Comprehension
A, S, M and D are functions of $x$ and $y$, and they are defined as follows:
$A(x, y) = x + y$
$S(x, y) = x - y$
$M(x, y) = xy$
$D(x, y) = \dfrac{x}{y}, \quad y \neq 0$
Question: 1
What is the value of M(M(A(M(x, y), S(y, x)), x), A(y, x)) for $x = 2, y = 3$?
What is the value of M(M(A(M(x, y), S(y, x)), x), A(y, x)) for $x = 2, y = 3$?
Show Hint
Evaluate inside-out for nested functions to avoid mistakes.
Updated On: Aug 6, 2025
- 60
- 140
- 25
- 70
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The Correct Option is D
Solution and Explanation
Given $A(x,y) = x+y$, $S(x,y) = x-y$, $M(x,y) = xy$.
Step 1: $M(x,y) = 2\times 3 = 6$.
Step 2: $S(y,x) = 3-2 = 1$.
Step 3: $A(M(x,y), S(y,x)) = A(6,1) = 6+1 = 7$.
Step 4: $M(7, x) = 7\times 2 = 14$.
Step 5: $A(y,x) = 3+2 = 5$.
Step 6: $M(14, 5) = 70$.
Step 1: $M(x,y) = 2\times 3 = 6$.
Step 2: $S(y,x) = 3-2 = 1$.
Step 3: $A(M(x,y), S(y,x)) = A(6,1) = 6+1 = 7$.
Step 4: $M(7, x) = 7\times 2 = 14$.
Step 5: $A(y,x) = 3+2 = 5$.
Step 6: $M(14, 5) = 70$.
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Question: 2
What is the value of $S[M(D(A(a, b), 2), D(A(a, b), 2)), M(D(S(a, b), 2), D(S(a, b), 2))]$?
What is the value of $S[M(D(A(a, b), 2), D(A(a, b), 2)), M(D(S(a, b), 2), D(S(a, b), 2))]$?
Show Hint
Recognize patterns like $(a+b)^2 - (a-b)^2 = 4ab$ to simplify faster.
Updated On: Aug 6, 2025
- $a^2 + b^2$
- $ab$
- $a^2 - b^2$
- $\frac{a}{b}$
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The Correct Option is C
Solution and Explanation
$A(a,b) = a+b$, $S(a,b) = a-b$, $D(x,y) = x/y$, $M(x,y) = xy$.
$D(A(a,b), 2) = (a+b)/2$. Multiplying same: $M((a+b)/2, (a+b)/2) = (a+b)^2 / 4$.
Similarly, $S(a,b) = a-b$, $D(S(a,b),2) = (a-b)/2$, multiply same: $(a-b)^2/4$.
$S[(a+b)^2/4, (a-b)^2/4] = \frac{(a+b)^2 - (a-b)^2}{4} = \frac{4ab}{4} = a^2 - b^2$.
$D(A(a,b), 2) = (a+b)/2$. Multiplying same: $M((a+b)/2, (a+b)/2) = (a+b)^2 / 4$.
Similarly, $S(a,b) = a-b$, $D(S(a,b),2) = (a-b)/2$, multiply same: $(a-b)^2/4$.
$S[(a+b)^2/4, (a-b)^2/4] = \frac{(a+b)^2 - (a-b)^2}{4} = \frac{4ab}{4} = a^2 - b^2$.
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