Mathematics-I (DI01000021) - Winter 2024 Solution

Q.1 [14 marks]

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

Q1.1 [1 mark]

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$\begin{vmatrix} 5 & 1 \ 2 & 3 \end{vmatrix} = $ _______

Answer: b. 13

Solution: For 2×2 determinant $\begin{vmatrix} a & b \ c & d \end{vmatrix} = ad - bc$

$\begin{vmatrix} 5 & 1 \ 2 & 3 \end{vmatrix} = (5 \times 3) - (1 \times 2) = 15 - 2 = 13$

Q1.2 [1 mark]

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If $\begin{vmatrix} x & 1 \ 2 & 1 \end{vmatrix} = 0$ then $x = $ _______

Answer: b. 2

Solution: $\begin{vmatrix} x & 1 \ 2 & 1 \end{vmatrix} = x \times 1 - 1 \times 2 = x - 2 = 0$

Therefore, $x = 2$

Q1.3 [1 mark]

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If $f(x) = x^2$ then $f(-1) = $ _______

Answer: a. 1

Solution: $f(x) = x^2$ $f(-1) = (-1)^2 = 1$

Q1.4 [1 mark]

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$\log_{10} 1 = $ _______

Answer: b. 0

Solution: By logarithm property: $\log_a 1 = 0$ for any base $a > 0$ Therefore, $\log_{10} 1 = 0$

Q1.5 [1 mark]

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$\sin \frac{\pi}{2} + \cos \frac{\pi}{2} = $ _______

Answer: c. 1

Solution: $\sin \frac{\pi}{2} = 1$ and $\cos \frac{\pi}{2} = 0$ Therefore, $\sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 + 0 = 1$

Q1.6 [1 mark]

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$\tan^{-1}(1) = $ _______

Answer: a. $\frac{\pi}{4}$

Solution: $\tan \frac{\pi}{4} = 1$ Therefore, $\tan^{-1}(1) = \frac{\pi}{4}$

Q1.7 [1 mark]

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$\frac{2\pi}{3}$ radian = _______ degree

Answer: d. 120

Solution: To convert radians to degrees: $\text{degrees} = \text{radians} \times \frac{180}{\pi}$ $\frac{2\pi}{3} \times \frac{180}{\pi} = \frac{2 \times 180}{3} = \frac{360}{3} = 120°$

Q1.8 [1 mark]

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$\hat{i} \times \hat{j} = $ _______

Answer: c. $\hat{k}$

Solution: By right-hand rule for cross product: $\hat{i} \times \hat{j} = \hat{k}$

Q1.9 [1 mark]

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$|\hat{i} + \hat{j} + \hat{k}| = $ _______

Answer: d. $\sqrt{3}$

Solution: $|\hat{i} + \hat{j} + \hat{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

Q1.10 [1 mark]

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Slope of line $2x + y - 3 = 0$ is _______

Answer: a. -2

Solution: Convert to slope-intercept form: $y = -2x + 3$ Slope = coefficient of $x = -2$

Q1.11 [1 mark]

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Radius of circle $x^2 + y^2 = 81$ is _______

Answer: b. 9

Solution: Standard form: $x^2 + y^2 = r^2$ Here, $r^2 = 81$, so $r = 9$

Q1.12 [1 mark]

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$\lim_{n \to \infty} \frac{1}{n} = $ _______

Answer: c. 0

Solution: As $n$ approaches infinity, $\frac{1}{n}$ approaches 0

Q1.13 [1 mark]

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$\lim_{x \to 1} (x^2 + x + 1) = $ _______

Answer: a. 3

Solution: Direct substitution: $(1)^2 + (1) + 1 = 1 + 1 + 1 = 3$

Q1.14 [1 mark]

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$\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = $ _______

Answer: b. 1

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

Q.2 (A) [6 marks]

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Attempt any two

Q2.1 [3 marks]

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Find the value of $\begin{vmatrix} 1 & 3 & 1 \ 2 & -1 & 0 \ 4 & -2 & 5 \end{vmatrix}$

Answer:

Solution: Using expansion along second row (has zero): $= -2\begin{vmatrix} 3 & 1 \ -2 & 5 \end{vmatrix} + (-1)\begin{vmatrix} 1 & 1 \ 4 & 5 \end{vmatrix} + 0$

$= -2(15 + 2) - 1(5 - 4)$ $= -2(17) - 1(1)$ $= -34 - 1 = -35$

Table:

Q2.2 [3 marks]

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If $f(x) = x^3 + 5$ then find $f(0)$, $f(1)$ and $f(-1)$

Answer:

Solution: Given: $f(x) = x^3 + 5$

$f(0) = (0)^3 + 5 = 0 + 5 = 5$ $f(1) = (1)^3 + 5 = 1 + 5 = 6$ $f(-1) = (-1)^3 + 5 = -1 + 5 = 4$

Table:

Q2.3 [3 marks]

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Prove that $\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{4}$

Answer:

Solution: Using formula: $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$

Let $a = \frac{1}{2}$, $b = \frac{1}{3}$

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

$= \tan^{-1}\left(\frac{\frac{5}{6}}{1 - \frac{1}{6}}\right) = \tan^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right) = \tan^{-1}(1) = \frac{\pi}{4}$

Hence proved.

Q.2 (B) [8 marks]

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Attempt any two

Q2.1 [4 marks]

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

Answer:

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

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

Now calculate $f(x) \cdot f(-x)$: $f(x) \cdot f(-x) = \frac{x-1}{x+1} \cdot \frac{x+1}{x-1} = \frac{(x-1)(x+1)}{(x+1)(x-1)} = 1$

Hence proved.

Q2.2 [4 marks]

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If $\log\left(\frac{x+y}{2}\right) = \frac{1}{2}(\log x + \log y)$ then prove that $x = y$

Answer:

Solution: Given: $\log\left(\frac{x+y}{2}\right) = \frac{1}{2}(\log x + \log y)$

Using logarithm properties: $\frac{1}{2}(\log x + \log y) = \frac{1}{2}\log(xy) = \log\sqrt{xy}$

So: $\log\left(\frac{x+y}{2}\right) = \log\sqrt{xy}$

Taking antilog: $\frac{x+y}{2} = \sqrt{xy}$

Squaring both sides: $\left(\frac{x+y}{2}\right)^2 = xy$

$\frac{(x+y)^2}{4} = xy$

$(x+y)^2 = 4xy$

$x^2 + 2xy + y^2 = 4xy$

$x^2 - 2xy + y^2 = 0$

$(x-y)^2 = 0$

Therefore, $x = y$. Hence proved.

Q2.3 [4 marks]

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Solve $\log(x+3) + \log(x-3) = \log 27$

Answer:

Solution: Given: $\log(x+3) + \log(x-3) = \log 27$

Using logarithm property: $\log a + \log b = \log(ab)$ $\log[(x+3)(x-3)] = \log 27$

Taking antilog: $(x+3)(x-3) = 27$

$x^2 - 9 = 27$

$x^2 = 36$

$x = \pm 6$

Check validity:

• For $x = 6$: $x+3 = 9 > 0$ and $x-3 = 3 > 0$ ✓
• For $x = -6$: $x+3 = -3 < 0$ (invalid for logarithm)

Therefore, $x = 6$

Q.3 (A) [6 marks]

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Attempt any two

Q3.1 [3 marks]

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Prove that $\frac{\sin\left(\frac{\pi}{2}+\theta\right)}{\cos(\pi-\theta)} + \frac{\tan(\pi-\theta)}{\cot\left(\frac{3\pi}{2}-\theta\right)} + \frac{\text{cosec}\left(\frac{\pi}{2}-\theta\right)}{\sec(\pi+\theta)} = -3$

Answer:

Solution: Using trigonometric identities:

$\sin\left(\frac{\pi}{2}+\theta\right) = \cos\theta$ $\cos(\pi-\theta) = -\cos\theta$ $\tan(\pi-\theta) = -\tan\theta$ $\cot\left(\frac{3\pi}{2}-\theta\right) = \tan\theta$ $\text{cosec}\left(\frac{\pi}{2}-\theta\right) = \sec\theta$ $\sec(\pi+\theta) = -\sec\theta$

Substituting: $\frac{\cos\theta}{-\cos\theta} + \frac{-\tan\theta}{\tan\theta} + \frac{\sec\theta}{-\sec\theta}$

$= -1 + (-1) + (-1) = -3$

Hence proved.

Q3.2 [3 marks]

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Prove that $\tan 55° = \frac{\cos 10° + \sin 10°}{\cos 10° - \sin 10°}$

Answer:

Solution: We know that $\tan 55° = \tan(45° + 10°)$

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

$\tan 55° = \frac{\tan 45° + \tan 10°}{1 - \tan 45° \tan 10°} = \frac{1 + \tan 10°}{1 - \tan 10°}$

Now, $\tan 10° = \frac{\sin 10°}{\cos 10°}$

$\tan 55° = \frac{1 + \frac{\sin 10°}{\cos 10°}}{1 - \frac{\sin 10°}{\cos 10°}} = \frac{\cos 10° + \sin 10°}{\cos 10° - \sin 10°}$

Hence proved.

Q3.3 [3 marks]

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

Answer:

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

$2\vec{a} = 2(2\hat{i} + 3\hat{j} + \hat{k}) = 4\hat{i} + 6\hat{j} + 2\hat{k}$

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

$= (4 + 1 - 3)\hat{i} + (6 + 1 - 1)\hat{j} + (2 + 1 - 1)\hat{k}$

$= 2\hat{i} + 6\hat{j} + 2\hat{k}$

Q.3 (B) [8 marks]

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Attempt any two

Q3.1 [4 marks]

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Prove that $\frac{\sin(x-y)}{\cos x \cos y} + \frac{\sin(y-z)}{\cos y \cos z} + \frac{\sin(z-x)}{\cos z \cos x} = 0$

Answer:

Solution: Using identity: $\sin(A-B) = \sin A \cos B - \cos A \sin B$

$\frac{\sin(x-y)}{\cos x \cos y} = \frac{\sin x \cos y - \cos x \sin y}{\cos x \cos y} = \tan x - \tan y$

Similarly: $\frac{\sin(y-z)}{\cos y \cos z} = \tan y - \tan z$ $\frac{\sin(z-x)}{\cos z \cos x} = \tan z - \tan x$

Adding all three: $(\tan x - \tan y) + (\tan y - \tan z) + (\tan z - \tan x) = 0$

Hence proved.

Q3.2 [4 marks]

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

Answer:

Solution:

Table of values:

Q3.3 [4 marks]

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Find equation of line passing through (1, 2) and (-3, 1)

Answer:

Solution: Given points: $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (-3, 1)$

Slope: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 2}{-3 - 1} = \frac{-1}{-4} = \frac{1}{4}$

Using point-slope form: $y - y_1 = m(x - x_1)$ $y - 2 = \frac{1}{4}(x - 1)$ $4(y - 2) = x - 1$ $4y - 8 = x - 1$ $x - 4y + 7 = 0$

Equation: $x - 4y + 7 = 0$

Q.4 (A) [6 marks]

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Attempt any two

Q4.1 [3 marks]

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Find unit vector perpendicular to $\vec{a} = \hat{i} - 3\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$

Answer:

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

$= \hat{i}[(-3)(2) - (1)(1)] - \hat{j}[(1)(2) - (1)(2)] + \hat{k}[(1)(1) - (-3)(2)]$ $= \hat{i}(-6-1) - \hat{j}(2-2) + \hat{k}(1+6)$ $= -7\hat{i} + 0\hat{j} + 7\hat{k}$

Magnitude: $|\vec{a} \times \vec{b}| = \sqrt{(-7)^2 + 0^2 + 7^2} = \sqrt{49 + 49} = 7\sqrt{2}$

Unit vector: $\hat{n} = \frac{-7\hat{i} + 7\hat{k}}{7\sqrt{2}} = \frac{-\hat{i} + \hat{k}}{\sqrt{2}}$

Q4.2 [3 marks]

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Forces (1, 2, 1) and (2, -1, 3) act on a particle and the particle moves from point (2, 3, 1) to (4, 6, 2). Find the work done.

Answer:

Solution: Resultant force: $\vec{F} = (1, 2, 1) + (2, -1, 3) = (3, 1, 4)$

Displacement: $\vec{s} = (4, 6, 2) - (2, 3, 1) = (2, 3, 1)$

Work done: $W = \vec{F} \cdot \vec{s} = (3)(2) + (1)(3) + (4)(1) = 6 + 3 + 4 = 13$ units

Q4.3 [3 marks]

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Show that lines $2x - 3y + 5 = 0$ and $8x - 12y - 3 = 0$ are parallel lines.

Answer:

Solution: For line $2x - 3y + 5 = 0$: slope $m_1 = \frac{2}{3}$ For line $8x - 12y - 3 = 0$: slope $m_2 = \frac{8}{12} = \frac{2}{3}$

Since $m_1 = m_2 = \frac{2}{3}$, the lines are parallel.

Table:

Q.4 (B) [8 marks]

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Attempt any two

Q4.1 [4 marks]

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Show that angle between $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$ is $\sin^{-1}\left(\frac{\sqrt{26}}{27}\right)$

Answer:

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

$|\vec{a}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$ $|\vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3$

$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{-1}{\sqrt{3} \times 3} = \frac{-1}{3\sqrt{3}}$

$\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{1}{27} = \frac{26}{27}$

Therefore, $\sin\theta = \frac{\sqrt{26}}{3\sqrt{3}} = \frac{\sqrt{26}}{\sqrt{27}}$

Hence, $\theta = \sin^{-1}\left(\frac{\sqrt{26}}{\sqrt{27}}\right)$

Q4.2 [4 marks]

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If $\vec{a} = (1, 1, 1)$, $\vec{b} = (2, 0, 1)$ and $\vec{c} = (-2, 1, 0)$ then find $\vec{a} \cdot (\vec{b} \times \vec{c})$

Answer:

Solution: First find $\vec{b} \times \vec{c}$: $\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 0 & 1 \ -2 & 1 & 0 \end{vmatrix}$

$= \hat{i}(0 \times 0 - 1 \times 1) - \hat{j}(2 \times 0 - 1 \times (-2)) + \hat{k}(2 \times 1 - 0 \times (-2))$ $= \hat{i}(-1) - \hat{j}(2) + \hat{k}(2)$ $= -\hat{i} - 2\hat{j} + 2\hat{k}$

Now find $\vec{a} \cdot (\vec{b} \times \vec{c})$: $\vec{a} \cdot (\vec{b} \times \vec{c}) = (1, 1, 1) \cdot (-1, -2, 2)$ $= (1)(-1) + (1)(-2) + (1)(2) = -1 - 2 + 2 = -1$

Q4.3 [4 marks]

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Evaluate $\lim_{\theta \to 0} \frac{\sin 4\theta}{\theta}$

Answer:

Solution: $\lim_{\theta \to 0} \frac{\sin 4\theta}{\theta} = \lim_{\theta \to 0} \frac{\sin 4\theta}{4\theta} \times 4$

Using standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$:

Let $u = 4\theta$, then as $\theta \to 0$, $u \to 0$

$\lim_{\theta \to 0} \frac{\sin 4\theta}{4\theta} = \lim_{u \to 0} \frac{\sin u}{u} = 1$

Therefore, $\lim_{\theta \to 0} \frac{\sin 4\theta}{\theta} = 4 \times 1 = 4$

Q.5 (A) [6 marks]

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Attempt any two

Q5.1 [3 marks]

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Evaluate $\lim_{x \to 9} \frac{x^2 - 81}{x - 9}$

Answer:

Solution: Direct substitution gives $\frac{0}{0}$ form.

Factor the numerator: $x^2 - 81 = (x-9)(x+9)$

$\lim_{x \to 9} \frac{x^2 - 81}{x - 9} = \lim_{x \to 9} \frac{(x-9)(x+9)}{x-9}$

$= \lim_{x \to 9} (x+9) = 9 + 9 = 18$

Q5.2 [3 marks]

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

Answer:

Solution: Let $y = \left(1 + \frac{3}{x}\right)^{2x}$

Taking natural logarithm: $\ln y = 2x \ln\left(1 + \frac{3}{x}\right)$

As $x \to \infty$, $\frac{3}{x} \to 0$

Using $\ln(1+u) \approx u$ for small $u$: $\ln y = 2x \times \frac{3}{x} = 6$

Therefore, $y = e^6$

Q5.3 [3 marks]

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Evaluate $\lim_{x \to 1} \frac{x - 1}{x^2 + x - 2}$

Answer:

Solution: Factor the denominator: $x^2 + x - 2 = (x+2)(x-1)$

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

$= \lim_{x \to 1} \frac{1}{x+2} = \frac{1}{1+2} = \frac{1}{3}$

Q.5 (B) [8 marks]

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Attempt any two

Q5.1 [4 marks]

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Find the equation of line passing through the point (2, -3) and having slope 4.

Answer:

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

Given: $(x_1, y_1) = (2, -3)$ and slope $m = 4$

$y - (-3) = 4(x - 2)$ $y + 3 = 4x - 8$ $y = 4x - 11$

Equation: $y = 4x - 11$ or $4x - y - 11 = 0

Q5.2 [4 marks]

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For what value of m, lines $7x + y - 1 = 0$ and $3x - my + 2 = 0$ are perpendicular to each other.

Answer:

Solution: For perpendicular lines, product of slopes = -1

For line $7x + y - 1 = 0$: slope $m_1 = -7$ For line $3x - my + 2 = 0$: slope $m_2 = \frac{3}{m}$

Condition: $m_1 \times m_2 = -1$ $(-7) \times \frac{3}{m} = -1$ $\frac{-21}{m} = -1$ $21 = m$

Therefore, $m = 21$

Table:

Verification: When $m = 21$, slopes are $-7$ and $\frac{3}{21} = \frac{1}{7}$ Product: $(-7) \times \frac{1}{7} = -1$ ✓

Q5.3 [4 marks]

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

Answer:

Solution: First, divide by 4 to get standard form: $x^2 + y^2 + 2x - 3y - \frac{3}{4} = 0$

Complete the square for x and y terms: $x^2 + 2x = (x+1)^2 - 1$ $y^2 - 3y = \left(y - \frac{3}{2}\right)^2 - \frac{9}{4}$

Substituting: $(x+1)^2 - 1 + \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{3}{4} = 0$

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

Centre: $(-1, \frac{3}{2})$ Radius: $r = \sqrt{4} = 2$

Table:

Formula Cheat Sheet

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Determinants

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

Functions & Logarithms

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• Basic: $\log_a 1 = 0$, $\log_a a = 1$
• Properties: $\log(ab) = \log a + \log b$, $\log\left(\frac{a}{b}\right) = \log a - \log b$

Trigonometry

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• Basic Values: $\sin 0° = 0$, $\sin 30° = \frac{1}{2}$, $\sin 45° = \frac{\sqrt{2}}{2}$, $\sin 60° = \frac{\sqrt{3}}{2}$, $\sin 90° = 1$
• Conversion: Radians to degrees: $\times \frac{180}{\pi}$
• Identities: $\sin^2\theta + \cos^2\theta = 1$
• Inverse: $\tan^{-1}(1) = \frac{\pi}{4}$

Vectors

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• Magnitude: $|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}$
• Dot Product: $\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z$
• Cross Product: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$
• Work Done: $W = \vec{F} \cdot \vec{s}$

Coordinate Geometry

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• Slope: $m = \frac{y_2 - y_1}{x_2 - x_1}$
• Point-Slope Form: $y - y_1 = m(x - x_1)$
• Parallel Lines: Same slope
• Perpendicular Lines: Product of slopes = -1
• Circle: $(x-h)^2 + (y-k)^2 = r^2$

Limits

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• Standard Limits: $\lim_{x \to 0} \frac{\sin x}{x} = 1$, $\lim_{x \to 0} \frac{\tan x}{x} = 1$
• Factorization: Use for $\frac{0}{0}$ forms
• L’Hôpital’s Rule: For indeterminate forms

Problem-Solving Strategies

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For Determinants:

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1. Choose the row/column with most zeros for expansion
2. Use cofactor expansion systematically
3. Check calculations by expanding along different rows

For Functions:

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1. Direct substitution first
2. Use function properties and definitions
3. Check domain restrictions

For Trigonometry:

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1. Convert all angles to same unit (degrees or radians)
2. Use standard angle values
3. Apply appropriate identities
4. Simplify step by step

For Vectors:

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1. Write components clearly
2. Use right-hand rule for cross products
3. Check units and directions
4. Verify with geometric interpretation

For Coordinate Geometry:

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1. Plot points when possible
2. Use appropriate formulas based on given information
3. Check parallel/perpendicular conditions
4. Complete the square for circles

For Limits:

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1. Try direct substitution first
2. Factor polynomials for $\frac{0}{0}$ forms
3. Use standard limit formulas
4. Apply L’Hôpital’s rule for indeterminate forms

Common Mistakes to Avoid

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Determinants:

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• ❌ Wrong sign in calculations
• ✅ Follow cofactor signs carefully: $(-1)^{i+j}$

Logarithms:

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• ❌ $\log(a+b) = \log a + \log b$ (WRONG)
• ✅ $\log(ab) = \log a + \log b$ (CORRECT)

Trigonometry:

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• ❌ Mixing degrees and radians
• ✅ Convert to same unit first

Vectors:

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• ❌ $\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$ (WRONG)
• ✅ $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ (CORRECT)

Slopes:

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• ❌ Confusing parallel and perpendicular conditions
• ✅ Parallel: same slope, Perpendicular: product = -1

Limits:

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• ❌ Direct substitution without checking indeterminate forms
• ✅ Check for $\frac{0}{0}$ or $\frac{\infty}{\infty}$ first

Exam Tips

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Time Management:

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• Spend 2 minutes per mark (14 marks = 28 minutes for Q1)
• Start with familiar questions
• Leave difficult problems for the end

Calculation Tips:

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• Show all steps clearly
• Use tables for organized presentation
• Double-check arithmetic
• Write final answers clearly

Writing Strategy:

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• Write given information first
• State formulas before using them
• Include units where applicable
• Box or underline final answers

Last-Minute Checks:

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• Verify all calculations
• Check if answers are reasonable
• Ensure all parts are attempted
• Review question requirements

Mnemonic for Standard Angles: “Some People Have Curly Brown Hair Through Proper Brushing”

• Sin 0° = 0, Pi/6 = 1/2, Half = √2/2, Cos complement, etc.

Remember: Mathematics is about understanding patterns, not memorizing formulas. Practice regularly and think step by step!

Quick Reference Table

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Final Tip: Keep practicing and stay confident! 🎯