1. **Find the modulus and argument of** $z = 3 + 4i$. - The modulus is given by $$|z| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$$ - The argument $\theta$ is $$\theta = \tan^{-1}\left(\frac{4}{3}\right).$$ 2. **Solve the system using determinants:** $$2x + 3y = 5,$$ $$3x - y = 4.$$ - Coefficient matrix determinant: $$D = \begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} = (2)(-1) - (3)(3) = -2 - 9 = -11.$$ - Determinant for $x$: $$D_x = \begin{vmatrix} 5 & 3 \\ 4 & -1 \end{vmatrix} = (5)(-1) - (3)(4) = -5 - 12 = -17.$$ - Determinant for $y$: $$D_y = \begin{vmatrix} 2 & 5 \\ 3 & 4 \end{vmatrix} = (2)(4) - (5)(3) = 8 - 15 = -7.$$ - Solutions: $$x = \frac{D_x}{D} = \frac{-17}{-11} = \frac{17}{11}, \quad y = \frac{D_y}{D} = \frac{-7}{-11} = \frac{7}{11}.$$ 3. **Prove:** $$(1 + \tan^2 A)(1 - \sin A) = 1 + \sin A.$$ - Recall identity: $$1 + \tan^2 A = \sec^2 A = \frac{1}{\cos^2 A}.$$ - Left side: $$\frac{1}{\cos^2 A}(1 - \sin A) = \frac{1 - \sin A}{\cos^2 A}.$$ - Multiply numerator and denominator by $1 + \sin A$: $$\frac{(1 - \sin A)(1 + \sin A)}{\cos^2 A (1 + \sin A)} = \frac{1 - \sin^2 A}{\cos^2 A (1 + \sin A)}.$$ - Since $1 - \sin^2 A = \cos^2 A$, this simplifies to: $$\frac{\cos^2 A}{\cos^2 A (1 + \sin A)} = \frac{1}{1 + \sin A}.$$ - But this contradicts the original statement, so re-check: actually, the original expression is $$(1 + \tan^2 A)(1 - \sin A) = \sec^2 A (1 - \sin A) = \frac{1 - \sin A}{\cos^2 A}.$$ - Multiply numerator and denominator by $1 + \sin A$: $$\frac{(1 - \sin A)(1 + \sin A)}{\cos^2 A (1 + \sin A)} = \frac{1 - \sin^2 A}{\cos^2 A (1 + \sin A)} = \frac{\cos^2 A}{\cos^2 A (1 + \sin A)} = \frac{1}{1 + \sin A}.$$ - The right side is $1 + \sin A$, so the original statement is incorrect as written. Possibly a typo; if the right side is $\frac{1}{1 + \sin A}$, the identity holds. 4. **Find 6th term and sum of first 6 terms of A.P. with** $a=2$, $d=3$. - $n$th term: $$a_n = a + (n-1)d = 2 + (6-1)3 = 2 + 15 = 17.$$ - Sum of first 6 terms: $$S_6 = \frac{6}{2} [2a + (6-1)d] = 3 [4 + 15] = 3 \times 19 = 57.$$ 5. **Find coefficient of** $x^4$ **in** $(2 - 3x)^6$. - Use binomial theorem: $$\binom{6}{4} (2)^{6-4} (-3x)^4 = \binom{6}{4} 2^2 (-3)^4 x^4.$$ - Calculate: $$\binom{6}{4} = 15, \quad 2^2 = 4, \quad (-3)^4 = 81.$$ - Coefficient: $$15 \times 4 \times 81 = 4860.$$ 6. **Find** $\det(A)$ **for** $A = \begin{bmatrix}1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0\end{bmatrix}$. - Use cofactor expansion along first row: $$\det(A) = 1 \times \begin{vmatrix}1 & 4 \\ 6 & 0\end{vmatrix} - 2 \times \begin{vmatrix}0 & 4 \\ 5 & 0\end{vmatrix} + 3 \times \begin{vmatrix}0 & 1 \\ 5 & 6\end{vmatrix}.$$ - 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. **Expand** $(1 - 2x)^3$ **up to** $x^3$. - Binomial expansion: $$(1 - 2x)^3 = \sum_{k=0}^3 \binom{3}{k} 1^{3-k} (-2x)^k = 1 - 6x + 12x^2 - 8x^3.$$ 8. **Find angle between vectors** $\mathbf{a} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$, $\mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}$. - Dot product: $$\mathbf{a} \cdot \mathbf{b} = 2 \times 1 + 1 \times (-1) + (-1) \times 1 = 2 - 1 - 1 = 0.$$ - Magnitudes: $$|\mathbf{a}| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6},$$ $$|\mathbf{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3}.$$ - Angle $\theta$: $$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{0}{\sqrt{6} \times \sqrt{3}} = 0 \implies \theta = 90^\circ.$$ 9. **Solve for** $\theta$: $$2 \sin^2 \theta - 3 \sin \theta + 1 = 0.$$ - Let $x = \sin \theta$, then $$2x^2 - 3x + 1 = 0.$$ - Solve quadratic: $$x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}.$$ - Solutions: $$x = 1, \quad x = \frac{1}{2}.$$ - So, $$\sin \theta = 1 \implies \theta = 90^\circ,$$ $$\sin \theta = \frac{1}{2} \implies \theta = 30^\circ \text{ or } 150^\circ.$$ 10. **In a G.P.,** $a=81$, $r=\frac{1}{3}$, **find sum to infinity**. - Since $|r| < 1$, sum to infinity is $$S_\infty = \frac{a}{1 - r} = \frac{81}{1 - \frac{1}{3}} = \frac{81}{\frac{2}{3}} = 81 \times \frac{3}{2} = 121.5.$$ 11. **If** $a,b,c$ **are in A.P., show that** $b = \frac{a + c}{2}$. - By definition of A.P., $$b - a = c - b \implies 2b = a + c \implies b = \frac{a + c}{2}.$$ 12. **Find area of triangle with sides 7, 8, 9 using Heron's formula.** - Semi-perimeter: $$s = \frac{7 + 8 + 9}{2} = 12.$$ - Area: $$\sqrt{s(s - a)(s - b)(s - c)} = \sqrt{12(12 - 7)(12 - 8)(12 - 9)} = \sqrt{12 \times 5 \times 4 \times 3} = \sqrt{720} = 12 \sqrt{5}.$$