Class 11

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 7 Linear Inequalities will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 7 Linear Inequalities

Inequality:
A statement with symbols like >, ≥, <, ≤ is an inequality.

Different types of inequality:

(a) Numerical inequality: It is an inequality involving numbers not variables.
(b) Literal inequality: It is the inequality involving literal numbers(variable).
(c) Strict inequality: An inequality with only > or < symbols is a strict inequality.
(d) Slack inequality: An inequality with only ≥ or ≤ symbols is a slack inequality.

Linear inequality:
An inequality involving variables in the first degree is called linear inequalities.
(a) General form of inequalities:
(i) In one variable: ax + b > or ≥ or < or ≤ 0
(ii) In two variables: ax + by + c > or ≥ or < or ≤ 0.

Intervals:

• Closed Interval: [a, b] = {x ∈ R: a ≤ x ≤ b}
• Open Interval: (a, b) = {x ∈ R: a < x < b}
• Semi-open or semi-closed interval:
  ⇒ [a, b) = {x ∈ R: a ≤ x < b}
  ⇒ (a, b] = {x ∈ R: a < x ≤ b}

Basic properties of inequalities:
(1) a > b, b > c ⇒ a > c
(2) a > b ⇒ a ± c > b ± c
(3) a > b

• m > 0 ⇒ am > bm, \(\frac{a}{m}>\frac{b}{m}\)
• m < 0 ⇒ am < bm, \(\frac{a}{m}<\frac{b}{m}\)

(4) If a > b > 0, then
a² > b², |a| > |b| and \(\frac{1}{a}>\frac{1}{b}\)
If a < b < 0, then
|a| > |b| and \(\frac{1}{a}>\frac{1}{b}\)

Graphical solution of linear inequalities in two variables:
Working rule:

Let the inequality is ax + by + c < or ≤ or > or ≥ 0

Step – 1: Consider the equation ax + by + c = 0 in place of the inequality and draw its graph (Draw a dotted line for > or < and a bold line for ≥ or ≤).
Step – 2: Take any point that does not lie on the graph, and put the coordinate in the inequality.
If you get true then the inequality is satisfied. Shade the half-plane containing that point otherwise the inequality is not satisfied. In this case shade the half plane region that does not contain the point.
Step – 3: The shaded region is the required solution.

Solution of a system of linear inequalities in two variables:

Step – 1: Draw the graph of all lines.
Step – 2: Shade the appropriate region for each inequality.
Step – 3: The common region is the required solution.

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 11 Straight Lines will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 11 Straight Lines

Distance formula:
Distance between two points A (x₁, y₁) and A (x₂, y₂) = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Section Formula:
If C(x, y) divides the join of A (x₁, y₁) and A (x₂, y₂) in the ratio m: n internally then, x = \(\frac{m x_2+n x_1}{m+n}\), y = \(\frac{m y_2+n y_1}{m+n}\)

Note:

• If the division is external then, x = \(\frac{m x_2-n x_1}{m-n}\), y = \(\frac{m y_2-n y_1}{m-n}\)
• If C(x, y) is the midpoint then x = \(\frac{x_1+x_2}{2}\), y = \(\frac{y_1+y_2}{2}\)

Area of triangle formula:
The area of triangle with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is given by  = \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Different points related to a triangle:
(a) Centroid of the triangle with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is = G \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)

(b) In centre of a triangle with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is = I \(\left(\frac{a x_1+b x_2+c x_3}{3}, \frac{a y_1+b y_2+c y_3}{3}\right)\)

Slope Of A Line:
(a) Angle of inclination: the angle θ made by a line with positive x-axis is the angle of inclination.
(b) Slope of a line: Slope of a line is the tangent of angle of inclination. i,.e m = tan θ.
(c) Slope of a line joining A(x₁, y₁), and B(x₂, y₂) = \(\frac{y_2-y_1}{x_2-x_1}\)

Note:

(i) Slope of x-axis = 0
Slope of any line parallel to x-axis = 0

(ii) Slope of y-axis  = ∞
Slope of any line parallel to y-axis = ∞

Angle Between two Lines:
Angle Φ between two lines with slope m₁ and m₂ is given by tan Φ = \(\pm \frac{\left(m_1-m_2\right)}{1+m_1 m_2}\)

Note:

• To find the acute angle between two lines use the formula. tan Φ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
• Two lines are parallel if m₁ = m₂
• Two lines are perpendicular if m₁m₂ = (-1).

Collinearity Of Three Points:
Three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are collinear if
(i) Sum of distances between two pairs of points = Distance between the 3rd pair.
Or, (ii) Area of Δ ABC = 0
Or, (iii) Let B(x₂, y₂) divides the join of AC in ratio k: 1
∴ \(x_2=\frac{k x_3+x_1}{k+1}, y_2=\frac{k y_3+y_1}{k+1}\)
The value of k obtained from two cases are equal.
Or, (iv) Slope of AB = Slope of AC.

Equation of a straight line:
Lines parallel to co-ordinate axes:
(i) Equation of any line parallel to x-axis is, y = k
⇒ Equation of x-axis is, y = 0

(ii) Equation of any line parallel to y-axis is, x = k
⇒ Equation of y-axis is, x = 0

Lines Not Parallel To Any Axes:
(i) Slope intercept form:
Equation of a line with slope ‘m’ and y-intercepts ‘c’ is: y = mx + c

(ii) Point slope form:
Equation of a line with slope ‘m’ and passing through a point A(x₁, y₁) is: y – y₁ = m(x – x₁)

(iii) Two point form:
Equation of the line passing through A(x₁, y₁) and B(x₂, y₂) is : \(\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}\)

(iv) Intercept form:
Equation of a line with x-intercept ‘a’ and y-intercept ‘b’ is \(\frac{x}{a}+\frac{y}{b}=1\)

(v) Normal form:
Equation of a line whose distance form origin is P and the perpendicular drawn form origin to the line makes an angle α with positive direction of x-axis is: x cos α + y sin α = P

(vi) Parameteric form or symmetric form:
Equation of the line passing through A(x₁, y₁) and making an angle θ with positive direction of x-axis is: \(\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}\) = r
Or, x = x₁ + r cos θ, y = y₁ + r sin θ
where r = The directed distance between points P(x, y) and A(x₁, y₁)

(vii) General form:
General equation of a straight line is Ax + By + C = 0

Note:

• Slope of this line = –\(\frac{\mathrm{A}}{\mathrm{B}}\)
• x-intercept = –\(\frac{\mathrm{C}}{\mathrm{A}}\)
• y-intercept = – \(\frac{\mathrm{C}}{\mathrm{B}}\)
• Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\) perpendicular if a₁a₂ + b₁b₂ = 0 and coincident if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Condition of concurrency of three lines:
Three lines a₁x + b₁y + c₁ = 0 a₂x + b₂y + c₂ = 0 and a₃x + b₃y + c₃ = 0 are concurrent if \(\left|\begin{array}{lll}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3
\end{array}\right|\) = 0

Family Of Lines:

(i) Equation of lines parallel to the line ax + by + c = 0 is given by: ax + by + λ = 0
(ii) Equation of lines perpendicular to the line ax + by + c = 0 is given by bx – ay + λ = 0
(iii) Equation of lines passing through the point of intersection of two lines.
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 is given by: (a₁x + b₁y + c₁) + λ(a₂x + b₂y + c₂)

Distance of a point from a line:
The perpendicular distance of A(x₁, y₁) from the line ax + by + c  = 0 is: d = \(\left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right|\)

Distance between two parallel lines:
ax + by + c₁ = 0 and  ax + by + c₂ = 0 is d = \(\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|\)

Position of a point with respect to a line:
A point A(x₁, y₁) lies
(i) above the line ax + by + c = 0 if \(\frac{a x_1+b y_1+c}{b}\) > 0
(ii) below the line ax + by + c = 0 if \(\frac{a x_1+b y_1+c}{b}\) < 0

Equation of bisectors of angle between two intersecting lines:
(i) Equation of angle bisector of two lines. a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 is given by \(\frac{a_1 x+b_1 y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm \frac{a_2 x+b_2 y+c_2}{\sqrt{a_2^2+b_2^2}}\)

Note:

Out of two bisector take one and find the angle between that bisector and one line. If the angle is less than 45° then that bisector is the bisector of acute angle, otherwise, the other bisector is the bisector of acute angle.

(ii) Bisector of angle containing a given point (h, k):

Step – 1: Check the sign of a₁h + b₁k + c₁  and a₂h + b₂k + c₂

• If they have same sign then the bisector of angle containing (h, k) is: \(\frac{a_1 x+b_1 y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{a_2 x+b_2 y+c_2}{\sqrt{a_2^2+b_2^2}}\)
• If they have opposite sign then the bisector of angle containing (h, k) is: \(\frac{a_1 x+b_1 y+c_1}{\sqrt{a_1^2+b_1^2}}=-\frac{a_2 x+b_2 y+c_2}{\sqrt{a_2^2+b_2^2}}\)

Change Of Axes (Shifting Of Origin):

(i) Translation of coordinate axes.
Let O'(h, k) is the origin of system S’ with respect to origin O(0, 0) of the system S. S’ is the translation of S. If (x, y) and (x’, y’) are the coordinate of a point P in the system S and S’ respectively then
x’ = x – h and y’ = y – k Or, x = x’ + h, y = y’ + k

(ii) Rotation of axes:
Let S’ is a rotation of S, α is the measure of rotation
If (x, y) and (x’, y’) are the coordinate of a point P with respect to S and S’ then x = x’ cos α – y’ sin α and y = x’ sin α + y’ cos α

(iii) Translation as well as a rotation:
If S’ is a combination of translation followed by a rotation then x = h + x’ cos α – y’ sin α, y = k + x’ sin α + y’ cos α

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 6 Complex Numbers and Quadratic Equations will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 6 Complex Numbers and Quadratic Equations

Unit imaginary number ‘i’.
The unit imaginary number i = √-1
i² = -1
i³ = -i
i⁴ = 1
In general (i)⁴ⁿ = 1, (i)⁴ⁿ⁺¹ = i, (i)⁴ⁿ⁺² = -1, and (i)⁴ⁿ⁺³ = -i.
⇒ If a and b are positive real numbers then
√-a × √-b = -√ab
√a × √b = √ab

Complex Number
General form: = z = a +ib

• a = Real part of (z)  = Re (z)
• b = Imaginary part of (z) = I_m(z)
• a + i0 is purely real and 0 + ib is purely imaginary .
• a + ib = c + id iff a = c and b = d

Complex Algebra
(a) Addition of complex numbers
If z₁ = a + ib and z₂ = c + id then z₁ + z₂ = (a + c) + i(b + d)

Properties:

• Addition is commutative: z₁ + z₂ = z₂ + z₁
• Addition is associative: (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃)
• 0 + i0 is the additive identity.
• -z is the additive inverse of z.

(b) Subtraction of complex numbers:
z₁ = a + ib and z₂ = c + id then z₁ – z₂ = (a – c) + i(b – d)

(c) Multiplication of complex numbers:
z₁ = a + ib and z₂ = c + id then z₁z₂ = (ac – bd) + i(bc + ad)

Properties:

• Multiplication is commutative: z₁z₂ = z₂z₁
• Multiplication is associative: z₁(z₂z₃) = z₁z₂(z₃)
• 1 = 1 + i0 is the multiplicative identity.
• If z = a + ib then the inverse of z.
  z^-1 = \(\frac{1}{a+i b}=\frac{a-i b}{(a+i b)(a-i b)}\)
  = \(\frac{a-i b}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{i b}{a^2+b^2}\)
• Multiplication is distributive over addition. z₁(z₂ + z₃) = z₁z₂ + z₁z₃

Conjugate and modulus of a complex number:
If  z = a + ib the conjugate of z is \(\bar{Z}\) = a – ib.
⇒ We get conjugate by replacing i by (-i) Modulus of z = a + ib is denoted by |z| and |z| = \(\sqrt{a^2+b^2}\)

Properties Of Conjugate:
(i) \((\overline{\bar{z}})\) = z
(ii) z + \(\bar{z}\) = 2 Re (z)
(iii) z – \(\bar{z}\) = 2i m̂ (z)
(iv) z – \(\bar{z}\) ⇔ z is purely real
(v) Conjugate of real number is itself.
(vi) z + \(\bar{z}\) = 0 ⇒ z is purely imaginary.
(vii) z. \(\bar{z}\) = [Re(z)]² + [m̂(z)]²
= a² + b²
= |z|²
(viii) \(\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\)
(ix) \(\overline{z_1-z_2}=\overline{z_1}-\overline{z_2}\)
(x) \(\overline{z_1z_2}=\overline{z_1}\overline{z_2}\)
(xi) \(\left(\overline{\frac{z_1}{z_2}}\right)=\frac{\overline{z_1}}{\overline{z_2}}\)

Properties of modulus:
(1) Order relations are not defined for complex numbers. i,e,. z₁ > z₂ or z₁ < z₂ has no meaning but |z₁| < |z₂| or |z₁| > |z₂| is meaningful because |z₁| and |z₂| are real numbers.
(2) |z|  = 0 ⇔ z = 0
(3) |z| = |\(\bar{z}\)| = |-z|
(4) |z| ≤ Re (z) ≤ |z| and -|z| ≤ m̂ (z) ≤ |z|
(5) |z₁z₂| = |z₁| |z₂|
(6) \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\)
(7) |z₁ ± z₂|² = |z₁|² + |z₂|² ± 2 Re (z₁\(\bar{z}_2\))
(8) |z₁ + z₂|² = |z₁ – z₂|² = 2(|z₁|² + |z₂|²)
(9) |z₁ + z₂|² ≤ |z₁| + |z₂|

Square Root Of Complex Number:
Let z = a + ib
Let √z = x + iy

If b > 0 then x and y are taken as same sign.
If b < 0 then x and y are of opposite sign.

Representation of a complex number:
We represent a complex number in different forms like
(i) Geometrical form
(ii) Vector form
(iii) Polar form
(iv) Eulerian form or Exponential form

(i) Geometrical form:
Geometrically z = x + iy = (x, y) represents a point in a coordinate plane known as Argand plane or Gaussian plane.

(ii) Vector form:
In vector form a complex number z = x + iy is the vector \(\overrightarrow{\mathrm{OP}}\) where p(x, y) is the point in the cartesian plane.

(iii) Polar form:
A complex number z = x + iy  in polar form can be written as z = r(cos θ + i sin θ) where r = \(\sqrt{x^2+y^2}\) = |z| and θ is called the argument and -π < θ ≤ π. Technique to write z = x + iy in polar form.
Step – 1: Find r = |z| = \(\sqrt{x^2+y^2}\)
Step – 2: Find α = tan^-1 \(\left|\frac{y}{x}\right|\)
Step – 3:
θ = α for x > 0, y > 0
θ = π – α for x > 0, y > 0
θ = -π + α for x > 0, y > 0
θ = -α for x > 0, y > 0
Step – 4: Write z = r(cos θ + i sin θ)

(iv) Eulerian form or Exponential form z = r e^iθ, because e^iθ = cos θ + i sin θ where θ is the argument and r is the modulus if z.

Note:
(1) |z₁ z₂ z₃ ….. z_n| = |z₁||z₂| …. |z_n|
(2) arg (z₁z₂ …. Z_n) = arg (z₁) + arg (z₂) + ….. + arg (z_n)
(3) arg \(\left(\frac{z_1}{z_2}\right)\) = arg (z₁) – arg (z₂)
(4) arg \((\bar{z})\) = -arg (z)

Cube Roots Of Unity:
Cube roots of unity are 1, ω, ω² where ω = \(\frac{-1 \pm i \sqrt{3}}{2}\)

Properties of Cube roots of unity:
(i) Cube roots of unity lie on unit circle |z| = 1
(ii) 1 + ω + ω² = 0
(iii) Cube roots of -1 are -1, -ω, -ω²
(iv) 1 + ωⁿ + ω²ⁿ \(=\left\{\begin{array}{l}
0 \text { if } n \text { is not a multiple of } 3 \\
3 \text { if } n \text { is a multiple of } 3
\end{array}\right.\)
(v) z³ + 1 = (z + 1) (z + ω) (z + ω²)
(vi) -ω and -ω² are roots of z² – z + 1  = 0.

De-moivre’s theorem:
(a) (De-moivre’s theorem for integral index)
(cos θ + i sin θ)ⁿ = cos (nθ) + i sin (nθ)

(b) (De-moivre’s theorem for rational index)
cos (nθ) + i sin (nθ) is one of the values of (cos θ + i sin θ)ⁿ

(c) nth roots of unity
nth roots of unity are 1, α, α², α³ …..α^n-1. where α = e^{i\(\frac{2 \pi}{n}\)} = cos \(\frac{2 \pi}{n}\) + i sin \(\frac{2 \pi}{n}\)

Properties:

• 1 + α + α² ….. + α^n-1 = 0
• 1 + α^p + α^2p + ….. + α^(n-1)p \(= \begin{cases}0 & \text { if } p \text { is not a multiple of } n \\ n & \text { if } p \text { is a multiple of } n\end{cases}\)
• 1. α. α² ….. α^n-1 = (-1)^n-1
• zⁿ – 1 = (z – 1) (z – α) (z – α²) …..(z – α^n-1)

Quadratic Equations:
The general form: ax² + bx + c = 0  …(i)
Solutions of quadratic equation(1) are
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
D = b² – 4ac is called the discrimination of a quadratic equation.
D > 0 ⇒ The equation has real and distinct roots.
D = 0 ⇒ The equation has real and equal roots.
D < 0 ⇒ The equation has complex roots.

Note:
In a quadratic equation with real coefficients, the complex roots occur in conjugate pairs.

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 9 Binomial Theorem will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 9 Binomial Theorem

Binomial Theorem For Positive Integral Index:
For any a,b ∈ R, and n ∈ N
(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ a^n-1b + ….. ⁿC_n bⁿ

Note:

(a) (a + b)ⁿ = aⁿ + na^n-1 b + \(\frac{n(n-1)}{2 !}\) a^n-2b² ….. + bⁿ
(b) (1 + x)ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x² + ….. + ⁿC_n xⁿ
(c) (a – b)ⁿ = ⁿC₀ aⁿ – ⁿC₁ a^n-1 b + ⁿC₂ a^n-2b² ….. + (-1)ⁿ bⁿ
(d) (1 – x)ⁿ = ⁿC₀ – ⁿC₁ x + ⁿC₂ x² ….. + (-1)ⁿ xⁿ

Some conclusions from the Binomial theorem:

• There are (n + 1) terms in the expansion of (a + b)ⁿ
• We can write (a + b)ⁿ = \(\sum_{r=0}^n{ }^n \mathrm{C}_r a^{n-r} b^r\) and (a – b)ⁿ = \(\sum_{r=0}^n(-1)^r{ }^n \mathrm{C}_r a^{n-r} b^r\)
• The sum of powers of a and b in each term = n
• As ⁿC_r = ⁿC_n-r (The coefficient of terms equidistant from the beginning and the end are equal).
• (r + 1)^th term (General term)
  = t_r+1 = ⁿC_r a^n-rb^r
• (a + b)ⁿ + (a – b)ⁿ = 2[ⁿC₀aⁿ + ⁿC₂ a^n-2b² + ….]
• (a + b)ⁿ – (a – b)ⁿ = 2[ⁿC₁ a^n-1b + ⁿC₃ a^n-3b³ + ….]
• (middle terms):
  ⇒ If n is even then the middle term = \(t_{\left(\frac{n+2}{2}\right)}=t_{\left(\frac{n}{2}+1\right)}\)
  ⇒ If n is odd there are two middle terms. They are = \(t_{\left(\frac{n+1}{2}\right)} \text { and } t_{\left(\frac{n+3}{2}\right)}\)
• t_r+1 from the end in the expansion of (a + b)ⁿ = t_r+1 from the beginning in the expansion of (b + a)ⁿ.

Binomial Theorem For Any Rational Index:
If n ∈ Q and x ∈ R such that |x| < 1 then (1 + x)ⁿ = 1 + nx + \(\frac{n(n-1)}{2 !} x^2\) + \(\frac{n(n-1)(n-2)}{3 !} x^3+\ldots .\)

Note:

(1) (1 + x)^-1 = 1 – x + x² – x³ + …..
(2) (1 – x)^-1 = 1 + x + x² + …..
(3) (1 + x)^-2 = 1 – 2x + 3x² – 4x³ + …..
(4) (1 – x)^-2 = 1 + 2x + 3x² + 4x³ + …..

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 8 Permutations And Combinations will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 8 Permutations And Combinations

Fundamental Principle Of Counting:
(a) Fundamental principle of Multiplication:
If we choose an element from set A with m element and then one element from set B  with n elements, then are total number of ways we can make a choice is exactly mn.

OR

If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of ways in which both the events can occur in succession in mn ways.

(b) Fundamental Principle of addition: If there are two events such that they can be performed independently in m and n different ways respectively, then either of two events can be performed in (m + n) ways.

Note:
(a) Use the multiplication principle if by doing one part of the job, the job remains incomplete.
(b) Use the addition principle if by doing one part of the job, the job is completed.

Factorial Notation:
If n ∈ N then the factorial of n, denoted by n! or ∠n is defined as
n! = n (n – 1). (n – 2) … 3.2.1.

Note:
0! = 1

Properties of Factorial:
(1) Factorial of negative integers is not defined
(2) n! = n(n – 1)!
= n(n – 1) (n – 2)!
= n(n – 1) (n – 2) (n – 3)!
(3) \(\frac{n !}{r !}\) = n(n – 1) (n – 2) ….. (r + 1)
(4) Exponent of a prime number p in n! denoted by
\(\mathrm{E}_p(n !)=\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\ldots \ldots\)

Permutation:
Each of the arrangements which can be made by taking some or all objects or things at a time is called a permutation.

(a) Permutation of n different objects:

• Number of permutations of n different objects have taken all at a time = \({ }^n \mathrm{P}_n\) = n!.
• Number of permutations of n different objects taken none at a time = \({ }^n \mathrm{P}_0\) = 1
• Number of permutations of n different objects taken r at a time = \({ }^n \mathrm{P}_r\) = P(n, r) = \(\frac{n !}{(n-r) !}\)

(b) Permutation ofnon-distinct objects:
(1) Number of permutations of n objects taken all at a time of which p objects are of same kind and others are distinct = \(\frac{n !}{p !}\)
(2) Number of permutations of n objects taken all at a time of which p objects are of one kind, q objects are of a second kind and other are distinct = \(\frac{n !}{p ! q !}\)
(3) Number of permutations of n objects taken all at a time in which p₁ objects are of one kind, p₂ are of second kind, p₃ are 3rd kind ….. and
p_n are of nth kind and other are distinct. = \(\frac{n !}{p_{1} ! p_{2} ! \cdots p_{n} !}\)

(c) Restricted permutations:

• Permutation of distinct objects with repetition: The number of permutations of n different things taken r at a time when each thing may be repeated any number of times = n^r
• Number of permutations of n different things taken r at a time when a particular thing is to be always included in each arrangement = r. ^n-1P_r-1.
• Number of permutations of n different things, taken r at the time when p particular are to be always included in each arrangement = P(r – (p – 1) ^n-pP_r-p.
• Number of permutations of n different things taken r at a time, when a particular thing is never taken in each arrangement = ^n-1P_r.
• Number of permutations of n different things taken r at a time, when p particular things never taken in each arrangement = ^n-pP_r.

(d) Circular permutation:
(1) When we do an arrangement of objects along a closed curve we call it the circular permutation.
(2) Number of circular permutations of n distinct objects taken all at a time = (n – 1)!, where clockwise and anti-clockwise orders are taken as different, as arrangements round a table.
(3) Number of circular permutations of n distinct objects taken all at a time, where clockwise and anti-clockwise orders make no difference as beads or flowers in a necklace or garland.
= \(\frac{(n-1) !}{2}\)
(4) Number of circular permutations of n different things taken r at a time where clockwise and anti-clockwise orders are different = \(\frac{\left({ }^n \mathrm{P}_r\right)}{r}\)
(5) Number of circular permutations of n different things taken r at a time where clockwise and anti-clockwise orders make no difference = \(\frac{\left({ }^n \mathrm{P}_r\right)}{2 r}\)

(e) Some more restricted permutations:

• Number of permutations of n different things taken all at a time, when m specified things come together = m!(n – m + 1)!.
• Number of permutations of n different things taken all at a time when m specified things never come together = n!  – m!(n – m + 1)!.

Combinations:
Each of the different selections made by taking some or all objects at a time irrespective of any order is called a combination.

(a) Difference between permutation and combination:

• A combination is a selection but a permutation is not a selection but an arrangement.
• In combination the order of appearance of objects is immaterial, whereas in a permutation the ordering is essential.
• Practically to find permutations of n different objects taken r at a time, we first select objects then we arrange them.
• One combination corresponds to many permutations.

(b) Combinations of n different things taken r at a time:
The number of combinations of n different things have taken r at a time ⁿc_r = C(n, r) = \(\left(\begin{array}{l}
n \\
r
\end{array}\right)=\frac{n !}{r !(n-r) !}\)

(c) Properties of ⁿc_r :
(1) ⁿc_r = ⁿC₀ = 1, ⁿC₁ = n
(2) ⁿC_r = ⁿC_n-r
(3) ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r (Euler’s formula)
(4) ⁿC_x = ⁿC_y ⇒ x = y or x + y = n
(5) n. ^n-1C_r-1 = (n – r + 1) ⁿC_r-1
(6) ⁿC_r = \(\frac{n}{r}{ }^{n-1} \mathrm{C}_{r-1}\)
(7) \(\frac{{ }^n \mathrm{C}_r}{{ }^n \mathrm{C}_{r-1}}=\frac{n-r+1}{r}\)
(8) If n is even then the greatest value of ⁿC_r is ⁿC_n/2.
⇒ If n is odd then the greatest value of ⁿC_r is \({ }^n \mathrm{C}_{\left(\frac{n+1}{2}\right)} \text { or }{ }^n \mathrm{C}_{\left(\frac{n-1}{2}\right)}\)

(d) Number of combinations of n different things taken r at a time, when k particular things always occur = ^n-kC_r-k

(e) The number of combinations of n different things, taken r at a time where k particular things never occur = ^n-kC_r

(f) The total number of combinations of n different things taken one or more at a time (or the number of ways of n different things selecting at least one of them) = ⁿC₁ + ⁿC₂ + ⁿC₃ + ….. + ⁿC_n = 2ⁿ -1

(g) The number of combinations of n identical things taken r at a time = 1.

(h) Number of ways of selecting r things out of n alike things where r = 0, 1, 2, 3 ….. n is (n+ 1).

(i) Division into groups:

• The number of ways in which (m + n) different things can be divided into two groups which contain m and n things respectively = \(\frac{(m+n) !}{m ! n !}\) for m ≠ n.
• If m-n then the groups are of equal size. Thus, division can be done in two ways as:
  ⇒ If order of groups is not important: In this case the number of ways = \(\frac{(2 n) !}{2 !(n !)^2}\)
  ⇒ If order of groups is important: In this case the number of ways = \(\frac{(2 n) !}{(n !)^2}\)

(j) Arrangement in groups:

1. The number of ways in which n different things can be arranged into r different groups = ^n+r-1P_n or n! ^n-1C_r-1
2. The number of ways in which n different things can be distributed into r different groups = rⁿ – ^rC₁(r – 1)ⁿ + ^rC₂(r – 2)ⁿ ….. + (-1)^r-1 . ^rC_r-1. (Blank groups are not allowed)
3. The number of ways in which n identical things can be distributed into r different groups where blank groups are allowed
   = ^(n+r-1)C_(r-1)
   = ^(n+r-1)C_n
4. Number of ways in which n identical things can be distributed into r different groups where blank groups are not allowed (each group receives at least one item) = ^n-1C_r-1

(k) Number of divisors:

Odisha State Board CHSE Odisha Class 12 Invitation to English 4 Solutions Grammar Additional Questions Textbook Activity Questions and Answers.

CHSE Odisha 11th Class English Grammar Additional Questions

A. Rewrite the passages after correcting all grammatical errors in it.

(1) There is great excitement in the planet of Venus this week. For the first time, Venusia scientists manage to land an unmanned spacecraft in the planet Venus, and it is sending back signals, as well as photographs, ever since. The craft directed into an area known as Gonebay.
Answer:
There was great excitement on the planet of Venus this week. For the first time, Venusian scientists managed to land an unmanned spacecraft on the planet Venus, and it has been sending back signals, as well as photographs ever since. The craft was directed into an area known as Gonebay.
(2) How did birds know when to flew south for winter? How long do a bear sleep in winter? Do a porcupine really shoot its quills at an enemy? How do squirrels know where he buries nuts?
Answer:
How do birds know when to fly south for the winter? How long does a bear sleep in winter? Does a porcupine really shoot his quills at an enemy? How does a squirrel know where he buried a nut?

B. Correct the errors.

(a) He is an European.
(b) I met the concerned clerk.
(c) It is high time you get up early.
(d) It has been five years since I last met you.
(e) I congratulate you for winning the prize.
Answer:
(a) He is a European.
(b) I met the clerk concerned.
(c) It is high time you got up early.
(d) It is five years since I last met you.
(e) I congratulate you on winning the prize.

C. Supply the correct tense of the verbs given in brackets.

1. Water always (freeze) at 0 degrees centigrade.
2. Students frequently (make) mistakes of tense usage when they do this exercise.
3. I (have) my hair cut whenever it gets too long.
4. I (take) my dog for a walk every evening before it died.
5. He (come) to my office whenever he needed money.
6. Last year she (wear) the same dress at every party.
7. Whenever I climb a hill, my ear (boil).
8. She (sing) very beautifully before she was married, but nowadays she (not sing) anymore.
9. I seldom (see) him at concerts these days. He (go) to them regularly before the war.
10. She cooks very well but her sister (cook) much better when I knew her.
11. Every time he opens his mouth, he (say) something foolish.
12. He occasionally makes a big effort, but usually he (not bother).
13. Whenever I (go) to see him, he was out.
14. In the past men frequently (fight) duels. Nowadays they seldom (do).
15. How often you (go) to the theatre when you were in London?
16. You (play) with dolls when you were a little girl?
17. The ancient Egyptians (build) pyramids as tombs for their kings.
18. When I was young, my father always (give) me some money on Saturdays.
19. If he is wise, a pianist (practise) four hours a day.
20. His parents don’t know what to do with their child. He (lie) habitually.
21. My aunt Jane (hate) girls who made up.
22. We all (study) Latin when we were at school.
23. Wood always (float).

Answer:
1. Water always freezes at 0 degrees centigrade.
2. Students frequently make mistakes of tense usage when they do this exercise.
3. I have my hair cut whenever it gets too long.
4. I took my dog for a walk every evening before it died.
5. He came to my office whenever he needed money.
6. Last year she wore the same dress at every party.
7. Whenever I climb a hill, my ear boils.
8. She sang very beautifully before she was married, but nowadays she does not sing anymore.
9. I seldom see him at concerts these days. He went to them regularly before the war.
10. She cooks very well but her sister cooked much better when I knew her.
11. Every time he opens his mouth, he says something foolish.
12. He occasionally makes a big effort, but usually, he does not bother.
13. Whenever I went to see him, he was out.
14. In the past men frequently fought duels. Nowadays they seldom do.
15. How often you go to. the theatre when you were in London?
16. Did you play with dolls when you were a little girl?
17. The ancient Egyptians built pyramids as tombs for their kings.
18. When I was young, my father always gave me some money on Saturdays.
19. If he is wise, a pianist practices four hours a day.
20. His parents don’t know what to do with their child. He lies habitually.
21. My aunt Jane hated girls who made up.
22. We all studied Latin when we were at school.
23. Wood always floats.

D. Put the verbs in brackets into the correct Present Tense, Continuous, or Simple.

1. Buses usually (run) along this street. but today they (not run) because it is under repair.
2. John (pass) the post office on his way to work every day.
3. She usually (sit) at the back of the class, but today she (sit) in the front row.
4. I rarely (carry) an umbrella, but I (carry) one now because it is raining.
5. What you generally (do) for a living?
6. You (enjoy) your English class today?
7. You (enjoy) washing dishes as a rule?
8. We nearly always (spend) our holidays at the seaside, but this year we are going to France.
9. Mr. Jones usually (sell) only newspapers, but this week he (sell) magazines as well.
10. You (wash) your hands before every meals.
11. Mary generally (begin) cooking at 11, but today she came home early and (cook) now, although il is only 10.30.
12. I’m sorry you can’t see her. She (sleep) still. She usually (wake) much earlier.
13. Why you (wear) a coat this morning? I never (wear) one till October.
14. Joan still (do) her homework. Her sister, who always (work) quicker, (play) already in the garden.
15. These builders generally (build) very rapidly. They (work) at present on two separate contracts.
16. What (do) you at this moment? If you (not do) anything, please help me.
17. John, who (study) medicine at present, hopes to go abroad after graduation.
18. He generally (come) to my office every clay, but today he (visit) his parents in the country.
19. You (watch) television often? The electrician (install) ours at this moment.
20. Mary usually (wear) a hat to go shopping. but today, as the sun (shine) she (not wear) one.

Answer:
1. Buses usually run along this Street, but today they are not running because it is under repair.
2. John passes the post office on his way to work every day.
3. She usually sits at the back of the class, but today she is sitting in the front row.
4. I rarely carry an umbrella, but I am carrying one now because it is raining.
5. What do you generally do for a living?
6. Are you enjoying your English class today?
7. Do you enjoy washing dishes as a rule?
8. We nearly always spend our holidays at the seaside, but this year we are going to France.
9. Mr. Jones usually sells only newspapers, but this week he is selling magazines as well.
10. Do you wash your hands before every meal?
11. Mary generally begins cooking at 11, but today she came home early and is cooking now, although it is only 10.30.
12. I’m sorry you can’t see her. She is still sleeping. She usually wakes much earlier.
13. Why are you wearing a coat this morning? I never wear one till October.
14. Joan is still doing her homework. Her sister, who always works quicker, is already playing already in the garden.
15. These builders generally build very rapidly. They are working at present on two separate contracts.
16. What are you doing at this moment? If you are not doing anything, please help me.
17. John, who is studying medicine at present, hopes to go abroad after graduation.
18. He generally comes to my office every day, but today he (is visiting) his parents in the country.
19. Do you often watch television? The electrician is installing ours at this moment.
20. Mary usually wears a hat to go shopping, but today, as the sun is shining she is not wearing one.

E. Put the verbs in brackets into the correct tense, Continuous or Simple.

1. You (see) the house on the corner? That is where I was born.
2. You (listen) to what I am saying You (understand) me?
3. I (notice) Mary (wear) a new hat today.
4. She (not understand) what you (mean).
5. I (need) a new suit. They (offer) special prices at the tailor this week.
6. You (smell) gas? I (think) the new stove is leaking.
7. Look at Mary! She (drink) up her medicine, but I can say that she (hate) it.
8. John (seem) rather tired today.
9. it still (rain), but it (look) as if it will soon stop.
10. You (mind) helping me a moment? I (try) to mend this table.
11. Ask him what he (want).
12. You (remember) the name of that girl who (walk) on the other side of the street?
13. ‘Will you have some tea?’ ‘I (prefer) coffee, please.’
14. I (suppose) I must go now. My wife (wait) for me at home.
15. You (see) this box? It (contain) matches.
16. These twins, who (resemble) one another so strongly, (study) art at present.
17. After what has happened, you really (mean) to say that you still (believe) him?
18. You (suppose) the children still (sleep)?
19. ‘The train still (stand) in the station. You (think) we can just catch it?
20. I (notice) you (possess) a copy of Waugh’s latest book. Will you lend it me?

Answer:
1. Do you see the house on the corner? That is where I was born.
2. Are you listening to what I am saying? Do you understand me?
3. I notice Mary is wearing a new hat today.
4. She does no: understand what you mean.
5. I need a new suit. They are offering special prices at the tailor this week.
6. Do you smell gás? I think the new stove is leaking.
7. Look at Mary! She is drinking up her medicine, but I can say that she hates it.
8. John seems rather tired today.
9. It is still raining, but it looks as if it will soon stop.
10. Do you mind helping me a moment? I urn Irving to mend this table.
11. Ask him what he wants.
12. Do you remember the name of that girl who is walking on the other side of the street?
13. ‘Will you have some tea?’ ‘I prefer coffee, please.’
14. 1 suppose I must go now. My wife is waiting for me at home.
15. Do you see this box? It contains matches.
16. These twins, who resemble one another so strongly, are studying art at present.
17. After what has happened, do you really mean to say that you still believe him?
18. Do you suppose the children still sleeping?
19. ‘The train is still standing in the station. Do you think we can just catch it
20. I notice you possess a copy of Waugh’s latest book. Will you lend it me?

F. Supply the correct form of the Present Perfect tense, Continuous or Simple in the place of the verbs in brackets.

1. They just (arrive) from New York.
2. They still (not succeed) in reaching the summit.
3. I this very minutes (receive) a telegram from my brother in India.
4. We already (have) breakfast.
5. I now (study) your proposals and regret I cannot accept them.
6. They (live) here since January.
7. We (wait) on the platform since three o’ clock.
8. She already (ring) the bell twice.
9. I see you just (have) your hair cut.
10. She (write) letters all morning, but I (not start) to write any yet.
11. The children (sleep) all this afternoon.
12. How long you (stay) in that old hotel?
13. They (work) in the same factory for twenty years now.
14. Since when you (have) that new car?
15. I (knock) on the door for ten minutes now without an answer.
16. They (build) that bridge for over a year and still it isn’t finished.
17. I (try) three times and (be) successful only once.
18. How many times you (be) to the cinema this week?
19. He (go) to the dentist off and on for six months.
20. He (take) the exam. three times and (fail) every time.
21. William (marry) the eldest Jones girl at last.
22. I (try) to get in touch with you for several days now.
23. She just (spend) three weeks at her grandmother’s.
24. He (work) hard on his book for some time and (finish) it at last.
25. You ever (read) ‘War and Peace’?

Answer:
1. They have just arrived from New York.
2. They have still not succeeded in reaching the summit.
3. I have this very minutes received a telegram from my brother in India.
4. We have already had breakfast.
5. I have now studied your proposals and regret I cannot accept them.
6. They have been living here since January.
7. We have been waiting on the platform since three o’ clock.
8. She has already rung the bell twice.
9. I see you have just had your hair cut.
10. She has been writing letters all morning, but I have no: starred to write any yet.
11. The children have been sleeping all this afternoon.
12. How long have you been staying in that old hotel?
13. They have been working in the same factory for twenty years now.
14. Since when have you had that new car?
15. I have been knocking on the door for ten minutes now without an answer.
16. They have been building that bridge for over a year and still it isn’t finished.
17. I have tried three times and have been successful only once.
18. How many times have you been to the cinema this week?
19. He has been going to the dentist off and on for six months.
20. He has taken the exam. three times and has failed every time.
21. William has married the eldest Jones girl at last.
22. I have been trying to get in touch with you for several days now.
23. She has just spent three weeks at her grandmother’s.
24. He has been working hard on his book for some time and has just finished it at last.
25. Have you ever read ‘War and Peace’?

G. Supply the missing prepositions using only to or at.

1. He is quite blind _____ her faults.
Answer:
to

2. He is extraordinarily clever _____ mimicking others.
Answer:
at

3. She, on the other hand, is very efficient _____ her work.
Answer:
at

4. He is an expert making himself understood in foreign languages.
Answer:
at

5. Contrary my expectations, I quite enjoyed myself at the party.
Answer:
to

6. She was standing too close _____ the fire and got burned.
Answer:
to

7. That fellow’s no good games at all.
Answer:
at

8. He carried out the project which had always been dear _____ his heart.
Answer:
to

9. We are all very indignant _____ the injustice done to him.
Answer:
at

10. I’m sorry! I’m very bad _____ explaining myself.
Answer:
at

11. He’s the one who’s so very lucky _____ cards.
Answer:
at

12. That device is entirely new _____ me.
Answer:
to

13 He’s not equal _____ the job they’ve given him.
Answer:
to

14. He remained faithful _____ his principles in spite of great pressure.
Answer:
to

15. The delay proved fatal _____ our plans.
Answer:
to

16. She was overjoyed _____ the prospect of meeting him again.
Answer:
at

17. His activities are very harmful _____ my interests.
Answer:
to

18. The government showed itself hostile _____ any progress.
Answer:
to

19. She’s no terribly cruel _____ her dog.
Answer:
to

20. I engaged him because he was so prompt in understanding my instructions.
Answer:
at

21. He was heartbroken _____ her indifference to him.
Answer:
at

22. He doesn’t like me although I’ve always been kind _____ him.
Answer:
to

23. Yes, he’s the kind of person who is always quick figures.
Answer:
at

24. This is much inferior _____ the one I bought last week.
Answer:
to

25. You will be liable _____ a heavy fine if you do that.
Answer:
to

26. This flower is not native _____ England.
Answer:
to

27. Naturally she was sad _____ the death of her parrot.
Answer:
at

28. He does his work carefully but he’s terribly slow ______ it.
Answer:
at

29. Who wouldn’t be triumphant _____ their success in the examination?
Answer:
at

30. A dutiful daughter and obedient _____ her parents.
Answer:
to

31. You shouldn’t be surprised _____ a thing like that.
Answer:
at

32. It should be obvious _____ the meanest intelligence.
Answer:
to

33. The seeds are peculiar ______ this genus of plant.
Answer:
to

34. Can’t you manage to be a little more polite your aunt?
Answer:
to

35. He works in a factory. Previous _____ that he was in a laundry.
Answer:
to

36. I have been truly astonished _____ the number of people who believe it.
Answer:
at

37. This is quite irrelevant ______ the matter we are discussing.
Answer:
to

38. They were shocked _____ his apparent lack of appreciation.
Answer:
at

39. He was so rude _____ her that she never spoke to him again.
Answer:
to

40. Sacred _____ the memory of Mary Jones.
Answer:
to

41. They are very sensitive _____ people’s opinion of them.
Answer:
to

42. I’ve got one similar _____ yours.
Answer:
to

43. Subject _____ the exigencies of the service.
Answer:
to

44. The people of this country are very skilful ______ making dolls.
Answer:
at

45. It’s useful _____ me to have him about the house.
Answer:
to

46. It’s vital _____ a proper understanding of the problem.
Answer:
to

H. Supply the missing preposition using with, for or of.

1. Don’t be afraid _____ the dog! It won’t bite you.
Answer:
of

2. I can’t be angry _____ him now that he’s apologized _____ what he has done.
Answer:
with, for

3. He’s far ahead _____ the others in arithmetic.
Answer:
of

4. You ought to be ashamed _____ yourself.
Answer:
of

5. Are you aware _____ the fact that it is half past ten.
Answer:
of

6. I’m sorry. They are simply not capable doing it.
Answer:
of

7. Don’t disturb him! He’s busy _____ his accounts.
Answer:
with

8. He’s ambitious and eager _____ honours.
Answer:
for

9. For goodness sake! Do be careful _____ that vase. You could easily drop it.
Answer:
with

10. Children must be taught to be careful _____ traffic.
Answer:
of

11. I am not at all certain _____ the date of his arrival.
Answer:
of

12. His explanation was not consistent _____ the facts.
Answer:
with

13. Monsieur X was famous _____ his collection of pictures.
Answer:
for

14. I was conscious _____ a feeling of uneasiness.
Answer:
of

15. Kingston lies due west _____ London.
Answer:
of

16. I’m sorry. I’m not content _____ your explanations.
Answer:
with

17. The soldier was pronounced fit _____ service.
Answer:
for

18. Mary was terribly envious _____ Joan’s new hat.
Answer:
of

19. We are all very fond _____ going to the theatre.
Answer:
of

20. John is very discontented _____ his salary.
Answer:
with

21. I shall be grateful any advice you can give me.
Answer:
for

22. This exercise is full ______ the most terrible mistakes.
Answer:
of

23. Are you familiar _____ the works of Milton?
Answer:
with

24. The manager is well qualified _____ his position.
Answer:
for

25. That is something I am profoundly glad.
Answer:
of

26. This chair is not identical _____ the one I bought last year.
Answer:
with

27. He’s a sporting fellow, and always ready for anything.
Answer:
for

28. That student is ignorant _____ the first rules of grammar.
Answer:
of

29. Your explanation is incompatible _____ the story I heard.
Answer:
with

30. John, you will be responsible for providing the drinks.
Answer:
for

31. Judges must be independent of

political pressure.
Answer:
of

32. I know he’s a difficult child, but you must be patient _____ him.
Answer:
with

33. I am extremely sorry _____ the delay, but I was held up.
Answer:
for

34. He was jealous _____ his brother’s good fortune.
Answer:
of

35. Comedians are always popular holiday crowds.
Answer:
with

36. I can’t bake a cake as we are short _____ eggs this week.
Answer:
of

37. His income is sufficient _____ his needs.
Answer:
for

38. He’s always very shy approaching his chief.
Answer:
of

39. Let us be thankful _____ small mercies.
Answer:
for

40. It is wise to be sure _____ your facts before you speak.
Answer:
of

41. One is generally tolerant _____ small faults.
Answer:
of

42. She is, unfortunately, devoid _____ a sense of humor.
Answer:
of

43. They gave him a visa valid _____ all countries in Europe.
Answer:
for

44. I’m tired _____ arguing with you.
Answer:
of

45. They proved themselves unworthy _____ the trust which was placed in them.
Answer:
of

K. Supply the missing prepositions, from, about, on or in.

1. Keep away _____ the machine while it is running.
Answer:
from

2. He wsa singularly fortunate _____ his choice of wallpaper.
Answer:
in

3. He is intent _____ attending the football match on Saturday.
Answer:
on

4. I am very dubious _____ your chances of passing the examination.
Answer:
about

5. They are proficient _____ the use of their fists.
Answer:
in

6. This is quite different _____ what I expected.
Answer:
from

7. The diet here is deficient _____ vitamins.
Answer:
in

8. That young man is very keen _____ cycling.
Answer:
on

9. We are all very enthusiastic ________ our next holiday.
Answer:
about

10. It was far ______ my intention to suggest that he was unintelligent.
Answer:
from

11. He was perfectly honest _____ his intentions to win the prize at all costs.
Answer:
about

12. The secretary was not well qualified _____ shorthand.
Answer:
in

13. Some people appear completely immune _____ this disease.
Answer:
from

14. I am very reluctant _____ asking him to do this.
Answer:
about

15. Our plans must remain dependent _____ the weather.
Answer:
on

16. Everybody was very uneasy _____ the outcome of the negotiations.
Answer:
about

17. Of course, you are quite right _____ that.
Answer:
about

18. He was involved in an accident, resulting _____ the slippery condition of the road.
Answer:
from

19. The enemy is weak _____ artillery.
Answer:
in

20. Entomologists are still curious _____ the life-cycle of that month.
Answer:
about

21. I am not very interested _____ the story of your life.
Answer:
in

22. Put that cake back in the cupboard. where it will be safe _____ the cat.
Answer:
from

23. I’m afraid he is quite wrong _____ the date of the invasion.
Answer:
about

24. His father was very sad _____ his son’s failure in his final exams.
Answer:
about

25. I am extremely doubtful _____ the wisdom of pursuing that course of action.
Answer:
about

L. Supply the prepositions at or to as appropriate.

1. What time did you arrive _____ your home?
Answer:
at

2. He finds it difficult to accustom himself _____ the climate.
Answer:
to

3. AIl the visitors exclaimed _____ the beauty of the place.
Answer:
at

4. Just glance _____ this for me, would you?
Answer:
at

5. His debts amount _____ a considerable sum.
Answer:
to

6 I can only guess _____ the extent of the damage.
Answer:
at

7. It is useless to appeal _____ his better nature.
Answer:
to

8. She hinted darkly _____ all sorts of wild actions in his youth.
Answer:
at

9. If you want permission. you must apply _____ the caretaker.
Answer:
to

10. He was attached _____ the French Army during the war.
Answer:
to

11. He is too sick to attend _____ his duties.
Answer:
to

12. _____ what do you attribute your success in life?
Answer:
to

13. I’m sure this one doesn’t belong _____ me.
Answer:
to

14. He challenged him _____ a game of chess.
Answer:
to

15. It is very unkind to joke _____ the expense of the disabled.
Answer:
at

16. Shall I compare thee _____ a summer’s day?
Answer:
to

17. If you want them to hear, you’ll have to knock a good deal harder _____ the door.
Answer:
at

18. The prisoner was condemned _____ penal servitude for life.
Answer:
to

19. The fire was confined _____ the kitchen regions.
Answer:
to

20. I will never consent _____ her marrying that man.
Answer:
to

21. Just have a look _____ this for me, would you?
Answer:
at

22. I have been entirely converted _____ the use of an electric razor.
Answer:
to

23. Employees who have twenty-five years’ service become entitled _____ a pension.
Answer:
to

24. You can safely entrust your little son _____ her care.
Answer:
to

25. People who heard her voice marvelled _____ it.
Answer:
at

26. Let’s invite them all _____ dinner.
Answer:
to

27. Listen _____ me!
Answer:
to

28. Do you object my smoking?
Answer:
to

29. He peered _____ the exhibit on account of his short-sightedness.
Answer:
at

30. Such an idea would never occur _____ me!
Answer:
to

31. The patient is reacting very unsatisfactorily _____ the drug.
Answer:
to

32. The children peeped _____ the guests as they were arriving.
Answer:
at

33. I have been reduced ______ using oil for lack of fat.
Answer:
to

34. The children are playing _____ Red Indians again.
Answer:
at

35. She has been forced to resort _____ all sorts of devices to avoid him.
Answer:
to

36. The patient has not responded _____ treatment.
Answer:
to

37. It is very rude to point people in the street.
Answer:
at

38. If you’ll bring the drinks, I’ll see _____ the food.
Answer:
to

39. They have been subjected _____ all sorts of indignities.
Answer:
to

40. I refuse to submit ______ that sort of treatment.
Answer:
to

41. Can you wonder _____ it, if they are reduced _____ begging.
Answer:
at, to

42. I’m sorry he finally succumbed _____ the temptation of stealing.
Answer:
to

43. If you want to pass your examination, you il have to work very hard _____ your Latin.
Answer:
at

44. We shall never surrender _____ that enemy.
Answer:
to

45. I don’t know him, but he has been starting _____ me for ten minutes.
Answer:
at

46. I have never subscribed _____ the general opinion of him.
Answer:
to

47. Turn _____ page 22 and start reading!
Answer:
to

48. The government has again yielded _____ the pressure from outside.
Answer:
to

49. She always trusts _____ her neighbors to help her.
Answer:
to

50. Would you please reply my question?
Answer:
to

M. Turn the following into passive.

1. The government has called out troops.
Answer:
Troops have been called out.

2. Fog held up the trains. (agent required)
Answer:
Trains were held up by fog.

3. You are to leave this here. Someone will call for it later on.
Answer:
This is to be left here. k will be called for.

4. We called in the police.
Answer:
Police were called in.

5. They didn’t look after the children properly.
Answer:
Children were not properly looked after.

6. They are flying in reinforcements.
Answer:
Reinforcements are being flown in.

7. Then they called up men of 28.
Answer:
Men of 28 were called up.

8. Everyone looked up to him. (agent required)
Answer:
He was looked up to by everyone.

9. All the ministers will see him off at the airport. (agent required)
Answer:
He will be seen off at the airport by all the ministers.

10. He hasn’t slept in his bed.
Answer:
Bed hasn’t been slept in.

11. We can build on more rooms.
Answer:
More rooms can be built on.

12. They threw him out.
Answer:
He was thrown out.

13. They will have to adopt a different attitude.
Answer:
Different attitude will have to be adopted.

14. He’s a dangerous maniac. They ought to lock him up.
Answer:
He ought to be locked up.

15. Her story didn’t take them in. (agent required)
Answer:
They weren’t taken in by her story.

16. Burglars broke into the house.
Answer:
House was broken into.

17. The manufacturers are giving away small plastic toys with each packet of cereal.
Answer:
Small plastic toys are being given away.

18. They took down the notice.
Answer:
Notice was taken down.

19. They frown on smoking here.
Answer:
Smoking is frown on.

20. After the government had spent a million pounds on the scheme they decided that it was impracticable and gave it up. (Make only the first and last verbs passive)
Answer:
After a million pounds had been spent, the scheme was given up.

21. When I returned I found that they had towed my car away. I asked why they had done this and they told me that it was because I had parked it under a No Parking sign. (four passives)
Answer:
My car had been towed away. I asked why this had been done and was told that it had been parked.

22. People must hand in their weapons.
Answer:
Weapons must be handed in.

23. The crowd shouted him down.
Answer:
He was shouted down.

24. People often take him for his brother.
Answer:
He is often taken for his brother.

25. No one has taken cut the cork.
Answer:
The cork hasn’t been taken out.

26. The film company were to have used the pool for aquatic displays, but now they have changed their minds about it and are filling it in. (Make the first and last verbs passive)
Answer:
Pool was to have been used it is being filled in.

27. This college is already full. We ne turning away students the whole time.
Answer:
Students are being turned away.

28. You will have to pull down this skyscraper as you have not complied with the town planning regulations.
Answer:
Skyscraper will have to be pulled down as the town planning regulations have not been complied with.

N. Put the following sentences into passive, using infinitive construction where possible.

1. We added up the money and found that it was correct.
Answer:
Money was added up and found to be correct.

2. I’m employing a man to tile the bathroom.
Answer:
I am having the bathroom tiled.

3. Someone seems to have made a terrible mistake.
Answer:
A terrible mistake seem to have been made.

4. It is your duty to make tea at eleven o’ clock. (Use suppose.)
Answer:
You are supposed to make tea.

5. People know that he is armed.
Answer:
He is known to be armed.

6. Someone saw him pick up the gun.
Answer:
He was seen to pick up?

7. We know that you were in town on the night of the crime.
Answer:
You are known to have been.

8. We believe that he has special knowledge which may be useful to the police. (one passive)
Answer:
He is believed to have special knowledge.

9. You needn’t have done this.
Answer:
This needn’t have been done.

10. It’s a little too loose: you had better ask your tailor to take it in. (one passive)
Answer:
You had better have it taken in.

11. He likes people to call him — ‘sir’.
Answer:
He likes to be called sir’.

12. Don’t touch this switch.
Answer:
This switch isn’t to be/mustn’t be touched.

13. You will have to get someone to see to it.
Answer:
You will have to have/get it seen to.
(Or) It will have to be seen to.

14. It is impossible to do this. (Use can’t)
Answer:
This can’t be done.

15. Someone is following us.
Answer:
We are being followed.

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 5 Principles of Mathematical Induction will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 5 Principles of Mathematical Induction

Principles Of Mathematical Induction

(i) Principle – 1
Let P(n) is a statement , n ∈ Z

Step – 1: Verification step:
verify that P(1) is true.

Step – 2: Induction step – 1:
Assume that P(k) is true for any arbitrary k ∈ N.

Step – 3: Induction step – 2:
prove that P_(k+1) is true using step – 1 and step – 2

Step – 4: Conclusion Step:
If P_(k+1) is true then take a conclusion that by Principle of mathematical induction P(n) is true for all n ∈ N.

(ii) Principle – 2
Let P(n) be a statement, n ∈ N.

Step – 1: Verification step:
verify the P(1) is true

Step – 2: Induction step – 1:
Assume that P(2), P(3),….. P(k) is true.

Step – 3: Induction step – 2:
Using step – 1 and step – 2 prove that P(k + 1) is true.

Step – 4: Conclusion step:
If P(k + 1) is true, then take a conclusion that by Principle of mathematical induction P(n) is true for all n ∈ N.

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 4 Trigonometric Functions will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 4 Trigonometric Functions

Angle:
If A, B, and C are three non-collinear points, then ∠ABC = \(\overrightarrow{\mathrm{BA}} \cup \overrightarrow{\mathrm{BC}}\)

\(\overrightarrow{\mathrm{BA}}\) is the initial side, \(\overrightarrow{\mathrm{BC}}\) is called the terminal side and B is called the vertex of the angle.

Positive and negative angles:
If the direction of rotation is anti-clockwise then the angle is positive and if the direction of rotation is clockwise then the angle is negative.

Measure of an angle:
(a) Sexagesimal system or English System (Degree measure):
1 degree = 1° = \(\left(\frac{1}{360}\right) \text { th }\) of revolution from initial side to terminal side.

• One revolution = 360°
• 1° = 60′ (sixty minute)
• 1′ = 60” (sixty seconds)

(b) Circular system(Radian measure):
One radian = 1^c = The angle at the centre of the circle by an arc where the arc length equals to

Note:
(i) θ = \(\frac{l}{r}=\frac{\text { arc }}{\text { radius }}\) where θ is an radian.
(ii) θ in radian is created as a real number.

(c) Relation between Degree and radian measure:

• 2π radians = 360°
  ⇒ π radian = 180°
• We can convert radian to degree or degree to radian by using the identity.
  \(\frac{\mathrm{D}}{180}=\frac{\mathrm{R}}{\pi}\) where D is the degree measure and R is the radian measure of an angle.

Trigonometry Functions:

(i) Sign of trigonometry functions:

(ii) \(\begin{array}{cccc}
\text { Add } & \text { Sugar } & \text { To } & \text { Coffee } \\
\downarrow & \downarrow & \downarrow & \downarrow \\
\text { all }+ & \sin + & \tan + & \cos +
\end{array}\)

(iii) Periodicity of trigonometry functions:

Trigonometric function	Period
sin x	2π
cos x	2π
tan x	π
cot x	π
sec x	2π
cosec x	2π
sin2 x or cos2 x	π
|sin x| or |cos x|	π

Trigonometric functions of some standard angles

Fundamental trigonometric identities:
(a) sin θ = \(\frac{1}{{cosec} \theta}\)
(b) cos θ = \(\frac{1}{{sec} \theta}\)
(c) tan θ = \(\frac{1}{{cot} \theta}\)
(d) sin² θ + cos² θ = 1
(e) sec² θ – tan² θ = 1
(f) cosec² θ – cot² θ = 1
(g) sin (-θ) = -sin (θ)
(h) cos (-θ) = -cos (θ)

Trigonometric functions of allied angles:
(a) sin \(\left((2 n+1) \frac{\pi}{2} \pm \theta\right)\) = (±) cos θ choose + or – in (±) by using ASTC rule

(b) cos \(\left((2 n+1) \frac{\pi}{2} \pm \theta\right)\) = (±) sin θ choose + or – in (±) by using ASIC rule
Similar technique can be used for other trigonometric functions.

(c) sin (nπ ± θ) = (±) sin θ
cos (nπ ± θ) = (±) cos θ
tan (nπ ± θ) = (±) tan θ
choose + or – in (±) by using ASTC rule.

Sum And Difference Formulae:
(a) sin(A + B) = sin A . cos b + cos A . sin B

(b) sin(A – B) = sin A . cos B – cos A . sin B

(c) cos(A + B) = sin A . cos B – cos A . sin B

(d) cos(A – B) = sin A . cos B + cos A . sin B

(e) tan(A + B) = \(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}\)

(f) tan(A – B) = \(\frac{\tan A-\tan B}{1+\tan A \cdot \tan B}\)

(g) cot(A + B) = \(\frac{\cot A \cdot \cot B-1}{\cot A+\cot B}\)

(h) cot(A – B) = \(\frac{\cot A \cdot \cot B+1}{\cot A-\cot B}\)

(i) sin(A + B) + sin (A – B) = 2 sin A . cos B

(j) sin (A + B) – sin(A – B) = 2 cos A . sin B

(k) cos(A + B) + cos(A – B) = 2 cos A . cos B

(l) cos(A + B) – cos(A – B) = -2sin A . sin B

(m) sin(A + B) sin(A – B) = sin² A – sin² B = cos² B – cos² A

(n) cos(A + B) cos(A – B) = cos² A – sin² B = cos² B – sin² A

(o) sin 2A = 2 sin A cos A = \(\frac{2 \tan \mathrm{A}}{1+\tan ^2 \mathrm{~A}}\)

(p) cos 2A = cos² A – sin² A
= 2 cos² A – 1
= 1 – 2 sin² A
= \(\frac{1-\tan ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}\)

(q) tan 2A = \(\frac{2 \tan A}{1-\tan ^2 A}\)

(r) tan(A + B + C) = \(\frac{\tan A+\tan B+\tan C-\tan n A \cdot \tan B \cdot \tan C}{1-\tan A \cdot \tan B-\tan B \cdot \tan C-\tan C \cdot \tan A}\)

(s) sin 3A = 3 sin A – 4 sin³ A
= 4 sin A sin(\(\frac{\pi}{3}\) – A) sin(\(\frac{\pi}{3}\) + A)

(t) cos 3A = 4 cos³ A – 3 cos A
= 4 cos A cos(\(\frac{\pi}{3}\) – A) cos(\(\frac{\pi}{3}\) + A)

(u) tan 3A = \(\frac{3 \tan A-\tan ^3 A}{1-3 \tan ^2 A}\)
= tan A. tan(\(\frac{\pi}{3}\) – A) tan(\(\frac{\pi}{3}\) + A)

Sum or Difference → Product:
(a) sin A + sin B = 2 sin(\(\frac{A+B}{2}\)) cos(\(\frac{A-B}{2}\))
(b) sin A – sin B = 2 cos(\(\frac{A+B}{2}\)) cos(\(\frac{A-B}{2}\))
(c) cos A + cos B = 2 cos(\(\frac{A+B}{2}\)) cos(\(\frac{A-B}{2}\))
(d) cos A – cos B = -2 sin(\(\frac{A+B}{2}\)) sin(\(\frac{A-B}{2}\))

Submultiple Arguments:
(a)

(b) 2 sin² \(\frac{\theta}{2}\) = 1 – cos θ
2 cos² \(\frac{\theta}{2}\) = 1 + cos θ

(c) tan \(\frac{\theta}{2}\) = \(\frac{\sin \theta}{1+\cos \theta}=\frac{1-\cos \theta}{\sin \theta}\)

(d) sin θ = 3 sin \(\frac{\theta}{2}\) – 4 sin³ \(\frac{\theta}{2}\)
cos θ = 4 cos³ \(\frac{\theta}{2}\) – 3 cos \(\frac{\theta}{2}\)

(e) tan θ = \(\frac{3 \tan \frac{\theta}{2}-\tan ^3 \frac{\theta}{2}}{1-3 \tan ^2 \frac{\theta}{2}}\)

Trigonometric Equations:
(a) Equation involving trigonometric equations of unknown angles are called trigonometric function.
(b) Principle solution: The solution ‘x’ of a trigonometric equation is said to be a principle solution if x ∈ (0, 2π)
(c) The solution considered over the entire set R are called the general solution.
(d) General solution of some standard trigonometric equations.

• sin x = 0 ⇒ x = nπ, n ∈ Z
• cos x = 0 ⇒ x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z
• tan x = 0 ⇒ x = nπ, n ∈ Z
• sin x = sin α ⇒ x = nπ + (-1)ⁿ, n ∈ Z
• cos x = cos α ⇒ x = 2nπ ± α, n ∈ Z
• tan x = tan α ⇒ x = nπ + α, n ∈ Z
• \(\left.\begin{array}{l}
  \sin ^2 x=\sin ^2 \alpha \\
  \cos ^2 x=\cos ^2 \alpha \\
  \tan ^2 x=\tan ^2 \alpha
  \end{array}\right]\) ⇒ x = nπ ± α
• \(\left.\begin{array}{l}
  \cos x=\cos \alpha \\
  \text { and } \sin x=\sin \alpha
  \end{array}\right]\) ⇒ x = nπ ± α, n ∈ Z

Sine Formula:
In any Δ ABC, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) or, \(\frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}=\frac{\sin \mathrm{C}}{c}\) = 2R
∴ a = 2R sin A, b = 2R sin B and c = 2R sin C
Also, sin A = \(\frac{a}{2R}\), sin B = \(\frac{b}{2R}\) and sin c = \(\frac{c}{2R}\)

Cosine fromulae:
In any Δ ABC,
(i) a² = b² + c² – 2bc cos A
(ii) b² = c² + a² – 2ca cos B
(iii) c² = a² + b² – 2ab cos C
or, → cos A = \(\frac{b^2+c^2-a^2}{2 b c}\)
→ cos B = \(\frac{c^2+a^2-a^2}{2 c a}\)
→ cos C = \(\frac{a^2+b^2-c^2}{2 a b}\)

Projection formulae:
In any Δ ABC,
(i) a = b sin C + c sin B
(ii) b = c cos A + a cos C
(iii) c = a cos B + b cos A

Tangent formulae (Napier’s Analogy);
In any Δ ABC

Area of Triangle (Heron’s formulae):
(i) Area of triangle ABC

(ii) Heron’s formulae:
In any Δ ABC Let 2S = a + b + c
Area of Δ ABC = Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
Δ = \(\frac{1}{2}\) bc sin A = \(\frac{1}{2}\) ca sin B
= \(\frac{1}{2}\) ab sin C, Δ = \(\frac{abc}{4R}\)

Semi-Angle Formulae:

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 1 Mathematical Reasoning will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 1 Mathematical Reasoning

Proposition: (Mathematically Acceptable)
A proposition (or mathematically acceptable statement) is a declarative sentence that is either true or false but not both.
(1) Thus a sentence will be a statement if

• It is declarative
• It has a truth value (either true (T) or false (F).

(2) A sentence cannot be a statement if it is
(i) A question
(ii) An order
(iii) An exclamation
(iv) A wish
(v) Advice or it involves

• variable time like ‘today’, ‘tomorrow’, ‘yesterday’ etc.
• Variable place like ‘here’, ‘there’, etc.
• pronouns like ‘he’, ‘she’, ‘they’ etc.
• Relative words/adjectives / undefined words like ‘good’, ‘bad’, ‘beautiful’, ‘wise’ etc
• Variable x, y, z, u, v….etc

(3) We denote statements by same letters are p, q, r, s, etc.

Negative (~): Denial of a statement is its negation.
Axiom of negation:
For any proposition p, if p is true, then ~p (Negation of p) is false and if p is false, then ~p is true,
Truth table of Negation:

p	~p
T	F
F	T

Logical Connectives:

• Two statements can be combined together by using the words like or, and, if, only if, if and only if etc. These are known as logical connectives.
• A proposition in which one or more connectives appear is known as a composite or compound proposition.

Conjunction (∧), (and):
Axiom: A conjunction p ∧ q is true if both ‘p’ and ‘q’ are both true and false otherwise.
Truth table:

p	q	p ∧ q
T	T	T
T	F	F
F	T	F
F	F	F

Disjunction (∨) (or):
Axiom: A disjunction p ∨ q is false only when both ‘p’ and ‘q’ are false and is true otherwise.
Truth table:

p	q	p ∨ q
T	T	T
T	F	T
F	T	T
F	F	F

• Inclusive and exclusive sense of ‘OR’

→ An employee either goes on leave or attends to his duty. (Exclusive)
→ In this restaurant you can order veg or non-veg items. (Inclusive)

Conditional (→)(if … then):
Axiom: A conditional p → q is false only when ‘p’ is true and ‘q’ is false in all other cases it is true.
Truth table:

p	q	p → q
T	T	T
T	F	F
F	T	T
F	F	T

Converse, Inverse and Contrapositive:

• Converse of p → q is q → p
• Inverse of p → q is ~p → ~q
• Contra positive of p → q is ~q → ~ p

Biconditional (p ↔ q)(p if and only if q):
Axiom: A biconditional p ↔ q is true if both ‘p’ and ‘q’ have same truth value and is false otherwise.
Truth table:

p	q	p ↔ q
T	T	T
T	F	F
F	T	F
F	F	T

Equivalent statements:
Two statements ‘p’ and ‘q’ are said to be equivalent statements if they have same truth values.

Tautology:
A statement is a tautology if it is always true:

Implication and double implication:

• If a conditional p → q is a tautology then we say p implies q and we write P ⇒ q
• If a biconditional p ↔ q is a tautology and we write p ⇒ q.

Contradiction:
A contradiction we mean a proposition that is false for all possible assignments of truth values to its prime components.

Logical Quantifiers:
Logical quantifiers are the words that associate a quantity to it. There are two types of logical quantifiers.
(i) Existential (There exists)
(ii) Universal (For all, for every).

Validity Of Statements
A statement is said to be valid if it is true.
Techniques to check the validity of a statement:

Validity Of Statements With ‘And’
To prove p ∧ q is true we follow the following steps :
Step – 1: Show that ‘p’ is true.
Step – 2: Show that ‘q’ is true.

Validity Of Statements With ‘ OR’
To prove p ∧ q is true we have to consider the following cases :
Case – 1: By assuming p is false, prove that q is true.
Case – 2: By assuming q is false, prove that p is true.

Validity Of Statements With ‘if … then’
To prove if ‘p’ then ‘q’ is true we can adopt any one of the following methods.

• Method – 1 (Direct Method):
  Assume ‘p’ is true and prove that ‘q’ is true (i.e. p ⇒ q)
• Method – 2 (Contrapositive Method):
  Assume ‘q’ is false and prove that ‘p’ is false. (i.e. ~ q ⇒ ~ p)
• Method – 3 (Contradiction Method):
  → Assume that p → q is false, i.e. p is true and q is false
  → Obtain an absurd result
  → This is due to our false assumption.
  → Thus by the method of contradiction p → q is true. i.e., the statement is valid.
• Method – 4 (By giving a counter-example):
  To prove a statement is false we give a single example where it is false.

Validity Of Statement With ‘if and only if’.
To prove ‘p’ if and only if ‘q’ is true we have to follow the following steps.
Step – 1: Take ‘p’ is true and prove that ‘q’ is true.
Step – 2: Take q is true and prove that ‘p’ is true.

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 3 Relations And Function will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 3 Relations And Function

Order Pairs
An ordered pair consists of a pair of objects, or elements or numbers or functions in order.
We denote order pairs as (a, b)

• An order pair is not a set of two objects.
• (a, b) = (c, d) ⇒ a = c and b = d
• (a, b) × (b, a)

Cartesian Product Of Sets:
If A and B are non-empty sets, then their Cartesian product, denoted by A × B and defined by A × B = {(a, b): a ∈ A, b ∈ B} = Set of all ordered pairs (a,b) where a ∈ A and b ∈ B
Note:
1. For finite sets A and B |A × B| = |A| . |B|
2. A × B = Φ ⇔ A = Φ or B = Φ
3. A² = A × A

Properties of Cartesian product:
1. A × B ≠ B × A (Cartesian product is non-commutative)
2. A × (B ∪ C) = (A × B) ∪ (A × C)
3. A × (B ∩ C) = (A × B) ∩ (A × C)
4. A × B = B × A ⇔ A = B
5. A × (B – C) = ( A × B) – (A × C)
6. A ⊂ B ⇒ A × A ⊂ (A × B) ∩ (B × A)
7. A ⊂ B ⇒ A × C ⊂ B × C
8. A ⊂ Band C ⊂ D ⇒ A × C ⊂ B × D
9. (A × B) ∩ (C × D) = ( A ∩ C) × (B ∩ D)

Relation
Let A and B be two arbitrary sets. A binary relation from A to B is a subset of A × B.
OR f is a relation from A to B if f ⊆ A × B
Note:

• If a of A is related to b of B by relation ‘f’ then we write (a,b) ∈ f or a f b
• As Φ ⊂ A × B we have Φ is a relation from A to B. This relation is known as a null of void or empty relation.
• As A × B ⊆ A × B, A × B is also a relation from A to B. This relation is known as universal relation.
• If |A| = m and |B| = n then number of relations from A to B is 2^mn

Domain, co-domain, and Range of a relation:
Let f is a relation from A to B. Domain of f = Dom (f) or D_f
={x ∈ A : (x, y) ∈ f for some y ∈ B) Co-domain of f = B
Range of ‘f’ = Rng (f) or R_f = {y ∈ B : (x, y) ∈ f for some x ∈ A}

Types Of Relation:
(a) One-many relation: A relation f from A to B is one many if (a, b) and (a, b’) ∈ f ⇒ b ≠ b’
(b) Many-one relations: A relation f from A to B is many-one if (a, b) and (a’, b) ∈ f ⇒ a ≠ a’
(c) One-one relation: A relation f from A to B is one-one if (a, b), (a, b’) ∈ f ⇒ b = b’ and (a, b), (a’, b) ∈ f ⇒ a = a’

Inverse of a relation: Let f is a relation from A to B. The inverse of f is denoted by f^-1 is a relation from B to A defined as f^-1 = {(b, a): (a, b) ∈ f}

Function:
A relation ‘f’ from X to Y is called a function if:
(a) D_f = Dom (f) = X and
(b) (x, y) and (x, z) ∈ f ⇒ y = z or A relation from A to B is a function
if ⇒ Domain of f = X i.e All elements of X is engaged in the relation and
⇒ f is not one many.

Note:
(1) If a relation f from X to Y becomes a function then we write f: X → Y.
(2) If f is a function from A to B i.e. f: X → Y and (x, y) ∈ f then we write y = f(x)
(3) Mapping, map, transformation, transform, operator, and correspondence are different synonym terms of function.
(4) If f: X → Y is defined as y = f(x), then

• y is called the value of the function at x or the image of x under f or the dependent variable.
• x is called the independent variable or pre-image of y under f.

Domain, Co-domain or Range of a function:
Let f: X → Y defined as y = f(x)
(a) Domain of ‘f’ = Dom f = D_f = {x ∈ X: y = f(x)}
(b) Range of f = Rng f = R_f = f(A) = {f(x) ∈ Y: x ∈ A } Clearly f(A) ⊆ y
(c) If |A| = m, |B| = n then number of functions from A to B = n^m

Real valued function :
A function f: A → B is a real-valued function if B ⊆ R.
→ f is a real function if A ⊆ R and B ⊆ R

Techniques to find Domain and Range of a Real function:
(a) Techniques to find Domain: Let the function is defined as y = f(x).
Step -1: Check the values of x for which f(x) is well defined.
Step -2: The set of all values obtained from step -1 is the domain of ‘f.

(b) Techniques to find range: Let the function is y = f(x)

• Method-1 (By inspection):
  → Step -1: Get values of y for all values x ∈ Dom f.
  → Step -2: Set of all these values of y = Rng f
• Method-2:
  → Step -1: Write x in terms of y
  → Step -2: Get values of y for which x is well defined in Dom f.
  → Step -3: Rng (f) = The set of all y obtained from step 2.

Some Real Functions:
(a) Constant function: A function f: A → R defined as f(x) = k, for some k ∈ R is called a constant function.

(b) Identity function: Let A ⊆ R. The function f: A → A defined as f(x) = x, x ∈ A is called the identity function on A. We denote it by I_A

(c) Polynomial function: A function f: A → R defined by f(x) = f(x) = a₀ + a₁x + a₂x² + a_nxⁿ where a₀, a₁, a₂, ….., a_n are real numbers and a_n ≠ 0 is called a polynomial function (polynomial) of degree n.

(d) Rational function: A function of form f(x) = \(\frac{\mathrm{P}(x)}{\mathrm{Q}(x)}\) where P(x) and Q(x) are polynomial functions of x is known as a rational function.

(e) Absolute value function OR modulus function: The function f: R → R defined as  f(x) = |x| = \(\begin{cases}x, & x \geq 0 \\ -x, & x<0\end{cases}\) is called as the modulus function.
→ Rng f = [ 0, ∞] = R⁺U {0}

Properties Of Modulus Function:
1. For any real number x, we have \(\sqrt{x^2}=|x|\)
2. If a and b are positive real numbers then

• x² ≤ a² ⇔ |x| ≤ a
• x² ≥ a² ⇔ |x| ≥ a
• a² ≤ x² ≤ b² ⇔ a ≤ |x| ≤ b ⇔ x ∈ [-b, – a] ∪ [a, b]

(f) Signum function: The function f: R → R defined as f(x) = \(\begin{cases}\frac{x}{|x|}, & x \neq 0 \\ 0, & x=0\end{cases}\) is called signum function.
→ We denote a signum function as f(x) = sgn(x)
→ Range of a signum function = {-1, 0, 1}

(g) Greatest integer function: The function f: R → R defined by f(x) = [x] is called the greatest integer function. [x] = The greatest among all integers ≤ x. OR [x] = n for n ≤ x < n + 1

Properties of the greatest integer function :
Let n is an integer and x is a real number between n and n + 1
(i) [-n]= -[n]
(ii) [x + k] = [x] + k (for an integer ‘k’)
(iii) [-x] = -[x] – 1
(iv) [x] + [-x] = \(\begin{cases}-1, & \mathrm{x} \notin \mathrm{Z} \\ 0, & \mathrm{x} \in \mathrm{Z}\end{cases}\)
(v) [x] – [-x] = \(\begin{cases}2[\mathrm{x}]+1, & \mathrm{x} \notin \mathrm{Z} \\ 2[\mathrm{x}], & \mathrm{x} \in \mathrm{Z}\end{cases}\)
(vi) [x] ≥ k ⇒ x ≥ k for k ∈ Z
(vii) [x] ≤ k ⇒ x < k +1 for k ∈ Z
(viii) [x] > k ⇒ x > k + 1 for k ∈ Z
(ix) [x] < k ⇒ x < k for k ∈ Z

(h) Exponential Function: A function f: R → R defined as f(x) = a^x where a > 0 and a ≠ 1 is called the exponential function.

Properties Of Exponential Function:
1. a^x+y =  a^x . a^y
2. (a^x)^y = a^xy
3. a^x = 1 if x = 0
4. If a > 1, a^x > a^y ⇒ x > y
5. If a < 1, a^x > a^y ⇒ x < y

Logarithmic Function:
Let a ≠ 1 is a positive real number. The function f: (0, ∞) → R defined by f(x) = log_a^x is called the logarithmic function, where y = log_a^x  ⇔ a^y = x
→ Domain of a logarithmic function = (0, ∞) and Range = R

Properties of logarithmic function:
1. log_a (xy) = log_ax + log_ay
2. log_a (x/y) = log_ax – log_ay
3. log_aa = 1
4. log_a(x)^y = y log_ax
5. log_a x = 0 ⇔ x = 1
6. log_ax = \(\frac{1}{\log _a{ }_a}\) , x ≠ 1
7. log_ab = \(\frac{\log _a b}{\log _c a}\)
8. \(\log _{a^n}\left(x^m\right)\) = \(\frac{m}{n}\) log_a|x|

Different Categories of function:
(a) Algebraic Function: A function that can be generated by a variable by a finite number of algebraic operations such as addition, subtraction, multiplication, division, square root, etc. is called an algebraic function.

(b) Transcendental function: A non-algebraic function is a transcendental function.
⇒ Trigonometric, trigonometric, Exponential, and logarithmic functions are transcendental functions.

Even And Odd Functions:
A function ‘f’ is an even function  if f(-x) = x and is an odd function
if f(-x) = x and is an odd function: if f(-x) = -f(x)
Note:
1. If ‘f’ is any function f(x) + f(-x) is always an even function and f(x) – f(-x) is an odd function.
2. Every function f(x) can be expressed as the sum of an even and an odd function as f(x) = g(x) + h(x), where
g(x) = \(\frac{f(x)+f(-x)}{2}\)
h(x) = \(\frac{f(x)-f(-x)}{2}\)

Periodic Function:
A function is called a periodic function with period k if f(x + k) = f(x) for some constant k ≠ 0. The least positive value of k for which f(x + k) = f(x) holds is called the fundamental period of f.

Properties of periodic function :
(1) If k is the period of f then any non-zero integral multiples of k is also a period of f.
(2) If k is the period of f(x) then f(ax + b) is also periodic with period \(\frac{k}{a}\)
(3) If f₁(x) + f₂(x) and f₃(x) are periodic functions with periods k₁, k₂, k₃, respectively then the function a₁f₁(x) + a₂f₂(x) + a₃f₃(x) is also a periodic function with period, LCM (k₁, k₂, k₃)

Algebra Of Real functions:
(a) Equality of two functions: Two functions f and g are equal iff ‘
(i) Dom f = Dom g
(ii) Co-Dom f = Co-Dom g
(iii) f(x) = g(x) for all x belonging to their common domain.

(b) Addition of two functions: Let f: D₁ → R and g: D₂ → R be two real functions.
The sum function f + g is defined by f + g: D₁ ∩ D₂ → R and (f + g)(x) = f(x) + g(x) ∀ x ∈ D₁ n D₂

(c) Subtraction of two functions: Let f: D₁ → R and g: D₂ → R. The difference function (f – g) is f – g: D₁ ∩ D₂) → R defined by (f – g) (x) = f(x) – g(x) ∀ x ∈ D₁ ∩ D₂

(d) Scalar multiplication: Let f: D → R and c is any scalar. The scalar multiple of f by the scalar c is cf: D → R defined as (cf)(x) = c. f(x) ∀ x ∈ D₁.

(e) Multiplication of two functions: Let f: D₁ → R and g: D₂ → R are two real functions. The product function (fg) is (fg): D₁ ∩ D₂ → R defined as (fg)(x) = f(x)g(x) ∀ ∈ D₁ ∩ D₂

(f) The quotient of two functions: Let f: D₁ → R and g: D₂ → R are two real functions. the quotient function (\(\frac{f}{g}\)) i,e,. \(\frac{f}{g}\): D₁ ∩ D₂ → R, defined by (\(\frac{f}{g}\))(x) = \(\frac{f(x)}{g(x)}\), ∀ x ∈ D₁ ∩ D₂

Odisha State Board CHSE Odisha Class 11 Math Notes Chapter 2 Sets will enable students to study smartly.

CHSE Odisha 11th Class Math Notes Chapter 2 Sets

Set:
Set is an undefined term in mathematics. But we understand set as “a collection of well-defined objects”.

• Set is a collection.
• The objects (called elements) in a set must be well-defined.

Set Notation:
We denote set as capital alphabets like A, B, C, D…..and the elements by the small alphabets like x, y, z ….

• If x is an element of set A we say “x belongs to A” and write ‘x ∈ A’.
• If x is not an element of set A we say “x does not belong to A” and we write ‘x ∉ A’.

Set Representation:
(a) Extension or tabular or Roster Method: In this method, we describe a set listing the elements, separated by commas within curly brackets.
Note: While listing out the elements the repetition of an object has no effect. Thus, we don’t do this.

(b) Intention or set builder or set selector method: In this method: a set is described by a characterizing property p(x) of element x. In this case, the set is described as {x : p(x) holds}

Types Of Set:
(a) Empty of full or void set: It is a set with no element.

• We denote empty set by ‘Φ’
• There is only one empty set.

(b) Singleton set: It is a set with only one element.

(c) Finite set: A set is finite if it has a finite number of elements.

(d) Infinite Set: A set that is not finite is called an infinite set.

(e) Equal sets: Two sets A and B are equal if they have the same elements. Two sets A and B are equal if all elements of A are also elements of B and all elements of B are also elements of A.

(f) Equivalent set: Two finite sets A and B are equivalent if they have the same number of elements.

Subsets: Let A and B be two sets. If every element of A is an element of B then A is called a. subset of B (we write A ⊂ B) and B is called a superset of A (We write B ⊃ A)
Thus A ⊂ B is x ∈ A ⇒ x ∈ B
Note.
(i) A set is a subset of itself.
(ii) Empty set Φ is a subset of every set.
(iii) A is called a proper subset of B if B contains at least one element that is not in A
(iv) If A has n elements then total number of subsets of A = 2ⁿ.

Universal set:
A set ‘U’ that contains all sets in a given context is called the universal set.

Power set:
Let A is any set. The collection (or set) of all subsets of A is called the power set of A. We denote it as P(A)
P(A) = { S: S ⊂ A }

Set Operations:
(a) Union of sets :
The union of two sets A and B is the set of all elements of A or B or both.
∴ A ∪ B = {x ∈ A or x ∈ B}

(b) Intersection of sets:
Intersection of two sets A and B is the set of all those elements that belong to both A and B . (or all common elements of A and B)
∴ A ∩ B = {x: x ∈ A and x ∈ B }
Two sets A and B are disjoint if A ∩ B = Φ. Otherwise, A and B are intersecting or overlapping sets.

(c) Difference of sets: The difference of two sets ‘A and B’ is the set of all elements of A which do not belong to B.
∴ A- B = {x: x ∈ A and x ∈ B)

(d) Symmetric difference of two sets: Symmetric difference of two sets A and B is the set (A – B) ∪ (B – A)
∴ A Δ B = (A – B) ∪ (B – A) = (A ∪ B) – (A ∩ B)

(e) Complement of a set: Let the complement of a set A (denoted as A’ or A^c) be defined as U – A

• A’ = {x ∈ U) : x ∉ A)
• x ∈ A’ ⇔ x ∉ A

Laws Of Set Algebra:
(a) Idempotent law: For any set A we have
(i) A ∪ A = A
(ii) A ∩ A = A

(b) Identity laws: For any set A we have
(i) A ∪ Φ = A and
(ii) A ∩ U = A

(c) Commutative laws: For any three sets A, B, and C
(i) A ∪ B = B ∪ A
(ii) A ∩ B = B ∩ A

(d) Associative laws: For any three sets A, B, and C
(i) A ∪ (B ∪ C) = (A ∪ B) ∪ C
(ii) A ∩ (B ∩ C) = (A ∩ B) ∩ C

(e) Distributive laws: For any three sets A, B, and C
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(f) De-morgans laws: For any two sets A and B
(i) (A ∪ B)’ = A’ ∩ B’
(ii) (A ∩ B)’ = A’ ∪ B’

Some more properties of sets: For any three sets A, B, and C
(a) A ⊂ (A ∪ B) and (A ∩ B) ⊂ A
(b) A ∪ B = B ⇔ A ⊂ B
(c) A ∩ B = A ⇔ A ⊂ B
(d) B ⊂ A and C ⊂ A ⇒ (B ∪ C ) ⊂ A and A ⊂ B, A ⊂ C ⇒ A ⊂ (B ∩ C)
(e) B ⊂ C ⇒ A ∪ B ⊂ A ∪ C and A ∩ B ⊂ A ∩ C
(f) A – B = A ∩ B’
(g) A – B = A ⇔ A ∩ B = Φ
(h) (A – B) ∪ B = A ∪ B and (A – B ) ∩ B = Φ
(i) A ⊆ B ⇔  B’ ⊆ A’
(j) A Δ B = B Δ A

Cardinality or order of a finite set: The cardinality or order of a finite set A (denoted as |A| or O(A) or n (A)) is the number of elements in ‘A’.

Some important results on the cardinality of finite sets and applications of set theory:
If A, B, and C are finite sets and ‘U’ is the finite universal set then a number of elements belonging to at least one of A or B.
(a) |A ∪ B| = |A| + |B| – |A ∩ B|
(b) |A ∪ B| = |A| + |B| for A ∩ B = Φ i.e. for two disjoint sets A and B
(c) Number of elements belonging to at least one of A, B, or C
= |A ∪ B ∪ C|
= |A| + |B| + |C| – |A ∩ B| – |B ∩ C| – |C ∩ A| + |A ∩ B ∩ C|
(d) Number of elements belonging to exactly two of the three sets A, B, and C = |A ∩ B| + |B ∩ C| + |C ∩ A| – 3 |A ∩ B ∩ C|
(e) Number of elements belonging to exactly one of the three sets A, B, and C = |A| + |B| + |C| – 2 |A ∩ B| -2 |B ∩ C| – 2 |C ∩ A| + 3 |A ∩ B ∩ C|
(f) Number of elements belonging to A but not B = |A – B| = |A| – |A ∩ B|
∴ |A| = |A – B| + |A ∩ B|
(g) Number of elements belonging to exactly one of A or B
= |A Δ B| = |A – B| + |B – A|
= |A| + |B| – 2 |A ∩ B|
(h) |A’ ∪ B’| = |U| – |A ∩ B|
(i) |A’ ∩ B’| = |U| – |A ∪ B| = Number of elements belonging to neither A nor B.