1. Problem: Find the value of $\vec{i} \cdot \vec{j}$ where $\vec{i}$ and $\vec{j}$ are standard unit vectors along X and Y axes respectively. Step 1: Recall that $\vec{i} = (1,0)$ and $\vec{j} = (0,1)$. Step 2: The dot product is $\vec{i} \cdot \vec{j} = 1 \times 0 + 0 \times 1 = 0$. Answer: $\boxed{0}$. 2. Problem: Find $Q(x)$ and $R$ such that $4x^{3} - 5x^{2} + 8 = (x + 2) \times Q(x) + R$ using synthetic division. Step 1: Use synthetic division with root $-2$. Coefficients: 4 (for $x^3$), -5 (for $x^2$), 0 (for $x$ term missing), 8 (constant). Step 2: Bring down 4. Multiply by -2: $4 \times -2 = -8$. Add to -5: $-5 + (-8) = -13$. Multiply by -2: $-13 \times -2 = 26$. Add to 0: $0 + 26 = 26$. Multiply by -2: $26 \times -2 = -52$. Add to 8: $8 + (-52) = -44$. Step 3: Quotient polynomial $Q(x) = 4x^{2} - 13x + 26$, remainder $R = -44$. Answer: $Q(x) = 4x^{2} - 13x + 26$, $R = -44$. 3. Problem: Given $A = \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 2 & 3 \\ 1 & x \end{pmatrix}$ and $|AB| = 5$, find $x$. Step 1: Recall $|AB| = |A||B|$. Step 2: Calculate $|A| = (1)(1) - (2)(3) = 1 - 6 = -5$. Step 3: Calculate $|B| = (2)(x) - (3)(1) = 2x - 3$. Step 4: Use $|AB| = |A||B|$ so $5 = (-5)(2x - 3)$. Step 5: Solve $5 = -10x + 15$. Step 6: Rearranged: $-10x = 5 - 15 = -10$. Step 7: $x = \frac{-10}{-10} = 1$. Answer: $x = 1$. 4. Problem: Find vertex coordinates of parabola $y = x^{2} - 4x + 5$. Step 1: Vertex formula $x = -\frac{b}{2a}$ for $y = ax^{2} + bx + c$. Here, $a=1$, $b=-4$, $c=5$. Step 2: Calculate $x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2$. Step 3: Find $y$ at $x=2$: $y = 2^{2} - 4 \times 2 + 5 = 4 - 8 + 5 = 1$. Answer: Vertex is at $\boxed{(2, 1)}$. 5. Problem: Find $p$ if line segment joining $(2, p)$ and $(3, 6)$ is parallel to line $x - \frac{y}{2} = 0$. Step 1: Rewrite line as $x = \frac{y}{2}$ or $y = 2x$. Step 2: Slope of given line is 2. Step 3: Slope of segment joining points is $\frac{6 - p}{3 - 2} = 6 - p$. Step 4: Set equal: $6 - p = 2$. Step 5: Solve $p = 6 - 2 = 4$. Answer: $p = 4$. 6. Problem: Prove $\frac{\cos 40^\circ - \cos 70^\circ}{\sin 70^\circ - \sin 40^\circ} = \tan 55^\circ$. Step 1: Use cosine difference formula: $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$. Step 2: Numerator: $$\cos 40^\circ - \cos 70^\circ = -2 \sin 55^\circ \sin (-15^\circ) = 2 \sin 55^\circ \sin 15^\circ$$ Step 3: Use sine difference formula: $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$. Step 4: Denominator: $$\sin 70^\circ - \sin 40^\circ = 2 \cos 55^\circ \sin 15^\circ$$ Step 5: Substitute: $$\frac{2 \sin 55^\circ \sin 15^\circ}{2 \cos 55^\circ \sin 15^\circ} = \frac{\sin 55^\circ}{\cos 55^\circ} = \tan 55^\circ$$ Answer: Proven. 7. Problem: Solve $\sin \theta - \cos \theta = 0$ for $0^\circ \leq \theta \leq 90^\circ$. Step 1: Rearrange: $\sin \theta = \cos \theta$. Step 2: Divide both sides by $\cos \theta$ (nonzero in interval): $\tan \theta = 1$. Step 3: $\theta = 45^\circ$. Answer: $\theta = 45^\circ$. 8. Problem: If $\vec{a} = 5\vec{i} + 3\vec{j}$ and $\vec{b} = 3\vec{i} - (1+m)\vec{j}$ are perpendicular, find $m$. Step 1: Perpendicular vectors satisfy $\vec{a} \cdot \vec{b} = 0$. Step 2: Compute dot product: $$5 \times 3 + 3 \times (-(1+m)) = 15 - 3 - 3m = 12 - 3m$$ Step 3: Set equal to zero: $$12 - 3m = 0 \Rightarrow 3m = 12 \Rightarrow m = 4$$ Answer: $m = 4$. 9. Problem: Given $N=20$, $\sum fm = 1000$, $\sum f |m - \overline{X}| = 100$, find coefficient of mean deviation. Step 1: Mean $\overline{X} = \frac{\sum fm}{N} = \frac{1000}{20} = 50$. Step 2: Mean deviation $= \frac{\sum f |m - \overline{X}|}{N} = \frac{100}{20} = 5$. Step 3: Coefficient of mean deviation $= \frac{\text{Mean deviation}}{\overline{X}} = \frac{5}{50} = 0.1$. Answer: Coefficient of mean deviation is $0.1$.