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

B = {1, 2, 3, 4 …}

f: A → B, f(x) = x²

∴ f(1) = 1² = 1

f(2) = 2² = 4

f(3) = 3² = 9

f(4) = 4² = 16

f(5) = 5² = 25

∴ Range of f = {1, 4, 9, 16, 25)

Elements in A have been different elements in B.

Therefore it is one-one function. But not all the elements in B have preimages in A. Therefore it is not on-to function.

(2) 11

All the images are same in a constant function.

(4) {3, 4, 5}

f(x) = x – 2

f(5) = 5 – 2 = 3

f(6) = 6 – 2 = 4

f(7) = 7 – 2 = 5

Range of f = {3, 4, 5}

(4) not a function

2 has two images 4 and 2.

It is not a function.

(i) 3,7,11,… up to 40 terms 

First term (a) = 3 

Common difference (d) = 7 – 3 = 4 

Number of terms (n) = 40 

S_n = (n/2) [2a + (n – 1) d] 

S₄₀ = (40/2) [6 + 39 x 4] = 20 [6 + 156] 

= 20 x 162 

S₄₀ = 3240

(ii) Here a = 102, d = 97 – 102 = -5 

n = 27 

S_n = (n/2) [2a + (n – 1)d] 

S₂₇ = (27/2) [2(102) + 26(-5)]

= (27/2) [204 – 130] 

= (27/2) x 74 

= 27 x 37 = 999 

S₂₇ = 999

(iii) Here a = 6, d = 13 – 6 = 7, l = 97 

n = (l - a)/d + 1 

= (97 - 6)/7 + 1

= 91/7 + 1

= 13 + 1 = 14 

S_n = (n/2) (a + l) 

S_n = (14/2) (a + l) 

S_n = (14/2) (6 + 97) 

= 7 x 103 

S_n = 721

(4) A is larger than B by 

A = 2⁶⁵

B = 2^{64 + 63} + 2⁶² + … + 2⁰ 

= 2 

= 1 + 2² + 2² + … + 2⁶⁴ 

a = 1, r = 2, n = 65 it is in G.P. 

S₆₅ = 1 (2⁶⁵ – 1) = 2⁶⁵ – 1 

A = 2⁶⁵ is larger than B

4800 + 4750 + 4700 + … 10 terms 

Here a = 4800 

(i) d = t₂ – t₁ = 4750 – 4800 = -50 

n = 10 

S_n = (n/2) (2a + (n – 1)d) 

S₁₀ = (10/2) (2 x 4800 + 9 x -50) 

= 5 (9600 – 450) 

= 5 x 9150 = 45750 

Total amount paid in 10 installments = Rs 45750. 

(ii) The extra amount he pays in installments 

= Rs 45750 – Rs 40,000 

= Rs 5750

Tabulate the given table

Monthly savings form an A.P. 

2000, 2600, 3200 … 

a = 2000; d = 2600 – 2000 = 600

Given t_n = 20,000 

t_n = a + (n – 1) d 

20000 = 2000 + (n – 1) 600 

20000 = 2000 + 600n – 600 

= 1400 + 600n 

20000 – 1400 = 600n 

18600 = 600n

n = 18600/600 = 31 

He will take 31 years to save Rs 20,000 per month.

a₁ = 1, a₂ = 1, a_n = 2a_{n - 1} + a_{n - 2} 

a₃ = 2a_{(3 - 1)} + a_{(3 - 2)} 

= 2a₂ + a₁

= 2 x 1 + 1 = 3 

a₄ = 2a_{(4 - 1)} + a_{(4 - 2)} 

= 2a₃ + a₂ 

= 2 x 3 + 1 = 7 

a₅ = 2a_{(5 - 1)} + a_{(5 - 2)} 

= 2a₄ + a₃ 

= 2 x 7 + 3 = 17 

a₆ = 2a_{(6 - 1)} + a_{(6 - 2)} 

= 2a₅ + a₄ 

= 2 x 17 + 7 

= 34 + 7 

= 41

∴ The first six terms of the sequence are 1, 1, 3, 7, 17, 41 …

Starting salary (a) = Rs 60000 

Increased salary = 5% of starting salary 

= 5/100 x 60000 = Rs 3000 

Starting salary for the 2nd year = 60000 + 3000 = Rs 63000 

Year increase = 5% of 63000 = 5/100 x 63000 = Rs 3150 

Starting salary for the 3^rd year = 63000 + 3150 = Rs 66150 

60000, 63000, 66150,… form a G.P.

a = 60000; r = 63000/60000 = 63/60 = 21/20

t_n = an^{n - 1} 

t₅ = (60000) (21/20)⁴ 

= 60000 x (21/20) x (21/20) x (21/20) x (21/20) 

= (6 x 21 x 21 x 21 x 21)/(2 x 2 x 2 x 2)

= 72930.38 

5% increase = (5/100) x 72930.38 

= Rs 3646.51 

Salary after 5 years = Rs 72930.38 + 3646.51

= Rs 76576.90 

= Rs 76577

Bennett travelled by aero-car which shot across space at a speed of about four hundred miles ‘ an hour. Within half an hour, he reached his works at Niagara.

The gigantic advertisement signs are reflected on the clouds, so large that they can be seen all over the country. From that gallery, a thousand projectors were unendingly employed in sending to the clouds, on which they were reproduced in colour, these exorbitant advertisements.

Penalty for the first day (a) = Rs 4000 

Increased rate for every day (d) = Rs 1000

Maximum amount of penalty (S_n) 

= Rs 1,65,000 

S_n = (n/2) [2a + (n - 1) 1000] 

165000 = (n/2) [2(4000) + (n – 1) 1000] 

= (n/2) [8000 + 1000 n – 1000] 

165000 = (n/2) (7000 + 1000n) 

330000 = 7000 n + 1000 n² 

0 = 1000 n² + 7000 n – 330000 ÷ 1000 on both sides, 

n² + 7n – 330 = 0 

(n + 22) (n – 15) = 0

n = -22 or n = 15 

n = -22 (not possible) 

∴ Maximum number of days for which the work can be delayed is 15 days.

(4) both 11 and 3

7³ⁿ – 4³ⁿ is of the form a²ⁿ – b²ⁿ which is divisible by both a – b and a + b. So 7³ⁿ – 4³ⁿ is divisible by both 7 – 4 = 3 and 7 + 4 = 11.

Let t_n be n^th term of an A.P. 

t_n = S_n – S_{n - 1} 

= 2n² – 3n – [2(n – 1)² – 3(n – 1)] 

= 2n² – 3n – [2(n² – 2n + 1) – 3n + 3] 

= 2n² – 3n – [2n² – 4n + 2 – 3n + 3] 

= 2n² – 3n – [2n² – 7n + 5] 

= 2n² – 3n – 2n² + 7n – 5 

t_n = 4n – 5 

t₁ = 4(1) – 5 = 4 – 5 = -1 

t₂ = 4(2) -5 = 8 – 5 = 3 

t₃ = 4(3) – 5 = 12 – 5 = 7 

t₄ = 4(4) – 5 = 16 – 5 = 11 

The A.P. is -1, 3, 7, 11,…

The common difference is 4 

∴ The series is an A.P.

1. Some pocket money is given to me by my parents.

2. New houses were promised to people by the Mayor.

People were promised new houses by the Mayor.

3. A holiday was declared by the school.

4. A doll was given by me to the baby. 

The baby was given a doll by me.

5. I was offered a better job by them. 

A better job was offered to me by them.

Jules Verne visualizes in 1889 the world in 2889, a thousand years later where the world is filled with technological advancements.

Jules Verne talks about the people of twenty-ninth century who live in fairyland.

(c) Jules Verne

(a) twenty-ninth century

Correct answer is (b) 25th July

At home through phonotelephote, vision and speech are transmitted. In two minutes, without the help of an attendant, the machine gets a person to be ready, to his office. Food can be served through pneumatic tubes.

Given, mt_m = nt_n 

m[a + (m – 1) d] = n [a + (n – 1) d] 

[we know that t_n = a + (n – 1)d]

m[a + md – d] = n[a + nd – d] 

ma + m²d – md = na + n²d – nd 

ma – na + m²d – n²d = md – nd

a (m – n) + d (m² – n²) = d(m – n) 

a (m – n) + d(m + n)(m – n) = d(m – n) ÷ by (m – n) on both sides, 

a + d (m + n) = d 

a + d(m + n) – d = 0 

a + d(m + n – 1) = 0 … (1) 

To prove, t_{m + n} = 0 

t_{m + n} = a + (m + n – 1)d 

t_{m + n} = 0 (from(1)) 

Hence it is proved.

c, e, b, d, a 

c. This story speaks about the people of the twenty-ninth century. 

e. The year is 2889, the date 25th July and the place is the office block. 

b. Francis Bennett is the Managing Editor of the Earth Herald. 

d. Earth Herald is the world’s largest newspaper. 

a. In this world of technological advancements newspapers are not printed but ‘spoken’.

(2) an A.P.

In an A.P, if each term is multiplied by a constant or added by a constant, the resulting sequence is an A.P.

Number of times the clock strikes each hour form an A.P. 

Number of times strike in 12 hours is 1 + 2 + 3 + … + 12 

Here, a = 1, d = 1, n = 12, l = 12 

S_n = (n/2) (a + l) = (12/2) (1 + 12) 

= 6 x 13 = 78 times 

∴ Number of times the clock strike in 24 hours

= 78 x 2 = 156 times.

107 = 4 x 26 + 3. 

This is of the form a = bq + r. 

Hence it is proved.

(c) Champs-Elysees

Jules Verne was bom in the city of Nantes, France on February 8th, 1828. His father was a lawyer, and at the beginning, Veme wanted to study law as well. When he was nineteen, – he started writing long pieces of literature, but his father wanted him to earn money as a lawyer, not as a writer. He decided to give up being a lawyer, and become a full-time professional writer instead. Veme married Aimee du Fraysse de Viane in January 1857 with his father’s blessing and continued to write until his death. On 24th of March 1905, Veme, who was sick with diabetes, died at his home in Amiens, France.

I do not believe that Max deserved a chance to accomplish a risky task. He was not intelligent enough to understand that he was being tricked by the detective. Also, his appearance did not startle the detective.

t_n = a + (n – 1)d 

Given t_{m + 1} = 2 t_{n + 1} 

a + (m + 1 – 1)d = 2[a + (n + 1 – 1)d] 

a + md = 2(a + nd) ⇒ a + md = 2a + 2nd 

md – 2nd = a 

d(m – 2n) = a … (1) 

To Prove t_{(3m + 1)} = 2(t_{m + n + 1}) 

L.H.S. = t_{3m + 1} 

= a + (3m + 1 – 1)d

= a + 3md 

= d(m – 2n) + 3md (from 1) 

= md – 2nd + 3md 

= 4md – 2nd 

= 2d (2m – n) 

R.H.S. = 2(t_{m + n + 1}) 

= 2 [a + (m + n + 1 – 1)d] 

= 2 [a + (m + n)d] 

= 2 [d (m – 2n) + md + nd)] (from 1) 

= 2 [dm – 2nd + md + nd] 

= 2 [2 md – nd] 

= 2d (2m – n) 

R.H.S = L.H.S 

∴ t_{(3m + 1)} = 2 t_{(m + n + 1)} 

Hence it is proved.

(i) As soon as 

(ii) to the house 

(iii) completed by the telephote

(iv) by the electric current 

(v) in the stellar world

1. Francis Bennett

2. The young man with broad brow

3. The young man with broad brow

4. The young man with broad brow

5. Francis Bennett

 (a) Francis Bennett was feeling lonely as his wife had been to France eight days ago and so he was in a bad temper.

(b) As soon as Francis woke up, he switched on the phonotelephote whose wires connected to his house in Champs Elysees where his wife was staying.

(c) The valuable discovery was the telephotic mirror where Francis could see his wife appear on screen apart from hearing his voice.

(d) When Francis heard his wife’s sweet voice say, ‘Francis…Francis’ , his mood changed into a happy feel.

(e) Telephone completed by telephote, resulting in phonotelephote is a Time’s conquest.

(a) commutators

Correct answer is (b) Jovians

1. A relative told Mohan that he needed to work very hard with the instrument.

2. The other relative told Mohan that if it were the sitar, he would have been in competition with his brother and sister.

The given A.P is -2, -4, -6, … -100 

d = -4 – (-2) = -4 + 2 = – 2 

Finding the 12 term from the last term 

a = -100, d = 2 (taking from the last term) 

n = 12

t_n = a + (n – 1)d 

t₁₂ = – 100 + 11 (2) 

= -100 + 22 = -78 

∴ The 12^th term of the A.P from the last term is – 78

Correct answer is (b) ten

1. Francis Bennett

2. Aero-coachman

3. Francis Bennett

4. Francis Bennett

5. Aero Car

1. It was important for Mohan that his instrument should not only produce good sound but also express his vision.

2. Pandit Vishwa Mohan Bhatt has won the Grammy Award.

3. Pandit Vishwa Mohan Bhatt created the Mohan Veena after untiring efforts and understanding of the ragas.

4. On winning the Grammy Award, Pandit Vishwa Mohan Bhatt said that he was successful because he just followed his heart’s passion.

1. This sentence was said by Mohan to himself.

2. This sentence was said by a family member to Mohan.

3. This was said by Mohan’s aunty to Mohan.

4. This sentence was said by Mohan’s mother to the family members.

1. After a long search, Mr Mehta has finally found a house that is suitable for his joint family.

2. Grandmother has expressed her wish to go to Haridwar.

3. Nita got a lot of appreciation for her wonderful collection of paintings.

4. The king, who was lost in the jungle, could hear a faint sound of flowing water of a stream.

5. Mihir looked depressed when his friend was declared the winner of the race.

6. It is because of his passion for music that he has been able to give the best tunes ever.

7. Our principal’s untiring efforts and guidance helped us to incorporate some s changes in the election process for Class Representatives.

8. Geeta has a good understanding of western music.

Obama’s daughter Malia was 10 years old and Sasha was 7 years old when Obama wrote this letter.

When Obama was campaigning for the elections his daughters when to picnics, parades and state fairs and ate all sorts of junk food. They had enough fun with their new puppy.

Going to picnics and parades and state fairs, eating all sorts of junk food were the fun that Obama’s daughters had when he was dn the campaign trial.

What Obama wanted for his daughters, he wanted them for every child in that nation, so he contested for the president’s post.

After the birth of his daughters, all Obama’s big plans for himself did not seen so important anymore.

As a young man, Obama thought life was all about him-about how he would make his way in the world and become successful and get the things he wanted.

I feel that Obama is a benevolent man who wishes that every child should have the same chances to learn and dream and grow and thrive that similar to which his daughters have. He is a great visionary.