# 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.

MATHS.

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 $\large \displaystyle \lim_{n\to \infty} 3a_{ n }$

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

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

4. Is it AP, GP, HP ?

$\quad \quad \displaystyle \sum_{ k=1 }^{ 2 }{ { cos }^{ 8 }\left( \dfrac { k\Pi }{ 5 } \right) , } \sum _{ k=1 }^{ 3 }{ { cos }^{ 8 }\left( \dfrac { k\Pi }{ 7 } \right) , } \sum _{ k=1 }^{ 4 }{ { cos }^{ 8 }\left( \dfrac { k\Pi }{ 9 } \right) , } \sum _{ k=1 }^{ 5 }{ { cos }^{ 8 }\left( \dfrac { k\Pi }{ 11 } \right) , \ldots}$

Thanks a lot Note by Ritu Roy
4 years, 2 months ago

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

- 4 years, 2 months ago

Here's what I did

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

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

2. $D\geq 0 \Rightarrow a \geq 6 \ \ or \ a\leq 0$

3. $-6<-(a-3)<1 \Rightarrow 2

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

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

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

Finally we get $a_{3}× h_{18}=12$

- 4 years, 2 months ago

Oh yes yes, thanks a lot @Shubhendra Singh

Can you have a look at others too?

- 4 years, 2 months ago

You're welcome.

- 4 years, 2 months ago

Note first that $\dfrac{k^{3} - 1}{k^{3} + 1} = \dfrac{(k - 1)(k^{2} + k + 1)}{(k + 1)(k^{2} - k + 1)}.$

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

$\dfrac{(k - 1)(k^{2} + k + 1)}{(k + 1)(k^{2} - k + 1)} * \dfrac{((k + 1) - 1)((k + 1)^{2} + (k + 1) + 1)}{((k + 1) + 1)((k + 1)^{2} - (k + 1) + 1)} * \dfrac{((k + 2) - 1)((k + 2)^{2} + (k + 2) + 1)}{((k + 2) + 1)((k + 2)^{2} - (k + 2) + 1)} =$

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

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 $k = 2$ to infinity the only terms that remain uncanceled are $\dfrac{(k - 1)k}{k^{2} - k + 1}$ for $k = 2,$ leaving us with $\lim_{n \rightarrow \infty} 3a_{n} = 3*\dfrac{(2 - 1)*2}{2^{2} - 2 + 1} * 3 = 3*\dfrac{2}{3} = \boxed{2}.$

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

- 4 years, 2 months ago

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

Is the answer to the second one 1.

- 4 years, 2 months ago

Can you elaborate on the same

- 4 years, 2 months ago