MCQ
Let $*$ be a binary operation defined on $Q^+$ by the rule $\text{a}*\text{b}=\frac{\text{ab}}3\forall\text{ a, b}\in \text{Q}^+$. The inverse of $4 * 6$ is:
  • $\frac{9}{8}$
  • B
    $\frac{2}3$
  • C
    $\frac{3}2$
  • D
    None of these.

Answer

Correct option: A.
$\frac{9}{8}$
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 $e * a = a$, $\forall\text{ a}\in\text{Q}^+$
Then,
$\frac{\text{ae}}{3}=\text{a}\text{ and }\frac{\text{ea}}{3}=\text{a},\forall\text{ a}\in\text{Q}^+$
$e = 3$, $\forall\text{ a}\in\text{Q}^+$
Thus, $3$ 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,
$a * b = e = b * a$
$a * b = e$ and $b * a = e$
$\therefore\ \frac{\text{ab}}3=3\text{ and }\frac{\text{ba}}3=3$
$\text{b}=\frac{9}{\text{a}}\in\text{Q}^+$
Thus, $\frac{9}{\text{a}}$ is the inverse of $\text{a}\in\text{Q}^+$.
Given: $\text{a}*\text{b}=\frac{\text{ab}}3$
$4*6=\frac{4\times6}3=8$
Now,
$\text{a}^{-1}=\frac{9}{\text{a}}$
$(4*6)^{-1}=8^{-1}$
$=\frac{9}8$

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