Question

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

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

Answer 1

Answer: (B)

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