Class 12 Maths — Important Board Questions with Answers

A relation R on a set A is reflexive if (a, a) belongs to R for every a in A. It is symmetric if (a, b) in R implies (b, a) in R. It is transitive if (a, b) in R and (b, c) in R together imply (a, c) in R. A relation that is reflexive, symmetric and transitive at the same time is called an equivalence relation.

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.

A function f from A to B is one-one (injective) if distinct elements of A have distinct images, that is, f(x1) = f(x2) implies x1 = x2. It is onto (surjective) if every element of B has at least one pre-image in A, that is, the range of f equals the codomain B. A function that is both one-one and onto is called bijective (a one-to-one correspondence).

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.

(g o f)(x) means g(f(x)). So (g o f)(x) = g(f(x)) = g(x + 1) = 2(x + 1) = 2x + 2. (f o g)(x) means f(g(x)). So (f o g)(x) = f(g(x)) = f(2x) = 2x + 1. Note that (g o f)(x) is not equal to (f o g)(x), so composition of functions is not commutative in general.

(g o f)(x) = g(f(x))

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

The principal value branch of sin inverse is [-pi/2, pi/2]. Since sin(pi/6) = 1/2, the principal value of sin inverse (1/2) = pi/6. The principal value branch of cos inverse is [0, pi]. Since cos(2pi/3) = -1/2, the principal value of cos inverse (-1/2) = 2pi/3.

The principal values are: tan inverse (1) = pi/4, cos inverse (1/2) = pi/3, and sin inverse (1/2) = pi/6. Adding them: pi/4 + pi/3 + pi/6. Taking the LCD 12: 3pi/12 + 4pi/12 + 2pi/12 = 9pi/12 = 3pi/4.

Let sin inverse (x) = theta, so sin theta = x, where theta lies in [-pi/2, pi/2]. Then x = sin theta = cos(pi/2 - theta). Since (pi/2 - theta) lies in [0, pi], we have cos inverse (x) = pi/2 - theta = pi/2 - sin inverse (x). Rearranging gives sin inverse (x) + cos inverse (x) = pi/2.

sin inverse (x) + cos inverse (x) = pi/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))

The principal value branch of cos inverse is [0, pi], but 7pi/6 does not lie in this interval. First find cos(7pi/6) = cos(pi + pi/6) = -cos(pi/6) = -sqrt(3)/2. Now cos inverse (-sqrt(3)/2) = 5pi/6, since 5pi/6 lies in [0, pi] and cos(5pi/6) = -sqrt(3)/2. Therefore cos inverse (cos(7pi/6)) = 5pi/6.

For sin inverse (x): the domain is [-1, 1] and the principal value range is [-pi/2, pi/2]. For tan inverse (x): the domain is all real numbers R and the principal value range is the open interval (-pi/2, pi/2). Inverse trigonometric functions are defined by restricting the domain of the corresponding trigonometric functions so that they become one-one.

A diagonal matrix is a square matrix in which all the non-diagonal (off-diagonal) elements are zero. A scalar matrix is a diagonal matrix in which all the diagonal elements are equal (to the same non-zero scalar). An identity (unit) matrix is a scalar matrix in which all the diagonal elements are equal to 1; it is denoted by I and acts as the multiplicative identity, since AI = IA = A.

Two matrices are equal when their corresponding elements are equal. Comparing: x + y = 6 and x - y = 2. Adding these two equations: 2x = 8, so x = 4. Subtracting the second from the first: 2y = 4, so y = 2. Hence x = 4 and y = 2.

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)

The transpose of a matrix A, denoted A' (or A^T), is the matrix obtained by interchanging its rows and columns. Two properties are: (1) (A')' = A, that is, the transpose of the transpose is the original matrix; and (2) (AB)' = B' A', the transpose of a product equals the product of the transposes in reverse order. Also (A + B)' = A' + B' and (kA)' = k A'.

(AB)' = B' A'

A square matrix A is symmetric if A' = A, that is, aij = aji for all i and j (it is unchanged on taking its transpose). A square matrix A is skew-symmetric if A' = -A, that is, aij = -aji for all i and j; this forces all its diagonal elements to be zero. Every square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix.

symmetric: A' = A; skew-symmetric: A' = -A

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')

If a matrix has 12 elements, then the order (number of rows x number of columns) must be a pair of factors whose product is 12. The possible orders are: 1 x 12, 12 x 1, 2 x 6, 6 x 2, 3 x 4 and 4 x 3. So there are 6 possible orders. (In general, the number of possible orders equals the number of ways of writing the total number of elements as a product of two natural numbers.)

First find 2A = [[2, 4],[6, 8]]. Then subtract B element by element: 2A - B = [[2 - 2, 4 - 1],[6 - 0, 8 - 5]] = [[0, 3],[6, 3]].

Expand along the first row: |A| = 1 x (1 x 0 - 4 x 6) - 2 x (0 x 0 - 4 x 5) + 3 x (0 x 6 - 1 x 5) = 1 x (0 - 24) - 2 x (0 - 20) + 3 x (0 - 5) = -24 + 40 - 15 = 1. So the value of the determinant is 1.

(1) The value of a determinant remains unchanged if its rows and columns are interchanged (that is, |A'| = |A|). (2) If any two rows (or two columns) of a determinant are interchanged, the sign of the determinant changes. (3) If any two rows (or columns) of a determinant are identical (or proportional), the value of the determinant is zero. Also, if each element of a row (or column) is multiplied by a constant k, the determinant is multiplied by k.

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)|

The minor of an element aij of a determinant is the value of the determinant obtained by deleting the ith row and the jth column in which that element lies; it is denoted by Mij. The cofactor of the element aij is Aij = (-1)^(i + j) x Mij, that is, the minor with a sign attached according to the position of the element. Cofactors are used to expand a determinant and to find the adjoint of a matrix.

Aij = (-1)^(i+j) Mij

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

A square matrix A is called singular if its determinant is zero (|A| = 0), and non-singular if its determinant is non-zero (|A| is not equal to 0). The condition for a square matrix to be invertible (to have an inverse) is that it must be non-singular, that is, |A| must not be zero, because the inverse A inverse = (1/|A|) adj A is defined only when |A| is not zero.

A inverse exists if and only if |A| is not 0

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 an n x n matrix, |adj A| = |A|^(n - 1). Here n = 3, so |adj A| = |A|^2 = 5^2 = 25. Also, for a scalar k, |kA| = k^n |A|. Here |2A| = 2^3 x |A| = 8 x 5 = 40. So |adj A| = 25 and |2A| = 40.

|adj A| = |A|^(n-1); |kA| = k^n |A|

A function f(x) is continuous at a point x = a if the following three conditions are all satisfied: (1) f(a) is defined (the function has a value at a); (2) the limit of f(x) as x approaches a exists (that is, the left-hand limit equals the right-hand limit); and (3) this limit is equal to the value of the function, that is, limit of f(x) as x approaches a = f(a).

limit(x->a) f(x) = f(a)

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)

If a function is differentiable at a point, then it is necessarily continuous at that point. However, the converse is not always true: a function that is continuous at a point need not be differentiable there. For example, f(x) = |x| is continuous at x = 0 but not differentiable at x = 0 (it has a sharp corner). So differentiability implies continuity, but continuity does not imply differentiability.

Let y = sin(x^2). Using the chain rule, dy/dx = cos(x^2) x d/dx(x^2) = cos(x^2) x 2x = 2x cos(x^2). The chain rule states that the derivative of a composite function f(g(x)) is f'(g(x)) x g'(x).

d/dx f(g(x)) = f'(g(x)) g'(x)

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)

The derivative of sin inverse (x) with respect to x is 1/sqrt(1 - x^2), valid for -1 < x < 1. The derivative of tan inverse (x) with respect to x is 1/(1 + x^2), valid for all real x. These standard results are obtained by implicit differentiation of y = sin inverse (x) and y = tan inverse (x).

d/dx(sin inverse x) = 1/sqrt(1 - x^2)

Differentiate each with respect to the parameter t: dx/dt = -a sin t and dy/dt = a cos t. Then dy/dx = (dy/dt)/(dx/dt) = (a cos t)/(-a sin t) = -cos t/sin t = -cot t. So dy/dx = -cot t.

dy/dx = (dy/dt)/(dx/dt)

Rolle's theorem: if a function f is continuous on the closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one point c in (a, b) such that f'(c) = 0. Lagrange's mean value theorem: if a function f is continuous on [a, b] and differentiable on (a, b), then there exists at least one point c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).

f'(c) = (f(b) - f(a))/(b - a)

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

The derivative dy/dx represents the instantaneous rate of change of y with respect to x. When a quantity changes with time, its derivative with respect to time gives the rate of change. The area of a circle is A = pi r^2, so the rate of change of area with respect to the radius is dA/dr = 2 pi r. If the radius changes with time, dA/dt = 2 pi r (dr/dt).

dA/dr = 2 pi r

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)

A differentiable function f is increasing on an interval if its derivative f'(x) is greater than or equal to 0 (strictly increasing if f'(x) > 0) for all x in that interval. It is decreasing if its derivative f'(x) is less than or equal to 0 (strictly decreasing if f'(x) < 0) for all x in that interval. So the sign of the first derivative tells us whether the function rises or falls.

f'(x) > 0 increasing; f'(x) < 0 decreasing

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 to a curve y = f(x) at a point is the value of the derivative dy/dx at that point. The normal is the line perpendicular to the tangent at that point, so its slope is the negative reciprocal of the tangent's slope, that is, slope of normal = -1/(dy/dx). Using the point-slope form with these slopes gives the equations of the tangent and the normal.

slope of normal = -1/(slope of tangent)

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 f be a function and c be a point where f'(c) = 0 (a critical point). By the second derivative test: if f''(c) < 0, then f has a local maximum at c; if f''(c) > 0, then f has a local minimum at c; and if f''(c) = 0, the test fails and we must use the first derivative test (or higher derivatives) to decide.

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.

Integration is the reverse (inverse) process of differentiation; it is used to find a function whose derivative is given. The indefinite integral of a function includes an arbitrary constant of integration C. The power rule for integration is: the integral of x^n dx = x^(n + 1)/(n + 1) + C, valid for n not equal to -1. For n = -1, the integral of (1/x) dx = ln|x| + C.

integral of x^n dx = x^(n+1)/(n+1) + C

Integrate term by term using the power rule. The integral of x^3 dx = x^4/4, the integral of 2x dx = x^2, and the integral of 1 dx = x. Adding a single constant of integration, the integral of (x^3 + 2x + 1) dx = x^4/4 + x^2 + x + C.

integral of x^n dx = x^(n+1)/(n+1) + C

The standard integrals are: the integral of sin x dx = -cos x + C; the integral of cos x dx = sin x + C; and the integral of e^x dx = e^x + C. Also, the integral of sec^2 x dx = tan x + C. These are obtained by reversing the corresponding derivative formulas.

integral of sin x dx = -cos x + C

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

Write 1/((x - 1)(x - 2)) = A/(x - 1) + B/(x - 2). Then 1 = A(x - 2) + B(x - 1). Putting x = 1: 1 = A(-1), so A = -1. Putting x = 2: 1 = B(1), so B = 1. Hence the integral = integral of [-1/(x - 1) + 1/(x - 2)] dx = -ln|x - 1| + ln|x - 2| + C = ln|(x - 2)/(x - 1)| + C.

(1) The integral from a to b of f(x) dx = - the integral from b to a of f(x) dx (interchanging the limits changes the sign). (2) The integral from a to b of f(x) dx = the integral from a to c of f(x) dx + the integral from c to b of f(x) dx (splitting the interval at a point c). (3) The integral from 0 to a of f(x) dx = the integral from 0 to a of f(a - x) dx. Also, for an odd function the integral from -a to a is 0, and for an even function it is twice the integral from 0 to a.

integral(a to b) f(x) dx = -integral(b to a) f(x) dx

First find the indefinite integral: the integral of x^2 dx = x^3/3. Now apply the limits using the fundamental theorem of calculus: the integral from 0 to 1 of x^2 dx = [x^3/3] evaluated from 0 to 1 = (1^3/3) - (0^3/3) = 1/3 - 0 = 1/3.

integral(a to b) f(x) dx = F(b) - F(a)

The integral of sin x dx = -cos x. Applying the limits from 0 to pi/2: the integral from 0 to pi/2 of sin x dx = [-cos x] from 0 to pi/2 = (-cos(pi/2)) - (-cos 0) = (-0) - (-1) = 0 + 1 = 1. So the value of the integral is 1.

integral of sin x dx = -cos x

Integrate term by term. The integral of (1/x) dx = ln|x| and the integral of e^x dx = e^x. Adding the constant of integration, the integral of (1/x + e^x) dx = ln|x| + e^x + C.

integral of (1/x) dx = ln|x| + C

The function f(x) = x^3 is an odd function, because f(-x) = (-x)^3 = -x^3 = -f(x). By the property of definite integrals, the integral from -a to a of an odd function is zero. Therefore the integral from -1 to 1 of x^3 dx = 0. (We can also verify: [x^4/4] from -1 to 1 = 1/4 - 1/4 = 0.)

integral(-a to a) of odd function = 0

The area bounded by the curve y = f(x), the x-axis, and the ordinates x = a and x = b (with f(x) >= 0) is given by the definite integral: Area = the integral from a to b of y dx = the integral from a to b of f(x) dx. The area between two curves y = f(x) (upper) and y = g(x) (lower) from x = a to x = b is Area = the integral from a to b of [f(x) - g(x)] dx.

Area = integral(a to b) f(x) dx

The area is given by the definite integral of y with respect to x from 0 to 3: Area = the integral from 0 to 3 of x^2 dx = [x^3/3] from 0 to 3 = (3^3/3) - (0) = 27/3 = 9. So the required area is 9 square units.

Area = integral(a to b) y dx

The area is the definite integral of y = 2x from 0 to 3: Area = the integral from 0 to 3 of 2x dx = [x^2] from 0 to 3 = (3^2) - (0^2) = 9. So the required area is 9 square units. (This agrees with the area of the triangle of base 3 and height 6, which is (1/2)(3)(6) = 9.)

Area = integral(a to b) y dx

The equation x^2 + y^2 = 16 represents a circle centred at the origin with radius r = 4 (since r^2 = 16). The area enclosed by a circle of radius r is pi r^2. So the area = pi (4)^2 = 16 pi square units. (By integration, the area = 4 times the integral from 0 to 4 of sqrt(16 - x^2) dx = 16 pi.)

Area of circle = pi r^2

If a region is symmetric about the x-axis, the y-axis or the origin, we can compute the area of one symmetric part and then multiply by the appropriate factor to get the total area. For example, a circle x^2 + y^2 = a^2 is symmetric about both axes, so its total area equals 4 times the area of the part lying in the first quadrant. Using symmetry reduces the amount of integration needed and simplifies the calculation.

A differential equation is an equation that involves an independent variable, a dependent variable and the derivatives of the dependent variable with respect to the independent variable. The order of a differential equation is the order of the highest derivative appearing in it. The degree of a differential equation is the power (exponent) of the highest order derivative, provided the equation is a polynomial in the derivatives (free from radicals and fractions in the derivatives).

The highest order derivative present in the equation is d2y/dx2 (the second derivative), so the order of the differential equation is 2. The power (exponent) of this highest order derivative is 3, and the equation is a polynomial in the derivatives, so the degree is 3. Hence the order is 2 and the degree is 3.

The general solution of a differential equation is the solution that contains as many arbitrary constants as the order of the differential equation; it represents a whole family of curves. A particular solution is obtained from the general solution by giving definite values to these arbitrary constants using given initial or boundary conditions; it represents one specific member of the family.

Separate the variables: dy/(1 + y^2) = dx/(1 + x^2). Now integrate both sides: the integral of dy/(1 + y^2) = the integral of dx/(1 + x^2). This gives tan inverse (y) = tan inverse (x) + C, where C is the constant of integration. This is the general solution of the differential equation.

integral dy/(1 + y^2) = tan inverse (y)

Separate the variables by cross-multiplying: y dy = x dx. Integrate both sides: the integral of y dy = the integral of x dx, giving y^2/2 = x^2/2 + C1. Multiplying by 2: y^2 = x^2 + C (where C = 2 C1). So the general solution is y^2 - x^2 = C, which represents a family of hyperbolas.

A differential equation of the form dy/dx = f(x, y) is called homogeneous if the function f(x, y) can be written as a function of the ratio y/x only (that is, f depends on x and y only through y/x). To solve it, we use the substitution y = v x (so v = y/x), which gives dy/dx = v + x (dv/dx). This substitution reduces the equation to a variable separable form in v and x, which can then be integrated.

y = v x, dy/dx = v + x(dv/dx)

A linear differential equation of the first order and first degree has the standard form dy/dx + P y = Q, where P and Q are functions of x only (or constants). It is solved using the integrating factor (IF) = e^(integral of P dx). Multiplying the equation by the integrating factor makes the left side an exact derivative, and the solution is y x (IF) = integral of [Q x (IF)] dx + C.

IF = e^(integral of P dx)

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

A scalar quantity is one that has only magnitude and no direction (for example, mass, length, temperature). A vector quantity is one that has both magnitude and direction and obeys the laws of vector addition (for example, displacement, velocity, force). The position vector of a point P with coordinates (x, y, z) with respect to the origin O is the vector OP = x i + y j + z k, and its magnitude is sqrt(x^2 + y^2 + z^2).

|OP| = sqrt(x^2 + y^2 + z^2)

The magnitude of a = 2 i + 3 j + 6 k is |a| = sqrt(2^2 + 3^2 + 6^2) = sqrt(4 + 9 + 36) = sqrt(49) = 7. A unit vector in the direction of a is obtained by dividing a by its magnitude: a hat = a/|a| = (1/7)(2 i + 3 j + 6 k) = (2/7) i + (3/7) j + (6/7) k.

a hat = a/|a|

A unit vector is a vector whose magnitude is 1; it is used to specify direction. A zero (null) vector is a vector whose magnitude is zero and whose direction is indeterminate; it is denoted by 0. Collinear vectors are two or more vectors that are parallel to the same line, irrespective of their magnitudes and directions; such vectors can be written as scalar multiples of one another.

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)

The scalar projection (component) of vector a on vector b is given by (a . b)/|b|. It represents the length of the shadow of a along the direction of b. If the dot product a . b = 0 (and neither vector is zero), then cos theta = 0, so theta = 90 degrees; this means the two vectors are perpendicular (orthogonal) to each other.

projection of a on b = (a . b)/|b|

Two vectors are perpendicular if their dot product is zero. So a . b = (2)(1) + (lambda)(-2) + (1)(3) = 2 - 2 lambda + 3 = 5 - 2 lambda. Setting this equal to zero: 5 - 2 lambda = 0, so 2 lambda = 5 and lambda = 5/2.

a . b = 0 for perpendicular vectors

The direction cosines of a line are the cosines of the angles that the line makes with the positive directions of the x-, y- and z-axes; they are usually denoted by l, m and n. The direction ratios are any set of numbers proportional to the direction cosines. The direction cosines always satisfy the relation l^2 + m^2 + n^2 = 1.

l^2 + m^2 + n^2 = 1

The direction cosines are obtained by dividing the direction ratios by the square root of the sum of their squares. Here the magnitude = sqrt(1^2 + 2^2 + 2^2) = sqrt(1 + 4 + 4) = sqrt(9) = 3. So the direction cosines are l = 1/3, m = 2/3 and n = 2/3. We can check l^2 + m^2 + n^2 = 1/9 + 4/9 + 4/9 = 9/9 = 1.

l = a/sqrt(a^2+b^2+c^2), etc.

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 general Cartesian equation of a plane is Ax + By + Cz + D = 0, where A, B, C are the direction ratios of the normal to the plane and are not all zero. The equation of a plane that makes intercepts a, b and c on the x-, y- and z-axes respectively is x/a + y/b + z/c = 1 (the intercept form). A plane is uniquely determined by three non-collinear points.

Ax + By + Cz + D = 0

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 angle between two planes is equal to the angle between their normals. If the planes are A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0, then cos theta = |A1 A2 + B1 B2 + C1 C2| / (sqrt(A1^2 + B1^2 + C1^2) x sqrt(A2^2 + B2^2 + C2^2)). The two planes are perpendicular if A1 A2 + B1 B2 + C1 C2 = 0, and parallel if A1/A2 = B1/B2 = C1/C2.

cos theta = |A1A2 + B1B2 + C1C2|/(|n1||n2|)

For two parallel planes Ax + By + Cz + D1 = 0 and Ax + By + Cz + D2 = 0, the distance between them is |D1 - D2|/sqrt(A^2 + B^2 + C^2). Here A = 2, B = -2, C = 1, D1 = 3, D2 = 9. Distance = |3 - 9|/sqrt(2^2 + (-2)^2 + 1^2) = 6/sqrt(4 + 4 + 1) = 6/sqrt(9) = 6/3 = 2 units.

d = |D1 - D2|/sqrt(A^2 + B^2 + C^2)

A linear programming problem (LPP) is a problem of maximising or minimising a linear function of variables subject to certain linear conditions. The objective function is the linear function Z = ax + by that is to be optimised (maximised or minimised). The constraints are the linear inequalities (or equations) in the variables that restrict the values the variables can take; together with the non-negativity restrictions (x >= 0, y >= 0) they define the region of possible solutions.

Z = ax + by (objective function)

The feasible region is the common region determined by all the constraints (including the non-negativity restrictions) of a linear programming problem; it is the set of all points that satisfy every constraint. A feasible solution is any point that lies in the feasible region. An optimal solution is a feasible solution that gives the maximum or minimum value of the objective function, whichever is required.

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.

A feasible region is said to be bounded if it can be enclosed within a circle (or a rectangle) of finite radius, that is, it does not extend to infinity. It is unbounded if it extends indefinitely (to infinity) in some direction and cannot be enclosed. For a bounded feasible region, the objective function always has both a maximum and a minimum value at the corner points. For an unbounded region, a maximum or a minimum may not exist and must be checked separately.

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).

Conditional probability is the probability of occurrence of an event A given that another event B has already occurred; it is written as P(A | B). Its formula is P(A | B) = P(A intersection B)/P(B), provided P(B) is not zero. Here P(A intersection B) is the probability that both A and B occur, and P(B) is the probability of B.

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

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)

Two events A and B are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other, that is, P(A | B) = P(A) and P(B | A) = P(B). For independent events, the multiplication rule takes the simple form P(A intersection B) = P(A) x P(B). This is used as the test for independence: if P(A intersection B) = P(A) P(B), the events are independent.

P(A intersection B) = P(A) x 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))

A random variable is a real-valued function whose value is determined by the outcome of a random experiment; it assigns a number to each outcome of the sample space (for example, the number of heads when two coins are tossed). A probability distribution of a random variable is a table (or description) that lists all the possible values of the random variable together with their corresponding probabilities; the sum of all these probabilities is always 1.

sum of all p(x) = 1

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)