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

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.

Since A cannot sit next to B, C, D, or E can be seated next to A. Therefore, C is a valid option.

Since W is to the immediate left of Y and Y is to the immediate right of Z, Y must be in the middle of the arrangement.

Since C is between A and D, and A is not next to B, E must be next to D.

Since C is between A and D, and A is not next to B, E must be next to B.

Arranging the friends based on the given information, we find that B is opposite to D.

The arrangement is A, C, B, D, E. C is in the middle.

P is between Q and R, which means Q is on one side of P and R on the other. Since S is opposite T, the only person next to R must be P.

Since R is between P and Q, we can place them in the circle. S will then be opposite T, as there are only five friends.

Since R is between P and Q, we can place them in a circle. S must be placed such that T is directly opposite S, as there are only five friends.

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.

Determinant of G = (1*-1) - (1*1) = -1 - 1 = -2.

The eigenvalues are found by solving the characteristic equation det(G - λI) = 0. The eigenvalues are λ = 1, 8.

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.