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

Answer: (B)

$63qf (0)-p^2 = 0$

Answer 2

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,
$\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} \frac{(3^7+3x)^{\frac{1}{7}}-3}{(729+qx)^{\frac{1}{3}}-9}$
$\lim_{{x \to 0}} \frac{3[(1+\frac{x}{3^6})^{\frac{1}{7}}-1]}{9[(1+\frac{q}{729})^{\frac{1}{3}}-1]}$
$= \frac{1}{3}.\frac{\frac{1}{7}.\frac{1}{3^6}}{\frac{1}{3}.\frac{q}{729}}$
$∴ f(0) = \frac{1}{7q}$