Reflexive: for any integer a, a - a = 0, which is divisible by 3, so (a, a) in R. Symmetric: if (a - b) is divisible by 3, then (b - a) = -(a - b) is also divisible by 3, so aRb implies bRa. Transitive: if (a - b) and (b - c) are both divisible by 3, then their sum (a - b) + (b - c) = (a - c) is also divisible by 3, so aRb and bRc imply aRc. Since R is reflexive, symmetric and transitive, it is an equivalence relation.

One-one: suppose f(x1) = f(x2). Then 2x1 + 3 = 2x2 + 3, so 2x1 = 2x2 and x1 = x2; hence f is one-one. Onto: let y be any real number in the codomain. Solving y = 2x + 3 gives x = (y - 3)/2, which is a real number, and f((y - 3)/2) = y. So every y has a pre-image, and f is onto. Being both one-one and onto, f is a bijection.

Since f is a bijection, its inverse exists. Let y = f(x) = 2x + 3. Solve for x in terms of y: y - 3 = 2x, so x = (y - 3)/2. Therefore the inverse function is f inverse (y) = (y - 3)/2, or writing in terms of x, f inverse (x) = (x - 3)/2. We can check that f(f inverse (x)) = x.

f inverse (x) = (x - 3)/2

First combine tan inverse (2) + tan inverse (3). Using tan inverse (x) + tan inverse (y) = pi + tan inverse ((x + y)/(1 - xy)) when xy > 1: here x = 2, y = 3, xy = 6 > 1, so tan inverse (2) + tan inverse (3) = pi + tan inverse ((5)/(1 - 6)) = pi + tan inverse (-1) = pi - pi/4 = 3pi/4. Adding tan inverse (1) = pi/4: total = pi/4 + 3pi/4 = pi. Hence proved.

tan inverse x + tan inverse y = tan inverse((x + y)/(1 - xy))

To multiply, each element of a row of A is multiplied by the corresponding element of a column of B and summed. AB = [[(1)(2) + (2)(1), (1)(0) + (2)(3)], [(3)(2) + (4)(1), (3)(0) + (4)(3)]] = [[2 + 2, 0 + 6], [6 + 4, 0 + 12]] = [[4, 6], [10, 12]].

(AB)ij = sum of (row i of A)(column j of B)

Write A = (1/2)(A + A') + (1/2)(A - A'). Let P = (1/2)(A + A') and Q = (1/2)(A - A'). Then P' = (1/2)(A + A')' = (1/2)(A' + A) = P, so P is symmetric. And Q' = (1/2)(A - A')' = (1/2)(A' - A) = -(1/2)(A - A') = -Q, so Q is skew-symmetric. Since A = P + Q, every square matrix is the sum of a symmetric matrix P and a skew-symmetric matrix Q.

A = (1/2)(A + A') + (1/2)(A - A')

The area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Substituting: area = (1/2)|1(0 - 3) + 6(3 - 0) + 4(0 - 0)| = (1/2)|-3 + 18 + 0| = (1/2)(15) = 7.5 square units.

Area = (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|

First find the determinant: |A| = (2)(4) - (3)(1) = 8 - 3 = 5, which is non-zero, so the inverse exists. For a 2 x 2 matrix [[a, b],[c, d]], the adjoint is [[d, -b],[-c, a]], so adj A = [[4, -3],[-1, 2]]. Then the inverse is A inverse = (1/|A|) adj A = (1/5)[[4, -3],[-1, 2]] = [[4/5, -3/5],[-1/5, 2/5]].

A inverse = (1/|A|) adj A

Write D = determinant of coefficients = |[[2, 3],[1, 2]]| = (2)(2) - (3)(1) = 4 - 3 = 1. Dx (replace first column by constants) = |[[8, 3],[5, 2]]| = (8)(2) - (3)(5) = 16 - 15 = 1. Dy (replace second column by constants) = |[[2, 8],[1, 5]]| = (2)(5) - (8)(1) = 10 - 8 = 2. By Cramer's rule, x = Dx/D = 1/1 = 1 and y = Dy/D = 2/1 = 2. So x = 1, y = 2.

x = Dx/D, y = Dy/D

For continuity at x = 5, the left-hand limit, the right-hand limit and f(5) must all be equal. Left-hand value (using kx + 1): f(5) = 5k + 1. Right-hand limit (using 3x - 5): as x approaches 5, 3(5) - 5 = 10. For continuity, 5k + 1 = 10, so 5k = 9, giving k = 9/5.

LHL = RHL = f(a)

Differentiate both sides with respect to x, treating y as a function of x. d/dx(x^2) + d/dx(y^2) = d/dx(25). This gives 2x + 2y (dy/dx) = 0. Solving for dy/dx: 2y (dy/dx) = -2x, so dy/dx = -2x/(2y) = -x/y.

dy/dx = -x/y

Take the natural logarithm of both sides: ln y = ln(x^x) = x ln x. Now differentiate both sides with respect to x. On the left: (1/y)(dy/dx). On the right, using the product rule on x ln x: (1)(ln x) + x(1/x) = ln x + 1. So (1/y)(dy/dx) = ln x + 1, which gives dy/dx = y(ln x + 1) = x^x (1 + ln x).

dy/dx = x^x (1 + ln x)

First derivative using the product rule: dy/dx = e^x sin x + e^x cos x = e^x(sin x + cos x). Differentiate again using the product rule: d2y/dx2 = e^x(sin x + cos x) + e^x(cos x - sin x) = e^x(sin x + cos x + cos x - sin x) = e^x(2 cos x) = 2 e^x cos x.

d2y/dx2 = 2 e^x cos x

Area A = pi r^2. Differentiate with respect to time t: dA/dt = 2 pi r (dr/dt). Given dr/dt = 3 cm/s and r = 10 cm, so dA/dt = 2 pi (10)(3) = 60 pi cm^2/s. Thus the area is increasing at the rate of 60 pi square cm per second (about 188.5 cm^2/s).

dA/dt = 2 pi r (dr/dt)

Find the derivative: f'(x) = 2x - 4 = 2(x - 2). Set f'(x) = 0 to get the critical point: x = 2. For x < 2, f'(x) < 0, so the function is decreasing on (-infinity, 2). For x > 2, f'(x) > 0, so the function is increasing on (2, infinity). Thus f is decreasing on (-infinity, 2) and increasing on (2, infinity).

The slope of the tangent is dy/dx = 2x. At the point (1, 1), the slope = 2(1) = 2. Using the point-slope form y - y1 = m(x - x1) with m = 2 and (x1, y1) = (1, 1): y - 1 = 2(x - 1), so y - 1 = 2x - 2, giving y = 2x - 1. This is the equation of the tangent.

y - y1 = m(x - x1)

Find the first derivative: f'(x) = 3x^2 - 3 = 3(x^2 - 1). Set f'(x) = 0: x^2 = 1, so x = 1 or x = -1. Second derivative: f''(x) = 6x. At x = 1, f''(1) = 6 > 0, so there is a local minimum; minimum value f(1) = 1 - 3 + 2 = 0. At x = -1, f''(-1) = -6 < 0, so there is a local maximum; maximum value f(-1) = -1 + 3 + 2 = 4.

f'(x) = 0; use f''(x) sign

Let the two numbers be x and 16 - x (since their sum is 16). Their product P = x(16 - x) = 16x - x^2. To maximise, differentiate: dP/dx = 16 - 2x, and set it to zero: 16 - 2x = 0, so x = 8. The second derivative d2P/dx2 = -2 < 0, confirming a maximum. So the numbers are x = 8 and 16 - 8 = 8, and the maximum product is 8 x 8 = 64.

Use the substitution t = 1 + x^2, so dt = 2x dx. The integral becomes the integral of (1/t) dt = ln|t| + C. Substituting back t = 1 + x^2, the answer is ln(1 + x^2) + C. (Note that 1 + x^2 is always positive, so the modulus sign is not needed.)

integral of f'(x)/f(x) dx = ln|f(x)| + C

Integration by parts uses the integral of u v dx = u (integral of v) - integral of (u' times integral of v). Take u = x (so u' = 1) and dv = e^x dx (so integral of dv = e^x). Then integral of x e^x dx = x e^x - integral of (1 x e^x) dx = x e^x - e^x + C = e^x(x - 1) + C.

integral u dv = u v - integral v du