Explore topic-wise InterviewSolutions in Current Affairs.

Conviction and willpower play a very prominent and positive role in a patient’s life. They, along with hope and confidence, raise a patient’s spirits and will help him/her to overcome the disease. Many prominent persons have successfully fought, even terminal diseases, with their will power. For a sound body, one needs a sound mind.

1. d. The trees seemed to run towards them but then stopped as the bus reached them and stood there helpless by the side of the road.

2. a. The meticulous way with which Valli planned her bus journey reflects her management skills.

3. b. The bus rolled on cutting across a bare landscape, rushing through a tiny hamlet or past an odd wayside shop.

4. c. The bus seemed on a point of gobbling up another vehicle that was coming towards them or a pedestrian crossing the road.

5. d. Trees stood there helpless for a moment by the side of the road.

The old and dirty buildings and apartments and the Bohemian life added to the interest of Greenwich village.

The author calls Johnsy ‘merciless’ because she wants to die and not to live in this beautiful world.

‘Gulmohar’ 

Ahmedabad 

17 January 2019 

Hi Dominic,

How are you? You have hardly written j to me since you went to London. Why the silence?

Yes, I know you are busy, but you can spare some time for an old friend, can’t you?

Here in Ahmedabad, the kite festival is just over. We had a wonderful time flying kites and cheering others. The school was closed for two days, and my friends and I spent one j day watching the competition at the Police Stadium and the other day at the ground near our house, flying an assortment of kites. Our kites were ordinary ones, but you should have seen the exotic kites being flown at the Police Stadium! They were from Japan, the U.S., U.K., Thailand, Korea and other places. They were in the shape of eagles, snakes, crocodiles, dragons, etc. Some had shining tails while some had bejewelled bodies. The competition was very stiff-but guess what! The first prize was won by the Indian team! Our own competition too was very thrilling. We cut so many kites! We were also awake the whole night shopping for kites and eating. It was really a wonderful time.

Well, I hope that next year you will be back in India and maybe you can come to Ahmedabad during this time. You will certainly enjoy the festival-and the eats! Hope you find time to reply.

Your friend,

Harsh.

a. Main Clause: There was a colony 

Subordinate Clause: where many struggling artists lived

Subordlnator: where

b. Main Clause: Johnsy showed no signs of improvement

Subordinate Clause: Though Sue looked after her very well

Subordlnator: Though

c. Main Clause: Sue thought 

Subordinate Clause: that she had fallen asleep

Subordinator: that

d. Main Clause: Old Behrman was a painter 

Subordinate Clause: who lived on the ground floor 

Subordlnator: who

e. Main Clause: Johnsy was sleeping 

Subordinate Clause: when the two went upstairs

Subordlnator: when

f. Main Clause: I want to see the last one 

Subordinate Clause: before It falls 

Subordlnator: before

Behrman always called himself as a watchdog and protector of the two artists. It was very difficult for him to understand the uncanny belief of Johnsy and he was very concerned. Outside, it was raining heavily and there was every possibility that he would get Pneumonia. Still, he risked his life because he wanted Johnsy to live.

O. Henry, the master storyteller, injects several themes into his beautiful story The Last leaf. It is a story of hope, optimism, love, friendship, and sacrifice. Hope is one of the major themes in The Last Leaf. In fact, the whole Universe runs with the idea of hope. When it is deserted, people die and when it is there, they live. The last leaf on the ivy vine signifies life. Leaf is the hope of life for a tree or plant. When Johnsy sees the painted leaf against the wall, she regains hope and says, “It is a sin to want to die” and recovers. The theme of optimism and love is symbolically presented in the story.

O. Henry tries to convey that optimistic attitudes will help us to live our life better. Johnsy embodies pessimism when she says,” when the last one falls I must go, too”. Sue consoles Johnsy and stands as symbol of optimism. It is her words full of optimism that keep Johnsy going. It is through them that O. Henry brings in the theme of friendship. Sue, as a friend, sacrifices her time and money for Johnsy. She says,” I’d rather stay here with you”. When her efforts don’t work, she seeks Behrman’s help. Sue is a symbol of true friendship.

It appears that O. Henry has a penchant for the theme ‘Sacrifice’. He brings in this theme in two of his most popular stories The Last Leaf and ‘The Gift of the Magi’. In The Last Leaf, it is Mr. Behrman who gives up his life to help Johnsy live. In’ The Gift of the Magi’, Della and Jim sacrifice their prized possessions for each other’s happiness. O. Henry seems to suggest we should make small sacrifices almost every day to make this world a better place to live in.

The Last Leaf can be regarded as O. Henry’s, masterpiece. He portrays characters that have true human strengths and weaknesses, characters we can easily identify with. Thus his world becomes our world and his stories become our stories.

It was a place frequented by artists and for the artists, old and dirty places have more artistic elements in them.

a. Right

b. Wrong

c. Wrong

d. Wrong

e. Right

f. Right

g. Wrong

The doctor told Sue that Johnsy had one chance in ten. That chance was for Johnsy to want to live, and Johnsy had made up her mind that she was not going to get well. He also told Sue that medicines alone would not help if the patient did not want to live.

Henry portrays Behrman in such a way to evoke sympathy for the character and to make him immortal.

“In November, a cold, unseen stranger whom the doctors called pneumonia, stalked about the city, touching one here and one there with his icy finger,” is an instance of personification. The personification makes pneumonia a living character, waiting to attack human beings.

1. d. The conclusion made by Valli on the basis of information was that the town was six miles from her village and the fare was thirty paise one way.

2. a. Neighbours and people who regularly travelled by bus were the sources of information.

3. b. She wanted to get all the details about the bus journey, including the distance, the fare and the total time required to reach the town and come back to the village.

4. c. The fare was thirty paise one way -"which is almost nothing at all".

5. d. The town was six miles from her village.

(c) Pneumonia’s

It tells us that their friendship is deep and strong.

No, Johnsy has already decided that she’s not going to get well. Medication will not help if we don’t have the mental strength and hope to get well.

(c) They came for making a living.

Yes, the painting was Behrman’s masterpiece. It was not a mere painting, but a life saving one as it transformed Johnsy to a woman who wants to live and not die. It is indeed a masterpiece as it gave life to someone.

She believed that she would die when the last leaf fell.

The author personifies pneumonia here. The figure of speech is called personification.

1. Sue found Behrman in his poorly-lighted studio.

2. The blank canvas

3. Johnsy’s fancy was when the last leaf of the ivy. vine fell, she would die too.

4. The author describes Johnsy as ‘light and fragile as a leaf.

5. The expression “Oh, that poor little Miss Johnsy

(b) a person who is interested in art and lives in a very informal way

To instill confidence in Johnsy. The message is, “You will be soon alright, Johnsy, be cheerful.”

It tells us that she is a creative artist and wants to reach lofty heights.

The word‘stalk

Mr. Behrman had saved Johnsy’s life. He had realized that Johnsy had a fancy that she would die after the last leaf; fell. He wanted to make certain that the last J leaf never fell, and hence he painted a false leaf on the wall. The sight of the persistent leaf, which Johnsy thought was real, made her come to her senses and want to live.

When a person is ill, he/she should have a positive attitude. He/She should have cheerful thoughts and courage and try to keep the mind occupied. Only then will medication help.

‘A Sunny Morning’ is a romantic comedy with only two main characters Don Gonzalo and Dona Laura who are in their seventies now. Their chance meeting on a sunny morning in a park in Madrid gives the characters an opportunity to relive the story of their love affair when they were living in Marfcela. The whole story is built up through witty, humorous, lively and spontaneous interaction between them.

When the play opens, we find Dona Laura happily speaking to the pigeons which were feeding on the bread crumbs. When Don Gonzalo comes there looking for a bench, Dona Laura accuses him of scaring away her birds. In reply, Don Gonzalo tells her that it is a public park and he doesn’t care about the birds. Dona Laura asks him why then did he complain that the priests had taken his seat. Later, when Don Gonzalo comes back to the same place to sit, she asks him why he was seen there again. But, when he tries to put her off saying that they had not met, she tells him that she was only responding to his gesture.

When he tells her that she ought to have only returned his greeting, she remarks that he should have taken her permission to sit on that bench. Finally, with her witty remarks, she makes him tongue-tied and helpless. When he mutters to himself, saying that she was a senile old lady and she ought to be at home knitting and counting her beads, she asks him not to grumble any more. When she finds him cleaning his shoes with his handkerchief, she taunts him asking whether one uses a handkerchief as a shoe brush. In reply, when Gonzalo asks her what right she has to criticize his actions, she answers playfully that it was her right as a neighbour.

When Gonzalo tries to put her off rudely saying that he does not care to listen to nonsense, she once again teases him remarking that he was very polite. When Gonzalo asks her apologetically not to interfere with what does not concern her, she again tells him stubbornly that she generally says what she thinks. From then on their spontaneous exchange takes a positive direction and soon they become friends.

1. He did not like her company.
2. She always comes late.
3. Did they set the house on fire?
4. He did not confess his guilt.
5. Did he not kill the snake?
6. They have not seen the Taj.
7. Did I lose my pen?
8. As soon as he saw us, he did not run away.
9. I am not senior to you.
10. She is not richer than I.

R.I. of

i. glass =  1.67

ii. water = 1.33

iii. glass w. r. t. water = 1.25

Correct option is (D) \(n\overline\upsilon\)
 

Correct option is (D) 1.5 x 10⁸ m/s

Area of equilateral triangle = \(81\sqrt{3}\) cm²

Area of equilateral triangle = \((\frac{\sqrt{3}}{4}\times a^2)\), where a is the length of the side

\(\Rightarrow 81\sqrt{3}=\frac{\sqrt{3}}{4}\times a^2\)

\(\Rightarrow\) 324 = a²

\(\Rightarrow\) a = 18 cm

Height of triangle = \(\frac{\sqrt{3}}{2}\times a^2\)

= \(\frac{\sqrt{3}}2{}\times18\)

= \(9\sqrt{3}\) cm

Sum of numbers = 994 + 996 + 998 + 1000 + 100 

= 4990 

Total counts = 5 

Therefore, Mean = (Sum of numbers)/(Total Counts) 

= \(\frac{4990}{5}\)

= 998 

Mean = 998

Answer: (a) – 3 ≤ x ≤ 3 

f(x) = \(\frac{1}{\sqrt{9-x^2}}\) will be defined if 9 – x² > 0 (∵ 9 – x² ≠0) 

⇒ x² – 9 < 0 ⇒ (x + 3) (x – 3) < 0 

⇒ x ∈ (– 3, 3)

Answer: (d) (3, 7) 

The function f will be defined if

\(\big(\sqrt{7-x} +\sqrt{x-3}\big)\) > 0

i.e., 7 – x > 0 and x – 3 > 0 

⇒ 7 > x and x > 3 

⇒ 3 < x < 7 

∴  Domain of the given function is (3, 7).

Answer : (c) (– ∞, 0) 

For the function y = \(\frac{1}{\sqrt{|x|-x}} \)  to be defined,

| x | – x > 0 as | x | – x ≠ 0 

⇒ | x | > x which is only possible when x < 0 as | x | = x for all x ≥ 0.

∴  Domain of y = \(\frac{1}{\sqrt{|x|-x}} \)  is x < 0 or (– ∞, 0).

Answer : (d) (1, \(\frac{3}{2}\))

Given f (x) = \(exp(\sqrt{5x-3-2x^2}) = exp(\sqrt {-(2x^2-5x+3)})\)

For f (x) to be defined 

– (2x² – 5x + 3) ≥ 0 

⇒ 2x² – 5x + 3 ≤ 0 

⇒ 2x² – 3x – 2x + 3 ≤ 0 

⇒ (2x – 3) (x – 1) ≤ 0  ( ∵ (x-a) (x-b) ≤ 0 ⇒ a ≤ x ≤ b where a < b)

⇒ \(x\leq \frac{3}{2}\) and x ≥ 1

⇒ 1 ≤ x ≤ \(\frac{3}{2}\)

Answer: (c) (– 3, – 2) ∪ (– 2, – 1) ∪ (1, ∞)

The function is defined, when x² – 1 > 0, 3 + x > 0 and 3 + x ≠ 1 

⇒ x < – 1 or x > 1 and x > – 3 and x ≠ – 2 

∴ Domain of the function = (– 3, – 2) ∪ (– 2, – 1) ∪ (1, ∞).

log_b a is defined for all a > 0, b > 0 and b ≠ 1 

∴ f (x) will be defined if log₅ (log₃ (18x – x² – 77)) > 0 

⇒ log₃ (18x – x² – 77) > 5⁰, i.e., 1 

⇒ 18x – x² – 77 > 3¹ 

⇒ 18x – x² – 80 > 0 

⇒ x² – 18x + 80 < 0 

⇒ (x – 8) (x – 10) < 0 

⇒ 8 < x < 10 

⇒ x ∈ (8, 10). 

∴ Domain of f (x) is x ∈ (8, 10).

Given R = {x, y): y = 2x; x, y ∈ N}

Domain of R = set of natural numbers Co-domain of 

R = set of natural number Range of 

R = set of even natural numbers 

Clearly, every natural number is related to unique image, so this relation is a function.

(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x² 

f(x) = x – 1 

g(x) = 3x + 1 

f(x) = x² 

(fog)oh = fo(goh) 

LHS = (fog)oh 

fog = f(g(x)) = f(3x + 1) = 3x + 1 – 1 = 3x 

(fog)oh = (fog)(h(x)) = (fog)(x²) = 3² … (1) 

RHS = fo(goh) 

goh = g(h(x)) = g(x²) = 3x² + 1

fo(goh) = f(3x² + 1) = 3x² + 1 – 1= 3x² … (2) 

LHS = RHS 

Hence it is verified.

(ii) f(x) = x², g(x) = 2x, h(x) = x + 4 

(fog)oh = fo(goh) 

LHS = (fog)oh 

fog = f(g(x)) = f(2x) = (2x)² = 4x²

(fog)oh = (fog) h(x) = (fog) (x + 4) 

= 4(x + 4)² = 4(x² + 8x + 16) 

= 4x² + 32x + 64 … (1) 

RHS = fo(goh) goh = g(h(x)) = g(x + 4) 

= 2(x + 4) = (2x + 8) 

fo(goh) = f(goh) = f(2x + 8) = (2x + 8)² 

= 4x² + 32x + 64 … (2) 

(1) = (2) 

LHS = RHS 

∴ (fog)oh = fo(goh) 

It is proved.

(iii) f(x) = x – 4, g(x) = x², h(x) = 3x – 5 

(fog)oh = fo(goh)

LHS = (fog)oh 

fog = f(g(x)) = f(x²) = x² – 4 

(fog)oh = (fog)(3x – 5) = (3x – 5)² – 4 

= 9x² – 30x + 25 - 4 

= 9x² – 30x + 21 … (1) 

∴ RHS = fo(goh) 

(goh) = g(h(x)) = g(3x – 5) = (3x – 5)² 

= 9x² – 30x + 25 

fo(goh) = f(9x² – 30 x + 25) 

= 9x² – 30x + 25 – 4 

= 9x² – 30x + 21 … (2) 

(1) = (2) 

LHS = RHS 

∴ (fog)oh = fo(goh) 

It is proved.

Given, 

R = {(a, b) : a, b ∈ Z and (a – b) is divisible by 5}

Reflexivity : ∀ a ∈ Z

a – a = 0 is divisible by 5

⇒ (a, a) ∈ R ∀ a ∈ Z

Hence, R is reflexive.

Symmetry : Let (a, b) ∈ R

⇒ a – b is divisible by 5

⇒ – (b – a) is divisible by 5

⇒ (b – a) is divisible by 5

⇒ (b, a) ∈ R

Hence, R is symmetric.

Transitivity : Let (a, b), (b, c) ∈ R

⇒ (a – b) and (b – c) are divisible by 5

⇒ (a – b + b – c) is divisible by 5

⇒ a – c is divisible by 5

⇒ (a, c) ∈ R

Thus, R is an equivalence relation.

f(x) = x²

g(x) = 3x

h(x) = x – 2

(fog)oh = x – 2

LHS = fo(goh)

fog = f(g(x)) = f(3x) = (3x)² = 9x

(fog)oh = (fog) h(x) = (fog) (x – 2)

= 9(x – 2)² = 9(x² – 4x + 4)

= 9x² – 36x + 36 … (1)

RHS = fo(goh)

(goh) = g(h(x)) = g(x – 2)

= 3(x – 2) = 3x – 6

fo(goh) = f(3x – 6) = (3x – 6)²

= 9x² – 36x + 36 … (2)

(1) = (2)

LHS = RHS

(fog)oh = fo(goh) is proved.

f(x) = 5x + 2; g(x) = 2x – 3; h(x) = 3x + 1

L.H.S. = fo(goh)

goh = g[h(x)]

= g(3x + 1)

= 2(3x + 1) – 3

= 6x – 1

fo(goh) = f[goh(x)]

= f(6x – 1)

= 5(6x – 1) + 2

= 30 x – 5 + 2

fo(goh) = 30x – 3 … (1)

R.H.S. = (fog)oh

fog = f[g(x)]

= f(2x – 3)

= 5(2x – 3) + 2

= 30x – 5 + 2

fo(goh) = 30x – 3 …..(1)

R.H.S. = (fog) oh

fog = f[g(x)]

= f(2x – 3)

= 5 (2x – 3)

= 5 (2x – 3) + 2

= 10x – 15 + 2

= 10x – 13

(fog) oh = fog [h(x)]

= fog (3x + 1)

= 10 (3x + 1) – 13

= 30x + 10 – 13

= 30x – 3 … (2)

From (1) and (2) we get L.H.S. = R.H.S.

fo(goh) = (fog)oh

(3) 2 – 4x

f(x – 1) = (x – 1)² – (x – 1)

= x² – 2x + 1 – x + 1

= x² – 3x + 2

f(x + 1) = (x + 1)² – (x + 1)

= x² + 2x + 1 – x – 1

= x² + x

f(x – 1) – f(x + 1) = x² – 3x + 2 – (x² + x)

= x² – 3x + 2 – x² – x

= -4x + 2 = 2 – 4x

(2) 1001 kms^-1

f(t) = at² – at + 1

m = 50(5)² – 50(5) + 1

= 1250 – 250 + 1

= 1001 kms^-1

f(x) = x² + 4 ; g(x) – 3x – 2; h(x) = x – 5

To prove fo (goh) = (fog) oh

L.H.S. fo (goh)

goh = g[h(x)]

= g(x – 5)

= 3(x – 5) – 2

= 3x – 15 – 2

goh = 3x – 17

fo (goh) = f [goh (x)]

= f(3x – 17)

= (3x – 17)² + 4

= 9x² + 289 – 102 x + 4

= 9x² – 102x + 293 … (1)

R.H.S. = (fog) oh

fog – f[g(x)]

= f(3x - 2)

= (3x – 2)² + 4

= 9x² + 4 – 12x + 4

= 9x² – 12x + 8

(fog) oh = fog [h(x)]

= fog (x – 5)

= 9(x – 5)² – 12 (x – 5) + 8

= 9(x² + 25 – 10x) – 12x + 60 + 8

= 9x² + 225 – 90x – 12x + 60 + 8

= 9x² – 102x + 293 ….(2)

From (1) and (2) we get fo (goh) = (fog) oh.

Composition of function is associative

(3) 21x – 73

K(x) = 3x – 9; L(x) = 7x – 10

LOK = L[K(x)]

= L(3x – 9)

= 7(3x – 9) – 10

= 21x – 63 – 10

= 21x – 73