quadratic inequality...

1) 377 i.e. (a,b)=(14,9)
3) 65847
4) p(1) = 4 i.e. P(x) = x²-2x+5 (this ones' the easiest)

\hspace{-16}$Here $\mathbf{x,y\in\mathbb{R}}$ and $\mathbf{x^2+y^2>0}$\\\\ Let $\mathbf{\frac{y}{x}=t,x\neq 0}$. Then $\mathbf{x^2+y^2>0\Leftrightarrow t^2+1>0\forall x\in\mathbb{R}}$\\\\ Now Let $\mathbf{\mathbb{K}=\frac{x^2+y^2}{x^2+xy+4y^2}=\frac{1+t^2}{1+t+4t^2}}$\\\\ $\mathbf{4\mathbb{K}t^2+\mathbb{K}t+\mathbb{K}=1+t^2}$\\\\ $\mathbf{(4\mathbb{K}-1).t^2+\mathbb{K}.t+(\mathbb{K}-1)=0}$\\\\ Now $\mathbf{(4\mathbb{K}-1)>0}$ for real and equal roots.\\\\ So its Discriminant must be greater then or equal to $\mathbf{\mathbb{Z}}$ero\\\\ $\mathbf{\mathbb{K}^2-4.(4\mathbb{K}-1).(\mathbb{K}-1)}\geq 0$\\\\ After rearranging, We Get\\\\ $\mathbf{15.\mathbb{K}^2-20.\mathbb{K}+4\leq 0}$\\\\ After Solving We Get\\\\ $\mathbf{\frac{10-2\sqrt{10}}{15}\leq \mathbb{K} \leq \frac{10+2\sqrt{10}}{15}}$\\\\ Now Average of Max. and Min. value of $\mathbf{\frac{10-2\sqrt{10}}{15}}$ and $\mathbf{\frac{10+2\sqrt{10}}{15}}$ is\\\\

\hspace{-16}\mathbf{\mathbb{A}=\frac{\frac{10-2\sqrt{10}}{15}+\frac{10+2\sqrt{10}}{15}}{2}=\frac{10}{15}=\frac{2}{3}}$\\\\ So $\mathbf{2007.\mathbb{A}=2007.\frac{2}{3}=1338}$

Nice, Aditya and Man111

@koni... in q1.if (a,b) =(14,9) then a² + b² is 277 and not 377... :P nd plz expain how u got the answer...
@aditya....question 3 answer is 86 and not 96 and also explain.....

\hspace{-16}$Let $\mathbf{\alpha,\beta,\gamma,\delta}$ be the roots of the equation. and Let \\\\ $\mathbf{p=-(\alpha+\beta)\;\;\;,\alpha.\beta=-32}$(bcz product of two roots)\\\\ $\mathbf{q=-(\gamma+\delta)\;\;\;,\gamma.\delta =62 }$(bcz $\mathbf{\alpha.\beta.\gamma.\delta = -1984}$)\\\\ So $\mathbf{x^4-18x^3+kx^2+200x-1984=\left\{x^2-(\alpha+\beta).x+\alpha.\beta\right\}.\left\{x^2-(\gamma+\delta).x+\gamma.\delta\right\}}$\\\\ $\mathbf{x^4-18x^3+kx^2+200x-1984=(x^2+px-32).(x^2+qx+62)}$\\\\ After Camparing Coefficients, We Get\\\\ $\begin{Bmatrix} \bold{p+q=-18 } \\\\ \bold{pq+30=k } \\\\ \bold{62p-32q=200 } \end{Bmatrix}$\\\\\\ after Solving For $\mathbf{p\;,q}\;,$ We Get \\\\ $\mathbf{p=-4\;,q=-14}$\\\\ So $\mathbf{k=pq+30=56+30=86}$\\\\ So $\mathbf{\boxed{\boxed{\bold{K=86}}}}$

@ketan,
1) Since we want -5,4 to be extremum values we need the equations f(x)=-5 and f(x)=4 to have double roots i.e. their discriminants must be zero each. using this we get (a,b) ≡ (-10,-15) and (14,9). As it is mentioned that a,b are naturals .:. (a,b) = (14,9)
and yeah it's 277 :P