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• The displacement of valence electrons resulting in polarization of an organic molecule is called electronic effect. 
• Polarization can be either due to the presence of an atom or substituent group, or due to the influence of certain atornattacking reagent or due to the certain structural feature present in the molecule. 
• Such polarization results in the formation of electrophilic or nucleophilic centre in the neutral organic molecule.

• Organic molecules contain covalent bonds, which are made of valence electrons of the constituent atoms.
• During an organic reaction, molecules of the reactant undergo change in their structure due to redistribution of valence electrons of constituent atoms.
• This results in the bond breaking or bond forming processes as organic reaction proceeds. However, these processes are usually not instantaneous.
• As a result of this, the overall organic reaction occurs by the formation of one or more unstable species called intermediates.

Thus, organic reactions are often a multi-step process.

Structural isomerism: When two or more compounds have same molecular formula but different structural formulae, they are said to be structural isomers of each other and the phenomenon is known as structural isomerism.

i. Isomerism: The phenomenon of existence of two or more compounds possessing the same molecular formula is known as isomerism.

ii. Isomers: Two or more compounds having the same molecular formula are called as isomers of each other. [Note: The isomers are different compounds having same molecular formula and therefore they exhibit different physical and chemical properties.]

• Identification of parent chain: The longest carbon chain containing the single or the principal functional group is identified as parent chain.

e.g. Ethers are named as alkoxyalkane. While naming it, the larger alkyl group is chosen as parent chain.

• Numbering of parent chain: It is done so as to give the lowest possible locant numbers to the carbon atom of this functional group.
• Suffix: The name of the parent hydrocarbon is modified adequately with appropriate suffix in accordance with the single/principal functional group.
• Names of the other functional groups (if any) are attached to this modified name as prefixes. The locant numbers of all the functional groups are indicated before the corresponding suffix/prefix.

[Note: The carbon atom in -COOR, -COCl, - CONH₂ , -CN and -CHO is C – 1 by rule and therefore, is not mentioned in the IUPAC name.]

General formula of homologous series of alcohols can be represented as, C_nH_{2n + 1} OH (where n = 1, 2, 3, …).

Homologous series:

• A series of compounds of the same family in which each member has the same type of carbon skeleton and functional group, and differs from the next member by a constant difference of one methylene group (-CH₂ -) in its molecular and structural formula is called as homologous series.
• The individual members of the series are called homologues and they can be represented by a same general formula.
• Two successive homologues differ by one – CH² (methylene) unit (i.e., molecular weight of each successive member differs by 14 units).
• Homologues show similar chemical properties.
• Physical properties (like melting point, boiling point, density, solubility, etc.) of the homologues show a gradual change with increase in the molecular weight of the member.

Note: Consider the homologous series of straight chain aldehydes. The boiling point increases down the series as molecular weight increases.

i. 5-Phenylpent-1-ene 

ii. 2-Hydroxybenzoic acid

Correct option is: (b) \(– CH_2 –\)

A member of a homologous series differs from immediate above or below member by \(– CH_2 –\) group

Correct option is (B) – CH₂ – 

(B) alicyclic

Correct option is: Correct option is (A) locant

In IUPAC nomenclature, the number which indicates the position of the substituent is called locant

Correct option is (A) locant

(B) homocyclic but not aromatic

1. b) ± \(\cfrac12\)

2. d) \(\cfrac{-1}{360}\)

3. b) y = 360x  – 717

4. c) 3588

5. b) (-1/2, 0) ꓴ (1/2, ∞)

7, 14, 21, 28,....

t₁ = 7, t₂ = 14, t₃ = 21, t₄ = 28

Here, \(\frac{t_2}{t_1}=2,\frac{t_3}{t_2}=\frac32,\frac{t_4}{t_3}=\frac43\) 

\(\because\) \(\frac{t_2}{t_1}\neq\frac{t_3}{t_2}\neq\frac{t_4}{t_3},\) the sequence is not a Geometric progression.

2, 6,18, 54,....

t₁ = 2, t₂ = 6, t₃ = 18, t₄ = 54,...

Here, \(\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}\) = 3

\(\because\) the ration of any two consecutive terms is a constant, the given sequence is a Geometric progression.

Here, a = 2, r = 3

t_n = ar^n-1

^{\(\therefore\)} t_n = 2(3^n-1)

1, -5, 25, -125,.....

t₁ = 1, t₂ = -5, t₃ = 25, t₄ = -125

Here, \(\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=5\)

\(\because\)  the ration of any two consecutive terms is a constant, the given sequence is a Geometric progression.

Here, a = 1, r = -5

t_n = ar^n-1

^{\(\therefore\)} t_n = (-5)^n-1

3, 4, 5, 6,....

t₁ = 3, t₂ = 4, t₃ = 5, t₄ = 6

Here, \(\frac{t_2}{t_1}=\frac43,\frac{t_3}{t_2}=\frac54,\frac{t_4}{t_3}=\frac65\)

\(\because\) \(\frac{t_2}{t_1}\neq\frac{t_3}{t_2}\neq\frac{t_4}{t_3},\) then given sequence is not a Geometric progression.

\(\sqrt5,\frac1{\sqrt5},\frac1{5\sqrt5},\frac1{25\sqrt5},.....\)

t₁ = √5, t₂ = 1/√5, t₃ = 1/5√5, t₄ = 1/25√5,.....

Here, \(\frac{t_2}{t_1}=\frac{t_3}{t_2}=\frac{t_4}{t_3}=\frac15\)

\(\because\) the ratio of any two consecutive terms is a constant, the given sequence is a Geometric progression.

Here, a = √5, r = 1/5

t_n = ar^n-1

\(\therefore\) t_n = √5\((\frac15)^{n-1}\)

1. The banner is about the celebrations of Telugu Language Day.

2. We celebrate such days in memory of great people / great events.

3. Gurajada Apparao, Tagore, Gurram Jashuva, Sarojini Naidu, Gidudu Rama Murthy, Sri Sri, etc.

4. Kanyasulkam by Cmrajada Apparao, Vemana Satakam by Vemana, Rajashekara Charitramu by Kandukur Veeresalingam, etc.

5. Srirangam Srinivasa Rao was the first true modern Telugu poet to write about contemporary issues that affected the day to day life of a common man. He wrote ‘Mahaprasthanam’.

The given equations are: 

217x + 131y = 913 …..(i) 

131x + 217y = 827 ……(ii) 

On adding (i) and (ii), we get: 

348x + 348y = 1740 

⇒348(x + y) = 1740 

⇒x + y = 5 ……(iii) 

On subtracting (ii) from (i), we get: 

86x – 86y = 86 

⇒86(x – y) = 86 

⇒x – y = 1 ……(iv) 

On adding (iii) from (i), we get: 

2x = 6 

⇒x = 3 

On substituting x = 3 in (iii), we get: 

3 + y = 5 

⇒y = 5 – 3 = 2 

Hence, the required solution is x = 3 and y = 2.

Given pair of linear equations is

217x + 131y = 913 …(i)

And 131x + 217y = 827 …(ii)

On multiplying Eq. (i) by 131 and Eq. (ii) by 217 to make the coefficients of x equal, we get the equation as

28427x + 17161y = 119603 …(iii)

28427x + 47089y = 179459 …(iv)

On subtracting Eq. (iii) from Eq. (iv), we get

⇒ 28427x + 47089y – 28427x – 17161y = 179459 – 119603

⇒ 47089y – 17161y = 179459 – 119603

⇒ 29928y = 59856

⇒ y = 59856/29928

⇒ y = 2

On putting y = 2 in Eq. (ii), we get

⇒ 131x + 217(2) = 827 ⇒ 131x + 434 = 827

⇒ 131x = 393

⇒ x = 393/131

⇒ x = 3

Hence, x = 3 and y = 2 , which is the required solution.

The given equations are: 

217x + 131y = 913 …..(i) 

131x + 217y = 827 ……(ii) 

On adding (i) and (ii), we get: 

348x + 348y = 1740 

⇒348(x + y) = 1740 

⇒x + y = 5 ……(iii) 

On subtracting (ii) from (i), we get: 

86x – 86y = 86 

⇒86(x – y) = 86 

⇒x – y = 1 ……(iv) 

On adding (iii) from (i), we get: 

2x = 6 

⇒x = 3 

On substituting x = 3 in (iii), we get: 

3 + y = 5 

⇒y = 5 – 3 = 2 

Hence, the required solution is x = 3 and y = 2.

(i) f = {( 1,2), (2, 3), (3, 4), (4, 5) (5, 6)}

Domain = {1,2, 3,4, 5}

Range = {2, 3, 4, 5, 6}

(ii) R = {(-2,4), (-1, 1),(2,4), (1,1) (-3,9)}

Domain = {-2, -1,2, 1,-3} (or)

= {-3,-2,-1, 1,2}

Range = {4, 1, 9} (or) {1, 4, 9}

(i) f(x) = (2x + 1)/(x - 9) 

If the denominator vanishes when x = 9

So f(x) is not defined at x = 9

∴ Domain is x ∈ [R – {9}]

(ii) p(x) = -5/(4x² + 1)

p(x) is defined for all values of x. So domain is x ∈ R.

(iii) g(x) = √(x - 2)

When x < 2 g(x) becomes complex. But given “g” is real valued function.

So x > 2

Domain x ∈ (2, α)

(iv) h (x) = x + 6 

For all values of x, h(x) is defined. Hence domain is x ∈ R.

a) Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24 

b) Factors of 15 = 1, 3, 5, 15 

c) Factors of 21 = 1, 3, 7, 21 

d) Factors of 27 = 1, 3, 9, 27 

e) Factors of 12 = 1, 2, 3, 4, 6, 12 

f) Factors of 20 = 1, 2, 4, 5, 10, 20 

g) Factors of 18 = 1, 2, 3, 6, 9, 18 

h) Factors of 23 = 1, 23

i) Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36

A = {1, 2, 3, 4, . . . 45}, A x A = {(1, 1), (2, 2) … (45, 45)} 

R – is square of’ 

R = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)} 

R ⊂ (A x A) 

Domain of R = {1, 2, 3, 4, 5, 6} 

Range of R = {1, 4, 9, 16, 25, 36}

R = {(x, x³): x is a prime number less than 10} The prime numbers less
than 10 are 2, 3, 5, and 7.
∴ R = {(2, 8), (3, 27), (5, 125), (7, 343)}

(i) N = (1, 2, 3, …} 

The relation from N to N is given by: {(1, 3), (2, 5), (3, 7), (4, 9), …} 

when, x = 1 then y = 3 

x = 2 then y = 5 

x = 3 then y = 7 

x = 4 then y = 9 

3, 5, 7, 9, … is an A.P. 

Hence, its nth term = a + (n – 1 ).d, 

where a is first term and d, is a common difference. 

T_n = 3 + (n – 1) × 2 = 3 + 2n – 2 = 2n + 1 

Here, we get the required rule by putting n = x and T_n = y

 {(x, y) | x, y ∈ N and y = 2x + 1}. 

(ii) Relation in N is expressed as : 

{(2, 3), (4, 2), (6, 1)} = {(6, 1), (4, 2), (2, 3)} 

Here, 6, 4, 2 are in an A.P. 

Its general term T_n = 6 + (n – 1) × (-2) 

T_n = 6 – 2n + 2 

T_n = 8 – 2n 

Here, we get the required rule by putting x = y and T_n = x 

{(x, y) | x, y ∈ N, x = 8 – 2y or x + 2y = 8} and y < 4

(iii) Relation in N is expressed as: 

{(2, 1), (3, 2), (4, 3), (5, 4), …} 

Here, 2, 3, 4, 5, … are in an A.P. 

So, nth term T_n = 2 + (n – 1) × 1 = 2 + n – 1 = n + 1

Here, by putting n = x and T_n = y 

Required rule = {(x, y) | x, y ∈ N, x = y + 1 or y = x – 1}

Given R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}. 

Domain of R = {0, 1, 2, 3,4, 5} 

Range of R = {5, 6, 7, 8, 9, 10}

All non-zero relation are:

{(1, 1)}, {(2, 2)}, {(1, 2)}, {(2, 1)}
{(1, 1), (1, 2)}, {(1, 1), (2, 1)}, {(2, 2), (1,2)}, {(2, 2), (2, 1)}, {(1, 2), (2, 1)}

{(1, 1), (2, 2), (1, 2)}, {(1, 1), (2, 2), (2, 1)}, {(1, 1), (1, 2), (2, 1)}, {(2, 2), (1, 2), (2, 1)}

{(1, 1), (1, 2), (2, 1), (2, 2)}

(i) Clearly, every element of domain is related to unique element of co-domain, so it is a function. 

Domain = {2, 5, 8, 11, 14, 17} 

Range = {1} 

(ii) Clearly, every element of domain is related to unique element of co-domain, so it is a function. 

Domain = {2,4, 6, 8,10,12,14} 

Range = {1,2, 3,4, 5, 6,7}

(iii) 1 is related to two elements of co-domain, namely 3 and 5, so it is not a function.

Given : Set X = {(a, b) : a, b ∈ I} 

where I is the set of integers. 

A relation R in X is defined as: 

(a, b) R(c, d) ⇔ b – d = a – c ∀ (a, b) (c, d) ∈ X 

To prove that R is equivalence relation, we have to prove that R is reflexive, symmetric and transitive

(i) Reflexivity: 

Let (a, b) ∈ X 

(a, b) ∈ X ⇒ (a, b) ∈ I 

⇒ b – b = a – a = 0 

⇒ (a, b) R(a, b) ∀ (a, b) ∈ X 

R is a reflexive relation. 

(ii) Symmetricity: 

Let (a, b), (c, d) ∈ X is in this way 

(a, b) R(c, d) 

(a, b) R(c, d) 

⇒ b – d = a – c 

⇒ -(d – b) = -(c – a) 

⇒ d – b = c – a 

⇒ (c.d) R(a.b) 

(a, b) R(c, d) ⇒ (cd) R(ab) ∀ (a, b), (c, d) ∈ X 

R is a symmetric relations. 

(iii) Transitivity: 

Let (a, b), (c, d), (e, f) ∈ X is in this way 

(a, b) R(c, d) and (c, d) R(e, f) (a, b) R(c, d) 

⇒ b – d = a – c …(1) 

(c, d) R(e, f) ⇒ d – f = c – e …(2) 

Adding equation (i) and (2), we have 

b – d + d – f = a – c + c – e 

⇒ b – f = a – e 

⇒ (a, b) R(e, f) 

So, (a, b) R(c, d) and (c, d) R(e, f) 

⇒ (a, b) R(e, f) ∀ (a, b), (c, d), (e, f) ∈ X 

R is a transitive relation. 

Hence, according to (i), (ii) and (iii), the given relation is equivalence relation. 

Hence Proved.

In the figure ∠5 = ∠3 (Alternate interior angles) 

Given ∠3 = 116° 

So, ∠5 = ∠3 = 116° 

∴ ∠5 = 116°

Given relation

R = {(x + 1, x + 5) : x ∈ (0, 1, 2, 3, 4, 5}

Then, domain of relation R {(x + 1) : x ∈ {0, 1, 2, 3, 4, 5}

Domain of R = {1, 2, 3, 4, 5, 6}

and Range of R = {(x + 5) : x ∈ (0, 1, 2, 3, 4, 5}

Range of R = {5, 6, 7, 8, 9, 10}

In the given figure 

∠7 = ∠1 (Alternative exterior angles) 

Given ∠1 = 123° 

So, ∠7 = ∠1 = 123° 

∴ ∠7 = 123° .

In the given figure ∠4 = ∠8 (corresponding angles) 

Given ∠8 = 80° So, ∠4 — ∠8 = 80° 

∴ ∠4 = 80°

Given, R = {(x,y) : x,y ∈ z, xy = 4} 

= {(– 4, – 1), (– 2, – 2), (– 1, – 4), (1, 4), 

    (2, 2), (4, 1)} 

∴ Domain of R = {– 4, – 2, – 1, 2, 4}

Given, n (A) = 5 and n (B) = 7 

We know that, if n (A) = P and n (B) = q, then number of functions from A to q^p . 

∴ Number of function from A to B = 7⁵

From the figure we know that ∠PRQ = x^o = 60^o as the vertically opposite angles are equal

We know that EF || GH and RQ is a transversal

From the figure we also know that ∠PRQ and ∠RQS are alternate angles

So we get

∠PRQ = ∠RQS

∠x = ∠y = 60^o

We know that AB || CD and PR is a transversal

From the figure we know that ∠PRD and ∠APR are alternate angles

So we get

∠PRD = ∠APR

It can be written as

∠PRQ + ∠QRD = ∠APR

By substituting the values we get

x + ∠QRD = 110^o

60^o + ∠QRD = 110^o

On further calculation

∠QRD = 110^o – 60^o

By subtraction

∠QRD = 50^o

According to the △ QRS

We can write

∠QRD + ∠QSR + ∠RQS = 180^o

By substituting the values

∠QRD + t^o + y^o = 180^o

50^o + t^o + 60^o = 180^o

On further calculation

t^o = 180^o – 50^o – 60^o

By subtraction

t^o = 180^o – 110^o

t^o = 70^o

We know that AB || CD and GH is a transversal

From the figure we know that z^o and t^o are alternate angles

So we get

z^o = t^o = 70^o

Therefore, the values of x, y, z and t are 60^o, 60^o, 70^o and 70^o.

Co-exterior angles are supplementary. 

That is ∠2 + ∠7 = 180° and ∠1 + ∠8 = 180°

In the given figure ∠7 = ∠3 (corresponding angles) 

Given ∠3 = 145° So, ∠7 = ∠3 = 145° 

∴ ∠7 – 145°

Given,\(f(x)\frac{1}{\sqrt{x-|x|}}\) 

We know that, 

|x| = \(\begin{cases}x, & \quad \text{when } x \geq0\\-x, & \quad \text{when } x<0\end{cases}\)

x - |x| = \( \begin{cases}x-x & \quad \text{when } x\geq0\\x+x, & \quad \text{when } x<0\end{cases}\)

x - |x| = \(\begin{cases}0, & \quad \text{when } x\geq0\\2x, & \quad \text{when } x <0\end{cases}\)

= x − |x| ≤ 0 For all x 

\(\frac{1}{\sqrt{x-|x|}}\) does not take real values for any x ∈ R 1 

⇒ f(x) is not defined for any x ∈ R Hence, Domain  \(f(x)=\phi\)

Given, f(x) = x³ 

At x = 5, f(5) = 5³ = 125 

At x = 1, f(1) = 1³= 1 

∴\(\frac{f(5)-f(1)}{5-1}=\frac{125-1}{5-1}=\frac{124}{4}\) = 31

The value of the function for every rational number is 1 and for every irrational number is – 1 

 \(\quad \therefore \frac{1}{2}\in Q \implies f(\frac{1}{2})=1 \quad\\\text{ And }\quad \pi \notin \text {Q }\implies f(\pi) = -1 \) 

In the given figure, BA ∥ CD and BC is transversal. 

∠C = x + 35° and ∠B = 60° 

∠C = ∠B (∵ alternate interior angles are equal)

x + 35° = 60° 

x – 60° – 35° 

∴ x = 25°

Given, f(x) =\(\frac{x^2-9}{x-3}\) 

Domain: Clearly, f(x) is defined for all x ∈ R expect x = 3 

∴ Domain of f = R – {3} = (−∞, 3),∪ (3, ∞) 

Range: Let y = f(x) 

∴ y = \(\frac{x^2-9}{x-3}\)⇒ = + 3 

It follow from the above relation that y takes all real values except 6 when takes values in the set 

R – {3} 

∴ Range of f = R – {6}

Given, f(x) = x² + 2 

Let y = f (x)

y =  x²+ 2 

x² = y – 2 

x = \(\sqrt{y-2}\) 

Clearly, x will take real values, if 

Y – 2 ≥ 0 Y ≥ 2 

∴ Range of y = [2, ∞)