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Lt.      3x³-4x²+6x-1/2x³+x²-5x+7

x-infinite

Posted by Mansi Saini (Feb 22, 2017 12:45 p.m.) (Question ID: 2443)

• As the Lim x is Infinite that means 1/x = 0.

So, let's divide the Nr. & Dr. wit the highest power of x i.e. 3

After this, one may get the equation as (3 - 4/x + 6/x2 - 1/x3)/ (2 + 1/x - 5/x2 + 7/x3) = (3 - 0 + 0 - 0) / (2 + 0 - 0 + 0)

Answered by Manish Gandhi (Feb 22, 2017 1:24 p.m.)
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Integrate:

1.  Sin-1x / (1-x2)3/2

2. log x / xn

3. ( Sin x +1)/sin x (1+cos x)

Posted by Supriya Kumari (Feb 22, 2017 10:19 a.m.) (Question ID: 2431)

Posted by Supriya Kumari (Feb 22, 2017 12:43 p.m.)
• 3)

Ans. $$\int {({sin x + 1})dx \over sinx (1+cos x)} = \int {sin x dx \over sinx (1+cos x)} + \int { dx \over sinx (1+cos x)}$$

Let $$I = I_1 + I_2$$

Where $$I= \int {({sin x + 1})dx \over sinx (1+cos x)}, I_1 = \int {sin x dx \over sinx (1+cos x)} \space and \space I_2= \int { dx \over sinx (1+cos x)}$$

=> $$I_1= \int {sin x dx \over sinx (1+cos x)} = \int { dx \over (1+cos x)}$$

=> $$\int { dx \over 2cos^2 {x \over 2}} = {1\over 2}\int {sec^2 {x \over 2}dx}$$

=> $${2\over 2}{tan^2 {x \over 2} + C} = {tan^2 {x \over 2} + C} \space \space \space \space \space \space \space (1)$$

Now,

$$I_2= \int { dx \over sinx (1+cos x)} = \int { dx \over {2tan{x\over 2}\over (1+tan^2{x\over2})} ({2cos^2{x\over 2}})}$$

=> $${1\over 4} \int {({1+tan^2{x\over2})} ({sec^2{x\over 2}}) \over {tan{x\over 2}}}dx$$

$$Put \space tan {x\over2} = t \space\space\space\space => {1\over 2} sec^2{x\over 2} dx = dt$$, Now

$$=> {1\over 2} \int {({1+t^2)} \over t}dt = {1\over 2} \int {{1} \over t}dt + {1\over 2} \int {{t^2} \over t}dt$$

=> $${1\over 2} \int {{1} \over t}dt + {1\over 2} \int {{t}.}dt$$

=> $${1\over 2} log\space t + {1\over 2} {{t^2} \over 2} = {1\over 2} log\space t + {1\over 4} {{t^2}}$$

=> $${1\over 2} log\space (tan {x\over 2}) + {1\over 4} {tan^2{x\over 2}} \space \space \space \space \space \space \space (2)$$

From (1) And (2), We get

I = $${tan^2 {x \over 2} } + {1\over 2} log\space (tan {x\over 2}) + {1\over 4} {tan^2{x\over 2}} + C$$

=> $$I = {1\over 2} log\space (tan {x\over 2}) + {5\over 4} {tan^2{x\over 2}} +C$$

Answered by Naveen Sharma (Feb 22, 2017 1:44 p.m.)
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• Ans. 1)

$$\int{ {sin ^{-1}x}dx \over ({{1-x^2})^{3\over2}}} => \int{ {sin ^{-1}x}dx \over {({1-x^2}) \sqrt {({1-x^2)}}}}$$

Put $$sin^{-1} x = t => {1\over \sqrt{({1-x^2})}}dx = dt$$

=> $$\int {t dt \over (1-sin^2t)} = \int {t dt \over cos^2t} = \int {t .sec^2tdt }$$

Using Integaration By Parts, We get

=> t tan t + log (cos t) +  C

=> $$sin ^{-1}x. tan (sin ^{-1}x) + log (cos (sin^{-1}x)) + C$$

2)$$\int {log \space x \space dx \over x^n}$$

Using Integration by Parts

=> $$log \space x .\int {1\over x^n}dx - \int[ { 1\over x} . \int {1\over x^n}dx]dx + C$$

=> $$log \space x .{(x^{-n+1})\over (-n+1)} - \int {1\over x}.{(x^{-n+1})\over (-n+1)} dx + C$$

=> $$log \space x .{(x^{-n+1})\over (-n+1)} - {1\over (1-n)}\int {1\over x^{n} }dx + C$$

=> $$log \space x .{(x^{-n+1})\over (-n+1)} - {1\over (1-n)^2} {1\over x^{n-1} } + C$$

Answered by Naveen Sharma (Feb 22, 2017 1:11 p.m.)
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Calculate the shortest wavelength in the balmer series of hydrogen atom .  In which region (infra-red,visible ,UV)of hydrogen spectrum does this wavelength lie?

Posted by Anu Shka (Feb 22, 2017 8:09 a.m.) (Question ID: 2428)

• Ans. Since, we know that for Belmer Series

For shortest wavelength in Balmer series, the spectral series is given by

=> 1/λ =  R/4

=> λ = 4\R

=> λ = 4/ (1.097 * 107)

=> λ = 3.64 * 10 -7

This wavelength is found in the visible part of the spectrum.

Answered by Naveen Sharma (Feb 22, 2017 12:48 p.m.)
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Write 2 factor which justify the need of modulating a low frequency signal into high frequencies before transmission

Posted by Anu Shka (Feb 21, 2017 9:51 p.m.) (Question ID: 2415)

• An un-modulated signal is having lower frequency range hence unable to travel a longer distance because of lower energies. Secondly since antenna length is inversely proportional to frequency, so a lower frequency signal will increase the antenna length beyond imagination. In modulation the characteristics such as amplitude, frequency or phase of a a higher frequency(carrier) signal is changed in according with the instantaneous value of the base band signal which can then be transmitted to a larger distance. Reverse process is done at the receiver to recover back the original signal

Answered by Shweta Gulati (Feb 21, 2017 10:51 p.m.)
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Report writing on demonitisation.Tell me what should I write about it.

Posted by Shikha Doda (Feb 21, 2017 8:27 p.m.) (Question ID: 2409)

• What is demonetisation ?The reason for demonetisation. Advantages and disadvantages. Problem faced by poors and rich people.

Answered by Amrit Joseph (Feb 21, 2017 9:56 p.m.)
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Can you tell me summary of the invisible man in hindi english both

Posted by Shikha Doda (Feb 21, 2017 8:20 p.m.) (Question ID: 2408)

Posted by Shikha Doda (Feb 21, 2017 8:18 p.m.) (Question ID: 2407)

Answered by Riya Girdhar (Feb 22, 2017 9:18 a.m.)
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Prove that

1/2tan-1x=cos-1√1+√1+x2/2√1+x2

Posted by Dishaant Pandit (Feb 21, 2017 4:59 p.m.) (Question ID: 2399)

• Ans. 

$${1\over 2}tan^{-1}x = cos^{-1} (\sqrt {1 +\sqrt{1+x^2}\over 2\sqrt{1+x^2}})$$

$$taking \space LHS, put \space x = tan \theta$$

=> $$cos^{-1} (\sqrt {1 +\sqrt{1+tan^2\theta}\over 2\sqrt{1+tan^2\theta}})$$

=> $$cos^{-1} (\sqrt {1 +\sqrt{sec^2\theta}\over 2\sqrt{sec^2\theta}})$$

=> $$cos^{-1} (\sqrt {1 +{sec\theta}\over 2{sec\theta}})$$

=> $$cos^{-1} ({\sqrt {1 +cos\theta\over 2}})$$

=> $$cos^{-1} ({\sqrt { cos^2{\theta\over 2}}})$$

=> $$cos^{-1} ({ cos{\theta\over 2}})$$

=> $$LHS = {\theta\over 2}$$

$$Taking \space RHS, put \space x = tan\theta$$

=> $${1\over 2} tan^{-1}({tan \theta })$$

=> $$RHS = {{\theta \over 2}}$$

RHS = LHS

Hence Proved

Answered by Naveen Sharma (Feb 21, 2017 6:12 p.m.)
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What are the expected questions for The Invisible Man?

Posted by Nirveksen Rajkhowa (Feb 21, 2017 1:06 p.m.) (Question ID: 2398)

•

First of all go through the character sketches of this novel and after that have look at all these questions given below. Hope you find them useful.

All the best.!

1. Illustrate whether the ending of the novel justified? What is your reaction when Griffin gets killed and Marvel gets to keep all the stolen money? Are you glad that the invisiblity fomula is hidden from Kemp, who could use it?

2. Justify the title ‘The Strange Man’s Arrival’.

3. What impressions do you form of the stranger by his strange appearance and the intermittent conversation with Mrs. Hall?

4. Mrs. Hall tried to prove the best of hosts but the regular snubbing on the part of the stranger frustrated all her efforts. Comment. OR

What different conversational advances did Mrs. Hall try to make? How did the stranger respond and Why did he do so?

5. What difference do you find in Mrs. Hall’s treatment of the stranger and her husband? How much importance did Mrs. Hall give to Mr. Hall’s observations, remarks and suggestions about the strange guest that she had taken into the “Coaches and Horses” inn? What traits of her character are highlighted by her approach?

6. Describe the stranger’s behaviour while unpacking the crates. What does it indicate about his personality?

7. What made Cuss interview the stranger? What kind of interview was it and what did come out of it?

8. There has been a shift in Mrs. Hall’s temperament and opinion about the stranger in sixth chapter. Explain the statement in context of the chapter. OR

What made Mrs. Hall change her opinion about the stranger and what change do you notice in her attidude towards the stranger?

9. Why did the invisible man choose Mr. Marvel as his helper and why did Mr. Marvel comply?

10. The unusually strange experience with the invisible man could not weaken the enthusiasm of the people for much awaited ‘Whit Monday’. Justify the statement.

11. Give pen portrait of Dr. Kemp and contrast it with that of Griffin.

12. Unseen fear of something strange often leads to puzzle senses of the people. Justify the statement in the light of people at ‘Jolly Cricketers’ and their handling the invisible man.

13. “All men, however highly educated retain some superstitious inkling”. Explain the line in context of Dr. Kemp's reaction at unusual sight in his house.

14. What strange thoughts or nightmare Griffin had when he was sleeping after he had executed his experiment on himself and why did he feel so? What do these suggest about his present state?

15. Once Griffin saw no drawbacks in the invisibility but when he became invisible, there came to his view thousands of disadvantages. Explain the statement in the general way when we have certain fantastic plans and the ensuing disappointment when we have realized them.

16. Sometimes the plan or the intention behind some ambitious innovation is so wicked that the accomplishment of it is either prevented from or if it is done, it is done with some irreparable faults. Justify the statement with Griffin's plans and intentions.

17. Often the innocent people fall the victim of the mad raid. Justify the statement with Griffin attacking and killing Wickstead desperately and cruelly. Why did he do so?

18. The entire course of Griffin’s life is that of wickedness, cruelty, rudeness and unscrupulous existence. However, at the end he ended with a tragic note. Explain the statement with Griffin’s chasing his ambition, getting deceived and finally leading to his catastrophic death.

Answered by Shweta Gulati (Feb 21, 2017 7:07 p.m.)
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What are the effect in market of minimum price ceilling

Posted by Pallavi Bisht (Feb 21, 2017 12:35 p.m.) (Question ID: 2394)

What is budget line ? How it will change ? Explain with schedule and curve

Posted by Pallavi Bisht (Feb 21, 2017 12:33 p.m.) (Question ID: 2393)

Integrate:

___Sinx+1___  dx

Sinx (1+ cosx)

Posted by PRINCE KUMAR (Feb 21, 2017 11:19 a.m.) (Question ID: 2391)

• Exactly this.

Posted by Supriya Kumari (Feb 22, 2017 8:16 a.m.)
• Yes

Posted by Supriya Kumari (Feb 22, 2017 8:16 a.m.)
• Is it $$\int{ ({sin x + 1})dx \over sinx (1+cos x)}$$?

Posted by Naveen Sharma (Feb 21, 2017 12:12 p.m.)
• Ans. $$\int {({sin x + 1})dx \over sinx (1+cos x)} = \int {sin x dx \over sinx (1+cos x)} + \int { dx \over sinx (1+cos x)}$$

Let $$I = I_1 + I_2$$

Where $$I= \int {({sin x + 1})dx \over sinx (1+cos x)}, I_1 = \int {sin x dx \over sinx (1+cos x)} \space and \space I_2= \int { dx \over sinx (1+cos x)}$$

=> $$I_1= \int {sin x dx \over sinx (1+cos x)} = \int { dx \over (1+cos x)}$$

=> $$\int { dx \over 2cos^2 {x \over 2}} = {1\over 2}\int {sec^2 {x \over 2}dx}$$

=> $${2\over 2}{tan^2 {x \over 2} + C} = {tan^2 {x \over 2} + C} \space \space \space \space \space \space \space (1)$$

Now,

$$I_2= \int { dx \over sinx (1+cos x)} = \int { dx \over {2tan{x\over 2}\over (1+tan^2{x\over2})} ({2cos^2{x\over 2}})}$$

=> $${1\over 4} \int {({1+tan^2{x\over2})} ({sec^2{x\over 2}}) \over {tan{x\over 2}}}dx$$

$$Put \space tan {x\over2} = t \space\space\space\space => {1\over 2} sec^2{x\over 2} dx = dt$$, Now

$$=> {1\over 2} \int {({1+t^2)} \over t}dt = {1\over 2} \int {{1} \over t}dt + {1\over 2} \int {{t^2} \over t}dt$$

=> $${1\over 2} \int {{1} \over t}dt + {1\over 2} \int {{t}.}dt$$

=> $${1\over 2} log\space t + {1\over 2} {{t^2} \over 2} = {1\over 2} log\space t + {1\over 4} {{t^2}}$$

=> $${1\over 2} log\space (tan {x\over 2}) + {1\over 4} {tan^2{x\over 2}} \space \space \space \space \space \space \space (2)$$

From (1) And (2), We get

I = $${tan^2 {x \over 2} } + {1\over 2} log\space (tan {x\over 2}) + {1\over 4} {tan^2{x\over 2}} + C$$

=> $$I = {1\over 2} log\space (tan {x\over 2}) + {5\over 4} {tan^2{x\over 2}} +C$$

Answered by Naveen Sharma (Feb 22, 2017 1:44 p.m.)
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What is the difference between bio _data and resume?

Posted by nishant Bhardwaj (Feb 21, 2017 8:17 a.m.) (Question ID: 2387)

• Ans.

• Bio Data is the short form for Biographical Data and is an archaic terminology for Resume or C.V. In a bio data, the focus is on personal particulars like date of birth, gender, religion, race, nationality, residence, marital status, and the like. A chronological listing of education and experience comes after that.
• A Resume is ideally a summary of one's education, skills and employment when applying for a new job. A resume does not list out all details of a profile, but only some specific skills customized to the target job profile.
Answered by Naveen Sharma (Feb 21, 2017 10:24 a.m.)
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Name the diagrams to be learnt for board exam.

Posted by Dinesh Kumar (Feb 21, 2017 8:11 a.m.) (Question ID: 2386)

• For biology

Posted by Dinesh Kumar (Feb 22, 2017 9:06 a.m.)
• Please mention the subjects for which you want to practice the diagrams. As such,there are no diagrams in English.

Answered by Shweta Gulati (Feb 21, 2017 7:27 p.m.)
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What are true and false fruits?

Posted by Dinesh Kumar (Feb 21, 2017 8:10 a.m.) (Question ID: 2385)

• Ans. True fruits are formed by the ovary, which is the lower region of the pistil and the female sex organ of the flower. e.g. Pea

Sometimes the bulk of the fruit is not derived from the ovary but from some other part(s) of the flower. Such fruits are termed false fruits or accessory fruits. Strawberry is a good example of this

Answered by Naveen Sharma (Feb 21, 2017 8:28 a.m.)
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What is the theme of the poem Aunt Jennifer's Tigers and why is Aunt Jennifer terrified?

Posted by Dinesh Kumar (Feb 21, 2017 8:03 a.m.) (Question ID: 2384)

• Ans. The central theme of "Aunt Jennifer's Tigers" by Adrienne Rich is how the power of the patriarchy controls women's bodies but not their minds. The poem makes this point by presenting the wild, exotic, powerful tigers embroidered by Aunt Jennifer and contrasting them with Aunt Jennifer herself.

Aunt Jennifer is, probably, terrified of the oppression of her chauvinist husband.  She lives her life under constant pressure of duties and responsibilities of a married lady. The image of the wedding ring, even after her death, suggests that there is no escape whatsoever from the conventions of her marriage and that she had to succumb to them.

Answered by Naveen Sharma (Feb 21, 2017 10:59 a.m.)
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Money had been the central force behind the happenings in the novel(the invisible man).How do you agree with the statement?Justify by providing suitable examples.

Posted by Dinesh Kumar (Feb 21, 2017 7:59 a.m.) (Question ID: 2383)

• The story of Griffin seems to be a parable depicting the power of money. Griffin’s invisibility resulted from a scientific discovery . The chief use Griffin makes of his invisibility is to rob people of their money. This is supported by the statement, “The story of the flying money was true…money had been quietly and dexterously making off that day in handfuls …floating quietly along by walls and shady places…”Moreover, the only thing that gets Griffin accepted in Iping is money. He gets a room at the inn because of his ability to ‘strike a bargain’. Even Mrs Hall does not bother to learn Griffin’s name as long as he pays his bills on time. Indeed money helps a lot in manipulation as he calms down an agitated Mrs Hall by saying “Put it down in the bill”. As Mrs Hall says , he may be a bit overbearing , but bills settled everything… Thus griffin needs money to survive and his crimes started with his stealing of money from his father. He was able to live among the people by virtue of the power of money.

Answered by Shweta Gulati (Feb 21, 2017 7:28 p.m.)
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A car fitted with a convex side view mirror of focal length 20 cm. A second car 2.8m behind the first car is overtaking the first car with a relative speed of 15 m/s .what will be the speed of second car with respect to first car?

Posted by Nilesh Tiwari (Feb 21, 2017 7:17 a.m.) (Question ID: 2381)

• Ans.

Object Distance (u) = -2.8 m = -280 cm

Focal Lenght of Mirror (f) = 20 cm

Velocity of Object = $$({du\over dt}) =$$  15 m/s =  1500cm m/s

We Know Mirror Formula :

$${1\over v } + {1\over u } = {1\over f}$$

=> $${1\over v } - {1\over 280 } = {1\over 20}$$

=> $${1\over v } = {1\over 20} + {1\over 280 }$$

=> $$v = {280\over 15}$$

Again

$${d\over dt}{({1\over v } + {1\over u }}) = {d\over dt}({1\over f})$$

=> $$-{1\over v^2}{dv\over dt} - {1\over u^2}{du\over dt} = 0$$

=> $${1\over v^2}{dv\over dt} =-{1\over u^2}{du\over dt}$$

=> $${dv\over dt} = -{v^2\over u^2}{du\over dt}$$

Put Values of All, We get

=> $${dv\over dt} = -{{({280\over 15})^2}\over{ 280^2}} \times 1500 = {-1500\over 225} = -100/15 \space cm/s$$

=> $${dv\over dt}= -{1\over 15} \space m/s$$

-ve sign implies that image appears to be opposite to that of the object.

So, velocity of Image =$$-{1\over 15}m/s$$

Answered by Naveen Sharma (Feb 21, 2017 3:46 p.m.)
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cheques are bank money or not?

Posted by Chirag Singla (Feb 21, 2017 5:53 a.m.) (Question ID: 2378)

(1_y2)+(x_etan-1y)dy/dx=0

Posted by Prem Khaturiya (Feb 20, 2017 10:02 p.m.) (Question ID: 2375)

• Ans. $$(1+y^2) + (x - e^{tan^{-1}y}) {dy\over dx} = 0$$

=> $$(1+y^2){dx} + (x - e^{tan^{-1}y}) {dy} = 0$$

=> $$(1+y^2){dx} = (e^{tan^{-1}y}-x) {dy}$$

$$=> \space \space (1+y^2){dx\over dy} = (e^{tan^{-1}y}-x)$$

$$=> \space \space (1+y^2){dx\over dy} +x = e^{tan^{-1}y}$$

$$=> \space \space {dx\over dy} +{x\over ({1+y^2})} = {e^{tan^{-1}y}\over ({1+y^2})} \space \space\space \space\space \space\space \space [Divide\space \space by\space \space ({1+y^2}) ]$$

$$on \space \space Comparing \space \space with \space \space {dx\over dy} + P_1.x = Q_1, \space \space We\space \space get$$

$$P_1 = {1\over ({1+y^2})} \space and \space Q_1= {e^{tan^{-1}y}\over ({1+y^2})}$$

=> $$I.F. = {e ^{\int({1\over (1+y^2})dy}} = {e^{tan^{-1}y}}$$

Solution :

$$x . (I.F) = {\int { (Q_1 \times I.F})dy }+ C$$

=> $$x. {e^{tan^{-1}y}} = {\int} {{{e^{tan^{-1}y}\over (1+y^2)}}.{e^{tan^{-1}y}} dy } \space \space + C$$

$$let \space {e^{tan^{-1}y}} = t => ({1\over (1+y^2})dy = dt$$

=> $$\int e^{2t} dt = {e^{2t}\over 2} ={e^{2tan^{-1}y}\over 2}$$

So,

$$=> x. {e^{tan^{-1}y}} = {{e^{2tan^{-1}y}}\over 2} \space \space + C$$

This is required Equation.

Answered by Naveen Sharma (Feb 21, 2017 12:10 p.m.)
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Why does the rectifier give the output in only forward bias?

Posted by Manish Sahu (Feb 20, 2017 9:14 p.m.) (Question ID: 2373)

Give me tips to attempt the paper

Posted by Dinesh Kumar (Feb 20, 2017 6:11 p.m.) (Question ID: 2366)

• I have xamidea book,tell me whether it is good or i have to buy evergreen

Posted by Dinesh Kumar (Feb 20, 2017 8:37 p.m.)
• Well, the most dreaded paper of board has one craziest thing, the methods of scoring high are like solving physics questions with limits and derivations.

you can go through on following steps

1. Be sure that you carry more than 2 pens with yourself. As the paper is lengthy, keep changing the pens in between, which will make you write better and faster.
2. Recommended book , evergreen's 100% success in english ( core or communication or whatever). This book completely defines what types of questions will be asked (very accurate too), and also shows the points, which must be covered in the answers. This helped me a lot. My novel questions, were exactly the same, as described by this book. Recommended for other subjects too.
3. Be sure to get insight of your NCERT textbook.
4. Now, while doing paper, if you encounter any question (usually in reading portion) having high level, then you must attempt it. It happened to me.  At my time, same thing occurred in reading portion, the CBSE agreed that the question was indeed of high level, and made a decision to give students grace, only those who have answered it, not left blank.
5. Be sure to get GOOD NIGHT sleep, that you wake up fresh and energetic.

Answered by imran khan (Feb 20, 2017 7:26 p.m.)
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• So now its the final time and board exams are just a couple of days away.

After studying for almost a year and practicing a lot of sample papers, the only thing left is to understand how to solve the paper.

See, in English paper start with Section - C i. e.  Literature part as it carries the maximum marks. Complete it in 1 hour.

Then give your 40 mins to Section-A Reading and attempt all the questions point to point.

After that give your left over time to Section-B and present your thoughts in a smarter way referring to the word-limit.

Make sure to have 10 mins left for your revision.

Do all the questions of a particular section at one place. Follow these points and you will easily score 90+ in exams.

All the very best :)

Answered by Shweta Gulati (Feb 20, 2017 7:25 p.m.)
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How to solved tha case studies of business study more efficiently?

Posted by Vipul Goyal (Feb 20, 2017 12:11 p.m.) (Question ID: 2359)

Government incurs expenditure to popularise yoga among the masses. Analyse its impact on GDP and welfare of the people.

Posted by Sarah Abdul Majid (Feb 20, 2017 12:39 a.m.) (Question ID: 2346)

• Ans. By trying to popularize yoga among the masses, the government is trying to create health awareness. It is also trying to get people to adopt a healthy lifestyle. This, in the long run, will benet both the society and economy. Adoption of healthy habits will raise the physical and emotional welfare of the people.

A healthy and t workforce will have a positive impact on the gross domestic product (GDP) of the country. Improvement in health and increase in tness levels will lead to improved productivity arid efciency of workforce along with greater stamina to work and reduced absenteeism. This in turn will increase the GDP of the country. i.e. higher availability of goods and services per person. This ll further increase the welfare of people.

Answered by Naveen Sharma (Feb 20, 2017 12:20 p.m.)
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What is the difference between parthenocarpy and parthenogenesis

Posted by Dinesh Kumar (Feb 19, 2017 4:47 p.m.) (Question ID: 2335)

• The development of seedless without fertilisation is called parthenocarpy and deveiopment of new orgnisms without fertilisation is called parthenogenesis.

Answered by RAJENDRA SINGH (Feb 19, 2017 6:44 p.m.)
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In group 15 trihalides,bond angle increases from fluoride to iodide

Posted by Dinesh Kumar (Feb 19, 2017 4:20 p.m.) (Question ID: 2333)

• All these hydrides are covalent in nature and have pyramidal structure. These involve sp3 hybridization of the central atom and one of the tetrahedral positions is occupied by a lone pair.Due to the presence of lone pair, the bond angle in NH3 is less than the normal tetrahedral angle. It has been found to be 107 degrees. Down the group the bond angle decreases as:

In all these hydrides, four electron pairs, three bond pairs and one lone pair surround the central atom. Now, as we move down the group from N to Bi, the size of the atom goes on increasing and its electro negativity decreases. Consequently, the position of bond pair shifts more and more away from the central atom in moving from NH3 to BiH3. For e.g., the bond pair in NH3 is close to N in N-H bond than the bond pair in P-H bond in PH3. As a result, the force of repulsion between the bonded pair of electrons in NH3 is more than in PH3. In general, the force of repulsion between bonded pairs of electrons decreases as we move from NH3 to BiH3 and therefore, the bond angle also decreases in the same order.

Answered by Shweta Gulati (Feb 20, 2017 1:14 a.m.)
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Why is N-N bond weaker than P-P bond

Posted by Dinesh Kumar (Feb 19, 2017 4:15 p.m.) (Question ID: 2332)

Posted by Dinesh Kumar (Feb 19, 2017 4:37 p.m.)
• Ans. Nitrogen does not have D orbitals whereas Phosphorus on the other hand can expand itself into the d-orbital and thus can share all 5 of it's outer shell electrons.Also Due to small size of Nitrogen, there is a lot of inter-electronic repulsion. So the N-N single bond is very weak.

Answered by Naveen Sharma (Feb 19, 2017 8:18 p.m.)
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Why  is globalisation of great social significance?

Posted by Puja Tiwari (Feb 19, 2017 8:39 a.m.) (Question ID: 2320)

• Globalisation is of great significance because in the process of globalisation there is a global interchange of customs and traditions and also a global flow of trade, capital & labour.

Answered by Vishu Tomar (Feb 19, 2017 10 a.m.)
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Intigrate : $$sinx/(cos^2x+1)(cos^2x+4)$$

Posted by KJ Nikhil (Feb 19, 2017 2:52 a.m.) (Question ID: 2313)

• Ans. $$\int{ {sin \space x }\over (cos^2 \space x+1 )(cos^2\space x+4)} dx$$

Put cos x = t

then -sin x dx = dt or sin x dx = -dt

$$\int{ {-dt }\over (t^2+1 )(t^2 +4)}$$

By Partial Fraction, We can Write

$${-1\over (t^2+1)(t^2+4)} = {-1\over 3 (t^2+1)} + {1\over 3 (t^2+4)}$$

So Now

$$\int {-1dt\over (t^2+1)(t^2+4)} ={\int {-1dt\over 3 (t^2+1)}} + {\int {1dt\over 3 (t^2+4)} }$$

=> $${-1\over 3}{\int {dt\over (t^2+1)}} + {{1\over 3}\int {dt\over (t^2+4)} }$$

=> $${-1\over 3}{tan^{-1}({t\over 1})} +{1\over 3}.{1\over 2}{tan^{-1}({t\over 2})} + C$$            [$$Applying \int {1dx\over (x^2 +a^2)} = { 1\over a}{tan^{-1} ({x\over a})}$$]

=> $${-1\over 3}{tan^{-1}({cos x\over 1})} +{1\over 6}{tan^{-1}({cosx\over 2})} + C$$

Answered by Naveen Sharma (Feb 20, 2017 1:15 p.m.)
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what is the impact of brexit on AD & AS of the Indian economy?

Posted by Sarah Abdul Majid (Feb 19, 2017 12:35 a.m.) (Question ID: 2312)