<p>Let \(f(x) = \begin{cases} \left(\dfrac{2^x + 3^x + 5^x}{3}\right)^{3/x}, & x \neq 0 \\ k, & x = 0 \end{cases}\). If \(f(x)\) is continuous then the value of \(k\) is equal to:</p>
Step-by-Step Solution
Key Concept: For continuity at x=0, evaluate lim(x→0) of the expression using the standard form (a+b+c)^(1/x) which requires recognizing this as a 1^∞ indeterminate form and applying logarithmic limits with L'Hôpital's rule or the exponential-logarithm technique.
<p><strong>Step 1:</strong> For continuity at x=0, we need lim(x→0) f(x) = k.</p><p><strong>Step 2:</strong> Note that as x→0: 2^x → 1, 3^x → 1, 5^x → 1, so the base → (1+1+1)/3 = 1, and exponent → ∞. This is a 1^∞ form.</p><p><strong>Step 3:</strong> Let L = lim(x→0) [(2^x + 3^x + 5^x)/3]^(3/x). Taking natural logarithm:</p><p>ln L = lim(x→0) (3/x)·ln[(2^x + 3^x + 5^x)/3]</p><p>= 3·lim(x→0) [ln(2^x + 3^x + 5^x) - ln 3]/x</p><p><strong>Step 4:</strong> This is 0/0 form. Apply L'Hôpital's rule:</p><p>= 3·lim(x→0) [(2^x·ln2 + 3^x·ln3 + 5^x·ln5)/(2^x + 3^x + 5^x)]/1</p><p><strong>Step 5:</strong> Substitute x=0:</p><p>= 3·[(ln2 + ln3 + ln5)/(1 + 1 + 1)]</p><p>= 3·[(ln(2·3·5))/3]</p><p>= ln 30</p><p><strong>Step 6:</strong> Therefore L = e^(ln 30) = 30</p><p>∴ k = 30 (Answer: D)</p>
Correct Answer: D