Question

The graph of $y=f(x)$ has a unique tangent of finite non zero slope at $\left({\pi }^3,0\right)$. If $A=\underset{x\to {\pi }^3}{\mathrm{Lim}}\frac{\ln (1+9f(x))-\sin (f(x))}{2f(x)}$, then $\sum _{n=1}^{\infty }{A}^{-n}$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. 4

Answer 1

Answer: (A)

$\underset{x\to {\pi }^3}{\mathrm{Lim}}\frac{\ln (1+9f(x))-\sin (f(x))}{2f(x)}\left(\frac{0}{0}\right)$
$=\underset{x\to {\pi }^3}{\mathrm{Lim}}\frac{\frac{9f^{\prime }(x)}{1+9f(x)}-\cos (f(x))\cdot f^{\prime }(x)}{2f^{\prime }(x)}=\frac{9f^{\prime }\left({\pi }^3\right)-f^{\prime }\left({\pi }^3\right)}{2f^{\prime }\left({\pi }^3\right)}=4=A$
$\therefore \sum _{n=1}^{\infty }{4}^{-n}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}$