Question:

<div class="solution"><p><strong>Step 1 — RHD at x=0:</strong><br><span class="math-block">\[f'(0^+)=\lim_{x\to 0^+}\frac{f(x)-f(0)}{x}=\lim_{x\to 0^+}\frac{\ln\cos x}{ax^2}\]</span>Using <span class="math-inline">\(\ln\cos x\approx -x^2/2\)</span>: <span class="math-inline">\(\text{RHD}=\frac{-1/2}{a}=-\frac{1}{2a}\)</span></p><p><strong>Step 2 — LHD at x=0:</strong><br><span class="math-block">\[f'(0^-)=\lim_{x\to 0^-}\frac{e^{x^2}-1}{bx^2}=\frac{1}{b}\]</span>(using <span class="math-inline">\(e^u-1\approx u\)</span>)</p><p><strong>Step 3:</strong> For f'(0) to exist: LHD=RHD=1/4.<br><span class="math-inline">\(-\frac{1}{2a}=\frac{1}{4}\implies a=-2\)</span><br><span class="math-inline">\(\frac{1}{b}=\frac{1}{4}\implies b=4\)</span></p><p><span class="math-inline">\(a+b=-2+4=2\)</span></p><p><strong>Answer: (A) a+b=2</strong></p><div class="trap-box"><strong>Trap:</strong> Using L'Hôpital instead of Taylor expansion — both work but Taylor is faster.</div><div class="key-concept"><strong>Key Concept:</strong> f'(0) via definition; use Taylor: ln cos x ≈ -x²/2 and eˣ-1≈x</div></div>