Question

Let $f$ be an invertible function defined on $R$ such that its inverse function $f ^{-1}( x )= x ^3+4 x ^2+6 x +1$, then the value of $\underset{x \rightarrow 1}{\text{Lim}} (f(x)+x)^{\frac{1}{x^2+5 x-6}}$ equals

• A. $\frac{1}{ e ^{1 / 6}}$
• B. $e ^{1 / 6}$
• C. $e ^6$
• D. $e ^{-6}$

Answer 1

Answer: (B)

$ f ^{-1}(0)=1 \Rightarrow f (1)=0$
$\therefore L=\underset{x \rightarrow 1}{\text{Lim}} (f(x)+x)^{\frac{1}{x^2+5 x-6}}=e^{\operatorname{Lim}_{x \rightarrow 1} \frac{f(x)+x-1}{x^2+5 x 6}}=e^{\frac{f^{\prime}(1)+1}{7}}$
Now, $f^{\prime}(1)=\frac{1}{\left(f^{-1}\right)^{\prime}(0)}=\frac{1}{6}$
$\Rightarrow L = e ^{1 / 6}$