Question

Find the value of: \[ \text{me}(a + \text{mo}(\text{le}(a, b)),\; \text{mo}(a + \text{me}(\text{mo}(a), \text{mo}(b)))) \] where $a = -2$ and $b = -3$.

• A. 1
• B. 0
• C. 5
• D. 3
• A. mo(le(a, b)) $\ge$ me(mo(a), mo(b))
• B. mo(le(a, b)) $>$ me(mo(a), mo(b))
• C. mo(le(a, b)) $\le$ me(mo(a), mo(b))
• D. mo(le(a, b)) = le(mo(a), mo(b))
• A. $a>3$
• B. $0<a<3$
• C. $a<0$
• D. $a = 3$
• A. $a>3$
• B. $0<a<3$
• C. $a<0$
• D. Both b and c

Answer 1

The Correct Option is A

Step 1: Evaluate $\text{le}(a, b)$ = least of $(-2, -3)$ = $-3$.
Step 2: mo$(\text{le}(a, b))$ = mo$(-3)$ = $|{-3}|$ = $3$.
Step 3: First argument of me = $a + \text{mo}(\text{le}(a, b))$ = $-2 + 3 = 1$. Step 4: mo$(a)$ = $|-2| = 2$, mo$(b)$ = $|-3| = 3$.
Step 5: $\text{me}(\text{mo}(a), \text{mo}(b))$ = me$(2, 3) = 3$.
Step 6: Second argument inside mo = $a + \text{me}(\text{mo}(a), \text{mo}(b))$ = $-2 + 3 = 1$.
Step 7: mo$(1) = |1| = 1$. Step 8: Final = me$(1, 1) = 1$.

Answer 2

Since $a, b>0$, le$(a, b)$ is simply the smaller of $a$ and $b$. - mo(le(a, b)) = $| \text{smaller of } a, b |$ = smaller of $a, b$. - me(mo(a), mo(b)) = max($a, b$) since $a, b$ are positive. Clearly, smaller $\le$ larger, so: \[ \text{mo(le(a, b))} \le \text{me(mo(a), mo(b))} \] is always true.

Answer 3

me$(X, Y)<0$ means both $X<0$ and $Y<0$. Here: $X = a^2 - 3a$, $Y = a - 3$. Step 1: $a - 3<0 \implies a<3$. Step 2: $a^2 - 3a<0 \implies a(a - 3)<0 \implies 0<a<3$. Both must hold simultaneously: $a<3$ and $0<a<3$ gives $0<a<3$. But careful — For $a<0$, - $a^2 - 3a>0$? Wait — check: if $a<0$, $a^2$ is positive, $-3a$ positive, so sum positive → contradiction. Actually the only valid range from intersection is $0<a<3$. Correction — answer is (b).

Answer 4

le$(X, Y)<0$ means the smaller of $X$ and $Y$ is negative. Case 1: $0<a<3$: - $a - 3<0$, so le$(X, Y)$ is negative → condition true. Case 2: $a<0$: - $a - 3<0$, so again le$(X, Y)$ is negative regardless of $X$. Thus, both $a<0$ and $0<a<3$ satisfy.