Question:

<div class="solution"><p>(A) f={2x²+3, x≤1; 3x+2, x>1}: at x=1: f(1⁻)=5, f(1⁺)=5 ✓ continuous. f'(1⁻)=4, f'(1⁺)=3. Not differentiable, but IS continuous. Statement (A) says "neither differentiable nor continuous" — FALSE.</p><p>(B) f(x)=x²|x|: f'(x)=3x|x|/... at x=0: f'(0)=lim h|h|/h=lim|h|=0. f''(0)=lim h|h|/h²=... wait, twice differentiable? f'(x)=x²·sign(x)+|x|·2x=3x|x|. Then f''(0)=lim 3h|h|/h = lim 3|h|=0. Yes, f is twice differentiable at x=0. (B) TRUE ✓</p><p>(C) f continuous at x=5, f(5)=2: lim_{x→2}f(4x²-11) = f(lim 4x²-11|_{x=2}) = f(5) = 2. TRUE ✓</p><p>(D) lim(f+g)=2, lim(f-g)=1: f→3/2, g→1/2. lim(fg)=3/4. Statement says "need not exist" — FALSE, it must equal 3/4.</p><p><strong>Answer: (B),(C)</strong></p><div class="key-concept"><strong>Key Concept:</strong> Twice differentiability requires both f' and f'' to exist at the point</div></div>