$\lim_{n \to \infty} a_n = \frac{1}{5} \neq 0$$

Conclusion

Since $\lim_{n \to \infty} \dfrac{n^2}{5n^2+4} = \dfrac{1}{5} \neq 0$, by the Divergence Test the series

$$\sum_{n=1}^{\infty} \frac{n^2}{5n^2+4} \quad \textbf{diverges.}$$

125 marksNumericalDot product and cross ProductAnswer

If $a = (4, 0, 3)$ and $b = (-2, 1, 5)$ , find $|a|$ , the vector $a - b$ and $2a + b$ . [5]

If $a = (4, 0, 3)$ and $b = (-2, 1, 5)$, find $|a|$, the vector $a - b$ and $2a + b$. [5]

$\vec{a} = (4, 0, 3) \qquad \vec{b} = (-2, 1, 5)$ --- $\vec{a} = \sqrt{4^2 + 0^2 + 3^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5$ $\boxed{\vec{a} = 5}$ --- $\vec{a} - \vec{b} = (4-(-2),\ 0-1,\ 3-5) = (6,\ -1,\ -2)$ $$\boxed{\vec{a} - ...

$$\vec{a} = (4, 0, 3) \qquad \vec{b} = (-2, 1, 5)$$ --- $$\vec{a} = \sqrt{4^2 + 0^2 + 3^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5$$ $$\boxed{\vec{a} = 5}$$ --- $$\vec{a} - \vec{b} = (4-(-2),\ 0-1,\ 3-5) = (6,\ -1,\ -2)$$ $$\boxed{\vec{a} - ...

135 marksNumericalPartial derivativesAnswer

Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ if z is defined as a function of x and y by the equation $x^3+y^3+z^3+6xyz=1$ . [5]

Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ if z is defined as a function of x and y by the equation $x^3+y^3+z^3+6xyz=1$. [5]

Equation defining $z$ implicitly as a function of $x$ and $y$ : $x^3 + y^3 + z^3 + 6xyz = 1$ Required: $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$ . Define $F(x,y,z) = x^3 + y^3 + z^3 + 6xyz - 1 = 0$ By the im...

Equation defining $z$ implicitly as a function of $x$ and $y$: $$x^3 + y^3 + z^3 + 6xyz = 1$$ Required: $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$. Define $$F(x,y,z) = x^3 + y^3 + z^3 + 6xyz - 1 = 0$$ By the im...

145 marksNumericalMaximum and minimum valuesAnswer

Find the extreme values of the function $f(x,y) = x^2 + 2y^2$ on the circle $x^2 + y^2 = 1$ . [5]

Find the extreme values of the function $f(x,y) = x^2 + 2y^2$ on the circle $x^2 + y^2 = 1$. [5]

Objective function: $f(x,y) = x^2 + 2y^2$ - Constraint: $g(x,y) = x^2 + y^2 - 1 = 0$ Set $\nabla f = \lambda \nabla g$ . $fx = 2x,\quad fy = 4y,\qquad gx = 2x,\quad gy = 2y$ System of equations: $$ 2x = \lambda(2x) ;\Rightarrow; 2x(...

Objective function: $f(x,y) = x^2 + 2y^2$ - Constraint: $g(x,y) = x^2 + y^2 - 1 = 0$ Set $\nabla f = \lambda \nabla g$. $$fx = 2x,\quad fy = 4y,\qquad gx = 2x,\quad gy = 2y$$ System of equations: $$ 2x = \lambda(2x) ;\Rightarrow; 2x(...

155 marksNumericalSecond order linear differential equationsAnswer

Find the solution of y′′+4y′+4=0y'' + 4y' + 4 = 0y′′+4y′+4=0. [5]

Differential equation: $y'' + 4y' + 4 = 0$ - Coefficients (interpreting as homogeneous constant-coefficient ODE): coefficient of $y''$ is $1$ , of $y'$ is $4$ , of $y$ is $4$ . Note on interpretation: As literally written,

Differential equation: $y'' + 4y' + 4 = 0$ - Coefficients (interpreting as homogeneous constant-coefficient ODE): coefficient of $y''$ is $1$, of $y'$ is $4$, of $y$ is $4$. Note on interpretation: As literally written,

Mathematics I 2075 question paper

Mean Value Theorem Verification for f(x)=x3−3x+2f(x) = x^3 - 3x + 2f(x)=x3−3x+2 on [−1,2][-1, 2][−1,2]

Given Data

Statement of MVT

Step 1: Verify Conditions

Step 2: Compute f(a)f(a)f(a) and f(b)f(b)f(b)

Step 3: Compute the Slope of the Chord

Step 4: Differentiate and Solve f′(c)=0f'(c) = 0f′(c)=0

Step 5: Select ccc in (−1,2)(-1, 2)(−1,2)

Conclusion

Mean Value Theorem Verification for $f(x) = x^3 - 3x + 2$ on $[-1, 2]$

Given Data

Statement of MVT

Step 1: Verify Conditions

Step 2: Compute $f(a)$ and $f(b)$

Step 3: Compute the Slope of the Chord

Step 4: Differentiate and Solve $f'(c) = 0$

Step 5: Select $c$ in $(-1, 2)$

Conclusion

Convergence/Divergence of ∑n=1∞n25n2+4\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}∑n=1∞​5n2+4n2​

Step 1 - Given Data

Step 2 - Solve

Conclusion

Convergence/Divergence of $\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}$

Step 1 - Given Data

Step 2 - Solve

Conclusion

Mean Value Theorem Verification for f(x)=x3−3x+2f(x) = x^3 - 3x + 2f(x)=x3−3x+2 on [−1,2][-1, 2][−1,2]

Given Data

Statement of MVT

Step 1: Verify Conditions

Step 2: Compute f(a)f(a)f(a) and f(b)f(b)f(b)

Step 3: Compute the Slope of the Chord

Step 4: Differentiate and Solve f′(c)=0f'(c) = 0f′(c)=0

Step 5: Select ccc in (−1,2)(-1, 2)(−1,2)

Conclusion

Convergence/Divergence of ∑n=1∞n25n2+4\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}∑n=1∞​5n2+4n2​

Step 1 - Given Data

Step 2 - Solve

Conclusion

Mean Value Theorem Verification for $f(x) = x^3 - 3x + 2$ on $[-1, 2]$

Step 2: Compute $f(a)$ and $f(b)$

Step 4: Differentiate and Solve $f'(c) = 0$

Step 5: Select $c$ in $(-1, 2)$

Convergence/Divergence of $\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}$

Mean Value Theorem Verification for $f(x) = x^3 - 3x + 2$ on $[-1, 2]$

Step 2: Compute $f(a)$ and $f(b)$

Step 4: Differentiate and Solve $f'(c) = 0$

Step 5: Select $c$ in $(-1, 2)$

Convergence/Divergence of $\sum_{n=1}^{\infty} \dfrac{n^2}{5n^2+4}$