Discussion
A function, \(\lambda\), is defined by
\[ \lambda(p,q) = \begin{cases} (p-q)^{2}, & \text{if } p \geq q, \\ p+q, & \text{if } p \lt q. \end{cases} \]
The value of the expression \(\frac{\lambda(-(-3+2),(-2+3))}{(-(-2+1))}\) is:
Answer: B
The value is 0.
Simplify the arguments first: -(-3 + 2) = 1 and (-2 + 3) = 1. So the numerator is \(\lambda(1, 1)\), and since p = q the first case applies:
\[ \lambda(1,1) = (1 - 1)^{2} = 0 \]
The denominator is -(-2 + 1) = 1, so the expression is 0/1 = 0.
The trap is the case split: with p = q the rule 'if p ≥ q' applies, so the square is used, not the sum. Using the sum by mistake would give 2, which is not offered, and none of -1, 16/3 or 16 can arise once both arguments are seen to equal 1.
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