9 algebra questions.

3) -1 ,0 ,2 are the only solutions i think

4)
(\sum{\frac{1}{r}})^{2} = \sum{\frac{1}{r^{2}}} + 2\sum{\frac{1}{rs}}

\sum{\frac{1}{r^{2}}} = (\sum{\frac{1}{r}})^{2} - \frac{2}{rstu}\sum{rs}

\sum{\frac{1}{r}} = \frac{1}{rstu}\sum{rst} = \frac{-3}{2}

\sum{rs} = 0

thus , \sum{\frac{1}{r^{2}}} = (\sum{\frac{1}{r}})^{2} = 9/4

5) S_{n} = \sum_{n=1}^{\propto }{\frac{(7n+32)(3/4)^{n}}{n(n+2)}}

S_{n} = \sum_{n=1}^{\propto }{\frac{(16n+32)(3/4)^{n}}{n(n+2)}} - \sum_{n=1}^{\propto }{\frac{(9n)(3/4)^{n}}{n(n+2)}}

S_{n} = 16\sum_{n=1}^{\propto }{\frac{(3/4)^{n}}{n}} - 16\sum_{n=1}^{\propto }{\frac{(9/16)(3/4)^{n}}{(n+2)}}

S_{n} = 16 \left[ \sum_{n=1}^{\propto }{\frac{(3/4)^{n}}{n}} - \sum_{n=1}^{\propto }{\frac{(3/4)^{n+2}}{(n+2)}} \right]

S_{n} = 16 \left[ \frac{(3/4)^{1}}{1} + {\frac{(3/4)^{2}}{(2)}} \right] = 16.5

8) P(x) = log₂x = a₀ +a₁x + .... + a_nxⁿ

now coeff. of x = a₁ = P ' (0)

P'(x) = ln2 .1x

P'(0) → ∞

thus a ₁ →∞

@aditya,
note that in Q8 its given that p is a polynomial of degree 2011, so it can't be log₂x.
maybe we could try something by taking a function
g(x) = p(x) - log₂x
but again thats not right..

yes rishab , i missed that .

using A.m ≥ H.m for the 2010 numbers apart from x,

[(2012-x)/2010] ≥ 2010/[2012-(1/x)]

(2012-x)(2012-1/x)≥(2010)^2

2012^2 - (x + 1/x) +1 ≥ 2010^2
so max value of x+1/x is 2012^2-2010^2 +1

Thanks Guys for Nice Discussion..

here is the answers....

\hspace{-16} \bf{(2)\;\; \left(-\frac{3}{5}\;,\frac{4}{5}\right)}$\\\\ $\bf{(3)\;\; x=-1\;,0\;,2}$\\\\ $\bf{(4)\;\; \frac{9}{4}}$\\\\ $\bf{(5)\;\; \frac{33}{2}}$\\\\ $\bf{(6)\;\; \left(-3^{1005}-1\right).x+\left(-2.3^{1005}-1\right)}$\\\\ $\bf{(7)\;\; \frac{8045}{2012}}$\\\\ $\bf{(8)\;\; 2-\frac{1}{2^{2010}}}$\\\\ $\bf{(9)\;\; -665}$\\\\ $\bf{(10)\;\; 10}$

Man111 please provide solutions for Qs 6 - 10