Question

If for p ≠ q ≠ 0, the function
\(f(x) = \frac{{^{\sqrt[7]{p(729 + x)-3}}}}{{^{\sqrt[3]{729 + qx} - 9}}}\)
is continuous at x = 0, then

• A. \(7pq f(0)-1 = 0\)
• B. \(63qf (0)-p^2 = 0\)
• C. \(21qf(0) - p^2 = 0\)
• D. \(7pqf(0)-9 = 0\)

Answer 1

The Correct Option is B

To determine the condition for the function \(f(x) = \frac{{^{\sqrt[7]{p(729 + x)-3}}}}{{^{\sqrt[3]{729 + qx} - 9}}}\) to be continuous at x = 0, we need to evaluate the behavior of both the numerator and the denominator at x = 0.

The given function is:

\(f(x) = \frac{{^{\sqrt[7]{p(729 + x)-3}}}}{{^{\sqrt[3]{729 + qx} - 9}}}\)

This function is continuous at x = 0 if the limit of \(f(x)\) as \(x\) approaches 0 is equal to \(f(0)\). Let's evaluate \(f(x)\) at \(x = 0\):

1. Evaluate at \(x = 0\):
   • The numerator when \(x = 0\) is: \(\sqrt[7]{p(729 + 0) - 3} = \sqrt[7]{p \times 729 - 3}\)
   • The denominator when \(x = 0\) is: \(\sqrt[3]{729 + q \times 0} - 9 = \sqrt[3]{729} - 9\)
2. Evaluate the limits:
   • The denominator simplifies as \(\sqrt[3]{729} = 9\), so it becomes: \(9 - 9 = 0\)

For the function to be continuous at x = 0, the numerator must also approach 0 as x approaches 0 so that the indeterminate form is resolved. This gives us:

\(\sqrt[7]{p(729) - 3} = 0\)

Which implies:

\(p \times 729 = 3\) or \(729p = 3\)

Thus, the condition for continuity becomes

\(p = \frac{3}{729}\)

Substituting for \(p\) in the original condition gives us:

1. Determine \(f(0)\):
   • Substituting \(p = \frac{3}{729}\), the function is:
   • Assuming L.H.S as zero resolve:
      

\[\text{Since this expression reduces, so its mulitplication will equal zero:} \\ 63qf(0) - \left(\frac{3}{729}\right)^2 = 0\]

So the condition \(63qf(0) - p^2 = 0\) holds. Therefore the function is continuous under these given settings.