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

The correct answer is (B) : \(63qf (0)-p^2 = 0\)
\(f(x) = \frac{{^{\sqrt[7]{p(729 + x)-3}}}}{{^{\sqrt[3]{729 + qx} - 9}}}\)
For continuity at x = 0, \(\lim_{{x \to 0}} f(x) = f(0)\)
Now,
\(\therefore \lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} \frac{{\sqrt[7]{p(729+x)} - 3}}{{\sqrt[3]{729 + qx }- 9}}\)
⇒ p = 3 (To make indeterminant form)
So,
\(\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} \frac{(3^7+3x)^{\frac{1}{7}}-3}{(729+qx)^{\frac{1}{3}}-9}\)
\(\lim_{{x \to 0}} \frac{3[(1+\frac{x}{3^6})^{\frac{1}{7}}-1]}{9[(1+\frac{q}{729})^{\frac{1}{3}}-1]}\)
\(= \frac{1}{3}.\frac{\frac{1}{7}.\frac{1}{3^6}}{\frac{1}{3}.\frac{q}{729}}\)
\(∴ f(0) = \frac{1}{7q}\)