Here P = 1/x and Q = x. The integrating factor IF = e^(integral of (1/x) dx) = e^(ln x) = x. Multiply through by x (or use the solution formula): y x (IF) = integral of [Q x (IF)] dx, that is, y x = integral of (x times x) dx = integral of x^2 dx = x^3/3 + C. Dividing by x gives the general solution y = x^2/3 + C/x.

y (IF) = integral of Q (IF) dx + C

The scalar (dot) product of two vectors is the sum of the products of their corresponding components: a . b = (1)(2) + (2)(-1) + (3)(1) = 2 - 2 + 3 = 3. The dot product is a scalar quantity and also equals |a||b| cos theta, where theta is the angle between the vectors.

a . b = a1 b1 + a2 b2 + a3 b3

The angle theta satisfies cos theta = (a . b)/(|a| |b|). Here a . b = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1. Also |a| = sqrt(1 + 1 + 1) = sqrt(3) and |b| = sqrt(1 + 1 + 1) = sqrt(3). So cos theta = 1/(sqrt(3) x sqrt(3)) = 1/3. Therefore theta = cos inverse (1/3).

cos theta = (a . b)/(|a| |b|)

The cross product is found by expanding the determinant with rows i, j, k; then the components of a; then the components of b. a x b = i[(1)(1) - (1)(-1)] - j[(1)(1) - (1)(1)] + k[(1)(-1) - (1)(1)] = i(1 + 1) - j(1 - 1) + k(-1 - 1) = 2 i - 0 j - 2 k = 2 i - 2 k. Its magnitude is |a x b| = sqrt(2^2 + 0^2 + (-2)^2) = sqrt(8) = 2 sqrt(2).

a x b = determinant of (i, j, k; a1, a2, a3; b1, b2, b3)

Vector form: the equation of a line passing through a point with position vector a and parallel to a vector b is r = a + lambda b, where lambda is a scalar parameter and r is the position vector of any point on the line. Cartesian form: if the line passes through the point (x1, y1, z1) and has direction ratios a, b, c, then its equation is (x - x1)/a = (y - y1)/b = (z - z1)/c.

r = a + lambda b

If two lines have direction ratios (a1, b1, c1) and (a2, b2, c2), the angle theta between them is given by cos theta = |a1 a2 + b1 b2 + c1 c2| / (sqrt(a1^2 + b1^2 + c1^2) x sqrt(a2^2 + b2^2 + c2^2)). The lines are perpendicular if a1 a2 + b1 b2 + c1 c2 = 0, and they are parallel if a1/a2 = b1/b2 = c1/c2 (their direction ratios are proportional).

cos theta = |a1a2 + b1b2 + c1c2|/(|d1||d2|)

The distance of a point (x1, y1, z1) from the plane Ax + By + Cz + D = 0 is |A x1 + B y1 + C z1 + D|/sqrt(A^2 + B^2 + C^2). Here (x1, y1, z1) = (0, 0, 0), A = 3, B = 4, C = 12, D = -26. Distance = |3(0) + 4(0) + 12(0) - 26|/sqrt(3^2 + 4^2 + 12^2) = |-26|/sqrt(9 + 16 + 144) = 26/sqrt(169) = 26/13 = 2 units.

d = |A x1 + B y1 + C z1 + D|/sqrt(A^2 + B^2 + C^2)

The corner point method is based on the fact that the optimal value of the objective function (if it exists) occurs at a corner point (vertex) of the feasible region. The steps are: (1) find the feasible region determined by the constraints and identify its corner points; (2) evaluate the objective function Z = ax + by at each corner point; (3) the largest of these values is the maximum and the smallest is the minimum. For an unbounded region, the optimal value must be checked for existence.

The feasible region is a triangle bounded by x + y = 4 and the coordinate axes. Its corner points are (0, 0), (4, 0) and (0, 4). Evaluate Z = 3x + 4y at each corner: at (0, 0), Z = 0; at (4, 0), Z = 12; at (0, 4), Z = 16. The maximum value of Z is 16, which occurs at the corner point (0, 4).

Using the conditional probability formula: P(A | B) = P(A intersection B)/P(B) = 0.3/0.6 = 0.5. Similarly, P(B | A) = P(A intersection B)/P(A) = 0.3/0.5 = 0.6. So P(A | B) = 0.5 and P(B | A) = 0.6.

P(A | B) = P(A intersection B)/P(B)

Since A and B are independent, P(A intersection B) = P(A) x P(B) = 0.3 x 0.4 = 0.12. Using the addition theorem, P(A union B) = P(A) + P(B) - P(A intersection B) = 0.3 + 0.4 - 0.12 = 0.58. So P(A intersection B) = 0.12 and P(A union B) = 0.58.

P(A union B) = P(A) + P(B) - P(A intersection B)

Let E1 and E2 be the events of choosing Bag I and Bag II, so P(E1) = P(E2) = 1/2. Let A be the event of drawing a red ball. Then P(A | E1) = 3/7 and P(A | E2) = 5/11. By Bayes' theorem, P(E1 | A) = [P(E1) P(A | E1)] / [P(E1) P(A | E1) + P(E2) P(A | E2)] = [(1/2)(3/7)] / [(1/2)(3/7) + (1/2)(5/11)] = (3/14) / (3/14 + 5/22). Converting to a common denominator (154): 3/14 = 33/154 and 5/22 = 35/154, so the sum = 68/154. Hence P(E1 | A) = (33/154)/(68/154) = 33/68.

P(E1 | A) = P(E1)P(A|E1)/(sum of P(Ei)P(A|Ei))

The mean (expectation) of a random variable is E(X) = sum of x times p(x). Here E(X) = 0 x (1/4) + 1 x (1/2) + 2 x (1/4) = 0 + 1/2 + 1/2 = 1. First we may check that the probabilities add to 1: 1/4 + 1/2 + 1/4 = 1. So the mean of X is 1.

E(X) = sum of x p(x)

This is a binomial experiment with n = 3 trials, probability of success (head) p = 1/2 and failure q = 1/2. The binomial probability of r successes is P(X = r) = nCr p^r q^(n - r). For exactly 2 heads (r = 2): P(X = 2) = 3C2 (1/2)^2 (1/2)^1 = 3 x (1/4)(1/2) = 3 x 1/8 = 3/8. So the probability of getting exactly 2 heads is 3/8.

P(X = r) = nCr p^r q^(n-r)

Microsporogenesis is the process of formation of microspores (pollen grains) from a microspore mother cell (pollen mother cell) through meiosis; each diploid mother cell gives four haploid microspores arranged in a tetrad. A mature pollen grain has a two-layered wall: an outer hard exine made of sporopollenin (very resistant) with germ pores, and an inner thin intine of cellulose and pectin. Inside, a mature pollen grain is two-celled, having a larger vegetative cell (with food and a tube nucleus) and a smaller generative cell that later forms two male gametes.

Double fertilization is a characteristic feature of flowering plants in which two fusion events take place inside the embryo sac using the two male gametes from one pollen tube. One male gamete fuses with the egg cell to form a diploid zygote; this is called syngamy (true fertilization). The other male gamete fuses with the two polar nuclei of the central cell to form a triploid (3n) primary endosperm nucleus; this is called triple fusion. Since two fusions (syngamy and triple fusion) occur, the process is called double fertilization.

Spermatogenesis is the process of formation of sperms (spermatozoa) from the germ cells in the testes, which begins at puberty. The diploid spermatogonia multiply by mitosis and some enlarge to form primary spermatocytes (2n). Each primary spermatocyte undergoes the first meiotic division to form two haploid secondary spermatocytes, which then undergo the second meiotic division to form four haploid spermatids. The spermatids are transformed into mature sperms by a process called spermiogenesis. It is stimulated by the hormones FSH and testosterone.

The menstrual cycle is the cyclic change in the reproductive system of a human female, of about 28 days. Its phases are: (1) Menstrual phase (days 1-5) - the uterine lining breaks down and is shed as menstrual flow. (2) Follicular (proliferative) phase (days 6-13) - under FSH the ovarian follicle matures and the uterine lining is rebuilt, and oestrogen rises. (3) Ovulation (about day 14) - a surge of LH causes the release of the ovum from the ovary. (4) Luteal (secretory) phase (days 15-28) - the corpus luteum secretes progesterone, which maintains the uterine lining; if fertilisation does not occur, it degenerates and the next cycle begins.

In a monohybrid cross, a pure tall plant (TT) is crossed with a pure dwarf plant (tt). All the F1 plants are Tt and are tall, because T (tall) is dominant over t (dwarf). When the F1 plants (Tt) are self-pollinated, the gametes T and t combine in all possible ways to give the F2 generation: TT, Tt, Tt and tt. This gives a phenotypic ratio of 3 tall : 1 dwarf and a genotypic ratio of 1 TT : 2 Tt : 1 tt.

In a dihybrid cross, two pairs of contrasting characters are studied together (for example, seed shape and seed colour in peas: round yellow RRYY crossed with wrinkled green rryy). All F1 plants are round and yellow (RrYy). When the F1 are self-pollinated, the F2 generation shows four phenotypes in the ratio 9 (round yellow) : 3 (round green) : 3 (wrinkled yellow) : 1 (wrinkled green). This 9 : 3 : 3 : 1 ratio illustrates Mendel's law of independent assortment.

F2 dihybrid ratio = 9 : 3 : 3 : 1