- $\text{a}$
- $\frac{1}{\text{a}}$
- $\frac{2}{\text{a}}$
- $\frac{4}{\text{a}}$
Solution:
Let e be the identity element in Q+ with respect to * such that
a * e = a = e * a, $\forall\text{ a}\in\text{Q}^+$
a * e = a and e * a = a,
$\forall\text{ a}\in\text{Q}^+$$\frac{\text{ae}}2=\text{a}$ and $\frac{\text{ea}}2=\text{a}$, $\forall\text{ a}\in\text{Q}^+$
$\text{e}=2\in\text{Q}^+, \forall\text{ a}\in\text{Q}^+$
Thus, 2 is the identity element in Q+ with respect to *.
Let $\text{ a}\in\text{Q}^+$ and $\text{ b}\in\text{Q}^+$ be the inverse of a.
Then,
a * e = a = e * a
a * b = e and b * a = e
$\frac{\text{ab}}2=2$ and $\frac{\text{ba}}2=2$
$\text{b}=\frac{4}{\text{a}}\in\text{Q}^+$
Thus, $\frac{4}{\text{a}}$ is the inverse of $\text{ a}\in\text{Q}^+$.
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$f(x)=\left\{\begin{array}{ll}\frac{x^{3}}{(1-\cos 2 x)^{2}} \log _{e}\left(\frac{1+2 x e^{-2 x}}{\left(1-x e^{-x}\right)^{2}}\right), & x \neq 0 \\ \,\alpha & , x=0\end{array}\right.$ If $\mathrm{f}$ is continuous at $\mathrm{x}=0$, then $\alpha$ is equal to :