Question:

<div class="solution"><p>(P): lim[f(3+h²)-f(3-h²)]/(2h²). Let u=h²: = f'(3)·2h²/(2h²) via L'H approach. More precisely: [f(3+h²)-f(3)]/(h²) + [f(3)-f(3-h²)]/(h²) → f'(3)+f'(3)=4. But divided by 2: = 2. Wait — actually the expression = [f(3+h²)-f(3-h²)]/(2h²) → f'(3)=2. So P→(3) 2.</p><p>(Q): f even, f'(0) exists. Since f(-x)=f(x), differentiating: -f'(-x)=f'(x). At x=0: -f'(0)=f'(0)⟹f'(0)=0. Q→(1) 0.</p><p>(R): Lf'(0)=lim(x→0⁻)[f(x)-f(0)]/x = lim x/(1+e^{1/x})/x = lim 1/(1+e^{1/x}). As x→0⁻: 1/x→-∞, e^{1/x}→0. So Lf'(0)=1/(1+0)=1. R→(2) 1.</p><p>(S): f=max{a-x, a+x, b}. The three functions intersect: a-x=a+x⟹x=0; a-x=b⟹x=a-b<0 (since b>a); a+x=b⟹x=b-a>0. So non-diff at x=0, x=a-b, x=b-a → 3 points. S→(4) 3.</p><p>Matching: P→3(2), Q→1(0), R→2(1), S→4(3). This corresponds to option (C): P→3, Q→1, R→2, S→3.</p><p><strong>Answer: (C)</strong></p><div class="key-concept"><strong>Key Concept:</strong> Symmetric/asymmetric derivatives, max function corners</div></div>