Question

If \(f(x) = \left\{ \begin{array}{cc}\frac{(8-2x)^{\frac{1}{3}}-2}{3-(243+5x)^{\frac{1}{5}}}, & \text{if }x\neq 0 \\ k, & \text{if }x = 0\end{array} \right.\) is continuous at \(x = 0\), then \(k =\)

• A. \(\frac{5}{2}\)
• B. \(-\frac{5}{2}\)
• C. \(\frac{27}{2}\)
• D. \(\frac{-27}{2}\)

Hint:

Both parts tend to 0, so use the derivative of each root expression at 0 and divide.

Answer 1

The Correct Option is C

Step 1: Continuity condition:
For \(f\) to be continuous at \(x=0\), \(k = \lim_{x\to0}\frac{(8-2x)^{1/3}-2}{3-(243+5x)^{1/5}}\). Both top and bottom go to \(0\), so this is a \(0/0\) form.

Step 2: Use the derivative idea:
Near \(x=0\), the top is \(g(x)-g(0)\) with \(g(x) = (8-2x)^{1/3}\), and the bottom is \(-(h(x)-h(0))\) with \(h(x) = (243+5x)^{1/5}\). Divide both by \(x\) and let \(x\to0\), so the limit becomes \(\frac{g'(0)}{-h'(0)}\).

Step 3: Find the derivatives:
\(g'(x) = \frac13(8-2x)^{-2/3}(-2)\), so \(g'(0) = \frac13\cdot\frac14\cdot(-2) = -\frac16\).
\(h'(x) = \frac15(243+5x)^{-4/5}(5)\), so \(h'(0) = 243^{-4/5} = \frac{1}{81}\).

Step 4: Combine:
\[ k = \frac{-1/6}{-1/81} = \frac{81}{6} = \frac{27}{2} \]

Step 5: Check the options:
The negative values come from losing one of the two minus signs. \(5/2\) would need wrong powers. Only \(27/2\) fits.

Final Answer:
\(k = \frac{27}{2}\), option (C). \[ \boxed{\frac{27}{2}} \]