Q.1 [14 marks]

Fill in the blanks using appropriate choice from the given options

Q1.1 [1 mark]

If $f(x) = \frac{1}{x}$ , then the value of $f(1)$ is __________

Answer: b. 1

Solution: $f(x) = \frac{1}{x}$ $f(1) = \frac{1}{1} = 1$

Q1.2 [1 mark]

$\log_b a \times \log_a b$ = __________

Answer: b. 1

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

Q1.3 [1 mark]

If $\begin{vmatrix} x & 3 \\ -2 & 2 \end{vmatrix} = 2$ then $x$ = _________

Answer: a. 2

Solution: $\begin{vmatrix} x & 3 \\ -2 & 2 \end{vmatrix} = x(2) - 3(-2) = 2x + 6$ Given: $2x + 6 = 2$ $2x = -4$ $x = -2$ Wait, let me recalculate: $2x + 6 = 2 \Rightarrow 2x = -4 \Rightarrow x = -2$ But -2 is option c, not a. Let me verify: If $x = 2$ : $2(2) + 6 = 10 \neq 2$ The correct answer should be c. -2

Q1.4 [1 mark]

Find the value: $\begin{vmatrix} 6 & 4 \\ 1 & 2 \end{vmatrix}$

Answer: a. 8

Solution: $\begin{vmatrix} 6 & 4 \\ 1 & 2 \end{vmatrix} = 6(2) - 4(1) = 12 - 4 = 8$

Q1.5 [1 mark]

$135° =$ __________ Radian

Answer: b. $\frac{3\pi}{4}$

Solution: $135° = 135 \times \frac{\pi}{180} = \frac{135\pi}{180} = \frac{3\pi}{4}$ radians

Q1.6 [1 mark]

$\sin 120° =$ _________

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

Solution: $120° = 180° - 60°$ $\sin 120° = \sin(180° - 60°) = \sin 60° = \frac{\sqrt{3}}{2}$

Q1.7 [1 mark]

$\sin(\frac{\pi}{2} + \theta) =$ __________

Answer: c. $\cos \theta$

Solution: Using the identity: $\sin(\frac{\pi}{2} + \theta) = \cos \theta$

Q1.8 [1 mark]

If $\vec{a} = (1,1,1)$ and $\vec{b} = (2,2,2)$ then $\vec{a} \times \vec{b} =$ _________

Answer: d. $(0,0,0)$

Solution: $\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{vmatrix}$

Since $\vec{b} = 2\vec{a}$ , they are parallel vectors, so their cross product is zero. $\vec{a} \times \vec{b} = (0,0,0)$

Q1.9 [1 mark]

$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ then $\vec{a} \cdot \vec{b} =$ _________

Answer: a. 2

Solution: $\vec{a} \cdot \vec{b} = (2)(1) + (-1)(1) + (1)(1) = 2 - 1 + 1 = 2$

Q1.10 [1 mark]

If lines $5x - py = 3$ and $2x + 3y = 4$ are parallel to each other then $p =$ ________

Answer: c. $-\frac{15}{2}$

Solution: For parallel lines, slopes must be equal. Line 1: $5x - py = 3 \Rightarrow y = \frac{5x - 3}{p}$ , slope = $\frac{5}{p}$ Line 2: $2x + 3y = 4 \Rightarrow y = \frac{-2x + 4}{3}$ , slope = $-\frac{2}{3}$

For parallel lines: $\frac{5}{p} = -\frac{2}{3}$ $5 \times 3 = -2p$ $15 = -2p$ $p = -\frac{15}{2}$

Q1.11 [1 mark]

The radius of the circle $x^2 + y^2 + 2x\cos\theta + 2y\sin\theta = 8$ is ________

Answer: d. 3

Solution: Rewriting: $x^2 + y^2 + 2x\cos\theta + 2y\sin\theta = 8$ $(x + \cos\theta)^2 + (y + \sin\theta)^2 = 8 + \cos^2\theta + \sin^2\theta$ $(x + \cos\theta)^2 + (y + \sin\theta)^2 = 8 + 1 = 9$

Radius = $\sqrt{9} = 3$

Q1.12 [1 mark]

$\lim_{x \to a} \frac{x^n - a^n}{x - a} =$ __________. $n \in \mathbb{R}$

Answer: a. $na^{n-1}$

Solution: This is the derivative of $x^n$ at $x = a$ . $\lim_{x \to a} \frac{x^n - a^n}{x - a} = \frac{d}{dx}(x^n)|_{x=a} = nx^{n-1}|_{x=a} = na^{n-1}$

Q1.13 [1 mark]

$\lim_{x \to 0} \frac{\sin x}{x} =$ __________

Answer: b. 1

Solution: This is a standard limit: $\lim_{x \to 0} \frac{\sin x}{x} = 1$

Q1.14 [1 mark]

Obtain the Limit of $\lim_{n \to \infty} (1 + \frac{1}{n})^n$

Answer: c. e

Solution: This is the definition of Euler's number: $\lim_{n \to \infty} (1 + \frac{1}{n})^n = e$

Q.2(A) [6 marks]

Attempt any two

Q2.1 [3 marks]

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

Solution: Expanding along the third row: $\begin{vmatrix} x-1 & 2 & 1 \\ x & 1 & x+1 \\ 1 & 1 & 0 \end{vmatrix} = 1 \cdot \begin{vmatrix} 2 & 1 \\ 1 & x+1 \end{vmatrix} - 1 \cdot \begin{vmatrix} x-1 & 1 \\ x & x+1 \end{vmatrix} + 0$

$= 1[2(x+1) - 1(1)] - 1[(x-1)(x+1) - x(1)]$ $= 2x + 2 - 1 - [(x-1)(x+1) - x]$ $= 2x + 1 - [x^2 - 1 - x]$ $= 2x + 1 - x^2 + 1 + x$ $= 3x + 2 - x^2$

Given: $3x + 2 - x^2 = 4$ $-x^2 + 3x - 2 = 0$ $x^2 - 3x + 2 = 0$ $(x-1)(x-2) = 0$

Therefore: $x = 1$ or $x = 2$

Q2.2 [3 marks]

If $\log(\frac{a+b}{2}) = \frac{1}{2}(\log a + \log b)$ then prove that $a = b$

Solution: Given: $\log(\frac{a+b}{2}) = \frac{1}{2}(\log a + \log b)$

RHS: $\frac{1}{2}(\log a + \log b) = \frac{1}{2}\log(ab) = \log(ab)^{1/2} = \log\sqrt{ab}$

So we have: $\log(\frac{a+b}{2}) = \log\sqrt{ab}$

Taking antilog: $\frac{a+b}{2} = \sqrt{ab}$

Squaring both sides: $(\frac{a+b}{2})^2 = ab$

$\frac{(a+b)^2}{4} = ab$

$(a+b)^2 = 4ab$

$a^2 + 2ab + b^2 = 4ab$

$a^2 - 2ab + b^2 = 0$

$(a-b)^2 = 0$

Therefore: $a = b$

Q2.3 [3 marks]

Obtain the value of $\tan 75°$ or obtain the value of $\tan \frac{5\pi}{12}$

Solution: $\tan 75° = \tan(45° + 30°)$

Using the formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$\tan 75° = \frac{\tan 45° + \tan 30°}{1 - \tan 45° \tan 30°}$

$= \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}}$

$= \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$

$= \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}}$

$= \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

Rationalizing: $= \frac{(\sqrt{3} + 1)^2}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$

Q.2(B) [8 marks]

Attempt any two

Q2.1 [4 marks]

If $\frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b}$ then prove that

(i) $xyz = 1$ (ii) $x^a y^b z^c = 1$

Solution: Let $\frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b} = k$ (say)

Then: $x = k(b-c)$ , $y = k(c-a)$ , $z = k(a-b)$

(i) Proving $xyz = 1$ :

We need to show: $x + y + z = 0$ first. $x + y + z = k(b-c) + k(c-a) + k(a-b) = k[(b-c) + (c-a) + (a-b)] = k[0] = 0$

Wait, this doesn't directly prove $xyz = 1$ . Let me reconsider.

Actually, we need additional conditions. The problem statement seems incomplete.

Let me assume the constraint: $x + y + z = 0$

From $x + y + z = 0$ and the given ratios: $k(b-c) + k(c-a) + k(a-b) = 0$ $k[(b-c) + (c-a) + (a-b)] = 0$ $k[0] = 0$ ✓

For part (ii), we need the constraint $a + b + c = 0$ or similar.

(ii) Proving $x^a y^b z^c = 1$ :

If $a + b + c = 0$ , then: $x^a y^b z^c = [k(b-c)]^a [k(c-a)]^b [k(a-b)]^c$ $= k^{a+b+c} (b-c)^a (c-a)^b (a-b)^c$ $= k^0 (b-c)^a (c-a)^b (a-b)^c = (b-c)^a (c-a)^b (a-b)^c$

With appropriate symmetry conditions, this equals 1.

Q2.2 [4 marks]

If $f(x) = \frac{1-x}{1+x}$ then prove that $f(f(x)) = x$

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

We need to find $f(f(x))$ :

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

Let $y = \frac{1-x}{1+x}$

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

Numerator: $1 - \frac{1-x}{1+x} = \frac{1+x-(1-x)}{1+x} = \frac{1+x-1+x}{1+x} = \frac{2x}{1+x}$

Denominator: $1 + \frac{1-x}{1+x} = \frac{1+x+(1-x)}{1+x} = \frac{1+x+1-x}{1+x} = \frac{2}{1+x}$

Therefore: $f(f(x)) = \frac{\frac{2x}{1+x}}{\frac{2}{1+x}} = \frac{2x}{1+x} \times \frac{1+x}{2} = x$

Hence proved: $f(f(x)) = x$

Q2.3 [4 marks]

If $\begin{vmatrix} a & b & b \\ b & a & b \\ b & b & a \end{vmatrix} = 0$ then prove that $a = b$ or $a = -2b$

Solution: Let $\Delta = \begin{vmatrix} a & b & b \\ b & a & b \\ b & b & a \end{vmatrix}$

Expanding along the first row: $\Delta = a\begin{vmatrix} a & b \\ b & a \end{vmatrix} - b\begin{vmatrix} b & b \\ b & a \end{vmatrix} + b\begin{vmatrix} b & a \\ b & b \end{vmatrix}$

$= a(a^2 - b^2) - b(ba - b^2) + b(b^2 - ab)$ $= a(a^2 - b^2) - b^2a + b^3 + b^3 - ab^2$ $= a^3 - ab^2 - ab^2 + b^3 + b^3 - ab^2$ $= a^3 - 3ab^2 + 2b^3$

Alternative method (easier): $\Delta = \begin{vmatrix} a & b & b \\ b & a & b \\ b & b & a \end{vmatrix}$

$R_1 \to R_1 + R_2 + R_3$ : $\Delta = \begin{vmatrix} a+2b & a+2b & a+2b \\ b & a & b \\ b & b & a \end{vmatrix}$

$= (a+2b)\begin{vmatrix} 1 & 1 & 1 \\ b & a & b \\ b & b & a \end{vmatrix}$

$C_2 \to C_2 - C_1, C_3 \to C_3 - C_1$ : $= (a+2b)\begin{vmatrix} 1 & 0 & 0 \\ b & a-b & 0 \\ b & 0 & a-b \end{vmatrix}$

$= (a+2b) \times 1 \times (a-b)(a-b) = (a+2b)(a-b)^2$

Given: $\Delta = 0$ $(a+2b)(a-b)^2 = 0$

Therefore: $a + 2b = 0$ or $(a-b)^2 = 0$ i.e., $a = -2b$ or $a = b$

Q.3(A) [6 marks]

Attempt any two

Q3.1 [3 marks]

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

Solution: Using sum-to-product formulas:

Numerator: $\sin A + \sin 2A + \sin 3A$ $= \sin 2A + (\sin A + \sin 3A)$ $= \sin 2A + 2\sin(\frac{A+3A}{2})\cos(\frac{3A-A}{2})$ $= \sin 2A + 2\sin(2A)\cos(A)$ $= \sin 2A(1 + 2\cos A)$

Denominator: $\cos A + \cos 2A + \cos 3A$ $= \cos 2A + (\cos A + \cos 3A)$ $= \cos 2A + 2\cos(\frac{A+3A}{2})\cos(\frac{3A-A}{2})$ $= \cos 2A + 2\cos(2A)\cos(A)$ $= \cos 2A(1 + 2\cos A)$

Therefore: $\frac{\sin A + \sin 2A + \sin 3A}{\cos A + \cos 2A + \cos 3A} = \frac{\sin 2A(1 + 2\cos A)}{\cos 2A(1 + 2\cos A)} = \frac{\sin 2A}{\cos 2A} = \tan 2A$

Q3.2 [3 marks]

Prove that $\frac{1 + \sin \theta + \cos \theta}{1 + \sin \theta - \cos \theta} = \cot \frac{\theta}{2}$

Solution: Using half-angle identities: $\sin \theta = 2\sin \frac{\theta}{2}\cos \frac{\theta}{2}$ $\cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2}$ $1 = \sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2}$

Numerator: $1 + \sin \theta + \cos \theta = \sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2} + 2\sin \frac{\theta}{2}\cos \frac{\theta}{2} + \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2}$ $= 2\cos^2 \frac{\theta}{2} + 2\sin \frac{\theta}{2}\cos \frac{\theta}{2}$ $= 2\cos \frac{\theta}{2}(\cos \frac{\theta}{2} + \sin \frac{\theta}{2})$

Denominator: $1 + \sin \theta - \cos \theta = \sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2} + 2\sin \frac{\theta}{2}\cos \frac{\theta}{2} - \cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2}$ $= 2\sin^2 \frac{\theta}{2} + 2\sin \frac{\theta}{2}\cos \frac{\theta}{2}$ $= 2\sin \frac{\theta}{2}(\sin \frac{\theta}{2} + \cos \frac{\theta}{2})$

Therefore: $\frac{1 + \sin \theta + \cos \theta}{1 + \sin \theta - \cos \theta} = \frac{2\cos \frac{\theta}{2}(\cos \frac{\theta}{2} + \sin \frac{\theta}{2})}{2\sin \frac{\theta}{2}(\sin \frac{\theta}{2} + \cos \frac{\theta}{2})} = \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}} = \cot \frac{\theta}{2}$

Q3.3 [3 marks]

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

Solution: First, divide by 2 to simplify: $x^2 + y^2 - 4x + 2y + 1 = 0$

Completing the square: $x^2 - 4x + y^2 + 2y = -1$ $(x^2 - 4x + 4) + (y^2 + 2y + 1) = -1 + 4 + 1$ $(x - 2)^2 + (y + 1)^2 = 4$

Table: Circle Properties

Property	Value
Center	$(2, -1)$
Radius	$\sqrt{4} = 2$

Mnemonic: "Complete the square to find the center's pair"

Q.3(B) [8 marks]

Attempt any two

Q3.1 [4 marks]

Plot the graph of $y = 2\sin \frac{x}{3}$ , $0 < x \leq 3\pi$

Solution: For the function $y = 2\sin \frac{x}{3}$ :

Table: Key Properties

Property	Value
Amplitude	$2$
Period	$2\pi \div \frac{1}{3} = 6\pi$
Frequency	$\frac{1}{3}$

Key Points Table:

$x$	$\frac{x}{3}$	$\sin \frac{x}{3}$	$y = 2\sin \frac{x}{3}$
$0$	$0$	$0$	$0$
$\frac{3\pi}{2}$	$\frac{\pi}{2}$	$1$	$2$
$3\pi$	$\pi$	$0$	$0$

      y
      |
    2 +     *
      |    / \
    1 +   /   \
      |  /     \
    0 +--+-----*--------> x
      0  π/2   π   3π/2  3π
     -1 +
      |
    -2 +

The graph shows one complete cycle from $0$ to $3\pi$ with amplitude 2.

Q3.2 [4 marks]

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

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

We need to prove: $\alpha + \beta + \gamma = \frac{\pi}{2}$

This is equivalent to proving: $\tan(\alpha + \beta + \gamma) = \infty$

Using the formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

First, find $\tan(\alpha + \beta)$ : $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\frac{2}{3} + \frac{10}{11}}{1 - \frac{2}{3} \cdot \frac{10}{11}}$

$= \frac{\frac{22 + 30}{33}}{1 - \frac{20}{33}} = \frac{\frac{52}{33}}{\frac{13}{33}} = \frac{52}{13} = 4$

Now find $\tan(\alpha + \beta + \gamma)$ : $\tan(\alpha + \beta + \gamma) = \frac{\tan(\alpha + \beta) + \tan \gamma}{1 - \tan(\alpha + \beta) \tan \gamma}$

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

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

Q3.3 [4 marks]

$\vec{a} = 2\hat{i} - \hat{j}$ and $\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$ then obtain $|(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|$

Answer:

Solution: Given: $\vec{a} = 2\hat{i} - \hat{j}$ , $\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$

First, let's complete $\vec{a}$ : $\vec{a} = 2\hat{i} - \hat{j} + 0\hat{k}$

$\vec{a} + \vec{b} = (2+1)\hat{i} + (-1+3)\hat{j} + (0-2)\hat{k} = 3\hat{i} + 2\hat{j} - 2\hat{k}$

$\vec{a} - \vec{b} = (2-1)\hat{i} + (-1-3)\hat{j} + (0+2)\hat{k} = \hat{i} - 4\hat{j} + 2\hat{k}$

Now, $(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})$ :

$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -2 \\ 1 & -4 & 2 \end{vmatrix}$

$= \hat{i}(2 \cdot 2 - (-2)(-4)) - \hat{j}(3 \cdot 2 - (-2)(1)) + \hat{k}(3(-4) - 2(1))$

$= \hat{i}(4 - 8) - \hat{j}(6 + 2) + \hat{k}(-12 - 2)$

$= -4\hat{i} - 8\hat{j} - 14\hat{k}$

$|(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})| = \sqrt{(-4)^2 + (-8)^2 + (-14)^2}$

$= \sqrt{16 + 64 + 196} = \sqrt{276} = 2\sqrt{69}$

Q.4(A) [6 marks]

Attempt any two

Q4.1 [3 marks]

Find $(10\hat{i} + 2\hat{j} + 3\hat{k}) \cdot [(\hat{i} - 2\hat{j} + 2\hat{k}) \times (3\hat{i} - 2\hat{j} - 2\hat{k})]$

Solution: Let $\vec{A} = 10\hat{i} + 2\hat{j} + 3\hat{k}$ Let $\vec{B} = \hat{i} - 2\hat{j} + 2\hat{k}$ Let $\vec{C} = 3\hat{i} - 2\hat{j} - 2\hat{k}$

We need to find $\vec{A} \cdot (\vec{B} \times \vec{C})$

This is a scalar triple product, which can be calculated as: $\vec{A} \cdot (\vec{B} \times \vec{C}) = \begin{vmatrix} 10 & 2 & 3 \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix}$

Expanding along the first row: $= 10\begin{vmatrix} -2 & 2 \\ -2 & -2 \end{vmatrix} - 2\begin{vmatrix} 1 & 2 \\ 3 & -2 \end{vmatrix} + 3\begin{vmatrix} 1 & -2 \\ 3 & -2 \end{vmatrix}$

$= 10[(-2)(-2) - (2)(-2)] - 2[(1)(-2) - (2)(3)] + 3[(1)(-2) - (-2)(3)]$

$= 10[4 + 4] - 2[-2 - 6] + 3[-2 + 6]$

$= 10(8) - 2(-8) + 3(4)$

$= 80 + 16 + 12 = 108$

Q4.2 [3 marks]

A particle under the constant forces $(1, 2, 3)$ and $(3, 1, 1)$ is displaced from point $(0, 1, -2)$ to point $(5, 1, 2)$ . Calculate the total work done by the particle

Solution: Work done = $\vec{F} \cdot \vec{d}$ where $\vec{F}$ is the resultant force and $\vec{d}$ is the displacement.

Step 1: Find resultant force $\vec{F_1} = 1\hat{i} + 2\hat{j} + 3\hat{k}$ $\vec{F_2} = 3\hat{i} + 1\hat{j} + 1\hat{k}$ $\vec{F_{resultant}} = \vec{F_1} + \vec{F_2} = 4\hat{i} + 3\hat{j} + 4\hat{k}$

Step 2: Find displacement Initial position: $(0, 1, -2)$ Final position: $(5, 1, 2)$ $\vec{d} = (5-0)\hat{i} + (1-1)\hat{j} + (2-(-2))\hat{k} = 5\hat{i} + 0\hat{j} + 4\hat{k}$

Step 3: Calculate work done $W = \vec{F_{resultant}} \cdot \vec{d} = (4\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (5\hat{i} + 0\hat{j} + 4\hat{k})$ $W = 4(5) + 3(0) + 4(4) = 20 + 0 + 16 = 36$ units

Table: Work Calculation

Component	Force	Displacement	Work
x	4	5	20
y	3	0	0
z	4	4	16
Total			36

Q4.3 [3 marks]

$5x + 6y + 3 = 0$ and $x - 11y + 7 = 0$ are two intersecting lines find the angle between them

Answer:

Solution: For lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ , the angle between them is: $\tan \theta = \left|\frac{a_1b_2 - a_2b_1}{a_1a_2 + b_1b_2}\right|$

Line 1: $5x + 6y + 3 = 0$ → $a_1 = 5, b_1 = 6$ Line 2: $x - 11y + 7 = 0$ → $a_2 = 1, b_2 = -11$

$\tan \theta = \left|\frac{5(-11) - 1(6)}{5(1) + 6(-11)}\right|$

$= \left|\frac{-55 - 6}{5 - 66}\right| = \left|\frac{-61}{-61}\right| = 1$

Therefore: $\theta = \tan^{-1}(1) = 45°$

Mnemonic: "Lines that intersect at forty-five, make slopes that multiply to negative one to stay alive"

Q.4(B) [8 marks]

Attempt any two

Q4.1 [4 marks]

Find the unit vector perpendicular to $\vec{a} = (1, -1, 1)$ and $\vec{b} = (2, 3, -1)$

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

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

$= \hat{i}[(-1)(-1) - (1)(3)] - \hat{j}[(1)(-1) - (1)(2)] + \hat{k}[(1)(3) - (-1)(2)]$

$= \hat{i}[1 - 3] - \hat{j}[-1 - 2] + \hat{k}[3 + 2]$

$= -2\hat{i} + 3\hat{j} + 5\hat{k}$

Magnitude: $|\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + 3^2 + 5^2} = \sqrt{4 + 9 + 25} = \sqrt{38}$

Unit vector: $\hat{n} = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{-2\hat{i} + 3\hat{j} + 5\hat{k}}{\sqrt{38}}$

$\hat{n} = \frac{-2}{\sqrt{38}}\hat{i} + \frac{3}{\sqrt{38}}\hat{j} + \frac{5}{\sqrt{38}}\hat{k}$

Q4.2 [4 marks]

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

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

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

Step 2: Calculate magnitudes $|\vec{A}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$ $|\vec{B}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$

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

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

Q4.3 [4 marks]

Find the Limit of $\lim_{x \to -1} \frac{2x^3 + 5x^2 + 4x + 1}{3x^3 + 5x^2 + x - 1}$

Solution: First, let's check if direct substitution works: At $x = -1$ : Numerator: $2(-1)^3 + 5(-1)^2 + 4(-1) + 1 = -2 + 5 - 4 + 1 = 0$ Denominator: $3(-1)^3 + 5(-1)^2 + (-1) - 1 = -3 + 5 - 1 - 1 = 0$

Since we get $\frac{0}{0}$ form, we need to factor both polynomials.

Factoring the numerator: $2x^3 + 5x^2 + 4x + 1$ Since $x = -1$ is a root, $(x + 1)$ is a factor. Using polynomial division: $2x^3 + 5x^2 + 4x + 1 = (x + 1)(2x^2 + 3x + 1)$ Further factoring: $2x^2 + 3x + 1 = (2x + 1)(x + 1)$ So: $2x^3 + 5x^2 + 4x + 1 = (x + 1)^2(2x + 1)$

Factoring the denominator: $3x^3 + 5x^2 + x - 1$ Since $x = -1$ is a root, $(x + 1)$ is a factor. Using polynomial division: $3x^3 + 5x^2 + x - 1 = (x + 1)(3x^2 + 2x - 1)$ Further factoring: $3x^2 + 2x - 1 = (3x - 1)(x + 1)$ So: $3x^3 + 5x^2 + x - 1 = (x + 1)^2(3x - 1)$

Therefore: $\lim_{x \to -1} \frac{2x^3 + 5x^2 + 4x + 1}{3x^3 + 5x^2 + x - 1} = \lim_{x \to -1} \frac{(x + 1)^2(2x + 1)}{(x + 1)^2(3x - 1)}$

$= \lim_{x \to -1} \frac{2x + 1}{3x - 1} = \frac{2(-1) + 1}{3(-1) - 1} = \frac{-1}{-4} = \frac{1}{4}$

Q.5(A) [6 marks]

Attempt any two

Q5.1 [3 marks]

Find the Limit of $\lim_{x \to 1} \frac{\sqrt{x+7} - \sqrt{3x+5}}{\sqrt{3x+5} - \sqrt{5x+3}}$

Solution: At $x = 1$ : Numerator: $\sqrt{1+7} - \sqrt{3+5} = \sqrt{8} - \sqrt{8} = 0$ Denominator: $\sqrt{3+5} - \sqrt{5+3} = \sqrt{8} - \sqrt{8} = 0$

We have $\frac{0}{0}$ form. We'll rationalize both numerator and denominator.

Rationalizing the numerator: $\sqrt{x+7} - \sqrt{3x+5} = \frac{(\sqrt{x+7} - \sqrt{3x+5})(\sqrt{x+7} + \sqrt{3x+5})}{\sqrt{x+7} + \sqrt{3x+5}}$

$= \frac{(x+7) - (3x+5)}{\sqrt{x+7} + \sqrt{3x+5}} = \frac{x + 7 - 3x - 5}{\sqrt{x+7} + \sqrt{3x+5}} = \frac{-2x + 2}{\sqrt{x+7} + \sqrt{3x+5}}$

Rationalizing the denominator: $\sqrt{3x+5} - \sqrt{5x+3} = \frac{(\sqrt{3x+5} - \sqrt{5x+3})(\sqrt{3x+5} + \sqrt{5x+3})}{\sqrt{3x+5} + \sqrt{5x+3}}$

$= \frac{(3x+5) - (5x+3)}{\sqrt{3x+5} + \sqrt{5x+3}} = \frac{3x + 5 - 5x - 3}{\sqrt{3x+5} + \sqrt{5x+3}} = \frac{-2x + 2}{\sqrt{3x+5} + \sqrt{5x+3}}$

Therefore: $\lim_{x \to 1} \frac{\sqrt{x+7} - \sqrt{3x+5}}{\sqrt{3x+5} - \sqrt{5x+3}} = \lim_{x \to 1} \frac{\frac{-2x + 2}{\sqrt{x+7} + \sqrt{3x+5}}}{\frac{-2x + 2}{\sqrt{3x+5} + \sqrt{5x+3}}}$

$= \lim_{x \to 1} \frac{-2x + 2}{\sqrt{x+7} + \sqrt{3x+5}} \times \frac{\sqrt{3x+5} + \sqrt{5x+3}}{-2x + 2}$

$= \lim_{x \to 1} \frac{\sqrt{3x+5} + \sqrt{5x+3}}{\sqrt{x+7} + \sqrt{3x+5}}$

Substituting $x = 1$ : $= \frac{\sqrt{8} + \sqrt{8}}{\sqrt{8} + \sqrt{8}} = \frac{2\sqrt{8}}{2\sqrt{8}} = 1$

Q5.2 [3 marks]

Find the Limit of $\lim_{x \to 0} \frac{\cos(ax) - \cos(bx)}{x^2}$

Solution: Using the identity: $\cos A - \cos B = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$

$\cos(ax) - \cos(bx) = -2\sin(\frac{ax + bx}{2})\sin(\frac{ax - bx}{2})$

$= -2\sin(\frac{(a+b)x}{2})\sin(\frac{(a-b)x}{2})$

Therefore: $\lim_{x \to 0} \frac{\cos(ax) - \cos(bx)}{x^2} = \lim_{x \to 0} \frac{-2\sin(\frac{(a+b)x}{2})\sin(\frac{(a-b)x}{2})}{x^2}$

$= -2 \lim_{x \to 0} \frac{\sin(\frac{(a+b)x}{2})}{x} \times \frac{\sin(\frac{(a-b)x}{2})}{x}$

$= -2 \lim_{x \to 0} \frac{\sin(\frac{(a+b)x}{2})}{\frac{(a+b)x}{2}} \times \frac{(a+b)}{2} \times \frac{\sin(\frac{(a-b)x}{2})}{\frac{(a-b)x}{2}} \times \frac{(a-b)}{2}$

Using $\lim_{u \to 0} \frac{\sin u}{u} = 1$ :

$= -2 \times 1 \times \frac{(a+b)}{2} \times 1 \times \frac{(a-b)}{2} = -2 \times \frac{(a+b)(a-b)}{4} = -\frac{(a^2 - b^2)}{2} = \frac{b^2 - a^2}{2}$

Q5.3 [3 marks]

Find the Limit of $\lim_{x \to 3} \frac{x^3 - 27}{\sqrt[3]{x} - \sqrt[3]{3}}$

Solution: Let $u = \sqrt[3]{x}$ , then $x = u^3$ and as $x \to 3$ , $u \to \sqrt[3]{3}$

$\lim_{x \to 3} \frac{x^3 - 27}{\sqrt[3]{x} - \sqrt[3]{3}} = \lim_{u \to \sqrt[3]{3}} \frac{(u^3)^3 - 27}{u - \sqrt[3]{3}} = \lim_{u \to \sqrt[3]{3}} \frac{u^9 - 27}{u - \sqrt[3]{3}}$

Since $27 = (\sqrt[3]{3})^9$ , we have: $\lim_{u \to \sqrt[3]{3}} \frac{u^9 - (\sqrt[3]{3})^9}{u - \sqrt[3]{3}}$

This is of the form $\frac{f(a) - f(b)}{a - b}$ where $f(u) = u^9$ , which gives us $f'(\sqrt[3]{3})$ .

$f'(u) = 9u^8$ $f'(\sqrt[3]{3}) = 9(\sqrt[3]{3})^8 = 9 \times 3^{8/3} = 9 \times 3^{8/3} = 9 \times (3^2)^{4/3} = 9 \times 9^{4/3} = 9 \times 9 \times 9^{1/3} = 81 \times \sqrt[3]{9}$

Alternative approach using direct factorization: $x^3 - 27 = x^3 - 3^3 = (x-3)(x^2 + 3x + 9)$

Let $y = \sqrt[3]{x}$ , then $x = y^3$ : $\sqrt[3]{x} - \sqrt[3]{3} = y - \sqrt[3]{3}$

Using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ : $x - 3 = y^3 - (\sqrt[3]{3})^3 = (y - \sqrt[3]{3})(y^2 + y\sqrt[3]{3} + (\sqrt[3]{3})^2)$

Therefore: $\lim_{x \to 3} \frac{x^3 - 27}{\sqrt[3]{x} - \sqrt[3]{3}} = \lim_{x \to 3} \frac{(x-3)(x^2 + 3x + 9)}{\sqrt[3]{x} - \sqrt[3]{3}}$

$= \lim_{x \to 3} \frac{(y - \sqrt[3]{3})(y^2 + y\sqrt[3]{3} + (\sqrt[3]{3})^2)(x^2 + 3x + 9)}{y - \sqrt[3]{3}}$

$= \lim_{x \to 3} (y^2 + y\sqrt[3]{3} + (\sqrt[3]{3})^2)(x^2 + 3x + 9)$

At $x = 3$ , $y = \sqrt[3]{3}$ : $= ((\sqrt[3]{3})^2 + \sqrt[3]{3} \cdot \sqrt[3]{3} + (\sqrt[3]{3})^2)(3^2 + 3 \cdot 3 + 9)$ $= (3^{2/3} + 3^{2/3} + 3^{2/3})(9 + 9 + 9)$ $= 3 \cdot 3^{2/3} \cdot 27 = 81 \cdot 3^{2/3} = 81\sqrt[3]{9}$

Q.5(B) [8 marks]

Attempt any two

Q5.1 [4 marks]

Find the equation of lines passing through point $A(3\sqrt{3}, 4)$ and making angle $\frac{\pi}{6}$ with line $\sqrt{3}x - 3y + 5 = 0$

Solution: Given line: $\sqrt{3}x - 3y + 5 = 0$ Rewriting in slope form: $3y = \sqrt{3}x + 5$ , so slope $m_1 = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$

Let the slope of required lines be $m_2$ .

The angle between two lines with slopes $m_1$ and $m_2$ is given by: $\tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$

Given $\theta = \frac{\pi}{6}$ , so $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$

$\frac{1}{\sqrt{3}} = \left|\frac{m_2 - \frac{1}{\sqrt{3}}}{1 + \frac{m_2}{\sqrt{3}}}\right|$

This gives us two cases:

Case 1: $\frac{1}{\sqrt{3}} = \frac{m_2 - \frac{1}{\sqrt{3}}}{1 + \frac{m_2}{\sqrt{3}}}$

$\frac{1}{\sqrt{3}}(1 + \frac{m_2}{\sqrt{3}}) = m_2 - \frac{1}{\sqrt{3}}$

$\frac{1}{\sqrt{3}} + \frac{m_2}{3} = m_2 - \frac{1}{\sqrt{3}}$

$\frac{2}{\sqrt{3}} = m_2 - \frac{m_2}{3} = \frac{2m_2}{3}$

$m_2 = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3}$

Case 2: $\frac{1}{\sqrt{3}} = -\frac{m_2 - \frac{1}{\sqrt{3}}}{1 + \frac{m_2}{\sqrt{3}}}$

Following similar steps: $m_2 = 0$

Equations of the lines: Using point-slope form with point $(3\sqrt{3}, 4)$ :

Line 1 (slope = $\sqrt{3}$ ): $y - 4 = \sqrt{3}(x - 3\sqrt{3})$ $y - 4 = \sqrt{3}x - 9$ $y = \sqrt{3}x - 5$ or $\sqrt{3}x - y - 5 = 0$

Line 2 (slope = $0$ ): $y - 4 = 0(x - 3\sqrt{3})$ $y = 4$

Q5.2 [4 marks]

Find the equation of circle passing through origin and point $(1,2)$ and whose center lies on the X-axis

Solution: Let the center of the circle be $(h, 0)$ since it lies on the X-axis. Let the radius be $r$ .

The general equation of circle with center $(h, k)$ and radius $r$ is: $(x - h)^2 + (y - k)^2 = r^2$

Since center is $(h, 0)$ : $(x - h)^2 + y^2 = r^2$

Condition 1: Circle passes through origin $(0, 0)$ $(0 - h)^2 + 0^2 = r^2$ $h^2 = r^2$ ... (1)

Condition 2: Circle passes through $(1, 2)$ $(1 - h)^2 + 2^2 = r^2$ $(1 - h)^2 + 4 = r^2$ ... (2)

From equations (1) and (2): $h^2 = (1 - h)^2 + 4$ $h^2 = 1 - 2h + h^2 + 4$ $0 = 5 - 2h$ $h = \frac{5}{2}$

From equation (1): $r^2 = h^2 = (\frac{5}{2})^2 = \frac{25}{4}$

Table: Circle Properties

Property	Value
Center	$(\frac{5}{2}, 0)$
Radius	$\frac{5}{2}$

Equation of circle: $(x - \frac{5}{2})^2 + y^2 = \frac{25}{4}$

Expanding: $x^2 - 5x + \frac{25}{4} + y^2 = \frac{25}{4}$ $x^2 + y^2 - 5x = 0$

Q5.3 [4 marks]

Find the equation of lines passing through point $A(-8, -10)$ and product of its intercepts on both axis is $-40$

Solution: Let the equation of line be $\frac{x}{a} + \frac{y}{b} = 1$ where $a$ and $b$ are x-intercept and y-intercept respectively.

Given conditions:

Line passes through $(-8, -10)$ : $\frac{-8}{a} + \frac{-10}{b} = 1$ ... (1)
Product of intercepts: $ab = -40$ ... (2)

From equation (2): $b = \frac{-40}{a}$

Substituting in equation (1): $\frac{-8}{a} + \frac{-10}{\frac{-40}{a}} = 1$

$\frac{-8}{a} + \frac{-10a}{-40} = 1$

$\frac{-8}{a} + \frac{a}{4} = 1$

Multiplying by $4a$ : $-32 + a^2 = 4a$ $a^2 - 4a - 32 = 0$ $(a - 8)(a + 4) = 0$

So $a = 8$ or $a = -4$

Case 1: $a = 8$ $b = \frac{-40}{8} = -5$ Equation: $\frac{x}{8} + \frac{y}{-5} = 1$ $\frac{x}{8} - \frac{y}{5} = 1$ $5x - 8y = 40$

Case 2: $a = -4$ $b = \frac{-40}{-4} = 10$ Equation: $\frac{x}{-4} + \frac{y}{10} = 1$ $\frac{-x}{4} + \frac{y}{10} = 1$ $-10x + 4y = 40$ $10x - 4y + 40 = 0$ $5x - 2y + 20 = 0$

The two equations are:

$5x - 8y - 40 = 0$
$5x - 2y + 20 = 0$

Mathematics Formula Cheat Sheet

Determinants

2×2 Matrix: $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$
3×3 Matrix: Expand along any row or column

Logarithms

$\log_a b \times \log_b a = 1$
$\log(xy) = \log x + \log y$
$\log(\frac{x}{y}) = \log x - \log y$
$\log(x^n) = n\log x$

Trigonometry

Basic Values:
- $\sin 30° = \frac{1}{2}$ , $\cos 30° = \frac{\sqrt{3}}{2}$ , $\tan 30° = \frac{1}{\sqrt{3}}$
- $\sin 60° = \frac{\sqrt{3}}{2}$ , $\cos 60° = \frac{1}{2}$ , $\tan 60° = \sqrt{3}$
- $\sin 45° = \cos 45° = \frac{1}{\sqrt{2}}$ , $\tan 45° = 1$
Compound Angles:
- $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
- $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$
- $\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$
Multiple Angles:
- $\sin 2A = 2\sin A \cos A$
- $\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$
- $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}$
Half Angles:
- $\sin \frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{2}}$
- $\cos \frac{A}{2} = \pm\sqrt{\frac{1 + \cos A}{2}}$
- $\tan \frac{A}{2} = \frac{1 - \cos A}{\sin A} = \frac{\sin A}{1 + \cos A}$
Sum-to-Product:
- $\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$
- $\sin A - \sin B = 2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$
- $\cos A + \cos B = 2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$
- $\cos A - \cos B = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$
Allied Angles:
- $\sin(90° - \theta) = \cos \theta$
- $\cos(90° - \theta) = \sin \theta$
- $\sin(90° + \theta) = \cos \theta$
- $\cos(90° + \theta) = -\sin \theta$
- $\sin(180° - \theta) = \sin \theta$
- $\cos(180° - \theta) = -\cos \theta$

Vectors

Dot Product: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos \theta = a_1b_1 + a_2b_2 + a_3b_3$
Cross Product: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Magnitude: $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$
Unit Vector: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$
Angle between vectors: $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
Scalar Triple Product: $\vec{a} \cdot (\vec{b} \times \vec{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$

Coordinate Geometry

Straight Lines

Slope: $m = \frac{y_2 - y_1}{x_2 - x_1}$
Point-Slope Form: $y - y_1 = m(x - x_1)$
Two-Point Form: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$
Slope-Intercept Form: $y = mx + c$
Intercept Form: $\frac{x}{a} + \frac{y}{b} = 1$
General Form: $Ax + By + C = 0$

Parallel and Perpendicular Lines

Parallel Lines: $m_1 = m_2$
Perpendicular Lines: $m_1 \times m_2 = -1$
Angle between lines: $\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|$

Circle

Standard Form: $(x - h)^2 + (y - k)^2 = r^2$
General Form: $x^2 + y^2 + 2gx + 2fy + c = 0$
Center: $(-g, -f)$
Radius: $\sqrt{g^2 + f^2 - c}$

Limits

Standard Limits:
- $\lim_{x \to 0} \frac{\sin x}{x} = 1$
- $\lim_{x \to 0} \frac{\tan x}{x} = 1$
- $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
- $\lim_{n \to \infty} (1 + \frac{1}{n})^n = e$
- $\lim_{x \to 0} (1 + x)^{1/x} = e$
L'Hôpital's Rule: If $\lim_{x \to a} \frac{f(x)}{g(x)}$ gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , then: $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$
Algebraic Limits: For polynomial $\frac{P(x)}{Q(x)}$ :
- If $P(a) \neq 0$ and $Q(a) \neq 0$ : Direct substitution
- If $P(a) = Q(a) = 0$ : Factor and cancel common factors
- For $\frac{\infty}{\infty}$ : Divide by highest power

Functions

Even Function: $f(-x) = f(x)$
Odd Function: $f(-x) = -f(x)$
Composite Function: $(f \circ g)(x) = f(g(x))$
Inverse Function: If $y = f(x)$ , then $x = f^{-1}(y)$

Useful Algebraic Identities

$(a + b)^2 = a^2 + 2ab + b^2$
$(a - b)^2 = a^2 - 2ab + b^2$
$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$
$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
$a^4 - b^4 = (a^2 + b^2)(a + b)(a - b)$

Conversion Formulas

Degrees to Radians: $\text{Radians} = \text{Degrees} \times \frac{\pi}{180}$
Radians to Degrees: $\text{Degrees} = \text{Radians} \times \frac{180}{\pi}$

Important Angles in Radians

Degrees	Radians
30°	$\frac{\pi}{6}$
45°	$\frac{\pi}{4}$
60°	$\frac{\pi}{3}$
90°	$\frac{\pi}{2}$
120°	$\frac{2\pi}{3}$
135°	$\frac{3\pi}{4}$
150°	$\frac{5\pi}{6}$
180°	$\pi$

Differentiation (Basic)

$\frac{d}{dx}(x^n) = nx^{n-1}$
$\frac{d}{dx}(\sin x) = \cos x$
$\frac{d}{dx}(\cos x) = -\sin x$
$\frac{d}{dx}(\tan x) = \sec^2 x$
$\frac{d}{dx}(e^x) = e^x$
$\frac{d}{dx}(\ln x) = \frac{1}{x}$

Problem-Solving Tips

For Determinants

Always expand along the row/column with most zeros
Factor out common terms first
Use row/column operations to create zeros

For Limits

Try direct substitution first
If you get $\frac{0}{0}$ , factor and cancel
For square roots, rationalize numerator/denominator
Use standard limit formulas

For Trigonometry

Convert everything to same angle measure (degrees or radians)
Use compound angle formulas for complex expressions
Check if angles are special angles (30°, 45°, 60°, etc.)

For Vectors

Write vectors in component form: $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$
For cross product, use determinant method
For dot product, multiply corresponding components and add

For Circle Problems

Complete the square to find center and radius
Use distance formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
Remember: All points on circle are equidistant from center

For Line Problems

Find slope first: $m = \frac{y_2-y_1}{x_2-x_1}$
Use point-slope form: $y - y_1 = m(x - x_1)$
For parallel lines: same slope
For perpendicular lines: product of slopes = -1

Memory Tips

SOHCAHTOA: Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent
CAST Rule: In quadrants I, II, III, IV - Cosine, All, Sine, Tangent are positive respectively
30-60-90 Triangle: Sides in ratio $1 : \sqrt{3} : 2$
45-45-90 Triangle: Sides in ratio $1 : 1 : \sqrt{2}$

Common Mistakes to Avoid

Sign errors in trigonometric identities
Forgetting to rationalize when dealing with surds in limits
Not checking domain for inverse trigonometric functions
Mixing up cross product and dot product formulas
Forgetting to complete the square properly in circle equations
Not factoring completely in limit problems

Quick Reference Values

$\sqrt{2} \approx 1.414$
$\sqrt{3} \approx 1.732$
$\pi \approx 3.14159$
$e \approx 2.718$

Final Tips for Exam Success

Time Management

Spend 2-3 minutes on each fill-in-the-blank question
Allocate 8-10 minutes per 3-mark question
Allow 12-15 minutes per 4-mark question
Reserve 20-25 minutes per 7-8 mark question

Question Selection Strategy

Read all options before selecting questions
Choose questions you're most confident about
Start with easier questions to build confidence

Presentation Tips

Show all working steps clearly
Draw diagrams where applicable
Use proper mathematical notation
Box your final answers

Common Topics That Appear Frequently

Trigonometric identities and compound angles
Limits involving rationalization
Vector operations (dot and cross products)
Circle and line equations
Determinant calculations

Best of luck with your exams! 🎯

Remember: Practice makes perfect. Work through similar problems multiple times to build speed and accuracy.