Q.1 [14 marks]

Fill in the blanks/MCQs using appropriate choice from the given options

Q1.1 [1 mark]

$\log_3 1 =$ ____

Answer: d. 0

Solution: For any base $a > 0, a \neq 1$ : $\log_a 1 = 0$ Therefore: $\log_3 1 = 0$

Q1.2 [1 mark]

If $f(x) = e^{x-1}$ then $f(1) =$ ____

Answer: c. 1

Solution: $f(x) = e^{x-1}$ $f(1) = e^{1-1} = e^0 = 1$

Q1.3 [1 mark]

$\log_5 125 =$ ____

Answer: b. 3

Solution: $\log_5 125 = \log_5 5^3 = 3$ Since $5^3 = 125$

Q1.4 [1 mark]

If $f(x) = x^3 - 7$ then $f(-2) =$ ____

Answer: c. -15

Solution: $f(x) = x^3 - 7$ $f(-2) = (-2)^3 - 7 = -8 - 7 = -15$

Q1.5 [1 mark]

Principal period of $\cos x$ is ____

Answer: c. $2\pi$

Solution: The cosine function repeats every $2\pi$ radians, so its principal period is $2\pi$ .

Q1.6 [1 mark]

$150° =$ ____

Answer: a. $\frac{5\pi}{6}$

Solution: Converting degrees to radians: $150° = 150 \times \frac{\pi}{180} = \frac{5\pi}{6}$

Q1.7 [1 mark]

$\sin^{-1}x + \cos^{-1}x =$ ____

Answer: a. $\frac{\pi}{2}$

Solution: This is a standard identity: $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$

Q1.8 [1 mark]

(1,0,0) × (1,0,0) = ____

Answer: d. (0,0,0)

Solution: Cross product of any vector with itself is zero vector: $(1,0,0) \times (1,0,0) = (0,0,0)$

Q1.9 [1 mark]

If $\vec{a} = 4\hat{i} - 3\hat{j}$ then $|\vec{a}| =$ ____

Answer: b. 5

Solution: $|\vec{a}| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Q1.10 [1 mark]

If a line makes an angle $45°$ with positive x-axis then slope of the line is ____

Answer: c. 1

Solution: Slope $m = \tan(45°) = 1$

Q1.11 [1 mark]

Radius of the circle $x^2 + y^2 = 4$ is ____

Answer: d. 2

Solution: Standard form: $x^2 + y^2 = r^2$ Comparing: $r^2 = 4$ , so $r = 2$

Q1.12 [1 mark]

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

Answer: a. 1

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

Q1.13 [1 mark]

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

Answer: d. 3

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

Q1.14 [1 mark]

$\lim_{n \to \infty} \frac{5n + 4}{4n + 5} =$ ____

Answer: c. 5/4

Solution: $\lim_{n \to \infty} \frac{5n + 4}{4n + 5} = \lim_{n \to \infty} \frac{5 + \frac{4}{n}}{4 + \frac{5}{n}} = \frac{5}{4}$

Q.2 (A) [6 marks]

Attempt any two

Q2(A).1 [3 marks]

Find value: $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$

Answer: 0

Solution: $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix} = 1(5 \times 9 - 6 \times 8) - 2(4 \times 9 - 6 \times 7) + 3(4 \times 8 - 5 \times 7)$

$= 1(45 - 48) - 2(36 - 42) + 3(32 - 35)$ $= 1(-3) - 2(-6) + 3(-3)$ $= -3 + 12 - 9 = 0$

Q2(A).2 [3 marks]

Prove that: $\log\left(\frac{x^p}{x^q}\right) + \log\left(\frac{x^q}{x^r}\right) + \log\left(\frac{x^r}{x^p}\right) = 0$

Solution: LHS = $\log\left(\frac{x^p}{x^q}\right) + \log\left(\frac{x^q}{x^r}\right) + \log\left(\frac{x^r}{x^p}\right)$

Using logarithm properties: $= \log(x^p) - \log(x^q) + \log(x^q) - \log(x^r) + \log(x^r) - \log(x^p)$ $= p\log x - q\log x + q\log x - r\log x + r\log x - p\log x$ $= 0$ = RHS

Q2(A).3 [3 marks]

Find value: $\tan(75°)$

Answer: $2 + \sqrt{3}$

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

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

$\tan(75°) = \frac{\tan 45° + \tan 30°}{1 - \tan 45° \tan 30°} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \times \frac{1}{\sqrt{3}}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$

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

Q.2 (B) [8 marks]

Attempt any two

Q2(B).1 [4 marks]

Prove that: $\frac{1}{\log_{12} 120} + \frac{1}{\log_2 120} + \frac{1}{\log_5 120} = 1$

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

LHS = $\log_{120} 12 + \log_{120} 2 + \log_{120} 5$

Using logarithm properties: $= \log_{120}(12 \times 2 \times 5) = \log_{120} 120 = 1$ = RHS

Q2(B).2 [4 marks]

Solve: $\begin{vmatrix} x & 1 & 1 \\ 1 & 2 & 1 \\ 0 & 0 & 3 \end{vmatrix} = 3$

Solution: Expanding along third row: $\begin{vmatrix} x & 1 & 1 \\ 1 & 2 & 1 \\ 0 & 0 & 3 \end{vmatrix} = 3 \begin{vmatrix} x & 1 \\ 1 & 2 \end{vmatrix}$

$= 3(2x - 1) = 6x - 3$

Given: $6x - 3 = 3$ $6x = 6$ $x = 1$

Q2(B).3 [4 marks]

If $f(x) = \frac{1-x}{1+x}$ prove that: (i) $f(x) + f\left(\frac{1}{x}\right) = 0$ (ii) $f(x) \times f(-x) = 1$

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

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

Therefore: $f(x) + f\left(\frac{1}{x}\right) = f(x) + (-f(x)) = 0$

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

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

Q.3 (A) [6 marks]

Attempt any two

Q3(A).1 [3 marks]

Prove that: $\frac{\sin(180° - x) + \cosec(180° - x) + \tan(180° + x)}{\cos(90° + x) + \sec(90° + x) + \cot(90° + x)} = -3$

Solution: Using trigonometric identities:

$\sin(180° - x) = \sin x$
$\cosec(180° - x) = \cosec x$
$\tan(180° + x) = \tan x$
$\cos(90° + x) = -\sin x$
$\sec(90° + x) = -\cosec x$
$\cot(90° + x) = -\tan x$

Numerator = $\sin x + \cosec x + \tan x$ Denominator = $-\sin x - \cosec x - \tan x = -(\sin x + \cosec x + \tan x)$

Therefore: $\frac{\sin x + \cosec x + \tan x}{-(\sin x + \cosec x + \tan x)} = -1 \neq -3$

Note: There appears to be an error in the problem statement or expected answer.

Q3(A).2 [3 marks]

Prove that: $\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right) = 45°$

Solution: Using $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$ :

$\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{2}\right) = \tan^{-1}\left(\frac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{3} \times \frac{1}{2}}\right)$

$= \tan^{-1}\left(\frac{\frac{5}{6}}{1 - \frac{1}{6}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) = 45°$

Q3(A).3 [3 marks]

Find out equation of the line whose X-intercept is 3 and Y-intercept is 2.

Solution: Using intercept form: $\frac{x}{a} + \frac{y}{b} = 1$

Where $a = 3$ (x-intercept) and $b = 2$ (y-intercept)

$\frac{x}{3} + \frac{y}{2} = 1$

Multiplying by 6: $2x + 3y = 6$

Q.3 (B) [8 marks]

Attempt any two

Q3(B).1 [4 marks]

Prove that: $\tan(70°) = \frac{\cos(25°) + \sin(25°)}{\cos(25°) - \sin(25°)}$

Solution: RHS = $\frac{\cos(25°) + \sin(25°)}{\cos(25°) - \sin(25°)}$

Dividing numerator and denominator by $\cos(25°)$ :

$= \frac{1 + \tan(25°)}{1 - \tan(25°)}$

Using $\tan(45° + θ) = \frac{1 + \tan θ}{1 - \tan θ}$ :

$= \tan(45° + 25°) = \tan(70°)$ = LHS

Q3(B).2 [4 marks]

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

Solution: Using sum-to-product formulas:

Numerator: $\sin θ + \sin 3θ + \sin 2θ = 2\sin 2θ \cos θ + \sin 2θ = \sin 2θ(2\cos θ + 1)$

Denominator: $\cos θ + \cos 3θ + \cos 2θ = 2\cos 2θ \cos θ + \cos 2θ = \cos 2θ(2\cos θ + 1)$

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

Q3(B).3 [4 marks]

If $\vec{a} = (1,2,3)$ , $\vec{b} = (4,0,0)$ and $\vec{c} = (2,0,1)$ find $2\vec{a} + 3\vec{b} - 5\vec{c}$

Solution: $2\vec{a} = 2(1,2,3) = (2,4,6)$ $3\vec{b} = 3(4,0,0) = (12,0,0)$ $5\vec{c} = 5(2,0,1) = (10,0,5)$

$2\vec{a} + 3\vec{b} - 5\vec{c} = (2,4,6) + (12,0,0) - (10,0,5)$ $= (2+12-10, 4+0-0, 6+0-5)$ $= (4,4,1)$

Q.4 (A) [6 marks]

Attempt any two

Q4(A).1 [3 marks]

If the vectors $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} + m\hat{j} - 4\hat{k}$ are perpendicular, find m.

Solution: For perpendicular vectors: $\vec{a} \cdot \vec{b} = 0$

$\vec{a} \cdot \vec{b} = (1)(2) + (-2)(m) + (3)(-4) = 2 - 2m - 12 = -10 - 2m$

Setting equal to zero: $-10 - 2m = 0$ $2m = -10$ $m = -5$

Q4(A).2 [3 marks]

Find the direction cosines and direction angles of the vector $\vec{a} = 5\hat{i} - 12\hat{k}$

Solution: $\vec{a} = 5\hat{i} + 0\hat{j} - 12\hat{k}$

Magnitude: $|\vec{a}| = \sqrt{5^2 + 0^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

Direction cosines:

$l = \frac{5}{13}$
$m = \frac{0}{13} = 0$
$n = \frac{-12}{13}$

Direction angles:

$α = \cos^{-1}\left(\frac{5}{13}\right)$
$β = \cos^{-1}(0) = 90°$
$γ = \cos^{-1}\left(\frac{-12}{13}\right)$

Q4(A).3 [3 marks]

Find out equation of the circle having center at $(2, -3)$ and radius 3.

Solution: Standard form: $(x - h)^2 + (y - k)^2 = r^2$

Where $(h, k) = (2, -3)$ and $r = 3$

$(x - 2)^2 + (y + 3)^2 = 9$

Expanding: $x^2 - 4x + 4 + y^2 + 6y + 9 = 9$ $x^2 + y^2 - 4x + 6y + 4 = 0$

Q.4 (B) [8 marks]

Attempt any two

Q4(B).1 [4 marks]

Show that the angle between vectors $\vec{a} = \hat{i} + 2\hat{j}$ and $\vec{b} = \hat{i} + \hat{j} + 3\hat{k}$ is $\sin^{-1}\sqrt{\frac{46}{55}}$

Solution: $\vec{a} \cdot \vec{b} = (1)(1) + (2)(1) + (0)(3) = 1 + 2 = 3$

$|\vec{a}| = \sqrt{1^2 + 2^2} = \sqrt{5}$ $|\vec{b}| = \sqrt{1^2 + 1^2 + 3^2} = \sqrt{11}$

$\cos θ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{3}{\sqrt{5}\sqrt{11}} = \frac{3}{\sqrt{55}}$

$\sin^2 θ = 1 - \cos^2 θ = 1 - \frac{9}{55} = \frac{46}{55}$

Therefore: $θ = \sin^{-1}\sqrt{\frac{46}{55}}$

Q4(B).2 [4 marks]

Under effect of the forces $2\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 3\hat{j} - \hat{k}$ a particle moves from the point $(1,2,-3)$ to the point $(5,3,7)$ . Find out work done.

Solution: Net force: $\vec{F} = (2\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 3\hat{j} - \hat{k}) = 3\hat{i} + 4\hat{j}$

Displacement: $\vec{s} = (5,3,7) - (1,2,-3) = (4,1,10)$

Work done: $W = \vec{F} \cdot \vec{s} = (3)(4) + (4)(1) + (0)(10) = 12 + 4 = 16$ units

Q4(B).3 [4 marks]

Evaluate: $\lim_{x \to 0} \frac{2^x - 5^x}{x}$

Solution: Using L'Hôpital's rule or the derivative definition:

$\lim_{x \to 0} \frac{2^x - 5^x}{x} = \lim_{x \to 0} \frac{2^x \ln 2 - 5^x \ln 5}{1}$

$= 2^0 \ln 2 - 5^0 \ln 5 = \ln 2 - \ln 5 = \ln\left(\frac{2}{5}\right)$

Q.5 (A) [6 marks]

Attempt any two

Q5(A).1 [3 marks]

Evaluate: $\lim_{x \to 0} \left(1 + \frac{3x}{7}\right)^{\frac{1}{x}}$

Solution: Let $y = \left(1 + \frac{3x}{7}\right)^{\frac{1}{x}}$

Taking natural log: $\ln y = \frac{1}{x} \ln\left(1 + \frac{3x}{7}\right)$

$\lim_{x \to 0} \ln y = \lim_{x \to 0} \frac{\ln\left(1 + \frac{3x}{7}\right)}{x}$

Using L'Hôpital's rule: $= \lim_{x \to 0} \frac{\frac{3/7}{1 + \frac{3x}{7}}}{1} = \frac{3}{7}$

Therefore: $\lim_{x \to 0} y = e^{3/7}$

Q5(A).2 [3 marks]

Evaluate: $\lim_{x \to 3} \frac{x^2 - 5x + 6}{x^2 - 9}$

Solution: Factoring numerator: $x^2 - 5x + 6 = (x-2)(x-3)$ Factoring denominator: $x^2 - 9 = (x-3)(x+3)$

$\lim_{x \to 3} \frac{x^2 - 5x + 6}{x^2 - 9} = \lim_{x \to 3} \frac{(x-2)(x-3)}{(x-3)(x+3)} = \lim_{x \to 3} \frac{x-2}{x+3} = \frac{3-2}{3+3} = \frac{1}{6}$

Q5(A).3 [3 marks]

Evaluate: $\lim_{x \to 0} \frac{\sqrt{4+x} - 2}{x}$

Solution: Rationalizing the numerator:

$\lim_{x \to 0} \frac{\sqrt{4+x} - 2}{x} \times \frac{\sqrt{4+x} + 2}{\sqrt{4+x} + 2}$

$= \lim_{x \to 0} \frac{(4+x) - 4}{x(\sqrt{4+x} + 2)} = \lim_{x \to 0} \frac{x}{x(\sqrt{4+x} + 2)} = \lim_{x \to 0} \frac{1}{\sqrt{4+x} + 2} = \frac{1}{2+2} = \frac{1}{4}$

Q.5 (B) [8 marks]

Attempt any two

Q5(B).1 [4 marks]

Find out equation of the line passing through points $(1,2)$ and $(2,1)$ .

Solution: Using two-point form: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$

$\frac{y - 2}{1 - 2} = \frac{x - 1}{2 - 1}$

$\frac{y - 2}{-1} = \frac{x - 1}{1}$

$y - 2 = -(x - 1) = -x + 1$

$x + y = 3$

Q5(B).2 [4 marks]

Find equation of the line that passes through $(-3, 2)$ and parallel to the line $x - 2y + 1 = 0$

Solution: The given line $x - 2y + 1 = 0$ has slope $m = \frac{1}{2}$

Since parallel lines have the same slope, required line has slope $m = \frac{1}{2}$

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

$y - 2 = \frac{1}{2}(x - (-3))$

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

$2y - 4 = x + 3$

$x - 2y + 7 = 0$

Q5(B).3 [4 marks]

Find out center and radius of the circle: $x^2 + y^2 + 6x - 4y - 3 = 0$

Solution: Completing the square:

$x^2 + 6x + y^2 - 4y = 3$

$(x^2 + 6x + 9) + (y^2 - 4y + 4) = 3 + 9 + 4$

$(x + 3)^2 + (y - 2)^2 = 16$

Center: $(-3, 2)$ Radius: $r = \sqrt{16} = 4$

Formula Cheat Sheet

Logarithms

$\log_a 1 = 0$
$\log_a a = 1$
$\log_a(xy) = \log_a x + \log_a y$
$\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$

Trigonometry

$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$
$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$
$\sin(180° - x) = \sin x$ , $\cos(90° + x) = -\sin x$

Vectors

$|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$
$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos θ$
For perpendicular vectors: $\vec{a} \cdot \vec{b} = 0$

Coordinate Geometry

Two-point form: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$
Circle: $(x - h)^2 + (y - k)^2 = r^2$
Parallel lines have equal slopes

Limits

$\lim_{x \to 0} \frac{\sin x}{x} = 1$
$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$
$\lim_{x \to \infty} \frac{ax + b}{cx + d} = \frac{a}{c}$

Problem-Solving Strategies

Logarithms: Use properties to simplify expressions
Trigonometry: Apply compound angle formulas and identities
Vectors: Remember dot and cross product properties

Common Mistakes to Avoid

Logarithms

Mistake: Confusing $\log_a b$ with $\log_b a$
Solution: Remember change of base: $\frac{1}{\log_a b} = \log_b a$

Trigonometry

Mistake: Wrong angle conversions between degrees and radians
Solution: Always use $180° = \pi$ radians for conversion

Vectors

Mistake: Confusing dot product with cross product
Solution: Dot product gives scalar, cross product gives vector

Limits

Mistake: Direct substitution in indeterminate forms
Solution: Use algebraic manipulation, L'Hôpital's rule, or standard limits

Determinants

Mistake: Sign errors in expansion
Solution: Follow the checkerboard pattern carefully

Exam Tips

Time Management

Q1 (14 marks): 20-25 minutes - Quick calculations
Q2-Q5: 35-40 minutes each - Show all steps clearly

Strategy

Read all questions first - Choose easier OR options
Start with Q1 - Build confidence with MCQs
Show work clearly - Partial credit is available
Use standard formulas - Don't derive unless asked

Key Points to Remember

Always write the final answer clearly
Use proper mathematical notation
Draw diagrams where helpful
Check units in physics-related problems (work, force)

Calculator Usage

Scientific calculator allowed
Use for complex arithmetic only
Show the setup before calculating
Round final answers appropriately

Common Formula Applications

Standard Limits (Memory aids)

lim(x→0) sin(x)/x = 1         "Sine over x is one"
lim(x→0) (e^x - 1)/x = 1      "e minus one over x is one"  
lim(x→0) (a^x - 1)/x = ln(a)  "General exponential form"

Trigonometric Identities (Quick Reference)

graph LR
    A[sin²θ + cos²θ = 1] --> B[1 + tan²θ = sec²θ]
    A --> C[1 + cot²θ = cosec²θ]
    D["sin(A±B)"] --> E[sin A cos B ± cos A sin B]
    F["cos(A±B)"] --> G[cos A cos B ∓ sin A sin B]

Vector Operations (Step-by-step)

Magnitude: $|\vec{a}| = \sqrt{sum \, of \, squares}$
Dot Product: $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$
Angle: $\cos θ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

Circle Equations (Forms)

Form	Equation	When to Use
Standard	$(x-h)² + (y-k)² = r²$	Given center and radius
General	$x² + y² + Dx + Ey + F = 0$	Need to find center/radius
Complete Square	$(x+D/2)² + (y+E/2)² = (D²+E²-4F)/4$	Converting general to standard

Problem-Specific Strategies

For Determinant Problems

Look for zeros to simplify expansion
Use row/column operations if allowed
Remember: if two rows/columns are proportional, determinant = 0

For Limit Problems

    Start with limit
         |
    Direct substitution?
      /        \
    Yes         No (0/0, ∞/∞, etc.)
     |           |
   Answer    Try factoring/
             rationalization
                 |
             Still indeterminate?
                 |
            L'Hôpital's Rule
                 |
             Find answer

For Vector Problems

Step 1: Write vectors in component form
Step 2: Apply required operation (dot/cross product)
Step 3: Simplify and find magnitude if needed
Step 4: Check perpendicularity condition ( $\vec{a} \cdot \vec{b} = 0$ )

For Coordinate Geometry

Line problems: Identify what's given (points, slope, parallel/perpendicular)
Circle problems: Identify center and radius from given information
Always check your equation by substituting known points

Memory Techniques

Logarithm Properties (MNEMONIC: "PLUS")

Product: $\log(ab) = \log a + \log b$
Limit: $\log_a 1 = 0$
Unity: $\log_a a = 1$
Subtraction: $\log(a/b) = \log a - \log b$

Trigonometric Values (30°, 45°, 60°)

Angle	sin	cos	tan
30°	1/2	√3/2	1/√3
45°	1/√2	1/√2	1
60°	√3/2	1/2	√3

Memory aid: "1, 2, 3" under square roots for sin values (30° to 60°)

Final Review Checklist

Before submitting your paper:

All questions attempted as required
Final answers clearly marked
Units included where applicable
No arithmetic errors in simple calculations
Proper mathematical notation used
Diagrams labeled clearly (if drawn)

Quick Problem Solving Guide

If you're stuck on a problem:

Read the problem again - Often missed details become clear
Try a different approach - Multiple methods usually exist
Work backwards - Start from what you want to prove/find
Use elimination - In MCQs, eliminate obviously wrong options
Move on and return - Don't spend too much time on one problem

Last 15 minutes strategy:

Focus on completing MCQs in Q1
Check arithmetic in longer problems
Ensure all final answers are clearly marked
Review any skipped parts of questions

Remember: This exam tests fundamental concepts. Focus on understanding rather than memorizing, and always show your reasoning clearly for maximum partial credit.