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Correct option is (3) \(-\frac\pi8\) 

\(tan^{-1} = \left(\frac{\frac1{\sqrt2}- 1}{\frac1{\sqrt2}}\right)\)

\(=tan^{-1}(1 - \sqrt2)\)

\(= - tan^{-1}(\sqrt 2 - 1)\)

\(= -\frac\pi8\)

f(x)=acosx+bxsinx+4x-3x^2-2x^3

=>f(3)=acos3+3bsin3+4*3-3*3^2-2*3^3

=>48=acos3+3bsin3-69

=>acos3+3bsin3=117

Now

f(-3)=acos(-3)+b(-3)sin(-3)+4*(-3)-3(-3)^2-2(-3)^3

f(-3)=acos3+3bsin3-12-27+54

=117-39+54=132

(26)ⁿ

unit digit of 26 is 6

∴ unit digit of (26)ⁿ is 6 ( ∵ 6ⁿ always ends with 6)

Hence, (26)ⁿ always ends with digit 6\(\forall n\in N\).

∴ (26)ⁿ never be end with digit 5

Or for no value of n, (26)ⁿ ends with digit 5.

y = sin(logx) 

y₁ = \(\frac{cos(logx)}{x}\)

xy₁ = cos(log x). 

Again differentiate w.r.t. x 

xy₂ + y₁¹ = \(\frac{-sin(logx)}{x}\)

x²y₂ + xy₁ = -y 

⇒ x²y₂ + xy₁ + y = 0.

(c) both commutative and associative

zw = (5 – 2i) (-1 + 3i) 

= -5 + 15i + 2i – 6i² 

= -5 + 17i + 6 

= 1 + 17i

Option : (b) one-one

Option : (d) At two corner points, value of Z are equal

Option : (a) 4

Option : (d) a - 8b = 0

Option : (c) 2π²r (2rh² - h³)

Option : (a) 8

a (iv) eating good helping of vegetables and fruits

b.(iii) glaucoma

c (i) at the time of birth

d.(ii) Cornea

(e) a. Eat vegetables and fruits. 

b. Omega 3 fatty acids. 

c. Vitamins A,C, E

(f) UV protection sunglasses.

(g) Computer Eye Syndrome

(h) Around the age of 40 almost everyone requires reading glasses this condition is known as 'presbyopia'.

(i) Lasers

(j) 'Cataract' is the loss of transparency in the natural lens of the eye.

(k) (i) afflicted 

(ii) inevitably

a. (ii) Tree blindness

b. (i) spiritual exercise

c. (iv) None of the above

d. (iii) Guerrilla Gardening

(e) The habit of people to not look at trees.

(f) Verhaen Khanna is educating Delhiites about climbing trees

(g) In Anuj Wadhwa’s opinion, spending time with nature and climbing trees is a spiritual exercise.

(h) Once you are in a tree you become part of its ecosystem, which includes birds, insects, fruits and flowers.

(i) Anyone can join NDNS. They teach people how to climb trees etc

(j) Hardy local tress like neem, babul, and jamun etc.

(k) i. ecosystem 

ii. jaunt

a. (ii) 14 percent

b. (ii) easier

c. (ii) United Nation Headquarters and Times Square

d. (ii) Sri Sri Ravi Shanker

(e) The weather forecast predicted heavy rains

(f) UN Secretary General and his wife, the President of the UN General Assembly and many ambassadors participated in the programme.

(g) Sri Sri Ravi Shanker is the founder of 'The Art of Living'.

(h) WHO has warned that depression will become the biggest killer after HIV/AIDS by 2030.

(i) Yoga can be used as a powerful tool in conflict resolution and trust building.

(j) The younger generation in the western world struggled to convince their parents to open up to the benefits of yoga.

(k) (i) remission 

(ii) arduous

Calculation:

Let the time taken by Sohan to complete the work = x days

Then, time is taken by Rohan to complete the work = 2x days and

Time is taken by Mohan to complete the work = 4x days

According to the question:  

⇒ \(12(\frac{1}{x}+{1}{2x}+{1}{4x})=1\)

⇒ \(\frac{4+2-1}{4x}=\frac{1}{12}\)

⇒ \(\frac{7}{4x}=\frac{1}{12}\)

⇒ x = 21 days

Time was taken by Sohan to complete the work = x = 21 days

Time taken by Rohan to complete the work = 2x = 2 x 21 = 42 days

Time taken by Mohan to complete the work = 4x = 4 x 21 = 84 days

Let the no of days taken by Rohan and Sohan to complete the work = p days

According to the question:

⇒ \(p(\frac{1}{21}+{1}{42})=1\)

⇒ \(p\frac{1+2}{42}=1\)

⇒ p = 14 days

∴ The number of days by Rohan and Sohan to complete the work is 14 days.

We have height of circular cone is h = 48 cm. 

And the area of the circular base of the cone is 616 cm². 

Let the radius of base of circular cone is r cm.

∴ πr² = 616

⇒ r² = \(\frac{616}{\pi}\) 

= \(\frac{616}{22}\) × 7 

= 28 × 7 

= 7² × 2² 

⇒ r = 7 × 2 = 14 cm.

∴ Radius of the base of cone is r = 14 cm.

The slant height of the cone is l = \(\sqrt{r^2+h^2}\)

 = \(\sqrt{14^2+48^2}\)

 = \(\sqrt{196+2304}\)

 = \(\sqrt{2500}\)

= 50cm.

Now,

The whole surface area of cone = πr(r + l)

= \(\frac{22}{7}\) × 14(14 + 50) 

= 22 × 2 × 64 

= 2816 cm².

∴ The whole surface area of the cone is 2816 cm².

Given:

A can fill up the tank in 60 min 

B can fill the tank in 75 min

C empties the tank in 100 min

Concept used:

Work = time × efficiency

Calculation:

Efficiency of C = -3 as it is emptying the pipe.

Total efficiency of A, B ,and C together = 5 + 4 - 3 = 6

Time to fill the tank = 300/6

⇒ 50 mins

∴ Time taken to fill the tank is 50 mins.

Given:

a : b = 3 : 2

 b : c = 1 / 4 : 1 / 5

 c : d = 1.5 : 3.5

Calculation:

a : b = 3 : 2      ----(1)

b : c = 1 / 4 : 1 / 5 

⇒ b ∶ c = 5 ∶ 4      ----(2)

c : d = 1.5 : 3.5

⇒ c : d = 3/2 : 7/2 = 3 ∶ 7      ----(3)

Let us make the value of c equal in (2) and (3),

(3) × 4 and (2) × 3 and combining both,

b ∶ c ∶ d = 15 ∶ 12 ∶ 28      ----(4)

Let us make the value of b equal in (1) and (4),

(1) × 15/2 and (4) and combining,

a ∶ b ∶ c ∶ d = 3 × 15/2 ∶ 2 × 15/2 ∶ 12 ∶ 28

⇒ a ∶ b ∶ c ∶ d = 45/2 ∶ 15 ∶ 12 ∶ 28

So, a ∶ d = 45/2 ∶ 28 

⇒ a ∶ d = 45 ∶ 56

∴ The value of a ∶ d is 45 ∶ 56.

Given:

3 Men can reap field in 43 days

4 women can reap field in 43 days

Concept used:

Work done = time × efficiency 

Calculation:

Let the efficiency of one man be 'm' and one women be 'w'.

Work done = 3 × m × 43 = 4 × w × 43

⇒ Ratio of efficiency of man to women, m ∶ w = 4 ∶ 3

Let the efficiency of man and women to be 4x and 3x respectively.

Number of days by which 7 men and 5 women can finish work = (3 × 4 × 43)/(7 × 4 + 5 × 3) = 12 days.

∴ The number of days required by 7 men and 5 women to finish work is 12 days.

Given:

Time taken by A to finish the work = 8 days

Time taken by B to finish the work = 10 days

Time taken by (A, B and C) to finish the work = 4 days

Total earnings = Rs. 1848

Concept used:

Efficiency = Total work/Time taken

The share will be proportional to work done

Calculation:

L.C.M of (8, 10 and 4) = 40 = Total work

Efficiency of A = 40/8 = 5 units/day

Efficiency of B = 40/10 = 4 units/day

Efficiency of (A, B and C) = 40/4 = 10 units/day

Efficiency of C = Efficiency of (A, B and C) – Efficiency of (A and B)

⇒ 10 – (5 + 4) units/day

⇒ (10 – 9) units/day

⇒ 1 unit/day

Share of C = (1/10) × Rs. 1848

⇒ Rs. 184.8

∴ The share of C is Rs. 184.8

Moles of compound (C₆H₅N₃O₆) = \(\frac{645}{215}\) = 3 mol 

moles of Nitrogen = 9 mole

No. of atoms of Nitrogen = 9 x 6.02 × 10²³

= 54.18 × 10²³

= 5418 × 10²¹

It has been said, “Never put off for tomorrow’ what you can do today”. Yet there are many people who enjoy postponing things. Such people do not realize the dangers of delaying. Work does not disappear if we postpone it. The difficulty of the work does not get reduced by postponing it. The more we postpone, the more the work piles up. Finally, the load of work seems too much. We then have to work for many days at a stretch under great strain and tension. Finally we do it hurriedly and in a careless manner. It is as bad as not doing the work at all.

Again, when some work is delayed, the time which would have been profitably used is wasted. Time wasted is time lost forever. Hence procrastination, that is the habit of postponing things, is rightly called the thief of time. Precious time wasted means opportunities lost, and lost forever. By the time we realize how much time we have wasted and how many opportunities we have missed by delaying word, it may be too late. And then nothing remains for us except to regret and repent. Sometimes a very heavy price has to be paid for delaying things.

The person who delays insuring his factory will regret it when the factory is gutted by a sudden fire. Delay in the treatment of a disease may make it worse, and may even result in death. There are other proverbs conveying similar meaning. Thus we say: “Time and tide wait for no man” and “A stitch in time saves nine”. All these proverbs warn us against the dangers of delay in actions, and stress the importance of timely action and punctuality.

Summary

No. of words given in the original passage: 369 No. of words to be written in the summary: 369/3 = 123 ±5

Rough Draft

The maintenance of the house reflects the personality of the people who live in that. So the distinctive decoration is as important as one attire in good clothes. A unity in the home can only be seen if the rooms in the house have a degree of harmony, colour and styling. Furniture is a working strategy for the pleasant living. If there is an expression of oneself then one will have a mental satisfaction every time one enter one’s home.

To attain such satisfaction one need to pore over plants, try colour schemes, window shopping to search the best thing for one’s home. Most of the home decorators plan extensively By trying colour schemes, finding ingenious ways to make the best of what you have. They shop around to search out the right purchases at prices you can afford to pay. There is a keen pleasure in striving for the perfect result, and great satisfaction in achieving it.

Fair Draft

Interior design of One’s Home The interior furnishings of’ a house reflect the personality of the people who live in it. It is as important as one dresses in good clothes. A unity in the home can only be seen if the rooms in the house have a degree of harmony, colour and styling. Furniture also enhances one’s perception on pleasant living. One will have a mental satisfaction every time one enter one’s home.

To attain such satisfaction one need to plan extensively, try colour schemes, window shopping to search the best thing for one’s home within their budget. Home decorators helps one decide on these matters. One can get more pleasure out of a house if they have a hobby connected with it like collecting antiques, gardening, art, cooking, decorating, flower arrangements, etc.

No. of words in the summary: 129

Notes
Title: Interior design of One’s Home
Home reflects:

• personality of house-owner
• unity & harmony bet. rooms
• colour & styling sh’d be uniform

Elements of decoration:

• selection of colour schemes
• draperies, rugs, upholstery, woods

Plan to enjoy the House:

• limit time, effort & money
• select furnish’gs which require little care
• hobby connected with house-great pleasure.

Choice of Colours:

• one colour sh’d predominate
• calm colours for restfulness; intense for liveliness
• colours sh’d harmonise with furniture, draperies, carpets

Abbreviations used: bet. – between; sh’d – should; fumish’gs – furnishings;

b. Both A & R are true, but R is not the correct explanation of A

Calculation:

Let the value of f(2) and f(3) be ‘x’ and ‘y’ respectively,

⇒ f(24) = f(8).f(3) = f(2).f(2).f(2).f(3) = 54

⇒ x³y = 54

⇒ x³y = 27 × 2

⇒ x³y = 3³ × 2

On comparing, x = 3 and y = 2

⇒ f(2) = 3

⇒ f(3) = 2

⇒ f(18) = f(9).f(2) = f(3).f(3).f(2)

⇒ f(18) = 2 × 2 × 3

⇒ f(18) = 12

∴ The value of f(18) equals 12.

The short story “After Twenty Years” takes place around 10 p.m. along a dark, windy New York City business avenue, mostly within the darkened doorway of a closed hardware store. This particular location had been a restaurant until five years ago. The plot begins with a policeman “on the beat” who discovers a man standing in the dark doorway. The man then proceeds to explain why he is there. He and his best friend, Jimmy Wells had parted exactly twenty years ago to make their fortunes and had promised to meet at that spot “After Twenty Years”. He had gone west and gotten rich and was sure his friend, Jimmy would meet him if he were alive. They talked a while and the policeman carried on. The man from the west wonders if his friend will come.

The drama increases in anticipation of the rendezvous. Twenty minutes later, another man, whom we assume is his long lost friend, greets him warmly and they walk arm in arm discussing careers until they come to a well-lit comer near a drug store. The man from the west gets a good look at his companion and discovers that he is not his friend, Jimmy. We are treated to several surprises for the man from the west is under arrest and secondly he is actually ‘Silky Bob’ , a gangster from Chicago and finally the stranger is a plainclothes policeman. However, it seems that these three surprises are not enough. We get the “real” surprise when Jimmy Wells, the original policeman didn’t have the heart to arrest Bob, because he was his friend.

Where there is a will, there is a way : Where there is a will, there is a way is an old saying in English which teaches us about the most significant subject of getting success in life. Reaching a nice target becomes our maxim. However achieving a target needs strong determination and commitment. People without having will power never achieve the victory and they curse their destiny. This common saying simply means that if the person is strong-minded enough to get something done, he/ she always will find a way to do that. Fortitude is necessary to get completion in the task as it helps a person to fight with all difficulties and get success.

Without grit and will power, we generally tend to give up very easily in the initial stages as we become desperate facing small teething issues. To learn or achieve something in life, will power acts as a channel and enhances the speed of action. It needs a firm and incessant exercise to deeply learn things. We can pass the exam by reading the lessons just a night before the exam. However, we cannot be ranked first in the class, district or state without hard work for the entire year. One needs to be systematic, meticulous and focused to with a strong will power to pavegood to have determination even to the smallest tasks in life and never yield to petty things easily. way for success. So, it is 

The young man who entered the coach gave out a smothered curse, he was engaged in searching something elusive angrily and uselessly. From time to time, he dug a six penny bit out of a waist coat pocket and stared at it sadly, then resumed his search. He voluntarily broke the silence. He exclaimed that Mr. Sletherby was going to Bill Manor. He introduced himself as Bertie, the younger son of Mrs. Saltpen-Jago. He admitted that he was away for about six months and had not seen his own mother.

Making use of the lucky coincidence that he was going to Brill Manor, he asked for a loan of three pounds as he had lost his sovereign purse and was desperately in need of help. He promised to meet him on the subsequent Monday. There is a dramatic irony when the son himself gives a different version of his mom’s appearance. This influences the decision of Mr. Sletherby in refusing to lend Bertie a loan of three pounds and mistaking the true son Bertie to be a fraud.

22nd July 2020

From

Vinitha

Railway Colony

Dindigu

To 

The Editor 

Daily Thanthi

Dindigul

Respected Sir,

Sub: Violence against the aged living alone at home.

I wish to draw your attention towards the security of the aged people living alone at home in Dindigul. Almost every day we find the newspapers full of such incidents. The miscreants rob the people and then murder them. Many times the servants of the house are found involved in these incidents. The aged are the gullible people and are easily made target.

I hope after publishing this letter in your esteemed newspaper, the police will take some necessaiy actions for their safety. Thanking you

Yours faithfully,

Vinitha

Address on the Envelope

To

The Editor

Indian Express

Chennai

\(\frac{dy}{dx}\) - y + tanx = cosx

I.F = e ^{∫-tanx dx}

= e^{-log secx} = \(\frac{1}{secx}=cosx\)

It's complete solution is

y x (I.F) = ∫(I.F) x 2 dx

\(\Rightarrow\) y. cosx = ∫cos²x dx

= \(\int \frac{1+cos2x}{2}\,dx\)

= \(\frac{x}{2}+\frac{sin\,2x}{4}+c\)

Hence, complete solution of given differential equation is y = \(\frac{x}{2}\) sec x + \(\frac{sin x}{2}+c\) sec x.

y" + y = x + cos x

It's auxiliary equation is

m² + 1 = 0

m = \(\pm i\)

\(\therefore\) C.F. = C₁ cos x + C₂ sin x

Let y₁ = cos x & y₂ = sin x are two solution of given differential equation

R = x + cos x

W(y₁, y₂) = \(\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}\) = \(\begin{vmatrix}cosx&sin x\\-sin x&cosx\end{vmatrix}\)

 = cos²x + sin²x = 1

\(\therefore\) u₁ = \(\int\frac{\begin{vmatrix}0&sin x\\R&cos x\end{vmatrix}}{W(y_1,y_2)}dx\) 

\(=\int\frac{\begin{vmatrix}0&sin x\\x+cos x&cos x\end{vmatrix}}{1}dx\) 

\(= \int(x sin x + sin x cos x)dx\)

\(=-\int xsin xdx-\frac12\int sin 2xdx\) 

\(=-[x\int sin x-\int(\frac{dx}{dx}\int sin xdx)dx)+\frac{cos 2x}4\) 

 = x cos x + \(\int-cos x dx+\frac{cos 2x}4\)

 = x cos x - sin x + \(\frac{cos 2x}4\)

u₂ = \(\int\frac{\begin{vmatrix}cos x&0\\-sin x&x+cos x\end{vmatrix}dx}{1}\)

\(=\int(xcosx+cos^2x)dx\) 

= \(\int\) x cos x dx + \(\int\)cos²x dx

 = x sin x - \(\int\) sin x dx + \(\int\) \(\frac{1+cos 2x}2dx\)

 = x sin x + cos x + x/2 + \(\frac{sin 2x}4\) 

\(\therefore\) Complete solution of given differential equation is 

y = u₁y₁ + u₂y₂ + C. F.

 = C₁ cos x + C₂ sin x + (x cos x - sin x + \(\frac{cos 2x}4\)) cos x +

 (x sin x + cos x + x/2 + \(\frac{sin 2x}4\)) sin x

 = C₁ cos x + C₂ sin x + x cos²x - sin x cos x + \(\frac{cos xcos 2x}4\) + x sin²x + sin x cos x + \(\frac{xsin x}2+\frac{sinx sin2x}4\)

 = C₁ cos x + C₂ sin x + x + \(\frac{xsin x}2+\frac14(sin x sin 2x + cosx cos2x)\)

First, differentiate the function: 8t³+9t²-1 which gives us the gradient at any point. 

When t=0, the function is 4 and when t=1, the function is 2+3-1+4=8 which confirms that the given points are on the curve.

We only need the t value to find the gradient at the given points, so at t=0 the gradient is -1 and at t=1 it's 8+9-1=16.

f(x) = (x+1)³ (x-1)³

f'(x) = 3(x+1)² (x-1)³ + 3(x+3)³(x-1)²

= 3(x+1)²(x-1)²(x-1+x+1)

= 6x(x+1)²(x-1)²

∵ f'(x) > 0 when x > 0 & x ≠ 1

& f'(x) < 0 when x < 0 & x ≠ -1

Hence,

f(x) is strictly increasing in (0,∞) - {1}

& stictly decreasing in (-∞,0) - {-1}

Correct option is A) to apply

correct option is A) to apply

Ram → R Shyam → S
No. of ways of arranging in which R & S together =4!×2!=48
No. of ways of arranging in which R & S not together = Total ways to arrange − No. of ways of arranging in which R & S together.
=5!−4!×2!
=120−48=72.