Let $\overrightarrow{\mathrm{a}}=-5 \hat{i}+\hat{j}-3 \hat{k}, \overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}-4 \hat{k}$ and

$\overrightarrow{\mathrm{c}}=(((\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \hat{i}) \times \hat{i}) \times \hat{i}$. Then $\vec{c} \cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to :

Let $\mathbf{S}=\left\{x \in \mathbf{R}:(\sqrt{3}+\sqrt{2})^x+(\sqrt{3}-\sqrt{2})^x=10\right\}$. Then the number of elements in $\mathrm{S}$ is :

The area enclosed by the curves $x y+4 y=16$ and $x+y=6$ is equal to :

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ and $g: \mathbf{R} \rightarrow \mathbf{R}$ be defined as

$f(x)=\left\{\begin{array}{ll}\log _{\mathrm{e}} x, & x>0 \\ \mathrm{e}^{-x}, & x \leq 0\end{array}\right.$ and

$g(x)=\left\{\begin{array}{ll}x, & x \geqslant 0 \\ \mathrm{e}^x, & x<0\end{array}\right.$. Then, gof : $\mathbf{R} \rightarrow \mathbf{R}$ is :