Let f: R rightarrow (0, infty) be a twice dif | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $L = \lim_{x \rightarrow 1} \log_{e} \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-1)^{2}}} = \lim_{x \rightarrow 1} \frac{18}{(x-1)^{2}} \

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Let $L = \lim_{x \rightarrow 1} \log_{e} \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-1)^{2}}} = \lim_{x \rightarrow 1} \frac{18}{(x-1)^{2}} \ln \left( \frac{f(x+2)}{f(3)} \right)$.Since $f(3) = 18$, we have $\lim_{x \rightarrow 1} \frac{f(x+2)}{f(3)} = \frac{f(3)}{f(3)} = 1$.This is a $\frac{0}{0}$ form. Using the Taylor expansion $f(x+2) \approx f(3) + f'(3)(x-1) + \frac{f''(3)}{2}(x-1)^2$ near $x=1$:$f(x+2) \approx 18 + 0(x-1) + \frac{4}{2}(x-1)^2 = 18 + 2(x-1)^2$.Thus, $\frac{f(x+2)}{f(3)} \approx 1 + \frac{2(x-1)^2}{18} = 1 + \frac{(x-1)^2}{9}$.Using $\ln(1+u) \approx u$ for small $u$, where $u = \frac{(x-1)^2}{9}$:$L = \lim_{x \rightarrow 1} \frac{18}{(x-1)^{2}} \cdot \frac{(x-1)^2}{9} = \frac{18}{9} = 2$.

List-$I$	List-$II$
$A. \frac{d}{dx}\left(\tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\right)$	$(i) \log(x+\sqrt{1+x^2})$
$B. \frac{d}{dx}\left(\frac{3+\|x-1\|}{3x+4}\right)$	$(ii) -\frac{4x}{(1+x^2)^2}$
$C. \sinh^{-1} x$	$(iii) \frac{1}{2}$
$D. \frac{d^2}{dx^2}\left(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right)$	$(iv) \frac{1}{\sqrt{1+x^2}}$
	$(v) \text{not differentiable at } x=1$

Let $f: R \rightarrow (0, \infty)$ be a twice differentiable function such that $f(3) = 18$, $f'(3) = 0$, and $f''(3) = 4$. Then $\lim_{x \rightarrow 1} \left( \log_{e} \left( \frac{f(x+2)}{f(3)} \right)^{\frac{18}{(x-1)^{2}}} \right)$ is equal to:

Similar Questions

Let $f(x)=x^2+a x+b$,where $a, b \in R$. If $f(x)=0$ has all its roots imaginary,then the roots of $f(x)+f^{\prime}(x)+f^{\prime \prime}(x)=0$ are

Let $a_1, a_2, \ldots, a_{100}$ be non-zero real numbers such that $a_1+a_2+\ldots+a_{100}=0$. Then,

Let $R$ be the set of all real numbers and $f(x) = \sin^{10} x (\cos^8 x + \cos^4 x + \cos^2 x + 1)$ for $x \in R$. Let $S = \{\lambda \in R : \text{there exists a point } c \in (0, 2\pi) \text{ with } f'(c) = \lambda f(c)\}$. Then,

At the point $x=1$,the function $f(x) = \begin{cases} x^3-1, & 1 < x < \infty \\ x-1, & -\infty < x \leq 1 \end{cases}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module