Maths - Help Required! (Nov 2015)

Hello Brilliantians

I have a few doubts in Maths which are listed below.... for which i didn't receive satisfactory answers from my teachers..

I request the readers help me in the concepts.

Thanks in advance for your valuable contribution.

MATHS.

  1. an=23123+1.33133+1.43143+1n31n3+1\large a_{ n }=\dfrac { { 2 }^{ 3 }-1 }{ { 2 }^{ 3 }+1 } .\dfrac { { 3 }^{ 3 }-1 }{ 3^{ 3 }+1 } .\dfrac { 4^{ 3 }-1 }{ 4^{ 3 }+1 } \ldots \dfrac { n^{ 3 }-1 }{ n^{ 3 }+1 }\quad Find limn3an\large \displaystyle \lim_{n\to \infty} 3a_{ n }

  2. limx3+[x2]2[x]3[x2]4[x]+3\large \displaystyle \lim_{x \rightarrow 3^+}\dfrac { \left[ { x }^{ 2 } \right] -2\left[ x \right] -3 }{ \left[ { x }^{ 2 } \right] -4\left[ x \right] +3 }

  3. For a constant a, let the roots of f(x)f(x) of the equation x2+2(a3)x+9=0{ x }^{ 2 }+2(a-3)x+9=0 lie between (6,1)(-6,1).
    2,h1,h2,h3,,h20,[a]2, { h }_{ 1 },{ h }_{ 2 },{ h }_{ 3 } , \ldots , { h }_{ 20 },\left[ a \right] are in HP.
    2,a1,a2,a3,,a20,[a]2, { a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 } , \ldots , {a}_{ 20 }, \left[ a \right] are in AP.
    Find the value of a3.h18{ a }_{ 3 }.{ h }_{ 18 }

  4. Is it AP, GP, HP ?

k=12cos8(kΠ5),k=13cos8(kΠ7),k=14cos8(kΠ9),k=15cos8(kΠ11),\quad \quad \displaystyle \sum_{ k=1 }^{ 2 }{ { cos }^{ 8 }\left( \dfrac { k\Pi }{ 5 } \right) , } \sum _{ k=1 }^{ 3 }{ { cos }^{ 8 }\left( \dfrac { k\Pi }{ 7 } \right) , } \sum _{ k=1 }^{ 4 }{ { cos }^{ 8 }\left( \dfrac { k\Pi }{ 9 } \right) , } \sum _{ k=1 }^{ 5 }{ { cos }^{ 8 }\left( \dfrac { k\Pi }{ 11 } \right) , \ldots}

Thanks a lot

Note by Ritu Roy
4 years ago

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1 vote

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I'm getting answer of question 3 as 12 is it correct??

Shubhendra Singh - 4 years ago

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Here's what I did

For roots lying in the range (-6,1) the following conditions must be satisfied

  1. f(6)>0f(-6)>0 and f(1)>0a<274 & a>2f(1)>0 \Rightarrow a <\dfrac{27}{4} \ \& \ a>-2

  2. D0a6  or a0D\geq 0 \Rightarrow a \geq 6 \ \ or \ a\leq 0

  3. 6<(a3)<12<a<9-6<-(a-3)<1 \Rightarrow 2<a<9

These all conditions give the final result that a[6,274)a \in [6,\dfrac{27}{4}). By this we get [a]=6[a]=6

Now the AP is is formed with first term 2 and 22nd22^{nd} term as 6 this gives a3=187a_{3}=\dfrac{18}{7}

For the HP part we have an AP with first term =12\dfrac{1}{2} and 22nd22^{nd} term =16\dfrac{1}{6} by this we get 1h18=942\dfrac{1}{h_{18}}=\dfrac{9}{42}

Finally we get a3×h18=12a_{3}× h_{18}=12

Shubhendra Singh - 4 years ago

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Oh yes yes, thanks a lot @Shubhendra Singh

Can you have a look at others too?

Ritu Roy - 4 years ago

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You're welcome.

Shubhendra Singh - 4 years ago

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A few observations to help you with 1.) .....

Note first that k31k3+1=(k1)(k2+k+1)(k+1)(k2k+1).\dfrac{k^{3} - 1}{k^{3} + 1} = \dfrac{(k - 1)(k^{2} + k + 1)}{(k + 1)(k^{2} - k + 1)}.

Next look at the sub-product of three consecutive terms in ana_{n}:

(k1)(k2+k+1)(k+1)(k2k+1)((k+1)1)((k+1)2+(k+1)+1)((k+1)+1)((k+1)2(k+1)+1)((k+2)1)((k+2)2+(k+2)+1)((k+2)+1)((k+2)2(k+2)+1)=\dfrac{(k - 1)(k^{2} + k + 1)}{(k + 1)(k^{2} - k + 1)} * \dfrac{((k + 1) - 1)((k + 1)^{2} + (k + 1) + 1)}{((k + 1) + 1)((k + 1)^{2} - (k + 1) + 1)} * \dfrac{((k + 2) - 1)((k + 2)^{2} + (k + 2) + 1)}{((k + 2) + 1)((k + 2)^{2} - (k + 2) + 1)} =

(k1)(k2+k+1)(k+1)(k2k+1)k(k2+3k+3)(k+2)(k2+k+1)(k+1)(k2+5k+7)(k+3)(k2+3k+3).\dfrac{(k - 1)(k^{2} + k + 1)}{(k + 1)(k^{2} - k + 1)}*\dfrac{k(k^{2} + 3k + 3)}{(k + 2)(k^{2} + k + 1)} * \dfrac{(k + 1)(k^{2} + 5k + 7)}{(k + 3)(k^{2} + 3k + 3)}.

One then notices how like terms in the numerator and denominator cancel, (both in successive terms and those with an intervening term). As the product stretches from k=2k = 2 to infinity the only terms that remain uncanceled are (k1)kk2k+1\dfrac{(k - 1)k}{k^{2} - k + 1} for k=2,k = 2, leaving us with limn3an=3(21)2222+13=323=2.\lim_{n \rightarrow \infty} 3a_{n} = 3*\dfrac{(2 - 1)*2}{2^{2} - 2 + 1} * 3 = 3*\dfrac{2}{3} = \boxed{2}.

Brian Charlesworth - 4 years ago

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Thanks a lot @Brian Charlesworth Sir. So glad to have the answer from you sir.

Ritu Roy - 4 years ago

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I'm not sure what specifically you want help with on most of those. For 3 I can tell you that the limit without absolute value doesn't exist because the function goes to -infinity is x approaches 1 from the left.

Cole Wyeth - 4 years ago

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Thanks for your answer.

While I am fairly able to solve many series,limits questions, perhaps the lack of clear concepts stops me from solving numerous questions . Some of them are posted above.

Ritu Roy - 4 years ago

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Please help for question 4!!

Ritu Roy - 4 years ago

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Please refrain from mass tagging.Tag at most 5 people if necessary.

Abdur Rehman Zahid - 4 years ago

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These are not specific doubts, it seems like you want us to do your homework. If you have some specific doubts about those problems I'm sure that the community will be glad to help you, but don't try to get us to do your homework.

Francisco Rodríguez - 4 years ago

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@Francisco Rodríguez

These are definitely not my homework problems, else I would have got the required help from my teachers.

For the questions above, I get stuck with the concepts required to solve the problems. For this I request the community for help.

...... If you are aware of the cbse system, these kind of questions are not found in our school texts. These type are found in competitive exams.

Ritu Roy - 4 years ago

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Is the answer to the second one 1.

kritarth lohomi - 4 years ago

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Can you elaborate on the same

Ritu Roy - 4 years ago

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