MTH117 · Exam intelligence

Mathematics I important questions

From 7 past TU papers: which questions keep coming back, how much they carry, and what is most likely to show up next. Every question links to a model answer.

Most likely Most repeated

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Ranked by how often a topic is asked, its marks weight, and whether it is due after skipping the 2081 paper. No guarantees; study the whole syllabus.

1asked 4xavg 36 marks · due (skipped 2081) · Linear mathematical model

Answer

As dry air moves upward, it expands and cools. If the ground temperature is $20°C$ and the temperature at height of 1 km is $10°C$ , express the temperature $T$ (in $°C$ ) as a function of the height $h$ (in kilometer), assuming that linear model is appropriate.

(a) Draw a graph of the function in part (b). What does the slope represent?

[5+5]

As dry air moves upward, it expands and cools. If the ground temperature is $20°C$ and the temperature at height of 1 km is $10°C$, express the temperature $T$ (in $°C$) as a function of the height $h$ (in kilometer), assuming that linear model is appropriate.

(a) Draw a graph of the function in part (b). What does the slope represent?

[5+5]

Temperature as a Linear Function of Height

STEP 1 - Given Data

Ground level: $h = 0$ km, $T = 20^{\circ}C$ → point $(0, 20)$
At $h = 1$ km: $T = 10^{\circ}C$ → point $(1, 10)$
Linear model assumed: $T = mh + b$
Required: temperature at $h = 2.5$ km

STEP 2 - Solution

Building the Linear Model

Slope:

$m = \frac{10 - 20}{1 - 0} = \frac{-10}{1} = -10$

Intercept (from point $(0,20)$ ):

$b = 20$

Therefore:

$\boxed{T(h) = 20 - 10h}$

with $T$ in $^{\circ}C$ and $h$ in km.

Part (a): Graph and Slope Meaning

The graph is a straight line through $(0,20)$ with negative slope $-10$ :

T (°C)
 20 |* (0,20)
    |  \
 10 |    * (1,10)
    |      \
  0 |________*___(2,0)_____ h (km)
    |          \
 -5 |            * (2.5,-5)
    +----+----+----+----+---
    0    1    2   2.5   3

Key points: $(0,20),\ (1,10),\ (2,0),\ (2.5,-5)$ .

Slope interpretation: The slope $m = -10\ ^{\circ}C/\text{km}$ is the rate of change of temperature with height. For each 1 km rise in altitude, the temperature falls by $10^{\circ}C$ (the lapse rate).

Part (c): Temperature at $h = 2.5$ km

$T(2.5) = 20 - 10(2.5) = 20 - 25 = -5$

$\boxed{T(2.5) = -5^{\circ}C}$

Final Results

$T(h) = 20 - 10h$
Slope $= -10^{\circ}C/\text{km}$ (rate of temperature decrease per km)
$T(2.5\text{ km}) = -5^{\circ}C$

Temperature as a Linear Function of Height

STEP 1 - Given Data

Ground level: $h = 0$ km, $T = 20^{\circ}C$ → point $(0, 20)$
At $h = 1$ km: $T = 10^{\circ}C$ → point $(1, 10)$
Linear model assumed: $T = mh + b$
Required: temperature at $h = 2.5$ km

STEP 2 - Solution

Building the Linear Model

Slope:

$$m = \frac{10 - 20}{1 - 0} = \frac{-10}{1} = -10$$

Intercept (from point $(0,20)$):

$$b = 20$$

Therefore:

$$\boxed{T(h) = 20 - 10h}$$

with $T$ in $^{\circ}C$ and $h$ in km.

Part (a): Graph and Slope Meaning

The graph is a straight line through $(0,20)$ with negative slope $-10$:

T (°C)
 20 |* (0,20)
    |  \
 10 |    * (1,10)
    |      \
  0 |________*___(2,0)_____ h (km)
    |          \
 -5 |            * (2.5,-5)
    +----+----+----+----+---
    0    1    2   2.5   3

Key points: $(0,20),\ (1,10),\ (2,0),\ (2.5,-5)$.

Part (c): Temperature at $h = 2.5$ km

$$T(2.5) = 20 - 10(2.5) = 20 - 25 = -5$$

$$\boxed{T(2.5) = -5^{\circ}C}$$

Final Results

$T(h) = 20 - 10h$
Slope $= -10^{\circ}C/\text{km}$ (rate of temperature decrease per km)
$T(2.5\text{ km}) = -5^{\circ}C$

2asked 5xavg 6 marks · due (skipped 2081) · Convergence tests and power series

Answer

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 $a_n = \dfrac{n^2}{5n^2+4}$
Task: determine convergence or divergence.

STEP 2 - Solve

Method: nth-Term Divergence Test.

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

Compute the limit of the general term. Divide numerator and denominator by $n^2$ :

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

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

\lim_{n\to\infty} a_n = \frac{1}{5+0} = \frac{1}{5}.

Apply the test.

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

Since the necessary condition for convergence ( $\lim a_n = 0$ ) fails, the series cannot converge.

Conclusion

\boxed{\sum_{n=1}^{\infty} \frac{n^2}{5n^2+4} \text{ diverges (by the nth-term / Divergence Test).}}

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 $a_n = \dfrac{n^2}{5n^2+4}$
Task: determine convergence or divergence.

STEP 2 - Solve

Method: nth-Term Divergence Test.

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

Compute the limit of the general term. Divide numerator and denominator by $n^2$:

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

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

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

Apply the test.

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

Since the necessary condition for convergence ($\lim a_n = 0$) fails, the series cannot converge.

Conclusion

$$ \boxed{\sum_{n=1}^{\infty} \frac{n^2}{5n^2+4} \text{ diverges (by the nth-term / Divergence Test).}} $$

3asked 5xavg 5 marks · due (skipped 2081) · Mean value theorem

Answer

State Rolle's theorem and verify the theorem for $f(x) = x^2 - 9$ , $x \in [-3,3]$ . [5]

State Rolle's theorem and verify the theorem for $f(x) = x^2 - 9$, $x \in [-3,3]$. [5]

Rolle's Theorem: Statement and Verification

Given Data

Function: $f(x) = x^2 - 9$
Interval: $[a, b] = [-3, 3]$

Statement of Rolle's Theorem

If a function $f$ satisfies:

$f$ is continuous on the closed interval $[a, b]$ ,
$f$ is differentiable on the open interval $(a, b)$ ,
$f(a) = f(b)$ ,

then there exists at least one point $c \in (a, b)$ such that

$f'(c) = 0.$

Verification for $f(x) = x^2 - 9$ on $[-3, 3]$

Condition 1: Continuity on $[-3, 3]$

$f(x) = x^2 - 9$ is a polynomial. Polynomials are continuous everywhere, hence $f$ is continuous on $[-3, 3]$ . ✓

Condition 2: Differentiability on $(-3, 3)$

$f'(x) = 2x$

This exists for all real $x$ , so $f$ is differentiable on $(-3, 3)$ . ✓

Condition 3: Equal endpoint values

$f(-3) = (-3)^2 - 9 = 9 - 9 = 0$ $f(3) = (3)^2 - 9 = 9 - 9 = 0$ $\therefore f(-3) = f(3) = 0 \checkmark$

All three conditions hold.

Finding $c$

Set $f'(c) = 0$ :

$2c = 0 \implies c = 0$

Check: $c = 0 \in (-3, 3)$ . ✓

Conclusion

Since all three conditions of Rolle's theorem are satisfied and there exists $c = 0 \in (-3, 3)$ with $f'(c) = 0$ , Rolle's theorem is verified for $f(x) = x^2 - 9$ on $[-3, 3]$ .

Rolle's Theorem: Statement and Verification

Given Data

Function: $f(x) = x^2 - 9$
Interval: $[a, b] = [-3, 3]$

Statement of Rolle's Theorem

If a function $f$ satisfies:

$f$ is continuous on the closed interval $[a, b]$,
$f$ is differentiable on the open interval $(a, b)$,
$f(a) = f(b)$,

then there exists at least one point $c \in (a, b)$ such that

$$f'(c) = 0.$$

Verification for $f(x) = x^2 - 9$ on $[-3, 3]$

Condition 1: Continuity on $[-3, 3]$

$f(x) = x^2 - 9$ is a polynomial. Polynomials are continuous everywhere, hence $f$ is continuous on $[-3, 3]$. ✓

Condition 2: Differentiability on $(-3, 3)$

$$f'(x) = 2x$$

This exists for all real $x$, so $f$ is differentiable on $(-3, 3)$. ✓

Condition 3: Equal endpoint values

$$f(-3) = (-3)^2 - 9 = 9 - 9 = 0$$ $$f(3) = (3)^2 - 9 = 9 - 9 = 0$$ $$\therefore f(-3) = f(3) = 0 \checkmark$$

All three conditions hold.

Finding $c$

Set $f'(c) = 0$:

$$2c = 0 \implies c = 0$$

Check: $c = 0 \in (-3, 3)$. ✓

Conclusion

Since all three conditions of Rolle's theorem are satisfied and there exists $c = 0 \in (-3, 3)$ with $f'(c) = 0$, Rolle's theorem is verified for $f(x) = x^2 - 9$ on $[-3, 3]$.

4asked 5xavg 5 marks · due (skipped 2081) · Dot product and cross Product

Answer

Find the angle between the vectors $a = (2, 2, -1)$ and $b = (1, 3, 2)$ . [5]

Find the angle between the vectors $a = (2, 2, -1)$ and $b = (1, 3, 2)$. [5]

$\mathbf{a} = (2, 2, -1), \qquad \mathbf{b} = (1, 3, 2)$ Formula for the angle $\theta$ : $\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a}\,\mathbf{b}}$ --- $$\mathbf{a} \cdot \mathbf{b} = (2)(1) + (2)(3) + (-1)(2) = 2 + 6 ...

$$\mathbf{a} = (2, 2, -1), \qquad \mathbf{b} = (1, 3, 2)$$ Formula for the angle $\theta$: $$\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{a},\mathbf{b}}$$ --- $$\mathbf{a} \cdot \mathbf{b} = (2)(1) + (2)(3) + (-1)(2) = 2 + 6 ...

5asked 6xavg 5 marks · Partial derivatives

Answer

If $f(x,y) = 2x^3 + x^2y^2 - y^4$ , find $f_x(1,-2)$ , $f_y(1,-1)$ and $f_{yx}(1,-1)$ . [5]

If $f(x,y) = 2x^3 + x^2y^2 - y^4$, find $f_x(1,-2)$, $f_y(1,-1)$ and $f_{yx}(1,-1)$. [5]

Given Data

$f(x, y) = 2x^3 + x^2y^2 - y^4$

Required:

$f_x(1, -2)$
$f_y(1, -1)$
$f_{yx}(1, -1)$

Step 1: Compute $f_x$

Differentiate with respect to $x$ (treat $y$ constant):

$f_x = 6x^2 + 2xy^2 - 0 = 6x^2 + 2xy^2$

Evaluate at $(1, -2)$ :

$f_x(1,-2) = 6(1)^2 + 2(1)(-2)^2 = 6 + 2(4) = 6 + 8 = 14$

Step 2: Compute $f_y$

Differentiate with respect to $y$ (treat $x$ constant):

$f_y = 0 + 2x^2y - 4y^3 = 2x^2y - 4y^3$

Evaluate at $(1, -1)$ :

$f_y(1,-1) = 2(1)^2(-1) - 4(-1)^3 = -2 - 4(-1) = -2 + 4 = 2$

Step 3: Compute $f_{yx}$

Differentiate $f_y = 2x^2y - 4y^3$ with respect to $x$ :

$f_{yx} = 4xy - 0 = 4xy$

Evaluate at $(1, -1)$ :

$f_{yx}(1,-1) = 4(1)(-1) = -4$

Summary

Expression	Value
$f_x(1,-2)$	$14$
$f_y(1,-1)$	$2$
$f_{yx}(1,-1)$	$-4$

All computations verified.

Given Data

$$f(x, y) = 2x^3 + x^2y^2 - y^4$$

Required:

$f_x(1, -2)$
$f_y(1, -1)$
$f_{yx}(1, -1)$

Step 1: Compute $f_x$

Differentiate with respect to $x$ (treat $y$ constant):

$$f_x = 6x^2 + 2xy^2 - 0 = 6x^2 + 2xy^2$$

Evaluate at $(1, -2)$:

$$f_x(1,-2) = 6(1)^2 + 2(1)(-2)^2 = 6 + 2(4) = 6 + 8 = 14$$

Step 2: Compute $f_y$

Differentiate with respect to $y$ (treat $x$ constant):

$$f_y = 0 + 2x^2y - 4y^3 = 2x^2y - 4y^3$$

Evaluate at $(1, -1)$:

$$f_y(1,-1) = 2(1)^2(-1) - 4(-1)^3 = -2 - 4(-1) = -2 + 4 = 2$$

Step 3: Compute $f_{yx}$

Differentiate $f_y = 2x^2y - 4y^3$ with respect to $x$:

$$f_{yx} = 4xy - 0 = 4xy$$

Evaluate at $(1, -1)$:

$$f_{yx}(1,-1) = 4(1)(-1) = -4$$

Summary

Expression	Value
$f_x(1,-2)$	$14$
$f_y(1,-1)$	$2$
$f_{yx}(1,-1)$	$-4$

All computations verified.

Most repeated questions

Topics asked at least twice, most-asked first.

asked 6xavg 5 marks · 2081, 2080, 2079, 2078, 2077...

Answer

If $f(x,y) = 2x^3 + x^2y^2 - y^4$ , find $f_x(1,-2)$ , $f_y(1,-1)$ and $f_{yx}(1,-1)$ . [5]

If $f(x,y) = 2x^3 + x^2y^2 - y^4$, find $f_x(1,-2)$, $f_y(1,-1)$ and $f_{yx}(1,-1)$. [5]

Given Data

$f(x, y) = 2x^3 + x^2y^2 - y^4$

Required:

$f_x(1, -2)$
$f_y(1, -1)$
$f_{yx}(1, -1)$

Step 1: Compute $f_x$

Differentiate with respect to $x$ (treat $y$ constant):

$f_x = 6x^2 + 2xy^2 - 0 = 6x^2 + 2xy^2$

Evaluate at $(1, -2)$ :

$f_x(1,-2) = 6(1)^2 + 2(1)(-2)^2 = 6 + 2(4) = 6 + 8 = 14$

Step 2: Compute $f_y$

Differentiate with respect to $y$ (treat $x$ constant):

$f_y = 0 + 2x^2y - 4y^3 = 2x^2y - 4y^3$

Evaluate at $(1, -1)$ :

$f_y(1,-1) = 2(1)^2(-1) - 4(-1)^3 = -2 - 4(-1) = -2 + 4 = 2$

Step 3: Compute $f_{yx}$

Differentiate $f_y = 2x^2y - 4y^3$ with respect to $x$ :

$f_{yx} = 4xy - 0 = 4xy$

Evaluate at $(1, -1)$ :

$f_{yx}(1,-1) = 4(1)(-1) = -4$

Summary

Expression	Value
$f_x(1,-2)$	$14$
$f_y(1,-1)$	$2$
$f_{yx}(1,-1)$	$-4$

All computations verified.

Given Data

$$f(x, y) = 2x^3 + x^2y^2 - y^4$$

Required:

$f_x(1, -2)$
$f_y(1, -1)$
$f_{yx}(1, -1)$

Step 1: Compute $f_x$

Differentiate with respect to $x$ (treat $y$ constant):

$$f_x = 6x^2 + 2xy^2 - 0 = 6x^2 + 2xy^2$$

Evaluate at $(1, -2)$:

$$f_x(1,-2) = 6(1)^2 + 2(1)(-2)^2 = 6 + 2(4) = 6 + 8 = 14$$

Step 2: Compute $f_y$

Differentiate with respect to $y$ (treat $x$ constant):

$$f_y = 0 + 2x^2y - 4y^3 = 2x^2y - 4y^3$$

Evaluate at $(1, -1)$:

$$f_y(1,-1) = 2(1)^2(-1) - 4(-1)^3 = -2 - 4(-1) = -2 + 4 = 2$$

Step 3: Compute $f_{yx}$

Differentiate $f_y = 2x^2y - 4y^3$ with respect to $x$:

$$f_{yx} = 4xy - 0 = 4xy$$

Evaluate at $(1, -1)$:

$$f_{yx}(1,-1) = 4(1)(-1) = -4$$

Summary

Expression	Value
$f_x(1,-2)$	$14$
$f_y(1,-1)$	$2$
$f_{yx}(1,-1)$	$-4$

All computations verified.

asked 5xavg 6 marks · 2080, 2077, 2075, 2074

Answer

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 $a_n = \dfrac{n^2}{5n^2+4}$
Task: determine convergence or divergence.

STEP 2 - Solve

Method: nth-Term Divergence Test.

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

Compute the limit of the general term. Divide numerator and denominator by $n^2$ :

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

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

\lim_{n\to\infty} a_n = \frac{1}{5+0} = \frac{1}{5}.

Apply the test.

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

Since the necessary condition for convergence ( $\lim a_n = 0$ ) fails, the series cannot converge.

Conclusion

\boxed{\sum_{n=1}^{\infty} \frac{n^2}{5n^2+4} \text{ diverges (by the nth-term / Divergence Test).}}

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 $a_n = \dfrac{n^2}{5n^2+4}$
Task: determine convergence or divergence.

STEP 2 - Solve

Method: nth-Term Divergence Test.

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

Compute the limit of the general term. Divide numerator and denominator by $n^2$:

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

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

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

Apply the test.

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

Since the necessary condition for convergence ($\lim a_n = 0$) fails, the series cannot converge.

Conclusion

$$ \boxed{\sum_{n=1}^{\infty} \frac{n^2}{5n^2+4} \text{ diverges (by the nth-term / Divergence Test).}} $$

asked 5xavg 5 marks · 2079, 2078, 2077, 2075, 2074

Answer

State Rolle's theorem and verify the theorem for $f(x) = x^2 - 9$ , $x \in [-3,3]$ . [5]

State Rolle's theorem and verify the theorem for $f(x) = x^2 - 9$, $x \in [-3,3]$. [5]

Rolle's Theorem: Statement and Verification

Given Data

Function: $f(x) = x^2 - 9$
Interval: $[a, b] = [-3, 3]$

Statement of Rolle's Theorem

If a function $f$ satisfies:

$f$ is continuous on the closed interval $[a, b]$ ,
$f$ is differentiable on the open interval $(a, b)$ ,
$f(a) = f(b)$ ,

then there exists at least one point $c \in (a, b)$ such that

$f'(c) = 0.$

Verification for $f(x) = x^2 - 9$ on $[-3, 3]$

Condition 1: Continuity on $[-3, 3]$

$f(x) = x^2 - 9$ is a polynomial. Polynomials are continuous everywhere, hence $f$ is continuous on $[-3, 3]$ . ✓

Condition 2: Differentiability on $(-3, 3)$

$f'(x) = 2x$

This exists for all real $x$ , so $f$ is differentiable on $(-3, 3)$ . ✓

Condition 3: Equal endpoint values

$f(-3) = (-3)^2 - 9 = 9 - 9 = 0$ $f(3) = (3)^2 - 9 = 9 - 9 = 0$ $\therefore f(-3) = f(3) = 0 \checkmark$

All three conditions hold.

Finding $c$

Set $f'(c) = 0$ :

$2c = 0 \implies c = 0$

Check: $c = 0 \in (-3, 3)$ . ✓

Conclusion

Since all three conditions of Rolle's theorem are satisfied and there exists $c = 0 \in (-3, 3)$ with $f'(c) = 0$ , Rolle's theorem is verified for $f(x) = x^2 - 9$ on $[-3, 3]$ .

Rolle's Theorem: Statement and Verification

Given Data

Function: $f(x) = x^2 - 9$
Interval: $[a, b] = [-3, 3]$

Statement of Rolle's Theorem

If a function $f$ satisfies:

$f$ is continuous on the closed interval $[a, b]$,
$f$ is differentiable on the open interval $(a, b)$,
$f(a) = f(b)$,

then there exists at least one point $c \in (a, b)$ such that

$$f'(c) = 0.$$

Verification for $f(x) = x^2 - 9$ on $[-3, 3]$

Condition 1: Continuity on $[-3, 3]$

$f(x) = x^2 - 9$ is a polynomial. Polynomials are continuous everywhere, hence $f$ is continuous on $[-3, 3]$. ✓

Condition 2: Differentiability on $(-3, 3)$

$$f'(x) = 2x$$

This exists for all real $x$, so $f$ is differentiable on $(-3, 3)$. ✓

Condition 3: Equal endpoint values

$$f(-3) = (-3)^2 - 9 = 9 - 9 = 0$$ $$f(3) = (3)^2 - 9 = 9 - 9 = 0$$ $$\therefore f(-3) = f(3) = 0 \checkmark$$

All three conditions hold.

Finding $c$

Set $f'(c) = 0$:

$$2c = 0 \implies c = 0$$

Check: $c = 0 \in (-3, 3)$. ✓

Conclusion

Since all three conditions of Rolle's theorem are satisfied and there exists $c = 0 \in (-3, 3)$ with $f'(c) = 0$, Rolle's theorem is verified for $f(x) = x^2 - 9$ on $[-3, 3]$.

asked 5xavg 5 marks · 2079, 2078, 2077, 2075, 2074

Answer

Find the angle between the vectors $a = (2, 2, -1)$ and $b = (1, 3, 2)$ . [5]

Find the angle between the vectors $a = (2, 2, -1)$ and $b = (1, 3, 2)$. [5]

asked 4xavg 36 marks · 2080, 2079, 2078, 2077

Answer

(a) Draw a graph of the function in part (b). What does the slope represent?

[5+5]

As dry air moves upward, it expands and cools. If the ground temperature is $20°C$ and the temperature at height of 1 km is $10°C$, express the temperature $T$ (in $°C$) as a function of the height $h$ (in kilometer), assuming that linear model is appropriate.

(a) Draw a graph of the function in part (b). What does the slope represent?

[5+5]

Temperature as a Linear Function of Height

STEP 1 - Given Data

Ground level: $h = 0$ km, $T = 20^{\circ}C$ → point $(0, 20)$
At $h = 1$ km: $T = 10^{\circ}C$ → point $(1, 10)$
Linear model assumed: $T = mh + b$
Required: temperature at $h = 2.5$ km

STEP 2 - Solution

Building the Linear Model

Slope:

$m = \frac{10 - 20}{1 - 0} = \frac{-10}{1} = -10$

Intercept (from point $(0,20)$ ):

$b = 20$

Therefore:

$\boxed{T(h) = 20 - 10h}$

with $T$ in $^{\circ}C$ and $h$ in km.

Part (a): Graph and Slope Meaning

The graph is a straight line through $(0,20)$ with negative slope $-10$ :

T (°C)
 20 |* (0,20)
    |  \
 10 |    * (1,10)
    |      \
  0 |________*___(2,0)_____ h (km)
    |          \
 -5 |            * (2.5,-5)
    +----+----+----+----+---
    0    1    2   2.5   3

Key points: $(0,20),\ (1,10),\ (2,0),\ (2.5,-5)$ .

Part (c): Temperature at $h = 2.5$ km

$T(2.5) = 20 - 10(2.5) = 20 - 25 = -5$

$\boxed{T(2.5) = -5^{\circ}C}$

Final Results

$T(h) = 20 - 10h$
Slope $= -10^{\circ}C/\text{km}$ (rate of temperature decrease per km)
$T(2.5\text{ km}) = -5^{\circ}C$

Temperature as a Linear Function of Height

STEP 1 - Given Data

Ground level: $h = 0$ km, $T = 20^{\circ}C$ → point $(0, 20)$
At $h = 1$ km: $T = 10^{\circ}C$ → point $(1, 10)$
Linear model assumed: $T = mh + b$
Required: temperature at $h = 2.5$ km

STEP 2 - Solution

Building the Linear Model

Slope:

$$m = \frac{10 - 20}{1 - 0} = \frac{-10}{1} = -10$$

Intercept (from point $(0,20)$):

$$b = 20$$

Therefore:

$$\boxed{T(h) = 20 - 10h}$$

with $T$ in $^{\circ}C$ and $h$ in km.

Part (a): Graph and Slope Meaning

The graph is a straight line through $(0,20)$ with negative slope $-10$:

T (°C)
 20 |* (0,20)
    |  \
 10 |    * (1,10)
    |      \
  0 |________*___(2,0)_____ h (km)
    |          \
 -5 |            * (2.5,-5)
    +----+----+----+----+---
    0    1    2   2.5   3

Key points: $(0,20),\ (1,10),\ (2,0),\ (2.5,-5)$.

Part (c): Temperature at $h = 2.5$ km

$$T(2.5) = 20 - 10(2.5) = 20 - 25 = -5$$

$$\boxed{T(2.5) = -5^{\circ}C}$$

Final Results

$T(h) = 20 - 10h$
Slope $= -10^{\circ}C/\text{km}$ (rate of temperature decrease per km)
$T(2.5\text{ km}) = -5^{\circ}C$

asked 4xavg 6 marks · 2079, 2077, 2075, 2074

Answer

Define initial value problem. Solve: $y'' + 4y' - 6y = 0$ , $y(0) = 1$ , $y'(0) = 0$ . Find the Taylor's series expansion for $\cos x$ at $x = 0$ . [10+0]

Define initial value problem. Solve: $y'' + 4y' - 6y = 0$, $y(0) = 1$, $y'(0) = 0$. Find the Taylor's series expansion for $\cos x$ at $x = 0$. [10+0]

Initial Value Problem: Definition and Solutions

Given Data

ODE: $y'' + 4y' - 6y = 0$
Initial conditions: $y(0) = 1$ , $y'(0) = 0$
Second task: Taylor (Maclaurin) series of $\cos x$ at $x = 0$

Definition of Initial Value Problem

An initial value problem (IVP) is a differential equation together with the values of the unknown function and its derivatives specified at a single point $x_0$ (the initial point). For a second-order ODE:

$y'' = f(x, y, y'), \quad y(x_0) = y_0, \quad y'(x_0) = y_1$

The constants $y_0, y_1$ are the initial conditions, and the aim is to find the particular solution satisfying both the ODE and these conditions.

Solving $y'' + 4y' - 6y = 0$

Step 1: Auxiliary Equation

Assume $y = e^{mx}$ :

$m^2 + 4m - 6 = 0$

Step 2: Roots

$m = \frac{-4 \pm \sqrt{16 + 24}}{2} = \frac{-4 \pm \sqrt{40}}{2} = -2 \pm \sqrt{10}$

Real, distinct roots: $m_1 = -2 + \sqrt{10}$ , $m_2 = -2 - \sqrt{10}$ .

Step 3: General Solution

$y = C_1 e^{(-2+\sqrt{10})x} + C_2 e^{(-2-\sqrt{10})x}$

Step 4: Apply Initial Conditions

$y(0) = 1$ : $C_1 + C_2 = 1 \quad (i)$

Differentiate: $y' = C_1(-2+\sqrt{10})e^{(-2+\sqrt{10})x} + C_2(-2-\sqrt{10})e^{(-2-\sqrt{10})x}$

$y'(0) = 0$ : $C_1(-2+\sqrt{10}) + C_2(-2-\sqrt{10}) = 0$ $-2(C_1+C_2) + \sqrt{10}(C_1 - C_2) = 0$

Using $(i)$ : $-2 + \sqrt{10}(C_1 - C_2) = 0 \;\Rightarrow\; C_1 - C_2 = \frac{2}{\sqrt{10}} = \frac{\sqrt{10}}{5} \quad (ii)$

Step 5: Solve

Adding $(i)$ and $(ii)$ : $2C_1 = 1 + \frac{\sqrt{10}}{5} = \frac{5+\sqrt{10}}{5} \;\Rightarrow\; C_1 = \frac{5+\sqrt{10}}{10}$

Subtracting: $2C_2 = 1 - \frac{\sqrt{10}}{5} \;\Rightarrow\; C_2 = \frac{5-\sqrt{10}}{10}$

Particular Solution

$\boxed{y = \frac{5+\sqrt{10}}{10}\, e^{(-2+\sqrt{10})x} + \frac{5-\sqrt{10}}{10}\, e^{(-2-\sqrt{10})x}}$

Taylor (Maclaurin) Series of $\cos x$ at $x = 0$

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$

$n$	$f^{(n)}(x)$	$f^{(n)}(0)$
0	$\cos x$	$1$
1	$-\sin x$	$0$
2	$-\cos x$	$-1$
3	$\sin x$	$0$
4	$\cos x$	$1$

Substituting:

$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$

$\boxed{\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\, x^{2n}}$

Verified results: roots $-2 \pm \sqrt{10}$ , constants $\frac{5\pm\sqrt{10}}{10}$ , and the standard cosine series.

Initial Value Problem: Definition and Solutions

Given Data

ODE: $y'' + 4y' - 6y = 0$
Initial conditions: $y(0) = 1$, $y'(0) = 0$
Second task: Taylor (Maclaurin) series of $\cos x$ at $x = 0$

Definition of Initial Value Problem

$$y'' = f(x, y, y'), \quad y(x_0) = y_0, \quad y'(x_0) = y_1$$

The constants $y_0, y_1$ are the initial conditions, and the aim is to find the particular solution satisfying both the ODE and these conditions.

Solving $y'' + 4y' - 6y = 0$

Step 1: Auxiliary Equation

Assume $y = e^{mx}$:

$$m^2 + 4m - 6 = 0$$

Step 2: Roots

$$m = \frac{-4 \pm \sqrt{16 + 24}}{2} = \frac{-4 \pm \sqrt{40}}{2} = -2 \pm \sqrt{10}$$

Real, distinct roots: $m_1 = -2 + \sqrt{10}$, $m_2 = -2 - \sqrt{10}$.

Step 3: General Solution

$$y = C_1 e^{(-2+\sqrt{10})x} + C_2 e^{(-2-\sqrt{10})x}$$

Step 4: Apply Initial Conditions

$y(0) = 1$: $$C_1 + C_2 = 1 \quad (i)$$

Differentiate: $$y' = C_1(-2+\sqrt{10})e^{(-2+\sqrt{10})x} + C_2(-2-\sqrt{10})e^{(-2-\sqrt{10})x}$$

$y'(0) = 0$: $$C_1(-2+\sqrt{10}) + C_2(-2-\sqrt{10}) = 0$$ $$-2(C_1+C_2) + \sqrt{10}(C_1 - C_2) = 0$$

Using $(i)$: $$-2 + \sqrt{10}(C_1 - C_2) = 0 ;\Rightarrow; C_1 - C_2 = \frac{2}{\sqrt{10}} = \frac{\sqrt{10}}{5} \quad (ii)$$

Step 5: Solve

Adding $(i)$ and $(ii)$: $$2C_1 = 1 + \frac{\sqrt{10}}{5} = \frac{5+\sqrt{10}}{5} ;\Rightarrow; C_1 = \frac{5+\sqrt{10}}{10}$$

Subtracting: $$2C_2 = 1 - \frac{\sqrt{10}}{5} ;\Rightarrow; C_2 = \frac{5-\sqrt{10}}{10}$$

Particular Solution

$$\boxed{y = \frac{5+\sqrt{10}}{10}, e^{(-2+\sqrt{10})x} + \frac{5-\sqrt{10}}{10}, e^{(-2-\sqrt{10})x}}$$

Taylor (Maclaurin) Series of $\cos x$ at $x = 0$

$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$

$n$	$f^{(n)}(x)$	$f^{(n)}(0)$
0	$\cos x$	$1$
1	$-\sin x$	$0$
2	$-\cos x$	$-1$
3	$\sin x$	$0$
4	$\cos x$	$1$

Substituting:

$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$

$$\boxed{\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}, x^{2n}}$$

Verified results: roots $-2 \pm \sqrt{10}$, constants $\frac{5\pm\sqrt{10}}{10}$, and the standard cosine series.

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Answer

Show the integral coverages $\int_0^3 \frac{dx}{x-1}$ . [5]

Show the integral coverages $\int_0^3 \frac{dx}{x-1}$. [5]

Integral: $\displaystyle\int0^3 \frac{dx}{x-1}$ - Integrand: $f(x) = \dfrac{1}{x-1}$ - Limits: lower $= 0$ , upper $= 3$ The denominator vanishes when $x - 1 = 0$ , i.e. at $x = 1$ . Since $1 \in (0,3)$ , the integrand has an infinite disc...

Integral: $\displaystyle\int0^3 \frac{dx}{x-1}$ - Integrand: $f(x) = \dfrac{1}{x-1}$ - Limits: lower $= 0$, upper $= 3$ The denominator vanishes when $x - 1 = 0$, i.e. at $x = 1$. Since $1 \in (0,3)$, the integrand has an infinite disc...

asked 4xavg 5 marks · 2080, 2077, 2075, 2074

Answer

Show that the function $f(x) = x^2 + \sqrt{7-x}$ is continuous at $x=4$ . [5]

Show that the function $f(x) = x^2 + \sqrt{7-x}$ is continuous at $x=4$. [5]

Function: $f(x) = x^2 + \sqrt{7 - x}$ - Point: $x = 4$ $f(x)$ is continuous at $x = a$ if all three hold: 1. $f(a)$ is defined 2. $\lim{x \to a} f(x)$ exists 3. $\lim{x \to a} f(x) = f(a)$ Here $a = 4$ . --- $$f(4) = 4^2 + \sqrt{7 - 4} ...

Function: $f(x) = x^2 + \sqrt{7 - x}$ - Point: $x = 4$ $f(x)$ is continuous at $x = a$ if all three hold: 1. $f(a)$ is defined 2. $\lim{x \to a} f(x)$ exists 3. $\lim{x \to a} f(x) = f(a)$ Here $a = 4$. --- $$f(4) = 4^2 + \sqrt{7 - 4} ...

asked 4xavg 5 marks · 2079, 2078, 2077, 2075

Answer

Starting with $x_1 = 1$ , find the third approximate $x_3$ to the root of the equation $x^3 - x - 5 = 0$ . [5]

Starting with $x_1 = 1$, find the third approximate $x_3$ to the root of the equation $x^3 - x - 5 = 0$. [5]

Newton-Raphson Method: Third Approximation for $x^3 - x - 5 = 0$

STEP 1 - EXTRACT (Given data)

Equation: $f(x) = x^3 - x - 5 = 0$
Starting value: $x_1 = 1$
Required: $x_3$ (third approximation)
Method implied: Newton-Raphson

STEP 2 - SOLVE

Iteration formula: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \qquad f'(x) = 3x^2 - 1$

Iteration 1: Find $x_2$

At $x_1 = 1$ : $f(1) = 1 - 1 - 5 = -5$ $f'(1) = 3(1)^2 - 1 = 2$

$x_2 = 1 - \frac{-5}{2} = 1 + 2.5 = 3.5$

Iteration 2: Find $x_3$

At $x_2 = 3.5$ : $f(3.5) = (3.5)^3 - 3.5 - 5 = 42.875 - 8.5 = 34.375$ $f'(3.5) = 3(3.5)^2 - 1 = 36.75 - 1 = 35.75$

$x_3 = 3.5 - \frac{34.375}{35.75} = 3.5 - 0.96154 = 2.53846$

Result

$\boxed{x_3 \approx 2.5385}$

Newton-Raphson Method: Third Approximation for $x^3 - x - 5 = 0$

STEP 1 - EXTRACT (Given data)

Equation: $f(x) = x^3 - x - 5 = 0$
Starting value: $x_1 = 1$
Required: $x_3$ (third approximation)
Method implied: Newton-Raphson

STEP 2 - SOLVE

Iteration formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}, \qquad f'(x) = 3x^2 - 1$$

Iteration 1: Find $x_2$

At $x_1 = 1$: $$f(1) = 1 - 1 - 5 = -5$$ $$f'(1) = 3(1)^2 - 1 = 2$$

$$x_2 = 1 - \frac{-5}{2} = 1 + 2.5 = 3.5$$

Iteration 2: Find $x_3$

At $x_2 = 3.5$: $$f(3.5) = (3.5)^3 - 3.5 - 5 = 42.875 - 8.5 = 34.375$$ $$f'(3.5) = 3(3.5)^2 - 1 = 36.75 - 1 = 35.75$$

$$x_3 = 3.5 - \frac{34.375}{35.75} = 3.5 - 0.96154 = 2.53846$$

Result

$$\boxed{x_3 \approx 2.5385}$$

asked 4xavg 6 marks · 2078, 2077, 2075, 2074

Answer

Question

If $f(x) = \sqrt{x}$ and $g(x) = \sqrt{3-x}$ , then find $f \circ g$ and its domain and range.

A rectangular storage container with an open top has a volume of $20 \text{ m}^3$ . The length of its base is twice its width. Material for the base costs Rs 10 per square meter; material for the sides costs Rs 4 per square meter. Express the cost of materials as a function of the width of the base.

[5+5]

Question

If $f(x) = \sqrt{x}$ and $g(x) = \sqrt{3-x}$, then find $f \circ g$ and its domain and range.

A rectangular storage container with an open top has a volume of $20 \text{ m}^3$. The length of its base is twice its width. Material for the base costs Rs 10 per square meter; material for the sides costs Rs 4 per square meter. Express the cost of materials as a function of the width of the base.

[5+5]

(a) Finding fog, its Domain and Range

Given data

$f(x) = \sqrt{x}, \qquad g(x) = \sqrt{3-x}$

Computing fog

$fog(x) = f(g(x)) = f\left(\sqrt{3-x}\right) = \sqrt{\sqrt{3-x}} = (3-x)^{1/4}$

$\boxed{fog(x) = (3-x)^{1/4}}$

Domain

Domain analysis via composition:

$g(x) = \sqrt{3-x}$ requires $3 - x \geq 0 \Rightarrow x \leq 3$ , so $D_g = (-\infty, 3]$ .
$f(x) = \sqrt{x}$ requires $x \geq 0$ , so $D_f = [0,\infty)$ .
For $fog$ we need $x \in D_g$ and $g(x) \in D_f$ . Since $g(x) = \sqrt{3-x} \geq 0$ wherever it is defined, the second condition is automatically satisfied.

$\therefore\ D_{fog} = (-\infty, 3]$

Range

On $(-\infty, 3]$ :

At $x = 3$ : $(3-3)^{1/4} = 0$ .
As $x \to -\infty$ : $(3-x)^{1/4} \to +\infty$ .

The function is continuous and decreasing, so it attains all values in $[0, \infty)$ .

$\therefore\ R_{fog} = [0, +\infty)$

(b) Cost as a Function of Width

Given data

Volume $V = 20\ \text{m}^3$
Length $= 2 \times$ width
Base cost $= \text{Rs } 10/\text{m}^2$
Side cost $= \text{Rs } 4/\text{m}^2$
Open top

Variables

Let width $= w$ , length $= 2w$ , height $= h$ .

Volume constraint

$V = (2w)(w)(h) = 2w^2 h = 20 \Rightarrow h = \frac{10}{w^2}$

Areas

Base area: $A_{\text{base}} = 2w \cdot w = 2w^2$

Sides (4 walls, open top):

Two of size $2w \times h$ : $2(2wh) = 4wh$
Two of size $w \times h$ : $2(wh) = 2wh$

$A_{\text{sides}} = 4wh + 2wh = 6wh$

Cost

$C = 10(2w^2) + 4(6wh) = 20w^2 + 24wh$

Substitute $h = \dfrac{10}{w^2}$ :

$C(w) = 20w^2 + 24w\cdot\frac{10}{w^2} = 20w^2 + \frac{240}{w}$

$\boxed{C(w) = 20w^2 + \frac{240}{w}, \quad w > 0}$

(a) Finding fog, its Domain and Range

Given data

$$f(x) = \sqrt{x}, \qquad g(x) = \sqrt{3-x}$$

Computing fog

$$fog(x) = f(g(x)) = f\left(\sqrt{3-x}\right) = \sqrt{\sqrt{3-x}} = (3-x)^{1/4}$$

$$\boxed{fog(x) = (3-x)^{1/4}}$$

Domain

Domain analysis via composition:

$g(x) = \sqrt{3-x}$ requires $3 - x \geq 0 \Rightarrow x \leq 3$, so $D_g = (-\infty, 3]$.
$f(x) = \sqrt{x}$ requires $x \geq 0$, so $D_f = [0,\infty)$.
For $fog$ we need $x \in D_g$ and $g(x) \in D_f$. Since $g(x) = \sqrt{3-x} \geq 0$ wherever it is defined, the second condition is automatically satisfied.

$$\therefore\ D_{fog} = (-\infty, 3]$$

Range

On $(-\infty, 3]$:

At $x = 3$: $(3-3)^{1/4} = 0$.
As $x \to -\infty$: $(3-x)^{1/4} \to +\infty$.

The function is continuous and decreasing, so it attains all values in $[0, \infty)$.

$$\therefore\ R_{fog} = [0, +\infty)$$

(b) Cost as a Function of Width

Given data

Volume $V = 20\ \text{m}^3$
Length $= 2 \times$ width
Base cost $= \text{Rs } 10/\text{m}^2$
Side cost $= \text{Rs } 4/\text{m}^2$
Open top

Variables

Let width $= w$, length $= 2w$, height $= h$.

Volume constraint

$$V = (2w)(w)(h) = 2w^2 h = 20 \Rightarrow h = \frac{10}{w^2}$$

Areas

Base area: $$A_{\text{base}} = 2w \cdot w = 2w^2$$

Sides (4 walls, open top):

Two of size $2w \times h$: $2(2wh) = 4wh$
Two of size $w \times h$: $2(wh) = 2wh$

$$A_{\text{sides}} = 4wh + 2wh = 6wh$$

Cost

$$C = 10(2w^2) + 4(6wh) = 20w^2 + 24wh$$

Substitute $h = \dfrac{10}{w^2}$:

$$C(w) = 20w^2 + 24w\cdot\frac{10}{w^2} = 20w^2 + \frac{240}{w}$$

$$\boxed{C(w) = 20w^2 + \frac{240}{w}, \quad w > 0}$$

asked 4xavg 5 marks · 2078, 2077, 2075, 2074

Answer

Find the local maximum and minimum values, saddle points of $f(x,y) = x^4 + y^4 - 4xy + 1$ . [5]

Find the local maximum and minimum values, saddle points of $f(x,y) = x^4 + y^4 - 4xy + 1$. [5]

Local Maxima, Minima, and Saddle Points of $f(x,y) = x^4 + y^4 - 4xy + 1$

Step 1: Given Data

Function: $f(x,y) = x^4 + y^4 - 4xy + 1$

Step 2: Find Critical Points

f_x = 4x^3 - 4y = 0 \Rightarrow y = x^3 \tag{1}

f_y = 4y^3 - 4x = 0 \Rightarrow x = y^3 \tag{2}

Substitute (1) into (2): $x = (x^3)^3 = x^9 \Rightarrow x^9 - x = 0 \Rightarrow x(x^8 - 1) = 0$

So $x = 0$ or $x = \pm 1$ .

$x = 0 \Rightarrow y = 0$ : point $(0,0)$
$x = 1 \Rightarrow y = 1$ : point $(1,1)$
$x = -1 \Rightarrow y = -1$ : point $(-1,-1)$

Step 3: Second Derivative Test

$f_{xx} = 12x^2, \quad f_{yy} = 12y^2, \quad f_{xy} = -4$ $D = f_{xx}f_{yy} - (f_{xy})^2 = 144x^2y^2 - 16$

At $(0,0)$ : $D = 144(0) - 16 = -16 < 0 \Rightarrow \text{Saddle point}$

At $(1,1)$ : $D = 144(1)(1) - 16 = 128 > 0, \quad f_{xx} = 12 > 0 \Rightarrow \text{Local minimum}$ $f(1,1) = 1 + 1 - 4 + 1 = -1$

At $(-1,-1)$ : $D = 144(1)(1) - 16 = 128 > 0, \quad f_{xx} = 12 > 0 \Rightarrow \text{Local minimum}$ $f(-1,-1) = 1 + 1 - 4(-1)(-1) + 1 = 1 + 1 - 4 + 1 = -1$

Summary

Point	$D$	$f_{xx}$	Conclusion
$(0,0)$	$-16$	--	Saddle point
$(1,1)$	$128$	$12$	Local minimum, $f = -1$
$(-1,-1)$	$128$	$12$	Local minimum, $f = -1$

Local minimum value: $-1$ at $(1,1)$ and $(-1,-1)$ Saddle point: $(0,0)$ No local maximum exists.

Local Maxima, Minima, and Saddle Points of $f(x,y) = x^4 + y^4 - 4xy + 1$

Step 1: Given Data

Function: $f(x,y) = x^4 + y^4 - 4xy + 1$

Step 2: Find Critical Points

$$ f_x = 4x^3 - 4y = 0 \Rightarrow y = x^3 \tag{1} $$

$$ f_y = 4y^3 - 4x = 0 \Rightarrow x = y^3 \tag{2} $$

Substitute (1) into (2): $$x = (x^3)^3 = x^9 \Rightarrow x^9 - x = 0 \Rightarrow x(x^8 - 1) = 0$$

So $x = 0$ or $x = \pm 1$.

$x = 0 \Rightarrow y = 0$: point $(0,0)$
$x = 1 \Rightarrow y = 1$: point $(1,1)$
$x = -1 \Rightarrow y = -1$: point $(-1,-1)$

Step 3: Second Derivative Test

$$f_{xx} = 12x^2, \quad f_{yy} = 12y^2, \quad f_{xy} = -4$$ $$D = f_{xx}f_{yy} - (f_{xy})^2 = 144x^2y^2 - 16$$

At $(0,0)$: $$D = 144(0) - 16 = -16 < 0 \Rightarrow \text{Saddle point}$$

At $(1,1)$: $$D = 144(1)(1) - 16 = 128 > 0, \quad f_{xx} = 12 > 0 \Rightarrow \text{Local minimum}$$ $$f(1,1) = 1 + 1 - 4 + 1 = -1$$

At $(-1,-1)$: $$D = 144(1)(1) - 16 = 128 > 0, \quad f_{xx} = 12 > 0 \Rightarrow \text{Local minimum}$$ $$f(-1,-1) = 1 + 1 - 4(-1)(-1) + 1 = 1 + 1 - 4 + 1 = -1$$

Summary

Point	$D$	$f_{xx}$	Conclusion
$(0,0)$	$-16$	--	Saddle point
$(1,1)$	$128$	$12$	Local minimum, $f = -1$
$(-1,-1)$	$128$	$12$	Local minimum, $f = -1$

Local minimum value: $-1$ at $(1,1)$ and $(-1,-1)$ Saddle point: $(0,0)$ No local maximum exists.

asked 4xavg 8 marks · 2081, 2080, 2075, 2074

Answer

Find the Maclaurin series expansion of $f(x) = \sin x$ for all x. [5]

The Maclaurin series of a function f(x) is the Taylor series expanded about x = 0, given by: $f(x) = f(0) + f'(0)\cdot x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots$ --- Let f(x) = sin x....

The Maclaurin series of a function f(x) is the Taylor series expanded about x = 0, given by: $$f(x) = f(0) + f'(0)\cdot x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots$$ --- Let f(x) = sin x....

asked 4xavg 5 marks · 2081, 2078, 2077, 2075

Answer

Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve $y = \sqrt{x}$ for $0$ to $1$ . [5]

Use cylindrical shells to find the volume of the solid obtained by rotating about the x-axis the region under the curve $y = \sqrt{x}$ for $0$ to $1$. [5]

Curve: $y = \sqrt{x}$ - Region: under the curve from $x = 0$ to $x = 1$ - Axis of rotation: the x-axis - Method required: cylindrical shells The method of cylindrical shells uses strips parallel to the axis of rotation. Since we rotate...

Curve: $y = \sqrt{x}$ - Region: under the curve from $x = 0$ to $x = 1$ - Axis of rotation: the x-axis - Method required: cylindrical shells The method of cylindrical shells uses strips parallel to the axis of rotation. Since we rotate...

asked 3xavg 10 marks · 2079, 2075, 2074

Answer

If a function is defined by $f(x) = \begin{cases} 1 + x, & x \leq -1 \\ x^2, & x > -1 \end{cases}$ , evaluate $f(-3)$ , $f(-1)$ and $f(0)$ and sketch the graph. Prove that $\lim_{x \to 0} \frac{|x|}{x}$ does not exist. [10+0]

If a function is defined by $f(x) = \begin{cases} 1 + x, & x \leq -1 \ x^2, & x > -1 \end{cases}$, evaluate $f(-3)$, $f(-1)$ and $f(0)$ and sketch the graph. Prove that $\lim_{x \to 0} \frac{|x|}{x}$ does not exist. [10+0]

Piecewise Function Evaluation, Graph, and Limit Proof

Given Data

$f(x) = \begin{cases} 1 + x, & x \leq -1 \\ x^2, & x > -1 \end{cases}$

Evaluate $f(-3)$ , $f(-1)$ , $f(0)$ ; sketch graph; prove $\lim_{x\to 0}\frac{|x|}{x}$ does not exist.

Part 1: Evaluations

$f(-3)$ : Since $-3 \leq -1$ , use $1+x$ : $f(-3) = 1 + (-3) = -2$

$f(-1)$ : Since $-1 \leq -1$ , use $1+x$ : $f(-1) = 1 + (-1) = 0$

$f(0)$ : Since $0 > -1$ , use $x^2$ : $f(0) = 0^2 = 0$

Part 2: Graph

Line $y = 1+x$ for $x \leq -1$ :

$x$	$-3$	$-2$	$-1$
$f(x)$	$-2$	$-1$	$0$ (closed)

Parabola $y = x^2$ for $x > -1$ :

$x$	$-1$	$0$	$1$	$2$
$f(x)$	$1$ (open)	$0$	$1$	$4$

At $x = -1$ : left branch gives closed point $(-1, 0)$ ; right branch approaches open point $(-1, 1)$ . The function is discontinuous at $x = -1$ (jump discontinuity of size 1).

f(x)
 4 |                          *
 3 |
 2 |
 1 |              o          *
 0 |            *       *
-1 |          * (line)
-2 |    *   /
   +--+--+--+--+--+--+--+---> x
     -3 -2 -1  0  1  2  3

Line drawn for $x \le -1$ , closed dot at $(-1,0)$
Parabola drawn for $x > -1$ , open dot at $(-1,1)$

Part 3: Prove $\lim_{x\to 0}\frac{|x|}{x}$ does not exist

Recall $|x| = \begin{cases} x, & x > 0 \\ -x, & x < 0 \end{cases}$

RHL ( $x \to 0^+$ ): here $|x| = x$ : $\lim_{x\to 0^+}\frac{|x|}{x} = \lim_{x\to 0^+}\frac{x}{x} = 1$

LHL ( $x \to 0^-$ ): here $|x| = -x$ : $\lim_{x\to 0^-}\frac{|x|}{x} = \lim_{x\to 0^-}\frac{-x}{x} = -1$

Since $\text{LHL} = -1 \neq 1 = \text{RHL}$ , the one-sided limits are unequal.

$\therefore \lim_{x\to 0}\frac{|x|}{x} \text{ does not exist.} \qquad \blacksquare$

Piecewise Function Evaluation, Graph, and Limit Proof

Given Data

$$f(x) = \begin{cases} 1 + x, & x \leq -1 \ x^2, & x > -1 \end{cases}$$

Evaluate $f(-3)$, $f(-1)$, $f(0)$; sketch graph; prove $\lim_{x\to 0}\frac{|x|}{x}$ does not exist.

Part 1: Evaluations

$f(-3)$: Since $-3 \leq -1$, use $1+x$: $$f(-3) = 1 + (-3) = -2$$

$f(-1)$: Since $-1 \leq -1$, use $1+x$: $$f(-1) = 1 + (-1) = 0$$

$f(0)$: Since $0 > -1$, use $x^2$: $$f(0) = 0^2 = 0$$

Part 2: Graph

Line $y = 1+x$ for $x \leq -1$:

$x$	$-3$	$-2$	$-1$
$f(x)$	$-2$	$-1$	$0$ (closed)

Parabola $y = x^2$ for $x > -1$:

$x$	$-1$	$0$	$1$	$2$
$f(x)$	$1$ (open)	$0$	$1$	$4$

At $x = -1$: left branch gives closed point $(-1, 0)$; right branch approaches open point $(-1, 1)$. The function is discontinuous at $x = -1$ (jump discontinuity of size 1).

f(x)
 4 |                          *
 3 |
 2 |
 1 |              o          *
 0 |            *       *
-1 |          * (line)
-2 |    *   /
   +--+--+--+--+--+--+--+---> x
     -3 -2 -1  0  1  2  3

Line drawn for $x \le -1$, closed dot at $(-1,0)$
Parabola drawn for $x > -1$, open dot at $(-1,1)$

Part 3: Prove $\lim_{x\to 0}\frac{|x|}{x}$ does not exist

Recall $|x| = \begin{cases} x, & x > 0 \ -x, & x < 0 \end{cases}$

RHL ($x \to 0^+$): here $|x| = x$: $$\lim_{x\to 0^+}\frac{|x|}{x} = \lim_{x\to 0^+}\frac{x}{x} = 1$$

LHL ($x \to 0^-$): here $|x| = -x$: $$\lim_{x\to 0^-}\frac{|x|}{x} = \lim_{x\to 0^-}\frac{-x}{x} = -1$$

Since $\text{LHL} = -1 \neq 1 = \text{RHL}$, the one-sided limits are unequal.

$$\therefore \lim_{x\to 0}\frac{|x|}{x} \text{ does not exist.} \qquad \blacksquare$$

asked 3xavg 8 marks · 2079, 2077, 2075

Answer

Solve $\int_0^3 \int_1^2 x^2y \, dx \, dy$ [5]

Solve $\int_0^3 \int_1^2 x^2y , dx , dy$ [5]

Integrand: $f(x,y) = x^2 y$ - Inner integral variable: $x$ , limits $[1, 2]$ - Outer integral variable: $y$ , limits $[0, 3]$ Treat $y$ as constant: $$\int1^2 x^2 y , dx = y \left[ \frac{x^3}{3} \right]1^2 = y\left(\frac{8}{3} - \frac{1...

Integrand: $f(x,y) = x^2 y$ - Inner integral variable: $x$, limits $[1, 2]$ - Outer integral variable: $y$, limits $[0, 3]$ Treat $y$ as constant: $$\int1^2 x^2 y , dx = y \left[ \frac{x^3}{3} \right]1^2 = y\left(\frac{8}{3} - \frac{1...

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Temperature as a Linear Function of Height

STEP 1 - Given Data

STEP 2 - Solution

Building the Linear Model

Part (a): Graph and Slope Meaning

Part (c): Temperature at h=2.5h = 2.5h=2.5 km

Final Results

Temperature as a Linear Function of Height

STEP 1 - Given Data

STEP 2 - Solution

Building the Linear Model

Part (a): Graph and Slope Meaning

Part (c): Temperature at $h = 2.5$ km

Final Results

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

Rolle's Theorem: Statement and Verification

Given Data

Statement of Rolle's Theorem

Verification for f(x)=x2−9f(x) = x^2 - 9f(x)=x2−9 on [−3,3][-3, 3][−3,3]

Condition 1: Continuity on [−3,3][-3, 3][−3,3]

Condition 2: Differentiability on (−3,3)(-3, 3)(−3,3)

Condition 3: Equal endpoint values

Finding ccc

Conclusion

Rolle's Theorem: Statement and Verification

Given Data

Statement of Rolle's Theorem

Verification for $f(x) = x^2 - 9$ on $[-3, 3]$

Condition 1: Continuity on $[-3, 3]$

Condition 2: Differentiability on $(-3, 3)$

Condition 3: Equal endpoint values

Finding $c$

Conclusion

Given Data

Step 1: Compute fxf_xfx​

Step 2: Compute fyf_yfy​

Step 3: Compute fyxf_{yx}fyx​

Summary

Given Data

Step 1: Compute $f_x$

Step 2: Compute $f_y$

Step 3: Compute $f_{yx}$

Summary

Most repeated questions

Given Data

Step 1: Compute fxf_xfx​

Step 2: Compute fyf_yfy​

Step 3: Compute fyxf_{yx}fyx​

Summary

Given Data

Step 1: Compute $f_x$

Step 2: Compute $f_y$

Step 3: Compute $f_{yx}$

Summary

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

Rolle's Theorem: Statement and Verification

Given Data

Statement of Rolle's Theorem

Verification for f(x)=x2−9f(x) = x^2 - 9f(x)=x2−9 on [−3,3][-3, 3][−3,3]

Condition 1: Continuity on [−3,3][-3, 3][−3,3]

Condition 2: Differentiability on (−3,3)(-3, 3)(−3,3)

Condition 3: Equal endpoint values

Finding ccc

Conclusion

Part (c): Temperature at $h = 2.5$ km

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Verification for $f(x) = x^2 - 9$ on $[-3, 3]$

Condition 1: Continuity on $[-3, 3]$

Condition 2: Differentiability on $(-3, 3)$

Finding $c$

Step 1: Compute $f_x$

Step 2: Compute $f_y$

Step 3: Compute $f_{yx}$

Step 1: Compute $f_x$

Step 2: Compute $f_y$

Step 3: Compute $f_{yx}$

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

Verification for $f(x) = x^2 - 9$ on $[-3, 3]$

Condition 1: Continuity on $[-3, 3]$

Condition 2: Differentiability on $(-3, 3)$

Finding $c$

Part (c): Temperature at $h = 2.5$ km

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Iteration 1: Find $x_2$

Iteration 2: Find $x_3$