Question

The sum of $A$ \& $B$ is given by:

• A. $\backslash(@(A,B), 2)$
• B. $@( \backslash(A,B), 2)$
• C. $@(x(A,B), 2)$
• D. None of these
• A. @ (/(@(/(B, A), 2), C), 3)
• B. X(@(/(@(B, A), 3), C), 2)
• C. /(X(@ (/(@(B, A), 2), C), 3), 2)
• D. /((X(@ (B, A), 2), C), 3)

Answer 1

The Correct Option is A

Given: - $@(A,B)$ = average of $A$ and $B$ = $\dfrac{A+B}{2}$. - $\backslash(P,Q)$ = product of $P$ and $Q$. The sum $A+B$ = $2 \times$ average$(A,B)$ = $2 \times @(A,B)$. In given notation, multiplying $@(A,B)$ by $2$ means: \[ \backslash(@(A,B), 2) \] Thus (1) is correct. \[ \boxed{\backslash(@(A,B), 2)} \]

Answer 2

Step 1: Recall the operations.
@(P, Q) is the average of P and Q, /(P, Q) is the product of P and Q, and X(P, Q) is P divided by Q. We need an expression equal to \(\dfrac{A+B+C}{3}\).

Step 2: Evaluate option (3) from the inside.
\(@(B,A)=\dfrac{A+B}{2}\)
\(/(@(B,A),2)=A+B\)
\(@(A+B,\,C)=\dfrac{A+B+C}{2}\)
\(X\!\left(\dfrac{A+B+C}{2},\,3\right)=\dfrac{A+B+C}{6}\)
\(/\!\left(\dfrac{A+B+C}{6},\,2\right)=\dfrac{A+B+C}{3}\), which is the required average.

Step 3: Rule out the others.
Option (1): \(/(B,A)=AB\), so the expression involves \(AB\) and cannot equal \(\dfrac{A+B+C}{3}\).
Option (2): \(/(@(B,A),3)=\dfrac{3(A+B)}{2}\), then \(@(\cdot,C)=\dfrac{3(A+B)/2+C}{2}\), divided by 2 again gives \(\dfrac{3(A+B)+2C}{8}\), not the average.
Option (4): as printed it is not a well-formed nesting; even read as /(/(X(@(B,A),2),C),3) it equals \(\dfrac{A+B}{4}\cdot C\cdot 3\), a product with C, not the average.

Final Answer:
The correct option is (3): /(X(@ (/(@(B, A), 2), C), 3), 2)