Exercise 1.13 [Section 1, BOMC-2011]

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Labib
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Exercise 1.13 [Section 1, BOMC-2011]

Unread post by Labib » Wed Oct 26, 2011 1:54 pm

For which real values of x the following inequality
holds:
$\frac{4x^2}{{(1 −\sqrt{1 + 2x)}}^2}< 2x + 9$?
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Re: Exercise 1.13 [Section 1, BOMC-2011]

Unread post by Labib » Wed Oct 26, 2011 1:58 pm

sourav das wrote:Hints
Ex:1.13 : $4n^2 < 4n^2 + n < 4n^2 + 4n + 1$
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Re: Exercise 1.13 [Section 1, BOMC-2011]

Unread post by sourav das » Wed Oct 26, 2011 6:24 pm

I follow the old book. So...
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
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Re: Exercise 1.13 [Section 1, BOMC-2011]

Unread post by Labib » Thu Oct 27, 2011 12:04 am

Please, some1 help me with this one!!
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Re: Exercise 1.13 [Section 1, BOMC-2011]

Unread post by *Mahi* » Thu Oct 27, 2011 12:36 am

Hint:
The left hand side is
$\frac{4x^2}{{(1 −\sqrt{1 + 2x)}}^2}=(1+ \sqrt {1+2x})^2$
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Re: Exercise 1.13 [Section 1, BOMC-2011]

Unread post by Labib » Thu Oct 27, 2011 1:57 am

Thanks for the help... :)
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Re: Exercise 1.13 [Section 1, BOMC-2011]

Unread post by sm.joty » Thu Oct 27, 2011 11:28 am

"For which real values of x the following inequality
holds"
মাহি ভাইয়া, দুটো বিষয় পরিস্কার করেন যে, এখানে কি x এর একটা মাত্র মান আসবে নাকি সেট কিংবা ব্যবধি ধরনের কিছু আসবে। কারন ভাইয়া x=-1/2 এবং x=4 দুটোর মানের জন্যই কিন্তু অসমতাটা সত্য। আর নিচের অংশটায় কোন ভুল করলাম কিনা দেখেনতো,
\[\frac{4x^2}{{(1 −\sqrt{1 + 2x)}}^2}< 2x + 9\]
if \[p =1+2x\]
then
\[[\frac{2x(1+\sqrt{p})}{1-p}]^2 < 2x+9 \]
\[\Leftrightarrow [-(1+\sqrt{p})]^2 < 2x+9\]
\[\Leftrightarrow 1+2\sqrt{p}+p < p+8\]
\[\Leftrightarrow \sqrt{p}< \frac{7}{2}\]
here p must be positive.Unless the inequality can't be define.(because inequality holds only for the real numbers) so,
\[x < \frac{45}{8}\]
but it's also true that from the first line of the inequality we see that \[1-\sqrt{p} \neq 0\]
so,\[x\neq 0\]
so the solution is
\[\frac{-1}{2}<x < \frac{45}{8}, x\neq 0\]
Last edited by sm.joty on Fri Oct 28, 2011 3:44 pm, edited 1 time in total.
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Re: Exercise 1.13 [Section 1, BOMC-2011]

Unread post by nafistiham » Thu Oct 27, 2011 3:00 pm

as you can see,here the problem says values.so, you got to include all the real values that satisfy the inequality.(i don't think there is any mistake. ;) )
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Exercise 1.13 [Section 1, BOMC-2011]

Unread post by *Mahi* » Thu Oct 27, 2011 9:37 pm

sm.joty wrote:\[[\frac{2x(1+\sqrt{p})}{1-p^2}]^2 < 2x+9 \]
This would be \[[\frac{2x(1+\sqrt{p})}{(1-p)}]^2 < 2x+9 \]
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Re: Exercise 1.13 [Section 1, BOMC-2011]

Unread post by sm.joty » Fri Oct 28, 2011 3:42 pm

Oups, that's a big mistake.I'm sorry for that. :oops: :oops: As soon as possible I will edit.

But I request all of our MENTORS & friends to inform me is my solution right or wrong.
I specially requeste the mentors to reply anything about my solution.

(It's very important for me because I already found another mistake in my solution. It's a problem from IMO . So it should be very hard to solve.So it is unbelievable for me to think that I can solve this!!! :!: Because I haven't a peculiar brain. :| )
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