$$ \text { If } y=\tan ^{-1}\left[\frac{\log _e\left(\frac{e}{x^2}\right)}{\log _e\left(e x^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log _e x}{1-6 \cdot \log _e x}\right] \text {, then } \frac{d^2 y}{d x^2}= $$
$$\lim _\limits{n \rightarrow \infty} \frac{1}{n^{k+1}}[2^k+4^k+6^k+\ldots .+(2 n)^k]=$$
The acceleration f $$\mathrm{ft} / \mathrm{sec}^2$$ of a particle after a time $$\mathrm{t}$$ sec starting from rest is given by $$\mathrm{f}=6-\sqrt{1.2 \mathrm{t}}$$. Then the maximum velocity $$\mathrm{v}$$ and time $$\mathrm{T}$$ to attend this velocity are
Let $$\Gamma$$ be the curve $$\mathrm{y}=\mathrm{be}^{-x / a}$$ & $$\mathrm{L}$$ be the straight line $$\frac{x}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$$ where $$\mathrm{a}, \mathrm{b} \in \mathbb{R}$$. Then