Class 11 Maths — Important Board Questions with Answers

The power set is the set of all subsets of A. P(A) = { {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} }. The number of elements n(P(A)) = 2^n = 2^3 = 8.

n(P(A)) = 2^n

Let n(C) = 24, n(F) = 16, n(C union F) = 35 (each likes at least one). By the formula n(C union F) = n(C) + n(F) - n(C intersection F), we get 35 = 24 + 16 - n(C intersection F), so n(C intersection F) = 40 - 35 = 5. Hence 5 students like both games.

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

n(T) = 150, n(C) = 225, n(T intersection C) = 100. Students drinking at least one = n(T union C) = n(T) + n(C) - n(T intersection C) = 150 + 225 - 100 = 275. Students drinking neither = total - n(T union C) = 600 - 275 = 325.

neither = n(U) - n(A union B)

Union (A union B) is the set of all elements that are in A or in B or in both. Intersection (A intersection B) is the set of elements common to both A and B. Difference (A - B) is the set of elements that are in A but not in B. Example: if A = {1,2,3} and B = {2,3,4}, then A union B = {1,2,3,4}, A intersection B = {2,3}, and A - B = {1}.

A union B = {2,3,4,5,6,7,8}, so (A union B)' = U - (A union B) = {1,9,10}. Now A' = {1,6,7,8,9,10} and B' = {1,2,3,9,10}, so A' intersection B' = {1,9,10}. Since (A union B)' = {1,9,10} = A' intersection B', De Morgan's law is verified.

(A union B)' = A' intersection B'

{x : -3 <= x < 5} includes -3 but not 5, so in interval notation it is [-3, 5). The interval (2, 9] excludes 2 and includes 9, so in set-builder form it is {x : x is a real number, 2 < x <= 9}.

The Cartesian product A x B is the set of all ordered pairs (a, b) with a in A and b in B. A x B = {(1,3), (1,4), (2,3), (2,4)}. The number of elements n(A x B) = n(A) x n(B) = 2 x 2 = 4.

n(A x B) = n(A) x n(B)

Two ordered pairs are equal when their corresponding components are equal. So x + 1 = 3 gives x = 2, and y - 2 = 1 gives y = 3. Hence x = 2 and y = 3.

(a, b) = (c, d) => a = c and b = d

The domain of a relation is the set of all first components (inputs), and the range is the set of all second components (outputs). For R = {(1,2), (2,4), (3,6)}, the domain = {1, 2, 3} and the range = {2, 4, 6}.

For f(x) to be real, the expression under the square root must be non-negative: 9 - x^2 >= 0, i.e. x^2 <= 9, so -3 <= x <= 3. Hence the domain is [-3, 3]. As x varies over this interval, 9 - x^2 ranges from 0 (at x = +/-3) to 9 (at x = 0), so sqrt(9 - x^2) ranges from 0 to 3. Hence the range is [0, 3].

sqrt() defined only for non-negative values

A function f from A to B is one-one (injective) if different elements of A have different images in B, i.e. f(x1) = f(x2) implies x1 = x2. A function f from A to B is onto (surjective) if every element of B is the image of at least one element of A, i.e. the range of f equals the codomain B.

f(2) = 2^2 + 1 = 4 + 1 = 5. f(-1) = (-1)^2 + 1 = 1 + 1 = 2. For f(x) = 10: x^2 + 1 = 10, so x^2 = 9, giving x = +3 or x = -3.

f(x) = x^2 + 1

Using pi radians = 180 degrees: 5pi/6 radians = (5/6) x 180 = 150 degrees. And 240 degrees = 240 x (pi/180) = 4pi/3 radians.

radians = degrees x (pi/180)

360 revolutions per minute = 360/60 = 6 revolutions per second. One revolution corresponds to an angle of 2pi radians. Therefore the angle turned in one second = 6 x 2pi = 12pi radians.

angle = number of revolutions x 2pi

Write 75 = 45 + 30. Using sin(A + B) = sin A cos B + cos A sin B: sin 75 = sin 45 cos 30 + cos 45 sin 30 = (1/sqrt2)(sqrt3/2) + (1/sqrt2)(1/2) = (sqrt3 + 1)/(2 sqrt2) = (sqrt6 + sqrt2)/4.

sin(A + B) = sin A cos B + cos A sin B

Expand each factor: cos(A + B) = cos A cos B - sin A sin B and cos(A - B) = cos A cos B + sin A sin B. Their product is of the form (p - q)(p + q) = p^2 - q^2 with p = cos A cos B and q = sin A sin B. So it equals cos^2 A cos^2 B - sin^2 A sin^2 B. Writing cos^2 B = 1 - sin^2 B and sin^2 A = 1 - cos^2 A: = cos^2 A (1 - sin^2 B) - (1 - cos^2 A) sin^2 B = cos^2 A - cos^2 A sin^2 B - sin^2 B + cos^2 A sin^2 B = cos^2 A - sin^2 B. Hence proved.

cos(A+B)cos(A-B) = cos^2 A - sin^2 B

Using sin^2 x + cos^2 x = 1: sin^2 x = 1 - (9/25) = 16/25, so sin x = +/- 4/5. In the second quadrant sine is positive, so sin x = 4/5. Then tan x = sin x / cos x = (4/5) / (-3/5) = -4/3.

sin^2 x + cos^2 x = 1

Start from the fundamental identity sin^2 x + cos^2 x = 1. Divide both sides by cos^2 x (cos x not equal to 0): sin^2 x/cos^2 x + cos^2 x/cos^2 x = 1/cos^2 x, which gives tan^2 x + 1 = sec^2 x. Hence 1 + tan^2 x = sec^2 x.

1 + tan^2 x = sec^2 x

The principal value for which sin x = 1/2 is x = pi/6. The general solution of sin x = sin a is x = n pi + (-1)^n a, where n is any integer. Here a = pi/6, so the general solution is x = n pi + (-1)^n (pi/6), n belongs to the set of integers.

sin x = sin a => x = n pi + (-1)^n a

The principal value for which tan x = sqrt(3) is x = pi/3. The general solution of tan x = tan a is x = n pi + a, where n is any integer. Here a = pi/3, so the general solution is x = n pi + pi/3, n belongs to the set of integers.

tan x = tan a => x = n pi + a

Write 15 = 45 - 30. Using sin(A - B) = sin A cos B - cos A sin B: sin 15 = sin 45 cos 30 - cos 45 sin 30 = (1/sqrt2)(sqrt3/2) - (1/sqrt2)(1/2) = (sqrt3 - 1)/(2 sqrt2) = (sqrt6 - sqrt2)/4.

sin(A - B) = sin A cos B - cos A sin B

Write sin 3x = sin(2x + x) = sin 2x cos x + cos 2x sin x. Substitute sin 2x = 2 sin x cos x and cos 2x = 1 - 2 sin^2 x: = (2 sin x cos x) cos x + (1 - 2 sin^2 x) sin x = 2 sin x cos^2 x + sin x - 2 sin^3 x. Replace cos^2 x = 1 - sin^2 x: = 2 sin x (1 - sin^2 x) + sin x - 2 sin^3 x = 2 sin x - 2 sin^3 x + sin x - 2 sin^3 x = 3 sin x - 4 sin^3 x. Hence proved.

sin 3x = 3 sin x - 4 sin^3 x

Write tan A = sin A/cos A and cot A = cos A/sin A. First term: cos A/(1 - sin A/cos A) = cos A/((cos A - sin A)/cos A) = cos^2 A/(cos A - sin A). Second term: sin A/(1 - cos A/sin A) = sin A/((sin A - cos A)/sin A) = sin^2 A/(sin A - cos A) = -sin^2 A/(cos A - sin A). Adding: [cos^2 A - sin^2 A]/(cos A - sin A) = [(cos A - sin A)(cos A + sin A)]/(cos A - sin A) = cos A + sin A. Hence proved.

cos^2 A - sin^2 A = (cosA - sinA)(cosA + sinA)

(1 + i)(2 - i) = 1 x 2 + 1 x (-i) + i x 2 + i x (-i) = 2 - i + 2i - i^2. Since i^2 = -1, this is 2 + i + 1 = 3 + i. So a = 3 and b = 1.

i^2 = -1

The multiplicative inverse of z is 1/z = conjugate(z)/|z|^2. Here conjugate of (4 - 3i) is (4 + 3i) and |z|^2 = 4^2 + (-3)^2 = 16 + 9 = 25. So the inverse = (4 + 3i)/25 = 4/25 + (3/25) i.

z^-1 = conjugate(z)/|z|^2

Modulus |z| = sqrt(1^2 + (sqrt3)^2) = sqrt(1 + 3) = sqrt4 = 2. For the argument, tan(theta) = sqrt3/1 = sqrt3, and since both real and imaginary parts are positive (first quadrant), theta = pi/3. Polar form: z = 2(cos(pi/3) + i sin(pi/3)).

z = r(cos theta + i sin theta)

Multiply numerator and denominator by the conjugate of the denominator (1 + 2i): [(3 + 4i)(1 + 2i)]/[(1 - 2i)(1 + 2i)]. Numerator = 3 + 6i + 4i + 8i^2 = 3 + 10i - 8 = -5 + 10i. Denominator = 1^2 + 2^2 = 5. So the result = (-5 + 10i)/5 = -1 + 2i.

multiply by conjugate of denominator

Using the quadratic formula x = [-b +/- sqrt(b^2 - 4ac)]/(2a) with a = 1, b = 1, c = 1: discriminant = 1 - 4 = -3. So x = [-1 +/- sqrt(-3)]/2 = [-1 +/- i sqrt3]/2. The roots are (-1 + i sqrt3)/2 and (-1 - i sqrt3)/2.

x = [-b +/- sqrt(b^2 - 4ac)]/(2a)

Here a = 1, b = -2, c = 5. Discriminant = b^2 - 4ac = 4 - 20 = -16. So x = [2 +/- sqrt(-16)]/2 = [2 +/- 4i]/2 = 1 +/- 2i. The roots are 1 + 2i and 1 - 2i.

x = [-b +/- sqrt(b^2 - 4ac)]/(2a)

Powers of i repeat with period 4 (i, -1, -i, 1). i^9 = i^(8+1) = (i^4)^2 x i = 1 x i = i. i^19 = i^(16+3) = (i^4)^4 x i^3 = 1 x (-i) = -i. So i^9 + i^19 = i + (-i) = 0.

i^4 = 1

Let sqrt(-5 + 12i) = a + i b. Squaring: a^2 - b^2 + 2ab i = -5 + 12i. So a^2 - b^2 = -5 and 2ab = 12 (ab = 6). Also |z| = sqrt((-5)^2 + 12^2) = sqrt(25 + 144) = 13, so a^2 + b^2 = 13. Adding a^2 - b^2 = -5 and a^2 + b^2 = 13 gives 2a^2 = 8, a^2 = 4, a = +/-2; then b^2 = 9, b = +/-3. Since ab = 6 is positive, a and b have the same sign. Hence sqrt(-5 + 12i) = +/- (2 + 3i).

(a + ib)^2 = a^2 - b^2 + 2ab i

3x - 5 < 7 gives 3x < 12, so x < 4. The solution set is {x : x < 4, x is real} = (-infinity, 4). On the number line, this is shown by an open circle at 4 with shading to the left (all values less than 4).

5x - 3 >= 3x - 5. Subtract 3x from both sides: 2x - 3 >= -5. Add 3: 2x >= -2. Divide by 2: x >= -1. The solution set is {x : x >= -1} = [-1, infinity). On the number line, a closed (filled) circle at -1 with shading to the right.

First inequality: 2x - 1 < 5 gives 2x < 6, so x < 3. Second inequality: 3x + 2 > -4 gives 3x > -6, so x > -2. The common solution is the intersection: -2 < x < 3, i.e. the interval (-2, 3).

Let the shortest side = x cm. Then the longest side = 3x and the third side = 3x - 2. Perimeter = x + 3x + (3x - 2) = 7x - 2. Given perimeter is at least 61: 7x - 2 >= 61, so 7x >= 63, giving x >= 9. Hence the minimum length of the shortest side is 9 cm.

perimeter = sum of sides

Write 8! = 8 x 7 x 6!. So 8!/(6! x 2!) = (8 x 7 x 6!)/(6! x 2!) = (8 x 7)/2! = 56/2 = 28.

n! = n x (n-1)!

The hundreds place can be filled in 9 ways (any of 1 to 9), the tens place in 8 ways (remaining digits) and the units place in 7 ways. By the multiplication principle, total numbers = 9 x 8 x 7 = 504.

total = product of choices at each place

MATHEMATICS has 11 letters in which M appears 2 times, A appears 2 times and T appears 2 times; the rest (H, E, I, C, S) are distinct. Number of distinct arrangements = 11! / (2! x 2! x 2!) = 39916800 / 8 = 4989600.

arrangements = n!/(p! q! r!)

nC2 = n(n-1)/2 = 15, so n(n-1) = 30. We need two consecutive integers whose product is 30: 6 x 5 = 30, so n = 6. (n = -5 is rejected since n must be a positive integer.)

nC2 = n(n-1)/2

Since order does not matter in a committee, we use combinations. Number of ways = 10C3 = 10!/(3! x 7!) = (10 x 9 x 8)/(3 x 2 x 1) = 720/6 = 120.

nCr = n!/(r!(n-r)!)

The thousands place cannot be 0, so it can be filled in 9 ways (digits 1 to 9). After using one digit, the hundreds place can be filled in 9 ways (including 0 now), the tens place in 8 ways and the units place in 7 ways. Total = 9 x 9 x 8 x 7 = 4536.

Choose 3 men from 7 in 7C3 ways and 2 women from 5 in 5C2 ways. 7C3 = (7 x 6 x 5)/(3 x 2 x 1) = 35 and 5C2 = (5 x 4)/(2 x 1) = 10. By the multiplication principle, total ways = 35 x 10 = 350.

total = 7C3 x 5C2

We use the property that nCa = nCb implies either a = b or a + b = n. Since r is not equal to r + 2, we must have r + (r + 2) = 10, so 2r + 2 = 10, giving 2r = 8 and r = 4.

nCa = nCb => a = b or a + b = n

The binomial theorem states that (a + b)^n = nC0 a^n + nC1 a^(n-1) b + nC2 a^(n-2) b^2 + ... + nCn b^n = sum over r from 0 to n of nCr a^(n-r) b^r. The general term (the (r+1)th term) is T(r+1) = nCr a^(n-r) b^r.

T(r+1) = nCr a^(n-r) b^r

(2x - 3)^4 = 4C0 (2x)^4 + 4C1 (2x)^3(-3) + 4C2 (2x)^2(-3)^2 + 4C3 (2x)(-3)^3 + 4C4 (-3)^4 = 16x^4 + 4(8x^3)(-3) + 6(4x^2)(9) + 4(2x)(-27) + 81 = 16x^4 - 96x^3 + 216x^2 - 216x + 81.

(a + b)^4 expansion

The general term is T(r+1) = 8Cr x^(8-r) 3^r. For the term in x^5, we need 8 - r = 5, so r = 3. The coefficient = 8C3 x 3^3 = 56 x 27 = 1512.

T(r+1) = nCr a^(n-r) b^r

General term T(r+1) = 6Cr (x^2)^(6-r) (-1/x)^r = 6Cr (-1)^r x^(12 - 2r) x^(-r) = 6Cr (-1)^r x^(12 - 3r). For the term independent of x, the power of x must be zero: 12 - 3r = 0, so r = 4. The term = 6C4 (-1)^4 = 15 x 1 = 15.

set power of x to 0

Here n = 6 is even, so there is one middle term, the (n/2 + 1) = 4th term. T4 = 6C3 x^3 (2y)^3 = 20 x^3 x 8 y^3 = 160 x^3 y^3.

middle term = (n/2 + 1)th term when n is even

For (a + b)^4 + (a - b)^4, the odd-power terms cancel and we get 2[a^4 + 6 a^2 b^2 + b^4]. With a = sqrt2 and b = 1: a^4 = (sqrt2)^4 = 4, a^2 b^2 = 2 x 1 = 2 so 6 a^2 b^2 = 12, and b^4 = 1. Sum = 2[4 + 12 + 1] = 2 x 17 = 34.

(a+b)^4 + (a-b)^4 = 2[a^4 + 6a^2 b^2 + b^4]

General term T(r+1) = 9Cr x^(9-r) (2y)^r = 9Cr 2^r x^(9-r) y^r. For x^6 y^3 we need 9 - r = 6 and r = 3, which is consistent (r = 3). Coefficient = 9C3 x 2^3 = 84 x 8 = 672.

T(r+1) = nCr a^(n-r) b^r

Here the first term a = 3 and common difference d = 7 - 3 = 4. The nth term is a_n = a + (n - 1)d. So a_20 = 3 + (20 - 1) x 4 = 3 + 76 = 79.

a_n = a + (n - 1)d

Here a = 3, d = 4, n = 20. The sum of n terms is S_n = (n/2)[2a + (n - 1)d]. So S_20 = (20/2)[2 x 3 + 19 x 4] = 10[6 + 76] = 10 x 82 = 820.

S_n = (n/2)[2a + (n - 1)d]

Let the three numbers be a - d, a, a + d. Their sum = 3a = 24, so a = 8. Their product = (a - d)(a)(a + d) = a(a^2 - d^2) = 440. So 8(64 - d^2) = 440, giving 64 - d^2 = 55, so d^2 = 9 and d = 3 (or -3). The numbers are 5, 8, 11.

AP terms: a - d, a, a + d

Here the first term a = 2 and common ratio r = 6/2 = 3. The nth term of a GP is a_n = a r^(n-1). So a_8 = 2 x 3^7 = 2 x 2187 = 4374.

a_n = a r^(n-1)

Here a = 5, r = 3, n = 6. Since r > 1, S_n = a(r^n - 1)/(r - 1). So S_6 = 5(3^6 - 1)/(3 - 1) = 5(729 - 1)/2 = 5 x 728/2 = 5 x 364 = 1820.

S_n = a(r^n - 1)/(r - 1), r > 1

Here a = 1 and r = 1/3. Since |r| < 1, the sum to infinity exists and is S_infinity = a/(1 - r) = 1/(1 - 1/3) = 1/(2/3) = 3/2.

S_infinity = a/(1 - r), |r| < 1

AM = (4 + 16)/2 = 20/2 = 10. GM = sqrt(4 x 16) = sqrt64 = 8. Since 10 > 8, we have AM > GM, which agrees with the general result AM >= GM for positive numbers.

AM = (a + b)/2; GM = sqrt(ab)

The sum of squares of the first n natural numbers is given by n(n + 1)(2n + 1)/6. For n = 10: sum = 10 x 11 x 21 / 6 = 2310/6 = 385.

sum of squares = n(n + 1)(2n + 1)/6

The rth term is a_r = r(r + 1) = r^2 + r. So the sum S = sum over r of (r^2 + r) = sum of r^2 + sum of r. Using sum of r^2 = n(n + 1)(2n + 1)/6 and sum of r = n(n + 1)/2: S = n(n + 1)(2n + 1)/6 + n(n + 1)/2. Take n(n + 1)/6 common: S = [n(n + 1)/6][(2n + 1) + 3] = [n(n + 1)/6](2n + 4) = [n(n + 1) x 2(n + 2)]/6 = n(n + 1)(n + 2)/3.

S = n(n + 1)(n + 2)/3

The slope of a line through (x1, y1) and (x2, y2) is m = (y2 - y1)/(x2 - x1). Here m = (9 - 3)/(5 - 2) = 6/3 = 2.

m = (y2 - y1)/(x2 - x1)

Using the point-slope form y - y1 = m(x - x1) with m = 2 and (x1, y1) = (1, -3): y - (-3) = 2(x - 1), so y + 3 = 2x - 2, giving y = 2x - 5 or 2x - y - 5 = 0.

y - y1 = m(x - x1)

Slope m = (6 - 2)/(3 - 1) = 4/2 = 2. Using point-slope form with point (1, 2): y - 2 = 2(x - 1), so y - 2 = 2x - 2, giving y = 2x, or 2x - y = 0.

y - y1 = m(x - x1)

Slope of first line m1 = -1/sqrt3; slope of second line m2 = -sqrt3. The angle between them satisfies tan(theta) = |(m1 - m2)/(1 + m1 m2)| = |(-1/sqrt3 + sqrt3)/(1 + (-1/sqrt3)(-sqrt3))| = |((-1 + 3)/sqrt3)/(1 + 1)| = |(2/sqrt3)/2| = 1/sqrt3. So theta = 30 degrees.

tan theta = |(m1 - m2)/(1 + m1 m2)|

The distance of point (x1, y1) from line Ax + By + C = 0 is d = |A x1 + B y1 + C|/sqrt(A^2 + B^2). Here A = 3, B = -4, C = -26 and (x1, y1) = (3, -5). d = |3(3) - 4(-5) - 26|/sqrt(9 + 16) = |9 + 20 - 26|/sqrt25 = |3|/5 = 3/5.

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

For parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0, the distance is d = |C1 - C2|/sqrt(A^2 + B^2). Here A = 3, B = -4, C1 = 7, C2 = 5. d = |7 - 5|/sqrt(9 + 16) = 2/5.

d = |C1 - C2|/sqrt(A^2 + B^2)

The intercept form of a line is x/a + y/b = 1, where a and b are the x- and y-intercepts. Here a = -3 and b = 2, so x/(-3) + y/2 = 1. Multiplying throughout by 6: -2x + 3y = 6, i.e. 2x - 3y + 6 = 0.

x/a + y/b = 1

Midpoint of the segment = ((1 + 3)/2, (1 + 5)/2) = (2, 3). Slope of the segment = (5 - 1)/(3 - 1) = 2, so the slope of the perpendicular bisector = -1/2. Using point-slope form through (2, 3): y - 3 = -1/2 (x - 2), so 2(y - 3) = -(x - 2), 2y - 6 = -x + 2, giving x + 2y - 8 = 0.

product of perpendicular slopes = -1

For the image (h, k) of point (x1, y1) in line Ax + By + C = 0, we use (h - x1)/A = (k - y1)/B = -2(A x1 + B y1 + C)/(A^2 + B^2). Here A = 1, B = 3, C = -7, (x1, y1) = (3, 8). A x1 + B y1 + C = 3 + 24 - 7 = 20, and A^2 + B^2 = 1 + 9 = 10, so the common ratio = -2(20)/10 = -4. Then h = 3 + 1(-4) = -1 and k = 8 + 3(-4) = 8 - 12 = -4. Hence the image is (-1, -4).

(h - x1)/A = (k - y1)/B = -2(A x1 + B y1 + C)/(A^2 + B^2)

Compare with x^2 + y^2 + 2gx + 2fy + c = 0: 2g = -4 so g = -2, 2f = 6 so f = 3, c = -12. Centre = (-g, -f) = (2, -3). Radius = sqrt(g^2 + f^2 - c) = sqrt(4 + 9 + 12) = sqrt25 = 5.

centre (-g, -f), radius sqrt(g^2 + f^2 - c)

The equation of a circle with centre (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2. Here (h, k) = (2, -3) and r = 5, so (x - 2)^2 + (y + 3)^2 = 25. Expanding: x^2 + y^2 - 4x + 6y - 12 = 0.

(x - h)^2 + (y - k)^2 = r^2

Compare y^2 = 12x with y^2 = 4ax: 4a = 12, so a = 3. The parabola opens to the right. Focus = (a, 0) = (3, 0). Directrix: x = -a, i.e. x = -3. Axis: the x-axis, i.e. y = 0. Length of latus rectum = 4a = 12.

y^2 = 4ax: focus (a,0), directrix x = -a

Here a^2 = 25 (a = 5) and b^2 = 9 (b = 3), with a > b so the major axis is along the x-axis. c = sqrt(a^2 - b^2) = sqrt(25 - 9) = 4. Vertices = (+/-5, 0); foci = (+/-4, 0); eccentricity e = c/a = 4/5; length of latus rectum = 2b^2/a = 2(9)/5 = 18/5.

e = c/a; latus rectum = 2b^2/a

Here a^2 = 16 (a = 4) and b^2 = 9 (b = 3). For a hyperbola, c = sqrt(a^2 + b^2) = sqrt(16 + 9) = sqrt25 = 5. Vertices = (+/-4, 0); foci = (+/-5, 0); eccentricity e = c/a = 5/4. (Length of latus rectum = 2b^2/a = 9/2.)

hyperbola: c = sqrt(a^2 + b^2), e = c/a

The focus is on the positive y-axis and the directrix is y = -3, so the parabola opens upwards with vertex at the origin and form x^2 = 4ay. Here a = 3 (distance from vertex to focus), so x^2 = 4(3)y = 12y.

x^2 = 4ay

The vertices and foci are on the x-axis, so the ellipse is x^2/a^2 + y^2/b^2 = 1 with a > b. From vertices, a = 5; from foci, c = 4. Then b^2 = a^2 - c^2 = 25 - 16 = 9. Hence the equation is x^2/25 + y^2/9 = 1.

b^2 = a^2 - c^2

Latus rectum = 2b^2/a and major axis = 2a. Given 2b^2/a = (1/2)(2a) = a, so 2b^2 = a^2. Since b^2 = a^2(1 - e^2), we get 2a^2(1 - e^2) = a^2, so 2(1 - e^2) = 1, giving 1 - e^2 = 1/2, e^2 = 1/2 and e = 1/sqrt2.

b^2 = a^2(1 - e^2)

The distance between (x1, y1, z1) and (x2, y2, z2) is sqrt[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]. Here = sqrt[(4 - 2)^2 + (3 - 3)^2 + (1 - 5)^2] = sqrt[4 + 0 + 16] = sqrt20 = 2 sqrt5.

d = sqrt[(x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2]

The distance of a point (x, y, z) from the origin is sqrt(x^2 + y^2 + z^2). Here = sqrt(1^2 + (-2)^2 + 3^2) = sqrt(1 + 4 + 9) = sqrt14 units.

d = sqrt(x^2 + y^2 + z^2)

By the section formula for internal division in ratio m : n, the coordinates are ((m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n), (m z2 + n z1)/(m + n)) with m = 2, n = 3. x = (2(3) + 3(1))/5 = 9/5; y = (2(4) + 3(-2))/5 = 2/5; z = (2(-5) + 3(3))/5 = -1/5. The point is (9/5, 2/5, -1/5).

((m x2 + n x1)/(m+n), ...)

The centroid of a triangle with vertices (x1,y1,z1), (x2,y2,z2), (x3,y3,z3) is ((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3). x = (1+3+4)/3 = 8/3; y = (2-1+0)/3 = 1/3; z = (3+5-2)/3 = 6/3 = 2. Centroid = (8/3, 1/3, 2).

G = ((x1+x2+x3)/3, (y1+y2+y3)/3, (z1+z2+z3)/3)

The midpoint of the segment joining (x1,y1,z1) and (x2,y2,z2) is ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2). = ((2+6)/2, (3+7)/2, (4+8)/2) = (4, 5, 6).

M = ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2)

Direct substitution gives 0/0, so factorise the numerator: x^2 - 4 = (x - 2)(x + 2). Then (x^2 - 4)/(x - 2) = x + 2 for x not equal to 2. Taking the limit as x approaches 2 gives 2 + 2 = 4.

a^2 - b^2 = (a - b)(a + b)

The standard limit is lim as x approaches 0 of (sin x)/x = 1 (with x in radians). This is a fundamental result used throughout limits and derivatives of trigonometric functions.

lim x->0 (sin x)/x = 1

Write (sin 3x)/(sin 5x) = [(sin 3x)/(3x) x 3x] / [(sin 5x)/(5x) x 5x]. As x approaches 0, (sin 3x)/(3x) -> 1 and (sin 5x)/(5x) -> 1, so the limit = (1 x 3x)/(1 x 5x) = 3/5.

lim x->0 sin(kx)/(kx) = 1

Direct substitution gives 0/0. Factorise: x^2 - 9 = (x - 3)(x + 3) and x^2 - x - 6 = (x - 3)(x + 2). Cancelling (x - 3): the expression becomes (x + 3)/(x + 2). Taking x approaching 3 gives (3 + 3)/(3 + 2) = 6/5.

Direct substitution gives 0/0. Rationalise by multiplying numerator and denominator by (sqrt(1 + x) + 1): [(sqrt(1+x) - 1)(sqrt(1+x) + 1)]/[x(sqrt(1+x) + 1)] = (1 + x - 1)/[x(sqrt(1+x) + 1)] = x/[x(sqrt(1+x) + 1)] = 1/(sqrt(1+x) + 1). As x approaches 0 this equals 1/(1 + 1) = 1/2.

(sqrt a - sqrt b)(sqrt a + sqrt b) = a - b

By definition f'(x) = lim as h approaches 0 of [f(x + h) - f(x)]/h = lim [ (x + h)^2 - x^2 ]/h = lim [x^2 + 2xh + h^2 - x^2]/h = lim [2xh + h^2]/h = lim (2x + h) = 2x. So f'(x) = 2x.

f'(x) = lim h->0 [f(x+h) - f(x)]/h

Using the power rule d/dx(x^n) = n x^(n-1) term by term: dy/dx = 3x^2 + 3(2x) - 5 + 0 = 3x^2 + 6x - 5.

d/dx(x^n) = n x^(n-1)

Let u = x^2 + 1 and v = x - 3, so u' = 2x and v' = 1. By the product rule dy/dx = u'v + uv' = (2x)(x - 3) + (x^2 + 1)(1) = 2x^2 - 6x + x^2 + 1 = 3x^2 - 6x + 1.

(uv)' = u'v + uv'

Let u = x + 1 and v = x - 1, so u' = 1 and v' = 1. By the quotient rule dy/dx = (u'v - uv')/v^2 = [1(x - 1) - (x + 1)(1)]/(x - 1)^2 = (x - 1 - x - 1)/(x - 1)^2 = -2/(x - 1)^2.

(u/v)' = (u'v - uv')/v^2

Using the product rule with u = sin x (u' = cos x) and v = cos x (v' = -sin x): dy/dx = cos x cos x + sin x (-sin x) = cos^2 x - sin^2 x = cos 2x. (Alternatively, y = (1/2) sin 2x, whose derivative is cos 2x.)

cos^2 x - sin^2 x = cos 2x

f'(x) = lim h->0 [sin(x + h) - sin x]/h. Expand sin(x + h) = sin x cos h + cos x sin h, so the numerator = sin x cos h + cos x sin h - sin x = sin x(cos h - 1) + cos x sin h. Thus f'(x) = sin x . lim h->0 (cos h - 1)/h + cos x . lim h->0 (sin h)/h. Since lim (cos h - 1)/h = 0 and lim (sin h)/h = 1, we get f'(x) = sin x . 0 + cos x . 1 = cos x.

d/dx(sin x) = cos x

The first 10 natural numbers are 1, 2, ..., 10. Their sum = n(n + 1)/2 = 10 x 11/2 = 55. Mean = sum/number of observations = 55/10 = 5.5.

mean = (sum of observations)/n

There are 8 observations. Mean = (6+7+10+12+13+4+8+12)/8 = 72/8 = 9. Absolute deviations from the mean: |6-9|=3, |7-9|=2, |10-9|=1, |12-9|=3, |13-9|=4, |4-9|=5, |8-9|=1, |12-9|=3. Sum of absolute deviations = 3+2+1+3+4+5+1+3 = 22. Mean deviation = 22/8 = 2.75.

MD = (sum of |xi - mean|)/n

Mean = (2+4+6+8+10)/5 = 30/5 = 6. Squared deviations: (2-6)^2=16, (4-6)^2=4, (6-6)^2=0, (8-6)^2=4, (10-6)^2=16; their sum = 40. Variance = 40/5 = 8. Standard deviation = sqrt(variance) = sqrt8 = 2 sqrt2 (approximately 2.83).

variance = (sum of (xi - mean)^2)/n; SD = sqrt(variance)

Variance is the arithmetic mean of the squares of the deviations of the observations from their mean; for n observations, variance = (1/n) sum of (xi - mean)^2. Standard deviation is the positive square root of the variance; it has the same units as the observations and measures the spread or dispersion of the data about the mean.

SD = sqrt(variance)

Coefficient of variation (CV) = (standard deviation/mean) x 100. Here SD = 2 sqrt2 which is approximately 2.83, and mean = 6. CV = (2.83/6) x 100, which is approximately 47.1 percent. A higher CV indicates greater relative variability.

CV = (SD/mean) x 100

The sample space is {1, 2, 3, 4, 5, 6}, so total outcomes = 6. The prime numbers on a die are 2, 3 and 5, which is 3 favourable outcomes. Probability = favourable/total = 3/6 = 1/2.

P(E) = favourable outcomes / total outcomes

When two dice are thrown, the total number of outcomes = 6 x 6 = 36. The outcomes giving a sum of 8 are (2,6), (3,5), (4,4), (5,3) and (6,2), which is 5 favourable outcomes. Probability = 5/36.

P(E) = favourable outcomes / total outcomes

Let A = the card is a king and B = the card is a heart. P(A) = 4/52, P(B) = 13/52, and P(A and B) = 1/52 (the king of hearts). By the addition theorem P(A or B) = P(A) + P(B) - P(A and B) = 4/52 + 13/52 - 1/52 = 16/52 = 4/13.

P(A or B) = P(A) + P(B) - P(A and B)