Question

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

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

Answer 1

Answer: (A)

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