Mathematics (4300001) - Winter 2022 Solution

Q.1 [14 marks]

Fill in the blanks using appropriate choice from the given options

Q1.1 [1 mark]

If $\begin{vmatrix} x & 8 \\ 2 & 4 \end{vmatrix} = 0$ then the value of $x$ is ____

Answer: c. 8

Solution: $\begin{vmatrix} x & 8 \\ 2 & 4 \end{vmatrix} = x(4) - 8(2) = 4x - 16$

Given: $4x - 16 = 0$ $4x = 16$ $x = 4$

Wait, let me recalculate: If the determinant is 0, then $4x - 16 = 0$, so $x = 4$. But 4 is option a, not c. Let me verify the options again... The answer should be a. 4

Q1.2 [1 mark]

$\begin{vmatrix} 2 & -9 & 1 \\ 5 & -8 & 4 \\ 0 & 3 & 0 \end{vmatrix} =$ ____

Answer: a. -9

Solution: Expanding along the third row (which has two zeros): $\begin{vmatrix} 2 & -9 & 1 \\ 5 & -8 & 4 \\ 0 & 3 & 0 \end{vmatrix} = 0 - 3 \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix} + 0$

$= -3(2 \times 4 - 1 \times 5) = -3(8 - 5) = -3(3) = -9$

Q1.3 [1 mark]

If $f(x) = \log x$ then $f(1) =$ ____

Answer: a. 0

Solution: $f(x) = \log x$ $f(1) = \log 1 = 0$

Q1.4 [1 mark]

$\log x + \log(\frac{1}{x}) =$ ____

Answer: a. 0

Solution: $\log x + \log(\frac{1}{x}) = \log x + \log x^{-1} = \log x + (-1)\log x = \log x - \log x = 0$

Q1.5 [1 mark]

$120° =$ _____ radian

Answer: b. $\frac{2\pi}{3}$

Solution: $120° = 120 \times \frac{\pi}{180} = \frac{120\pi}{180} = \frac{2\pi}{3}$ radians

Q1.6 [1 mark]

$\sin^{-1}(\sin \frac{\pi}{6}) =$ _____

Answer: c. $\frac{\pi}{6}$

Solution: Since $\frac{\pi}{6}$ lies in the principal range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ of $\sin^{-1}$: $\sin^{-1}(\sin \frac{\pi}{6}) = \frac{\pi}{6}$

Q1.7 [1 mark]

The principal period of $\tan \theta$ is _____

Answer: b. $\pi$

Solution: The principal period of $\tan \theta$ is $\pi$.

Q1.8 [1 mark]

$|2i - j + 2k| =$ ____

Answer: a. 3

Solution: $|2i - j + 2k| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

Q1.9 [1 mark]

$i \cdot i =$ ____

Answer: a. 1

Solution: The dot product of a unit vector with itself: $i \cdot i = |i|^2 = 1^2 = 1$

Q1.10 [1 mark]

The slope of line $x - 4 = 0$ is ______

Answer: d. Not Defined

Solution: The line $x - 4 = 0$ or $x = 4$ is a vertical line. The slope of a vertical line is undefined (not defined).

Q1.11 [1 mark]

The center of circle $x^2 + y^2 = 4$ is

Answer: c. $(0,0)$

Solution: Comparing with standard form $(x - h)^2 + (y - k)^2 = r^2$: $x^2 + y^2 = 4$ has center $(0, 0)$ and radius $2$.

Q1.12 [1 mark]

$\lim_{x \to 2} \frac{x^4 - 16}{x - 2} =$ ____

Answer: c. 32

Solution: $\lim_{x \to 2} \frac{x^4 - 16}{x - 2} = \lim_{x \to 2} \frac{x^4 - 2^4}{x - 2}$

This is of the form $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$

$= 4 \times 2^3 = 4 \times 8 = 32$

Q1.13 [1 mark]

$\lim_{n \to 0} (1 + n)^{\frac{1}{n}} =$ ____

Answer: d. $e$

Solution: This is the definition of $e$: $\lim_{n \to 0} (1 + n)^{\frac{1}{n}} = e$

Q1.14 [1 mark]

$\lim_{x \to 0} \frac{\sin 6x}{3x} =$ ____

Answer: c. 2

Solution: $\lim_{x \to 0} \frac{\sin 6x}{3x} = \lim_{x \to 0} \frac{\sin 6x}{6x} \times \frac{6x}{3x} = 1 \times 2 = 2$

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Q.2(A) [6 marks]

Attempt any two

Q2.1 [3 marks]

If $\begin{vmatrix} 2 & 6 & 4 \\ -1 & x & 0 \\ 5 & 9 & -2 \end{vmatrix} = 0$ then find $x$

Answer:

Solution: Expanding along the second row: $\begin{vmatrix} 2 & 6 & 4 \\ -1 & x & 0 \\ 5 & 9 & -2 \end{vmatrix} = -(-1) \begin{vmatrix} 6 & 4 \\ 9 & -2 \end{vmatrix} - x \begin{vmatrix} 2 & 4 \\ 5 & -2 \end{vmatrix} + 0$

$= 1(6 \times (-2) - 4 \times 9) - x(2 \times (-2) - 4 \times 5)$ $= 1(-12 - 36) - x(-4 - 20)$ $= -48 - x(-24)$ $= -48 + 24x$

Given: $-48 + 24x = 0$ $24x = 48$ $x = 2$

Q2.2 [3 marks]

If $f(x) = \tan x$ then prove that (i) $f(x+y) = \frac{f(x) + f(y)}{1 - f(x)f(y)}$, (ii) $f(2x) = \frac{2f(x)}{1 - [f(x)]^2}$

Answer:

Solution: Given: $f(x) = \tan x$

(i) Prove $f(x+y) = \frac{f(x) + f(y)}{1 - f(x)f(y)}$

LHS: $f(x+y) = \tan(x+y)$

Using the tangent addition formula: $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} = \frac{f(x) + f(y)}{1 - f(x)f(y)}$ = RHS

(ii) Prove $f(2x) = \frac{2f(x)}{1 - [f(x)]^2}$

LHS: $f(2x) = \tan(2x)$

Using the double angle formula: $\tan(2x) = \frac{2\tan x}{1 - \tan^2 x} = \frac{2f(x)}{1 - [f(x)]^2}$ = RHS

Q2.3 [3 marks]

Prove that $\frac{\sin 3A - \cos 3A}{\sin A - \cos A} = 2$

Answer:

Solution: Using the identities: $\sin 3A = 3\sin A - 4\sin^3 A = \sin A(3 - 4\sin^2 A)$ $\cos 3A = 4\cos^3 A - 3\cos A = \cos A(4\cos^2 A - 3)$

$\frac{\sin 3A - \cos 3A}{\sin A - \cos A} = \frac{\sin A(3 - 4\sin^2 A) - \cos A(4\cos^2 A - 3)}{\sin A - \cos A}$

$= \frac{3\sin A - 4\sin^3 A - 4\cos^3 A + 3\cos A}{\sin A - \cos A}$

$= \frac{3(\sin A + \cos A) - 4(\sin^3 A + \cos^3 A)}{\sin A - \cos A}$

Using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$: $\sin^3 A + \cos^3 A = (\sin A + \cos A)(\sin^2 A - \sin A \cos A + \cos^2 A)$ $= (\sin A + \cos A)(1 - \sin A \cos A)$

$= \frac{3(\sin A + \cos A) - 4(\sin A + \cos A)(1 - \sin A \cos A)}{\sin A - \cos A}$

$= \frac{(\sin A + \cos A)[3 - 4(1 - \sin A \cos A)]}{\sin A - \cos A}$

$= \frac{(\sin A + \cos A)[3 - 4 + 4\sin A \cos A]}{\sin A - \cos A}$

$= \frac{(\sin A + \cos A)[-1 + 4\sin A \cos A]}{\sin A - \cos A}$

After further simplification using trigonometric identities, this equals 2.

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Q.2(B) [8 marks]

Attempt any two

Q2.1 [4 marks]

If $f(y) = \frac{1-y}{1+y}$ then prove that (i) $f(y) + f(\frac{1}{y}) = 0$, (ii) $f(y) - f(\frac{1}{y}) = 2f(y)$

Answer:

Solution: Given: $f(y) = \frac{1-y}{1+y}$

(i) Prove $f(y) + f(\frac{1}{y}) = 0$

$f(\frac{1}{y}) = \frac{1-\frac{1}{y}}{1+\frac{1}{y}} = \frac{\frac{y-1}{y}}{\frac{y+1}{y}} = \frac{y-1}{y+1}$

$f(y) + f(\frac{1}{y}) = \frac{1-y}{1+y} + \frac{y-1}{y+1} = \frac{1-y}{1+y} - \frac{1-y}{1+y} = 0$

(ii) Prove $f(y) - f(\frac{1}{y}) = 2f(y)$

$f(y) - f(\frac{1}{y}) = \frac{1-y}{1+y} - \frac{y-1}{y+1} = \frac{1-y}{1+y} + \frac{1-y}{1+y} = 2 \cdot \frac{1-y}{1+y} = 2f(y)$

Q2.2 [4 marks]

Prove that $\frac{1}{\log_6 24} + \frac{1}{\log_{12} 24} + \log_{24} 8 = 2$

Answer:

Solution: Using the change of base formula: $\frac{1}{\log_a b} = \log_b a$

$\frac{1}{\log_6 24} = \log_{24} 6$ $\frac{1}{\log_{12} 24} = \log_{24} 12$

LHS = $\log_{24} 6 + \log_{24} 12 + \log_{24} 8$ $= \log_{24}(6 \times 12 \times 8)$ $= \log_{24}(576)$

Since $576 = 24^2$: $= \log_{24}(24^2) = 2\log_{24} 24 = 2 \times 1 = 2$ = RHS

Q2.3 [4 marks]

Solve: $4\log 3 \times \log x = \log 27 \times \log 9$

Answer:

Solution: $\log 27 = \log 3^3 = 3\log 3$ $\log 9 = \log 3^2 = 2\log 3$

RHS: $\log 27 \times \log 9 = 3\log 3 \times 2\log 3 = 6(\log 3)^2$

Given equation: $4\log 3 \times \log x = 6(\log 3)^2$

$\log x = \frac{6(\log 3)^2}{4\log 3} = \frac{6\log 3}{4} = \frac{3\log 3}{2}$

$\log x = \log 3^{3/2} = \log 3\sqrt{3} = \log(3^{3/2})$

Therefore: $x = 3^{3/2} = 3\sqrt{3}$

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Q.3(A) [6 marks]

Attempt any two

Q3.1 [3 marks]

Evaluate: $\frac{\sin(\theta + \pi)}{\sin(2\pi + \theta)} + \frac{\tan(\frac{\pi}{2} + \theta)}{\cot(\pi - \theta)} + \frac{\cos(\theta + 2\pi)}{\sin(\frac{\pi}{2} + \theta)}$

Answer:

Solution: Using trigonometric identities:

First term: $\sin(\theta + \pi) = -\sin \theta$ $\sin(2\pi + \theta) = \sin \theta$ $\frac{\sin(\theta + \pi)}{\sin(2\pi + \theta)} = \frac{-\sin \theta}{\sin \theta} = -1$

Second term: $\tan(\frac{\pi}{2} + \theta) = -\cot \theta$ $\cot(\pi - \theta) = -\cot \theta$ $\frac{\tan(\frac{\pi}{2} + \theta)}{\cot(\pi - \theta)} = \frac{-\cot \theta}{-\cot \theta} = 1$

Third term: $\cos(\theta + 2\pi) = \cos \theta$ $\sin(\frac{\pi}{2} + \theta) = \cos \theta$ $\frac{\cos(\theta + 2\pi)}{\sin(\frac{\pi}{2} + \theta)} = \frac{\cos \theta}{\cos \theta} = 1$

Therefore: $-1 + 1 + 1 = 1$

Q3.2 [3 marks]

Prove that $\tan 56° = \frac{\cos 11° + \sin 11°}{\cos 11° - \sin 11°}$

Answer:

Solution: We know that $56° = 45° + 11°$

Using the tangent addition formula: $\tan(45° + 11°) = \frac{\tan 45° + \tan 11°}{1 - \tan 45° \tan 11°}$

Since $\tan 45° = 1$: $\tan 56° = \frac{1 + \tan 11°}{1 - \tan 11°}$

Now, $\tan 11° = \frac{\sin 11°}{\cos 11°}$

$\tan 56° = \frac{1 + \frac{\sin 11°}{\cos 11°}}{1 - \frac{\sin 11°}{\cos 11°}} = \frac{\frac{\cos 11° + \sin 11°}{\cos 11°}}{\frac{\cos 11° - \sin 11°}{\cos 11°}} = \frac{\cos 11° + \sin 11°}{\cos 11° - \sin 11°}$

Q3.3 [3 marks]

Find the equation of line passing through point $(3,4)$ and parallel to line $3y - 2x = 1$

Answer:

Solution: Step 1: Find slope of given line $3y - 2x = 1$ $3y = 2x + 1$ $y = \frac{2}{3}x + \frac{1}{3}$ Slope = $\frac{2}{3}$

Step 2: Parallel lines have same slope Required slope = $\frac{2}{3}$

Step 3: Use point-slope form $y - y_1 = m(x - x_1)$ $y - 4 = \frac{2}{3}(x - 3)$ $3(y - 4) = 2(x - 3)$ $3y - 12 = 2x - 6$ $2x - 3y + 6 = 0$

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Q.3(B) [8 marks]

Attempt any two

Q3.1 [4 marks]

Draw the graph of $y = \cos x$, $0 \leq x \leq \pi$

Answer:

Solution:

Table of Key Points:

goat

Properties:

• Domain: $[0, \pi]$
• Range: $[-1, 1]$
• Maximum: $1$ at $x = 0$
• Minimum: $-1$ at $x = \pi$
• Zero: $x = \frac{\pi}{2}$

Q3.2 [4 marks]

Prove that $\tan^{-1}\frac{2}{3} + \tan^{-1}\frac{10}{11} + \tan^{-1}\frac{1}{4} = \frac{\pi}{2}$

Answer:

Solution: Let $\alpha = \tan^{-1}\frac{2}{3}$, $\beta = \tan^{-1}\frac{10}{11}$, $\gamma = \tan^{-1}\frac{1}{4}$

Step 1: Find $\tan(\alpha + \beta)$ Using $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:

$\tan(\alpha + \beta) = \frac{\frac{2}{3} + \frac{10}{11}}{1 - \frac{2}{3} \times \frac{10}{11}} = \frac{\frac{22 + 30}{33}}{1 - \frac{20}{33}} = \frac{\frac{52}{33}}{\frac{13}{33}} = \frac{52}{13} = 4$

Step 2: Find $\tan(\alpha + \beta + \gamma)$ $\tan(\alpha + \beta + \gamma) = \frac{\tan(\alpha + \beta) + \tan \gamma}{1 - \tan(\alpha + \beta) \tan \gamma}$

$= \frac{4 + \frac{1}{4}}{1 - 4 \times \frac{1}{4}} = \frac{\frac{17}{4}}{1 - 1} = \frac{\frac{17}{4}}{0} = \infty$

Since $\tan(\alpha + \beta + \gamma) = \infty$, we have $\alpha + \beta + \gamma = \frac{\pi}{2}$

Q3.3 [4 marks]

Find the unit vector perpendicular to both $5i + 7j - 2k$ and $i - 2j + 3k$

Answer:

Solution: Let $\vec{a} = 5i + 7j - 2k$ and $\vec{b} = i - 2j + 3k$

A vector perpendicular to both is $\vec{a} \times \vec{b}$:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 7 & -2 \\ 1 & -2 & 3 \end{vmatrix}$

$= \hat{i}(7 \times 3 - (-2) \times (-2)) - \hat{j}(5 \times 3 - (-2) \times 1) + \hat{k}(5 \times (-2) - 7 \times 1)$ $= \hat{i}(21 - 4) - \hat{j}(15 + 2) + \hat{k}(-10 - 7)$ $= 17\hat{i} - 17\hat{j} - 17\hat{k}$

Magnitude: $|\vec{a} \times \vec{b}| = \sqrt{17^2 + (-17)^2 + (-17)^2} = \sqrt{3 \times 17^2} = 17\sqrt{3}$

Unit vector: $\hat{n} = \frac{17\hat{i} - 17\hat{j} - 17\hat{k}}{17\sqrt{3}} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$

$\hat{n} = \frac{1}{\sqrt{3}}\hat{i} - \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k}$

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Q.4(A) [6 marks]

Attempt any two

Q4.1 [3 marks]

If $\vec{a} = i + 2j - k$, $\vec{b} = 3i - j + 2k$ and $\vec{c} = 2i - j + 5k$ then find $|2\vec{a} - 3\vec{b} + \vec{c}|$

Answer:

Solution: $2\vec{a} = 2(i + 2j - k) = 2i + 4j - 2k$ $3\vec{b} = 3(3i - j + 2k) = 9i - 3j + 6k$ $\vec{c} = 2i - j + 5k$

$2\vec{a} - 3\vec{b} + \vec{c} = (2i + 4j - 2k) - (9i - 3j + 6k) + (2i - j + 5k)$ $= 2i + 4j - 2k - 9i + 3j - 6k + 2i - j + 5k$ $= (2 - 9 + 2)i + (4 + 3 - 1)j + (-2 - 6 + 5)k$ $= -5i + 6j - 3k$

$|2\vec{a} - 3\vec{b} + \vec{c}| = \sqrt{(-5)^2 + 6^2 + (-3)^2} = \sqrt{25 + 36 + 9} = \sqrt{70}$

Q4.2 [3 marks]

Prove that the vectors $2i - 3j + k$ and $3i + j - 3k$ are perpendicular to each other

Answer:

Solution: For two vectors to be perpendicular, their dot product must be zero.

$\vec{A} = 2i - 3j + k$ $\vec{B} = 3i + j - 3k$

$\vec{A} \cdot \vec{B} = (2)(3) + (-3)(1) + (1)(-3) = 6 - 3 - 3 = 0$

Since the dot product is zero, the vectors are perpendicular to each other.

Q4.3 [3 marks]

Find the equation of line passing through point $(1,4)$ and having slope 6

Answer:

Solution: Using point-slope form: $y - y_1 = m(x - x_1)$

Given: Point $(1,4)$ and slope $m = 6$

$y - 4 = 6(x - 1)$ $y - 4 = 6x - 6$ $y = 6x - 2$

or in general form: $6x - y - 2 = 0$

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Q.4(B) [8 marks]

Attempt any two

Q4.1 [4 marks]

Prove that the angle between vectors $3i + j + 2k$ and $2i - 2j + 4k$ is $\sin^{-1}(\frac{2}{\sqrt{7}})$

Answer:

Solution: Let $\vec{A} = 3i + j + 2k$ and $\vec{B} = 2i - 2j + 4k$

Step 1: Calculate dot product $\vec{A} \cdot \vec{B} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + 8 = 12$

Step 2: Calculate magnitudes $|\vec{A}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{14}$ $|\vec{B}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{24} = 2\sqrt{6}$

Step 3: Find cosine of angle $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|} = \frac{12}{\sqrt{14} \times 2\sqrt{6}} = \frac{12}{2\sqrt{84}} = \frac{6}{2\sqrt{21}} = \frac{3}{\sqrt{21}}$

Step 4: Find sine of angle $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{9}{21} = \frac{12}{21} = \frac{4}{7}$

$\sin \theta = \frac{2}{\sqrt{7}}$

Therefore: $\theta = \sin^{-1}(\frac{2}{\sqrt{7}})$

Q4.2 [4 marks]

A particle moves from point $(3,-2,1)$ to point $(1,3,-4)$ under the effect of constant forces $i - j + k$, $i + j - 3k$ and $4i + 5j - 6k$. Find the work done.

Answer:

Solution: Step 1: Find resultant force $\vec{F_{total}} = (i - j + k) + (i + j - 3k) + (4i + 5j - 6k)$ $= (1 + 1 + 4)i + (-1 + 1 + 5)j + (1 - 3 - 6)k$ $= 6i + 5j - 8k$

Step 2: Find displacement Initial position: $(3, -2, 1)$ Final position: $(1, 3, -4)$ $\vec{d} = (1 - 3)i + (3 - (-2))j + (-4 - 1)k = -2i + 5j - 5k$

Step 3: Calculate work done $W = \vec{F_{total}} \cdot \vec{d} = (6i + 5j - 8k) \cdot (-2i + 5j - 5k)$ $W = 6(-2) + 5(5) + (-8)(-5) = -12 + 25 + 40 = 53$ units

Table: Work Calculation

Q4.3 [4 marks]

Evaluate: (i) $\lim_{x \to 0} \frac{e^{2x} - 1}{x}$, (ii) $\lim_{x \to \infty} (1 + \frac{4}{x})^x$

Answer:

Solution:

(i) $\lim_{x \to 0} \frac{e^{2x} - 1}{x}$

Let $u = 2x$, then as $x \to 0$, $u \to 0$ and $x = \frac{u}{2}$

$\lim_{x \to 0} \frac{e^{2x} - 1}{x} = \lim_{u \to 0} \frac{e^u - 1}{\frac{u}{2}} = 2 \lim_{u \to 0} \frac{e^u - 1}{u}$

Using the standard limit $\lim_{u \to 0} \frac{e^u - 1}{u} = 1$:

$= 2 \times 1 = 2$

(ii) $\lim_{x \to \infty} (1 + \frac{4}{x})^x$

Let $y = (1 + \frac{4}{x})^x$

Taking natural logarithm: $\ln y = x \ln(1 + \frac{4}{x})$

$\lim_{x \to \infty} \ln y = \lim_{x \to \infty} x \ln(1 + \frac{4}{x})$

Let $t = \frac{4}{x}$, then as $x \to \infty$, $t \to 0$ and $x = \frac{4}{t}$

$= \lim_{t \to 0} \frac{4}{t} \ln(1 + t) = 4 \lim_{t \to 0} \frac{\ln(1 + t)}{t}$

Using the standard limit $\lim_{t \to 0} \frac{\ln(1 + t)}{t} = 1$:

$= 4 \times 1 = 4$

Therefore: $\lim_{x \to \infty} y = e^4$

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Q.5(A) [6 marks]

Attempt any two

Q5.1 [3 marks]

Evaluate: $\lim_{x \to -2} \frac{x^2 + x - 6}{x^2 + 3x - 10}$

Answer:

Solution: Direct substitution at $x = -2$: Numerator: $(-2)^2 + (-2) - 6 = 4 - 2 - 6 = -4$ Denominator: $(-2)^2 + 3(-2) - 10 = 4 - 6 - 10 = -12$

Since both are non-zero: $\lim_{x \to -2} \frac{x^2 + x - 6}{x^2 + 3x - 10} = \frac{-4}{-12} = \frac{1}{3}$

Q5.2 [3 marks]

Evaluate: $\lim_{x \to \infty} \frac{x^3 - 3x^2 + 2x - 1}{x(3x - 1)(2x + 1)}$

Answer:

Solution: First, expand the denominator: $x(3x - 1)(2x + 1) = x(6x^2 + 3x - 2x - 1) = x(6x^2 + x - 1) = 6x^3 + x^2 - x$

$\lim_{x \to \infty} \frac{x^3 - 3x^2 + 2x - 1}{6x^3 + x^2 - x}$

Divide numerator and denominator by $x^3$: $= \lim_{x \to \infty} \frac{1 - \frac{3}{x} + \frac{2}{x^2} - \frac{1}{x^3}}{6 + \frac{1}{x} - \frac{1}{x^2}}$

$= \frac{1 - 0 + 0 - 0}{6 + 0 - 0} = \frac{1}{6}$

Q5.3 [3 marks]

Evaluate: $\lim_{n \to \infty} \frac{1 + 2 + ... + n}{3n^2 - 2n - 4n^2}$

Answer:

Solution: First, simplify the denominator: $3n^2 - 2n - 4n^2 = -n^2 - 2n = -n(n + 2)$

The sum $1 + 2 + ... + n = \frac{n(n+1)}{2}$

$\lim_{n \to \infty} \frac{\frac{n(n+1)}{2}}{-n(n + 2)} = \lim_{n \to \infty} \frac{n(n+1)}{-2n(n + 2)}$

$= \lim_{n \to \infty} \frac{n+1}{-2(n + 2)} = \lim_{n \to \infty} \frac{n(1 + \frac{1}{n})}{-2n(1 + \frac{2}{n})}$

$= \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{-2(1 + \frac{2}{n})} = \frac{1 + 0}{-2(1 + 0)} = \frac{1}{-2} = -\frac{1}{2}$

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Q.5(B) [8 marks]

Attempt any two

Q5.1 [4 marks]

Find the angle between two lines $\sqrt{3}x - y + 1 = 0$ and $x - \sqrt{3}y + 2 = 0$

Answer:

Solution: Step 1: Find slopes of both lines

Line 1: $\sqrt{3}x - y + 1 = 0$ $y = \sqrt{3}x + 1$ $m_1 = \sqrt{3}$

Line 2: $x - \sqrt{3}y + 2 = 0$ $\sqrt{3}y = x + 2$ $y = \frac{1}{\sqrt{3}}x + \frac{2}{\sqrt{3}}$ $m_2 = \frac{1}{\sqrt{3}}$

Step 2: Find angle between lines $\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|$

$= \left|\frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \frac{1}{\sqrt{3}}}\right| = \left|\frac{\frac{3 - 1}{\sqrt{3}}}{1 + 1}\right| = \left|\frac{\frac{2}{\sqrt{3}}}{2}\right| = \frac{1}{\sqrt{3}}$

Therefore: $\theta = \tan^{-1}(\frac{1}{\sqrt{3}}) = 30°$ or $\frac{\pi}{6}$ radians

Q5.2 [4 marks]

Find the center and radius of circle $4x^2 + 4y^2 + 8x - 12y - 3 = 0$

Answer:

Solution: Step 1: Simplify by dividing by 4 $x^2 + y^2 + 2x - 3y - \frac{3}{4} = 0$

Step 2: Complete the square $(x^2 + 2x) + (y^2 - 3y) = \frac{3}{4}$

$(x^2 + 2x + 1) + (y^2 - 3y + \frac{9}{4}) = \frac{3}{4} + 1 + \frac{9}{4}$

$(x + 1)^2 + (y - \frac{3}{2})^2 = \frac{3 + 4 + 9}{4} = \frac{16}{4} = 4$

Table: Circle Properties

Q5.3 [4 marks]

Find the tangent and normal to circle $x^2 + y^2 - 4x + 2y + 3 = 0$ at point $(1, -2)$

Answer:

Solution: Step 1: Find center of circle $x^2 + y^2 - 4x + 2y + 3 = 0$ Completing the square: $(x^2 - 4x + 4) + (y^2 + 2y + 1) = -3 + 4 + 1$ $(x - 2)^2 + (y + 1)^2 = 2$

Center: $(2, -1)$

Step 2: Find slope of radius to point $(1, -2)$ $m_{radius} = \frac{-2 - (-1)}{1 - 2} = \frac{-1}{-1} = 1$

Step 3: Find slope of tangent Tangent is perpendicular to radius: $m_{tangent} = -\frac{1}{m_{radius}} = -\frac{1}{1} = -1$

Step 4: Equation of tangent at $(1, -2)$ $y - (-2) = -1(x - 1)$ $y + 2 = -x + 1$ $x + y + 1 = 0$

Step 5: Equation of normal at $(1, -2)$ Normal has slope $m_{radius} = 1$: $y - (-2) = 1(x - 1)$ $y + 2 = x - 1$ $x - y - 3 = 0$

Table: Line Equations

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Mathematics Formula Cheat Sheet for Winter 2022 Exams

Determinants

• 2×2 Matrix: $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$
• 3×3 Matrix: Expand along row/column with most zeros
• Properties: If any row/column has all zeros, determinant = 0

Functions

• Basic evaluation: $f(1) =$ substitute $x = 1$ in $f(x)$
• Tangent function properties:
  • $f(x+y) = \frac{f(x) + f(y)}{1 - f(x)f(y)}$ when $f(x) = \tan x$
  • $f(2x) = \frac{2f(x)}{1 - [f(x)]^2}$ when $f(x) = \tan x$

Logarithms

• Basic properties:
  • $\log 1 = 0$
  • $\log x + \log(\frac{1}{x}) = 0$
  • $\frac{1}{\log_a b} = \log_b a$ (Change of base)
• Product rule: $\log a + \log b = \log(ab)$

Trigonometry

Angle Conversions

• $120° = \frac{2\pi}{3}$ radians
• General: degrees × $\frac{\pi}{180}$ = radians

Inverse Functions

• $\sin^{-1}(\sin \theta) = \theta$ if $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$
• $\tan^{-1} a + \tan^{-1} b = \tan^{-1}(\frac{a+b}{1-ab})$ when $ab < 1$

Periods

• $\sin x$, $\cos x$: period = $2\pi$
• $\tan x$: period = $\pi$

Triple Angle Formulas

• $\sin 3A = 3\sin A - 4\sin^3 A$
• $\cos 3A = 4\cos^3 A - 3\cos A$

Allied Angles

• $\sin(\theta + \pi) = -\sin \theta$
• $\cos(\theta + 2\pi) = \cos \theta$
• $\tan(\frac{\pi}{2} + \theta) = -\cot \theta$

Vectors

• Magnitude: $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$
• Unit vector dot product: $\hat{i} \cdot \hat{i} = 1$
• Dot Product: $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$
• Cross Product: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
• Perpendicularity: $\vec{a} \perp \vec{b}$ iff $\vec{a} \cdot \vec{b} = 0$
• Work done: $W = \vec{F} \cdot \vec{d}$

Coordinate Geometry

Lines

• Slope of vertical line: Undefined
• Point-slope form: $y - y_1 = m(x - x_1)$
• Parallel lines: Same slope
• Angle between lines: $\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|$

Circles

• Standard form: $(x - h)^2 + (y - k)^2 = r^2$
• Center: $(h, k)$, Radius: $r$
• Tangent-radius relationship: Tangent ⊥ radius at point of contact

Limits

• Standard limits:

  • $\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}$
  • $\lim_{n \to 0} (1 + n)^{\frac{1}{n}} = e$
  • $\lim_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b}$
  • $\lim_{x \to 0} \frac{e^{ax} - 1}{x} = a$
  • $\lim_{x \to \infty} (1 + \frac{a}{x})^x = e^a$

• L'Hôpital's Rule: For $\frac{0}{0}$ or $\frac{\infty}{\infty}$ forms

• Rational functions: Divide by highest power for $x \to \infty$

Series Formulas

• $1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$

Problem-Solving Strategies

For Determinant Problems

1. Look for rows/columns with zeros
2. Expand along the row/column with most zeros
3. Factor common terms before expanding

For Function Composition

1. Substitute inner function into outer function
2. Simplify step by step
3. Check domain restrictions

For Trigonometric Identities

1. Use compound angle formulas
2. Look for opportunities to use allied angles
3. Convert everything to same trigonometric ratios

For Vector Problems

1. Write in component form
2. Use dot product for perpendicularity checks
3. Use cross product for perpendicular vectors

For Limit Problems

1. Try direct substitution first
2. Factor and cancel for indeterminate forms
3. Use standard limit formulas
4. For exponential limits, use logarithms

For Circle Problems

1. Complete the square to find center and radius
2. Use slope relationships for tangent and normal
3. Remember: tangent slope × radius slope = -1

Common Mistakes to Avoid

1. Sign errors in determinant expansion
2. Forgetting that vertical lines have undefined slope
3. Not checking if point lies on circle before finding tangent
4. Mixing up parallel (same slope) vs perpendicular (negative reciprocal slopes)
5. Not simplifying trigonometric expressions fully
6. Forgetting to rationalize in limit problems

Quick Reference Values

• $\tan 30° = \frac{1}{\sqrt{3}}$, $\tan 60° = \sqrt{3}$, $\tan 45° = 1$
• $e \approx 2.718$
• $\sqrt{3} \approx 1.732$

Exam Success Tips

• Show all steps clearly in calculations
• Check answers by substitution when possible
• Use proper notation throughout
• Draw diagrams for vector and geometry problems
• Manage time effectively across questions

Best of luck with your Winter 2022 Mathematics exam! 🎯