1. Problem: Given $z = 3 + 4i$, find the modulus $|z|$ and the argument of $z$. Step 1: The modulus of a complex number $z = a + bi$ is given by $$|z| = \sqrt{a^2 + b^2}.$$ Here, $a=3$ and $b=4$. Step 2: Calculate modulus: $$|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$$ Step 3: The argument $\theta$ is given by $$\theta = \tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{4}{3}\right).$$ Step 4: So, $$\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ.$$ 2. Problem: Solve the system using determinants: $$2x + 3y = 5,$$ $$3x - y = 4.$$ Step 1: Write coefficient matrix $A$ and constants vector $B$: $$A = \begin{bmatrix}2 & 3 \\ 3 & -1\end{bmatrix}, \quad B = \begin{bmatrix}5 \\ 4\end{bmatrix}.$$ Step 2: Calculate determinant of $A$: $$\det(A) = (2)(-1) - (3)(3) = -2 - 9 = -11.$$ Step 3: Calculate determinants for $x$ and $y$: $$\det(A_x) = \begin{vmatrix}5 & 3 \\ 4 & -1\end{vmatrix} = (5)(-1) - (3)(4) = -5 - 12 = -17,$$ $$\det(A_y) = \begin{vmatrix}2 & 5 \\ 3 & 4\end{vmatrix} = (2)(4) - (5)(3) = 8 - 15 = -7.$$ Step 4: Solve for $x$ and $y$: $$x = \frac{\det(A_x)}{\det(A)} = \frac{-17}{-11} = \frac{17}{11},$$ $$y = \frac{\det(A_y)}{\det(A)} = \frac{-7}{-11} = \frac{7}{11}.$$ 3. Problem: Prove $$(1 + \tan^2 A)(1 - \sin A) = 1 + \sin A.$$ Step 1: Recall the identity $$1 + \tan^2 A = \sec^2 A = \frac{1}{\cos^2 A}.$$ Step 2: Substitute: $$(1 + \tan^2 A)(1 - \sin A) = \frac{1}{\cos^2 A}(1 - \sin A).$$ Step 3: Multiply numerator and denominator by $(1 + \sin A)$: $$= \frac{(1 - \sin A)(1 + \sin A)}{\cos^2 A (1 + \sin A)} = \frac{1 - \sin^2 A}{\cos^2 A}(1 + \sin A).$$ Step 4: Use $1 - \sin^2 A = \cos^2 A$: $$= \frac{\cos^2 A}{\cos^2 A}(1 + \sin A) = 1 + \sin A.$$ Hence proved. 4. Problem: Find the 6th term and sum of first 6 terms of A.P. with $a=2$, $d=3$. Step 1: The $n$th term of A.P. is $$a_n = a + (n-1)d.$$ Step 2: Calculate 6th term: $$a_6 = 2 + (6-1) \times 3 = 2 + 15 = 17.$$ Step 3: Sum of first $n$ terms: $$S_n = \frac{n}{2}[2a + (n-1)d].$$ Step 4: Calculate sum of first 6 terms: $$S_6 = \frac{6}{2}[2 \times 2 + (6-1) \times 3] = 3[4 + 15] = 3 \times 19 = 57.$$ 5. Problem: Find the coefficient of $x^2$ in $(2 - 3x)^4$. Step 1: Use binomial expansion: $$(a - b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} (-b)^k.$$ Step 2: Here, $a=2$, $b=3x$, $n=4$. The term containing $x^2$ corresponds to $k=2$: $$T_{k+1} = \binom{4}{2} 2^{4-2} (-3x)^2 = \binom{4}{2} 2^2 9 x^2.$$ Step 3: Calculate: $$\binom{4}{2} = 6, \quad 2^2 = 4,$$ $$\Rightarrow T_3 = 6 \times 4 \times 9 x^2 = 216 x^2.$$ Step 4: Coefficient of $x^2$ is $216$. 6. Problem: Find determinant of matrix $$A = \begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0\end{bmatrix}.$$ Step 1: Use cofactor expansion along first row: $$\det(A) = 1 \times \begin{vmatrix}1 & 4 \\ 6 & 0\end{vmatrix} - 2 \times \begin{vmatrix}0 & 4 \\ 5 & 0\end{vmatrix} + 3 \times \begin{vmatrix}0 & 1 \\ 5 & 6\end{vmatrix}.$$ Step 2: Calculate minors: $$= 1(1 \times 0 - 4 \times 6) - 2(0 \times 0 - 4 \times 5) + 3(0 \times 6 - 1 \times 5)$$ $$= 1(0 - 24) - 2(0 - 20) + 3(0 - 5)$$ $$= -24 + 40 - 15 = 1.$$ 7. Problem: Expand $(1 - 2x)^3$ up to $x^3$. Step 1: Use binomial theorem: $$(1 - 2x)^3 = \sum_{k=0}^3 \binom{3}{k} 1^{3-k} (-2x)^k.$$ Step 2: Calculate terms: - $k=0$: $\binom{3}{0} 1^3 (-2x)^0 = 1$ - $k=1$: $\binom{3}{1} 1^2 (-2x)^1 = 3 \times (-2x) = -6x$ - $k=2$: $\binom{3}{2} 1^1 (-2x)^2 = 3 \times 4x^2 = 12x^2$ - $k=3$: $\binom{3}{3} 1^0 (-2x)^3 = 1 \times (-8x^3) = -8x^3$ Step 3: Combine: $$(1 - 2x)^3 = 1 - 6x + 12x^2 - 8x^3.$$ 8. Problem: Find angle between vectors $\mathbf{a} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$ and $\mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}$. Step 1: Use formula: $$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}.$$ Step 2: Calculate dot product: $$\mathbf{a} \cdot \mathbf{b} = 2 \times 1 + 1 \times (-1) + (-1) \times 1 = 2 - 1 - 1 = 0.$$ Step 3: Since dot product is zero, vectors are perpendicular. Step 4: Therefore, $$\theta = 90^\circ.$$ 9. Problem: Solve for $\theta$: $$2 \sin^2 \theta - 3 \sin \theta + 1 = 0.$$ Step 1: Let $x = \sin \theta$, then $$2x^2 - 3x + 1 = 0.$$ Step 2: Solve quadratic: $$x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}.$$ Step 3: Roots: $$x_1 = \frac{4}{4} = 1, \quad x_2 = \frac{2}{4} = \frac{1}{2}.$$ Step 4: Find $\theta$: - For $\sin \theta = 1$, $$\theta = 90^\circ + 360^\circ k.$$ - For $\sin \theta = \frac{1}{2}$, $$\theta = 30^\circ, 150^\circ + 360^\circ k.$$ 10. Problem: In G.P., $a=81$, $r=\frac{1}{3}$, find sum to infinity. Step 1: Sum to infinity formula for $|r|<1$: $$S_\infty = \frac{a}{1-r}.$$ Step 2: Calculate: $$S_\infty = \frac{81}{1 - \frac{1}{3}} = \frac{81}{\frac{2}{3}} = 81 \times \frac{3}{2} = 121.5.$$ 11. Problem: If $a,b,c$ are in A.P., show that $$b = \frac{a + c}{2}.$$ Step 1: By definition of A.P., the difference between consecutive terms is constant: $$b - a = c - b.$$ Step 2: Rearrange: $$2b = a + c.$$ Step 3: Divide both sides by 2: $$b = \frac{a + c}{2}.$$ 12. Problem: Find area of triangle with sides 7, 8, 9 using Heron's formula. Step 1: Calculate semi-perimeter: $$s = \frac{7 + 8 + 9}{2} = 12.$$ Step 2: Use Heron's formula: $$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12(12-7)(12-8)(12-9)}.$$ Step 3: Calculate: $$= \sqrt{12 \times 5 \times 4 \times 3} = \sqrt{720} = 12 \sqrt{5}.$$ Final answer: Area = $12 \sqrt{5}$ square units.