Question:

<div class="solution"><p><strong>Analysis:</strong> Note <span class="math-inline">$\sin x - [\sin x]=\{\sin x\}$</span> (fractional part).</p><p><strong>At x=π/2:</strong> <span class="math-inline">$\sin(\pi/2)=1$</span>, so <span class="math-inline">$[\sin x]=1$</span> and <span class="math-inline">$\{\sin x\}=0$</span>. Near <span class="math-inline">$x=\pi/2$</span>, <span class="math-inline">$\sin x$</span> approaches 1 from below (for <span class="math-inline">$x\ne\pi/2$</span>), so <span class="math-inline">$[\sin x]=0$</span> and <span class="math-inline">$\{\sin x\}=\sin x$</span>. At x=π/2 exactly, f=1+0=1. From left: f→0+1=1. <strong>Continuous</strong> at π/2.</p><p>LHD at π/2: derivative of <span class="math-inline">$\sqrt{\sin x}$</span> requires <span class="math-inline">$\frac{\cos x}{2\sqrt{\sin x}}\to 0/0$</span> — exists. RHD: <span class="math-inline">$f=1+0=1$</span> const, derivative=0. Check: both sides give 0. <strong>Differentiable</strong> at π/2.</p><p><strong>At x=π:</strong> <span class="math-inline">$\sin\pi=0$</span>. Near π, <span class="math-inline">$\sin x$</span> changes sign. For x just less than π: sin x>0, so [sin x]=0, f=√(sin x). For x just greater than π: sin x<0, so [sin x]=-1, f=-1+√(sin x+1) — but sin x+1 near π: sin x≈ sin(π+ε)=-sin ε≈-ε, so sin x+1≈1-ε, f≈-1+1=0. LHD and RHD both give derivative 0 via L'Hôpital. Actually: LHD=cos(π)/(2√(sin x))|... this is -1/(2√0)→-∞. <strong>Not differentiable at x=π</strong>. Also discontinuous? Check: from left f→0, from right f→0, f(π)=-1+1=0. Continuous. But LHD→-∞. <strong>Not differentiable at π but continuous</strong>.</p><p><strong>Answer: (B) differentiable at x=π/2 but not at x=π</strong></p><div class="trap-box"><strong>Trap:</strong> Not carefully checking fractional part behaviour as sin x approaches an integer from different sides.</div><div class="key-concept"><strong>Key Concept:</strong> Fractional part {sin x} has corners wherever sin x is an integer</div></div>