Let $f: \mathbf{R} \rightarrow(0, \infty)$ be a twice differentiable function such that $f(3)=18, f^{\prime}(3)=0$ and $f^{\prime \prime}(3)=4$.

Then $\lim\limits _{x \rightarrow 1}\left(\log _e\left(\frac{f(2+x)}{f(3)}\right)^{\frac{18}{(x-1)^2}}\right)$ is equal to :

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that the quadratic equation $f(x) \mathrm{m}^2-2 f^{\prime}(x) \mathrm{m}+f^{\prime \prime}(x)=0$ in m , has two equal roots for every $x \in \mathbf{R}$. If $f(0)=1, f^{\prime}(0)=2$, and ( $\alpha, \beta$ ) is the largest interval in which the function $f\left(\log _{\mathrm{e}} x-x\right)$ is increasing, then $\alpha+\beta$ is equal to

$\_\_\_\_$ .

Let $S=\{(m, n): m, n \in\{1,2,3, \ldots . ., 50\}\}$. If the number of elements $(m, n)$ in $S$ such that $6^m+9^n$ is a multiple of 5 is $p$ and the number of elements ( $m, n$ ) in $S$ such that $m+n$ is a square of a prime number is q , then $\mathrm{p}+\mathrm{q}$ is equal to $\_\_\_\_$ .

Let $a_1=1$ and for $n \geqslant 1, a_{n+1}=\frac{1}{2} a_n+\frac{n^2-2 n-1}{n^2(n+1)^2}$. Then $\left|\sum_{n=1}^{\infty}\left(a_n-\frac{2}{n^2}\right)\right|$ is equal to $\_\_\_\_$ .