For continuity at x = 0, we need 1 = 3 and 1 = -1 + 3k, solving gives k = 1.

At x = 0, we use the second piece: f(0) = 2(0) + 3 = 3.

Setting the derivatives equal at x = 0 gives k = 0.

To ensure continuity at x = 0, we set k(0) + 1 = 0^2, leading to k = 2.

Setting the two pieces equal at x = 1 gives 1 = k + 1, hence k = 0.

For continuity at x = 2, f(2) must equal the limit from both sides, which is 4.

For continuity at x = 3, we need limit as x approaches 3 from left (9) to equal f(3) = k, thus k = 9.

For continuity at x = 3, we need k to equal the limit from both sides, which is 9.

f(2) = |2 - 2| = |0| = 0.

The left and right derivatives at x = 0 do not match, hence f is not differentiable at that point.

f(-3) = |-3| = 3.

By the Intermediate Value Theorem, a continuous function on a closed interval takes every value between f(a) and f(b).

The number of distinct functions from set A to set B is given by |B|^|A|. Here, |B| = 3 and |A| = 5, so the maximum number of distinct functions is 3^5 = 243.

The maximum number of distinct functions is |B|^|A| = 3^5 = 243.

G · H = 1*0 + 0*1 + 1*0 = 0 + 0 + 0 = 0.

G · H = 1*1 + 1*(-1) + 1*1 = 1 - 1 + 1 = 1.

G · H = 1*2 + 1*2 + 1*2 = 2 + 2 + 2 = 6.

G · H = 2*3 + 2*(-1) = 6 - 2 = 4.

The number of ways to choose 3 elements from 5 is given by the combination formula C(5, 3) = 10.

The number of subsets of a set with n elements is 2^n. Here, n = 5, so 2^5 = 32.

The subsets containing '1' can be formed by including '1' and choosing from the remaining elements {2, 3}. There are 2^2 = 4 subsets, but we need to exclude the empty subset, so there are 4 - 1 = 3 subsets containing '1'.

The subsets with exactly 2 elements are {1, 2}, {1, 3}, and {2, 3}. So, there are 3 such subsets.

The subsets containing exactly 2 elements are {x, y}, {x, z}, and {y, z}. So, there are 3 such subsets.

The number of subsets of a set with n elements is 2^n. Here, n=2, so the number of subsets is 2^2 = 4.

g(-1) = 3(-1) + 2 = -3 + 2 = -1.

The number of ways to choose 2 elements from 3 is given by the combination formula C(3, 2) = 3.

The total number of subsets is 2^3 = 8. Subtracting the empty set gives 8 - 1 = 7.

The power set of H is {∅, {x}, {y}, {x, y}}. The subsets of H are {∅, {x}, {y}}.

h'(x) = 3x^2 - 3 = 0 gives x^2 = 1, so x = 1 and x = -1 are critical points.

h(1) = 1^3 - 3(1) + 2 = 1 - 3 + 2 = 0.