# Maths doubt

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## Maths doubt

 A rectangle has its lower left hand corner at the origin and its upper right hand corner on the graph of f(x)=x^2+x^(-2).For which x is the area of the rectangle minimised? a. x=0 b. x=infinity c. x=(1/3)^(1/4) d. x=2^(1/3)
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## Re: Maths doubt

 This post was updated on . Area = x(x^2 + x^-2). Minimize A accordingly. I am getting x = (1/3)^(1/4) “Operator! Give me the number for 911!”
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## Re: Maths doubt

 Seems legit..thank you :)
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## Re: Maths doubt

 Q.1 Two women and four men are to be seated randomly around a circular table.find the probability that the women are not seated next to each other. a. 1/2 b. 1/3 c. 2/5 d. 3/5
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## Re: Maths doubt

 In reply to this post by Noel @Noel Total no of ways =5! Lets make the men seat first.4 men can be seated in 4! ways.Now remaining 2 women can be seated in 5P2 ways. therfore prob is (4!*5P2)/5!=2/5
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## Re: Maths doubt

 @ron thank you very much but there seems to be an error as (4!*5P2)/5! is equal to 4
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## Re: Maths doubt

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## Re: Maths doubt

 In reply to this post by Noel @Noel i made a silly error...here it goes Total=5!ways 4 men can be seated in 3! ways.now there are 4 gaps and these 4 gaps can be filled by women in 4P2 ways. Therefore prob is (3!*4P2)/5! =3/5
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## Re: Maths doubt

 Thanks ben10 @ron thanks again..i'm a bit weak at probability..can you please explain how the men can be seated in 3! ways and then how can there be 4 places for the women after you've seated all the men?.thank you
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## Re: Maths doubt

 In reply to this post by Homer Simpson can you please elaborate ?
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## Re: Maths doubt

 In reply to this post by Noel hi Noel, since its circular arrangement if u assume first the 4 men are seated, they can be seated in 3! ways. The two women cannot sit together, they have to be seated between men. There are 4 possible places(Imagine a chair between two men). The two women can sit in any of this 4 places in 4C2*2! ways. You get total possible ways as 3!*4P2
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## Re: Maths doubt

 In reply to this post by Noel Hi,Just need to confirm as ans coming out to be c.as y=x^2+1/(x^2).and we need to maximize area = xy=x^3+1/(x).thus ans c. Let me know if there's any mistake.Regards,AdityaOn Fri, Apr 11, 2014 at 12:20 AM, Noel [via Discussion forum] wrote: A rectangle has its lower left hand corner at the origin and its upper right hand corner on the graph of f(x)=x^2+x^(-2).For which x is the area of the rectangle minimised? a. x=0 b. x=infinity c. x=(1/3)^(1/4) d. x=2^(1/3) If you reply to this email, your message will be added to the discussion below: http://discussion-forum.2150183.n2.nabble.com/Maths-doubt-tp7586601.html To start a new topic under General Discussions, email [hidden email] To unsubscribe from General Discussions, click here. NAML "Woh mara papad wale ko!"
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## Re: Maths doubt

 In reply to this post by Noel Here Noel...check out
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## Re: Maths doubt

 In reply to this post by riyaf @Riyaf, this is the idea. “Operator! Give me the number for 911!”
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## Re: Maths doubt

 In reply to this post by The Villain Thank you so much ron and ben10..i got it now :) @aditya..you're correct
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## Re: Maths doubt

 Q.2 Suppose X1,X2,...,Xn are observed completion times of an experiment with values in [0,1].Each of these random variables is uniformly distributed on [0,1].If Y is the maximum observed completion time,then the mean of Y is a. [n/(n+1)]^2 b. n/[2*(n+1)] c. n/(n+1) d. 2n/(n+1)
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## Re: Maths doubt

 In reply to this post by Noel Is the ans c? MA Economics DSE 2014-16
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## Re: Maths doubt

 @sonia i don't have the answer key but please explain how you got it
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## Re: Maths doubt

 Q. The coefficient of x^2 in the polynomial (1-x)*(1+2x)*(1-3x)...(1+14x)*(1-15x) is a -121 b -191 c -255 d -291