<p>Let \(f(x) = \begin{cases} \left[1 + \ln(c^2 + c + 1)\tan^2(x-1)\right]^{\frac{1}{(\ln x)^2}}, & x \neq 1 \\ 3c, & x = 1 \end{cases}\), where \(c \in R\).</p><p>If \(\lim_{x \to 1} f(x)\) exists but \(f(x)\) is discontinuous at \(x = 1\), then \(c\) can take the value:</p>
Step-by-Step Solution
Key Concept: For the limit to exist but f be discontinuous at x=1, we need lim(x→1) f(x) to exist and be finite, but ≠ f(1)=3c. The limit has form 1^∞, so we must make the exponent→0 while the base→1, which requires c²+c+1=1.
<p><strong>Step 1:</strong> As x→1, we have tan(x-1)→0 and ln(x)→0. The expression has form 1^∞.</p><p><strong>Step 2:</strong> For the limit to exist and be finite, we need the exponent ln(x)² to approach 0 while keeping 1 + ln(c²+c+1)tan²(x-1) approaching 1. This requires: c² + c + 1 = 1, so c² + c = 0, giving c = 0 or c = -1.</p><p><strong>Step 3:</strong> When c²+c+1=1, the base becomes 1+0·tan²(x-1)=1 for all x near 1. Thus lim(x→1) f(x) = 1^∞ = 1 (using standard form).</p><p><strong>Step 4:</strong> For f to be discontinuous at x=1, we need f(1) = 3c ≠ 1. Since c=0 gives f(1)=0≠1, or c=-1 gives f(1)=-3≠1, both work. Typically c=0 is the answer.</p><p><strong>Step 5:</strong> Verify: lim(x→1) f(x) = 1 but f(1) = 0, so f is discontinuous (removable type) at x=1.</p><p>∴ Answer: A (c = 0 or c = -1)</p>
Correct Answer: A