1. Problem (6b): Find the maxima and minima of the function $$f(x) = x^3 - 12x^2 + 36x + 17$$ on the interval $$[1, 10]$$. 2. To find maxima and minima, we use the first derivative test. The critical points occur where $$f'(x) = 0$$ or is undefined. 3. Compute the first derivative: $$f'(x) = 3x^2 - 24x + 36$$ 4. Set the derivative equal to zero to find critical points: $$3x^2 - 24x + 36 = 0$$ Divide both sides by 3: $$x^2 - 8x + 12 = 0$$ 5. Solve the quadratic equation: $$x = \frac{8 \pm \sqrt{64 - 48}}{2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2}$$ 6. So the critical points are: $$x = \frac{8 + 4}{2} = 6$$ $$x = \frac{8 - 4}{2} = 2$$ 7. Evaluate the function at critical points and endpoints: $$f(1) = 1 - 12 + 36 + 17 = 42$$ $$f(2) = 8 - 48 + 72 + 17 = 49$$ $$f(6) = 216 - 432 + 216 + 17 = 17$$ $$f(10) = 1000 - 1200 + 360 + 17 = 177$$ 8. Determine maxima and minima by comparing values: - Minimum value is $$17$$ at $$x=6$$. - Maximum value is $$177$$ at $$x=10$$. --- 9. Problem (6a): Expand $$\log_e x$$ in powers of $$(x-1)$$ and evaluate $$\log_e 1.1$$ to 4 decimal places. 10. The Taylor series expansion of $$\log_e x$$ about $$x=1$$ is: $$\log_e x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots$$ 11. Substitute $$x=1.1$$: $$\log_e 1.1 \approx 0.1 - \frac{0.1^2}{2} + \frac{0.1^3}{3} - \frac{0.1^4}{4}$$ 12. Calculate each term: $$0.1 = 0.1$$ $$- \frac{0.01}{2} = -0.005$$ $$+ \frac{0.001}{3} \approx +0.0003333$$ $$- \frac{0.0001}{4} = -0.000025$$ 13. Sum the terms: $$0.1 - 0.005 + 0.0003333 - 0.000025 = 0.0953083$$ 14. Rounded to 4 decimal places: $$\log_e 1.1 \approx 0.0953$$ --- 15. Problem (6b): Find the equation of the evolute of the curve $$y^2 = 4ax$$. 16. The curve is a parabola. The evolute is the locus of centers of curvature. 17. Parametrize the curve: $$x = \frac{t^2}{4a}, \quad y = t$$ 18. The formula for the evolute of $$y^2 = 4ax$$ is: $$x = a - \frac{y^2}{4a}, \quad y = y$$ 19. More precisely, the evolute of $$y^2 = 4ax$$ is: $$x = a - \frac{t^2}{2}, \quad y = \frac{t^3}{2a}$$ 20. Eliminating parameter $$t$$ gives the evolute equation: $$27a y^2 = 4(a - x)^3$$ --- 21. Problem (7a): Show vectors $$v_1 = (1, -1, 0), v_2 = (1, 1, 0), v_3 = (0, 0, 1)$$ form a basis of $$\mathbb{R}^3$$ and find its dimension. 22. To form a basis, vectors must be linearly independent and span $$\mathbb{R}^3$$. 23. Check linear independence by setting: $$c_1 v_1 + c_2 v_2 + c_3 v_3 = 0$$ 24. This gives system: $$c_1 + c_2 = 0$$ $$-c_1 + c_2 = 0$$ $$c_3 = 0$$ 25. Solving, we get $$c_1 = c_2 = c_3 = 0$$, so vectors are linearly independent. 26. Since there are 3 vectors in $$\mathbb{R}^3$$ and they are independent, they form a basis. 27. Dimension of $$\mathbb{R}^3$$ is 3. --- 28. Problem (7b): State and prove rank-nullity theorem. 29. Theorem: For a linear transformation $$T: V \to W$$, $$\text{dim} (\text{Ker} T) + \text{dim} (\text{Im} T) = \text{dim} V$$ 30. Proof: Let $$\{v_1, ..., v_k\}$$ be a basis of $$\text{Ker} T$$. Extend it to a basis $$\{v_1, ..., v_k, v_{k+1}, ..., v_n\}$$ of $$V$$. 31. Then $$\{T(v_{k+1}), ..., T(v_n)\}$$ spans $$\text{Im} T$$ and is linearly independent. 32. So $$\text{dim} (\text{Im} T) = n - k$$. 33. Hence, $$\text{dim} (\text{Ker} T) + \text{dim} (\text{Im} T) = k + (n - k) = n = \text{dim} V$$. --- 34. Problem (8a): Show linear transformation $$T: \mathbb{R}^3 \to \mathbb{R}^3$$ defined by $$T(x,y,z) = (3x, x - y, 2x + y + z)$$ is one-one and onto. 35. To show one-one, check if $$T(v) = 0$$ implies $$v=0$$. 36. Solve: $$3x = 0$$ $$x - y = 0$$ $$2x + y + z = 0$$ 37. From first, $$x=0$$. From second, $$0 - y = 0 \Rightarrow y=0$$. From third, $$0 + 0 + z = 0 \Rightarrow z=0$$. 38. So kernel is zero vector only, hence $$T$$ is one-one. 39. To show onto, for any $$w = (a,b,c)$$ in $$\mathbb{R}^3$$, solve for $$v=(x,y,z)$$ such that $$T(v) = w$$: $$3x = a$$ $$x - y = b$$ $$2x + y + z = c$$ 40. From first, $$x = \frac{a}{3}$$. From second, $$y = x - b = \frac{a}{3} - b$$. From third, $$z = c - 2x - y = c - 2\frac{a}{3} - (\frac{a}{3} - b) = c - \frac{2a}{3} - \frac{a}{3} + b = c - a + b$$. 41. So for any $$w$$, there exists $$v$$, hence $$T$$ is onto. --- 42. Problem (8b): Find linear transformation mapping $$(1,1,1), (1,1,0), (1,0,0)$$ in $$\mathbb{R}^3$$ to $$(2,1), (2,1), (2,1)$$ in $$\mathbb{R}^2$$. 43. Since all images are the same vector $$(2,1)$$, the transformation maps the span of these vectors to $$(2,1)$$. 44. The transformation $$T$$ is defined by: $$T(x,y,z) = (2,1)$$ for all $$x,y,z$$ in the span of given vectors. 45. This is a constant transformation on that subspace. --- 46. Problem (9a): Show vectors $$x_1 = (1,2,4), x_2 = (2,-1,3), x_3 = (0,1,2), x_4 = (-3,7,2)$$ are linearly dependent. 47. Since there are 4 vectors in $$\mathbb{R}^3$$, which has dimension 3, they must be linearly dependent. --- 48. Problem (9b): Reduce matrix $$\begin{bmatrix}-1 & 2 & -2 \\ 1 & 2 & 1 \\ -1 & -1 & 0\end{bmatrix}$$ to diagonal form. 49. Perform row operations: - Add row 1 to row 3: $$R_3 = R_3 + R_1 = (-1 + -1, -1 + 2, 0 + -2) = (-2, 1, -2)$$ - Swap rows if needed and continue elimination to get diagonal matrix. 50. Final diagonal matrix is: $$\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$ --- 51. Problem (9c): Expand $$e^{a \sin^{-1} x}$$ in ascending powers of $$x$$. 52. Use series expansion of $$\sin^{-1} x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots$$ 53. Then, $$e^{a \sin^{-1} x} = e^{a(x + \frac{x^3}{6} + \cdots)} = e^{ax} e^{a \frac{x^3}{6}} \cdots$$ 54. Expand using series for $$e^{ax}$$ and multiply terms to get ascending powers. --- 55. Problem (9d): Linear transformations and inverse with example. 56. A linear transformation $$T: V \to W$$ satisfies: $$T(u+v) = T(u) + T(v)$$ and $$T(cv) = cT(v)$$. 57. The inverse $$T^{-1}$$ exists if $$T$$ is bijective. 58. Example: $$T: \mathbb{R}^2 \to \mathbb{R}^2$$ defined by $$T(x,y) = (2x, 3y)$$. 59. Its inverse is $$T^{-1}(a,b) = (\frac{a}{2}, \frac{b}{3})$$.