The solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=(x-y)^2$ when $y(1)=1$ is

If $x_0$ is the point of local minima of $f(x)=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})$ where $\overline{\mathrm{a}}=x \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$, $\overline{\mathrm{b}}=-2 \hat{\mathrm{i}}+x \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overline{\mathrm{c}}=7 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+x \hat{\mathrm{k}}$, then value of $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}$ at $x=x_0$ is

$\hat{a}, \hat{b}$, and $\hat{c}$ are three unit vectors such that $\hat{a} \times(\hat{b} \times \hat{c})=\frac{\sqrt{3}}{2}(\hat{b}+\hat{c})$. If $\dot{b}$ is not parallel to $\hat{c}$, then the angle between $\hat{a}$ and $\hat{b}$ is

For a suitable chosen real constant a, let a function $f: \mathbb{R}-\{-\mathrm{a}\} \rightarrow \mathbb{R}$ be defined by $f(x)=\frac{a-x}{a+x}$. Further suppose that for any real number $x \neq-\mathrm{a}$ and $\mathrm{f}(x) \neq-\mathrm{a}$, (fof) $(x)=x$. Then $f\left(-\frac{1}{5}\right)$ is equal to