In the interval $[0, 2\pi]$,the slope of the tangent to the function $f(x) = e^x \sin x$ is maximum at $x = \dots$

Let $x_0$ be the point of local minima of $f(x) = \overline{a} \cdot (\overline{b} \times \overline{c})$ where $\overline{a} = x \hat{i} - 2 \hat{j} + 3 \hat{k}$,$\overline{b} = -2 \hat{i} + x \hat{j} - \hat{k}$,and $\overline{c} = 7 \hat{i} - 2 \hat{j} + x \hat{k}$. Then the value of $\overline{a} \cdot \overline{b}$ at $x = x_0$ is:

Let $x=-1$ and $x=2$ be the critical points of the function $f(x)=x^3+ax^2+b \ln|x|+1, x \neq 0$. Let $m$ and $M$ respectively be the absolute minimum and the absolute maximum values of $f$ in the interval $\left[-2, -\frac{1}{2}\right]$. Then $|M+m|$ is equal to (Take $\ln 2 \approx 0.7$):

If $m$ and $M$ respectively denote the minimum and maximum of $f(x)=(x-1)^2+3$ for $x \in [-3, 1]$, then the ordered pair $(m, M)$ is equal to

Let $f(x) = \begin{cases} 3 - x, & 0 \le x < 1 \\ x^2 + \log_e b, & x \ge 1 \end{cases}$. The set of values of $b$ such that $f(x)$ has a local minimum at $x = 1$ is

Let $f(x)=x^2+\frac{1}{x^2}$ and $g(x)=x-\frac{1}{x}$ for $x \in R-\{-1,0,1\}$,then the local minimum of $\frac{f(x)}{g(x)}$ is