Question

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

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

Answer 1

Answer: (B)

$f(0)=\underset{x\to 0}{\lim }f(x)$
Limit should be $\frac{0}{0}$ form
So, $\sqrt[7]{p.729}-3=0\Rightarrow$ p. $3^6=3^7\Rightarrow p=3$
Now, $f(0)=\underset{x\to 0}{\lim }\frac{\sqrt[7]{3\left(3^6+x\right)}-3}{\sqrt[3]{3^6+qx}-9}$
$=\underset{x\to 0}{\lim }\frac{3\left[{\left(1+\frac{x}{3^6}\right)}^{1/7}-1\right]}{9\left[{\left(1+\frac{qx}{3^6}\right)}^{1/3}-1\right]}=\frac{3}{9}\times \frac{\frac{1}{7.3^6}}{\frac{q}{3.3^6}}$
$\Rightarrow f(0)=\frac{1}{3}\times \frac{3}{7q}=\frac{1}{7q}$
$\Rightarrow 7qf(0)-1=0$
$\Rightarrow 7\cdot p^2\cdot qf(0)-p^2=0\text{ (for option) }$
$\Rightarrow 63qf(0)-p^2=0$