Draw the truth table for each of the propositions p q and p q and show that they are identical. Also, given that ^-1(x) = alpha and ^-1(x) = beta, show that (alpha + beta) = 1.

Welcome back LA — been a while! Let's pick up where you left off. Here are the solutions to the problems: 3. (i) Draw the truth table for each of the propositions p q and p q and show that they are identical. Step 1: Create a truth table for p and q. Step 2: Determine the truth values for p q. Step 3: Determine the truth values for p. Step 4: Determine the truth values for p q. Step 5: Compare the columns for p q and p q. |c|c|c|c|c| p & q & p q & p & p q \\ T & T & T & F & T \\ T & F & F & F & F \\ F & T & T & T & T \\ F & F & T & T & T \\ The columns for p q and p q are identical, which shows that the two propositions are equivalent. 3. (ii) Given that ^-1(x) = and ^-1(x) = , show that ( + ) = 1. Step 1: Use the property of inverse trigonometric functions. We know that for any x [-1, 1], the sum of ^-1(x) and ^-1(x) is ()/(2). Step 2: Substitute the given values into the identity. Given ^-1(x) = and ^-1(x) = , we have: + = ()/(2) Step 3: Calculate ( + ). ( + ) = (()/(2)) Step 4: Evaluate the sine function. (()/(2)) = 1 Thus, ( + ) = 1. 4. (i) The function f is given by f(x) = (x+1)/((x-1)(x^2+1)). Express f(x) in partial fractions. Hence find f(x)\,dx. Part 1: Partial Fractions Step 1: Set up the partial fraction decomposition. (x+1)/((x-1)(x^2+1)) = (A)/(x-1) + (Bx+C)/(x^2+1) Step 2: Multiply by the common denominator (x-1)(x^2+1). x+1 = A(x^2+1) + (Bx+C)(x-1) Step 3: Solve for A, B, and C. Substitute x=1: 1+1 = A(1^2+1) + (B(1)+C)(1-1) 2 = 2A A = 1 Expand the equation: x+1 = Ax^2+A + Bx^2-Bx+Cx-C x+1 = (A+B)x^2 + (-B+C)x + (A-C) Equate coefficients: Coefficient of x^2: 0 = A+B 0 = 1+B B = -1 Coefficient of x: 1 = -B+C 1 = -(-1)+C 1 = 1+C C = 0 Constant term: 1 = A-C 1 = 1-0 1 = 1 (consistent) Step 4: Write the partial fraction decomposition. f(x) = (1)/(x-1) + (-x+0)/(x^2+1) = (1)/(x-1) - (x)/(x^2+1) Part 2: Integration Step 5: Integrate each term. f(x)\,dx = ((1)/(x-1) - (x)/(x^2+1))\,dx (1)/(x-1)\,dx - (x)/(x^2+1)\,dx The first integral is |x-1|. For the second integral, let u = x^2+1, then du = 2x\,dx, so x\,dx = (1)/(2)\,du. (x)/(x^2+1)\,dx = (1)/(u) · (1)/(2)\,du = (1)/(2) (1)/(u)\,du = (1)/(2)|u| + C = (1)/(2)(x^2+1) + C Step 6: Combine the results. f(x)\,dx = |x-1| - (1)/(2)(x^2+1) + C The final answer is |x-1| - (1)/(2)(x^2+1) + C. 4. (ii) By using the substitution u = x, find (( x)/(1+^2 x))\,dx. Step 1: Define the substitution. Let u = x. Step 2: Find du. Differentiating u with respect to x gives du = x\,dx. Step 3: Substitute u and du into the integral. ( x)/(1+^2 x)\,dx = (1)/(1+u^2)\,du Step 4: Evaluate the resulting integral. This is a standard integral: (1)/(1+u^2)\,du = (u) + C Step 5: Substitute back u = x. ( x) + C The final answer is ( x) + C. 4. (iii) Find (x+2)e^3x\,dx. Step 1: Use integration by parts, u\,dv = uv - v\,du. Step 2: Choose u and dv. Let u = x+2 and dv = e^3x\,dx. Step 3: Find du and v. Differentiate u: du = 1\,dx. Integrate dv: v = e^3x\,dx = (1)/(3)e^3x. Step 4: Apply the integration by parts formula. (x+2)e^3x\,dx = (x+2)((1)/(3)e^3x) - ((1)/(3)e^3x)(1)\,dx = (1)/(3)(x+2)e^3x - (1)/(3) e^3x\,dx Step 5: Evaluate the remaining integral. (1)/(3)(x+2)e^3x - (1)/(3)((1)/(3)e^3x) + C Step 6: Simplify the expression. = (1)/(3)(x+2)e^3x - (1)/(9)e^3x + C You can also factor out e^3x: = e^3x((1)/(3)(x+2) - (1)/(9)) + C = e^3x((3(x+2) - 1)/(9)) + C = e^3x((3x+6 - 1)/(9)) + C = e^3x(3x+5)9 + C The final answer is e^3x(3x+5)9 + C. Send me the next one 📸