1. **Problem 1: Survey of students studying sciences** Given: - Total students = 100 - Studying Mathematics (M) = 60 - Studying Physics (P) = 45 - Studying Chemistry (C) = 55 - Studying both M and P = 27 - Studying both M and C = 30 - Studying both P and C = 28 - Studying all three (M, P, C) = 15 **a) Find the number of students not studying any of the three subjects.** Use the principle of inclusion-exclusion: $$|M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |M \cap C| - |P \cap C| + |M \cap P \cap C|$$ Calculate: $$|M \cup P \cup C| = 60 + 45 + 55 - 27 - 30 - 28 + 15 = 160 - 85 + 15 = 90$$ Number not studying any subject: $$100 - 90 = 10$$ **b) Find the number of students studying at least two subjects.** Number studying exactly two subjects is: $$|M \cap P| + |M \cap C| + |P \cap C| - 3 \times |M \cap P \cap C| = 27 + 30 + 28 - 3 \times 15 = 85 - 45 = 40$$ Number studying at least two subjects includes those studying exactly two and those studying all three: $$40 + 15 = 55$$ --- 2. **Problem 2: Find the mixed second partial derivative** Given: $$f(x,y) = 2xy^2 + (xy)^2 - y^3$$ First, find the first partial derivative with respect to $y$: $$\frac{\partial f}{\partial y} = 2x \cdot 2y + 2xy \cdot x - 3y^2 = 4xy + 2x^2 y - 3y^2$$ Now, differentiate this with respect to $x$: $$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x}(4xy + 2x^2 y - 3y^2) = 4y + 4xy - 0 = 4y + 4xy = 4y(1 + x)$$ --- 3. **Problem 3: Equation of tangent line to $f'(x) = \frac{x}{x-5}$ at $x = -2$** Find $f''(x)$ to get slope of tangent to $f'(x)$: $$f'(x) = \frac{x}{x-5}$$ Rewrite: $$f'(x) = \frac{x}{x-5}$$ Differentiate using quotient rule: $$f''(x) = \frac{(1)(x-5) - x(1)}{(x-5)^2} = \frac{x - 5 - x}{(x-5)^2} = \frac{-5}{(x-5)^2}$$ Evaluate at $x = -2$: $$f''(-2) = \frac{-5}{(-2 - 5)^2} = \frac{-5}{(-7)^2} = \frac{-5}{49}$$ Find $f'(-2)$: $$f'(-2) = \frac{-2}{-2 - 5} = \frac{-2}{-7} = \frac{2}{7}$$ Equation of tangent line at $x = -2$: $$y - f'(-2) = f''(-2)(x + 2)$$ $$y - \frac{2}{7} = -\frac{5}{49}(x + 2)$$ Simplify: $$y = -\frac{5}{49}x - \frac{10}{49} + \frac{2}{7} = -\frac{5}{49}x - \frac{10}{49} + \frac{14}{49} = -\frac{5}{49}x + \frac{4}{49}$$ --- 4. **Problem 4: Compute $A^2 - B$** Given matrices: $$A = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 2 & -3 \\ 1 & 2 & -2 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & -2 & 3 \\ -1 & -4 & 6 \\ 0 & -5 & 5 \end{bmatrix}$$ Calculate $A^2 = A \times A$: Row 1: - $(2)(2) + (-1)(1) + (0)(1) = 4 - 1 + 0 = 3$ - $(2)(-1) + (-1)(2) + (0)(2) = -2 - 2 + 0 = -4$ - $(2)(0) + (-1)(-3) + (0)(-2) = 0 + 3 + 0 = 3$ Row 2: - $(1)(2) + (2)(1) + (-3)(1) = 2 + 2 - 3 = 1$ - $(1)(-1) + (2)(2) + (-3)(2) = -1 + 4 - 6 = -3$ - $(1)(0) + (2)(-3) + (-3)(-2) = 0 - 6 + 6 = 0$ Row 3: - $(1)(2) + (2)(1) + (-2)(1) = 2 + 2 - 2 = 2$ - $(1)(-1) + (2)(2) + (-2)(2) = -1 + 4 - 4 = -1$ - $(1)(0) + (2)(-3) + (-2)(-2) = 0 - 6 + 4 = -2$ So, $$A^2 = \begin{bmatrix} 3 & -4 & 3 \\ 1 & -3 & 0 \\ 2 & -1 & -2 \end{bmatrix}$$ Now compute $A^2 - B$: $$\begin{bmatrix} 3-2 & -4-(-2) & 3-3 \\ 1-(-1) & -3-(-4) & 0-6 \\ 2-0 & -1-(-5) & -2-5 \end{bmatrix} = \begin{bmatrix} 1 & -2 & 0 \\ 2 & 1 & -6 \\ 2 & 4 & -7 \end{bmatrix}$$ --- 5. **Problem 5: Functions $g(x) = (x-1)^2$ and $f(x) = \frac{3-x}{1-x}$** **i. Find $x$ if $(f \circ g)(x) = 4$** Calculate: $$(f \circ g)(x) = f(g(x)) = f((x-1)^2) = \frac{3 - (x-1)^2}{1 - (x-1)^2} = 4$$ Set equation: $$\frac{3 - (x-1)^2}{1 - (x-1)^2} = 4$$ Multiply both sides by denominator: $$3 - (x-1)^2 = 4(1 - (x-1)^2)$$ Expand right side: $$3 - (x-1)^2 = 4 - 4(x-1)^2$$ Bring all terms to one side: $$3 - (x-1)^2 - 4 + 4(x-1)^2 = 0$$ $$-1 + 3(x-1)^2 = 0$$ Solve: $$3(x-1)^2 = 1$$ $$(x-1)^2 = \frac{1}{3}$$ Take square root: $$x - 1 = \pm \frac{1}{\sqrt{3}}$$ So, $$x = 1 \pm \frac{1}{\sqrt{3}}$$ **ii. Find $g^{-1}(x)$** Given: $$y = (x-1)^2$$ Solve for $x$: $$x - 1 = \pm \sqrt{y}$$ $$x = 1 \pm \sqrt{y}$$ So, $$g^{-1}(x) = 1 \pm \sqrt{x}$$ **iii. Find domain of $f(x) = \frac{3-x}{1-x}$** Denominator cannot be zero: $$1 - x \neq 0 \implies x \neq 1$$ Domain: $$(-\infty, 1) \cup (1, \infty)$$ **iv. For $g(x) = (x-1)^2$, find intercepts and vertex** - Vertex is at $x=1$, value: $$g(1) = (1-1)^2 = 0$$ - $y$-intercept at $x=0$: $$g(0) = (0-1)^2 = 1$$ - $x$-intercept(s) where $g(x) = 0$: $$ (x-1)^2 = 0 \implies x=1$$ **Summary:** - Vertex: $(1,0)$ - $y$-intercept: $(0,1)$ - $x$-intercept: $(1,0)$ The graph is a parabola opening upwards with vertex at $(1,0)$.