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#
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.