Let f(1) = g(1) = k and their n^{th} derivati | vedclass

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

Question

(A) Given the limit: $\lim _{x \rightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$. Since $f(1) = g(1) = k$,the expression bec

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given the limit: $\lim _{x ightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$. Since $f(1) = g(1) = k$,the expression becomes $\lim _{x ightarrow 1} \frac{k g(x) - k - k f(x) + k}{g(x) - f(x)} = \lim _{x ightarrow 1} \frac{k(g(x) - f(x))}{g(x) - f(x)}$. As $x ightarrow 1$,$g(x) - f(x) ightarrow g(1) - f(1) = k - k = 0$,which is an indeterminate form of type $\frac{0}{0}$. Applying $L'	ext{Hospital's Rule}$: $\lim _{x ightarrow 1} \frac{k g'(x) - k f'(x)}{g'(x) - f'(x)} = 4$. $\lim _{x ightarrow 1} \frac{k(g'(x) - f'(x))}{g'(x) - f'(x)} = 4$. Since $g'(1) 
eq f'(1)$ (as given that derivatives are not equal),we can cancel the term $(g'(x) - f'(x))$. Thus,$k = 4$.

Let $f(1) = g(1) = k$ and their $n^{th}$ derivatives $f^{(n)}(1), g^{(n)}(1)$ exist and are not equal for some $n$. If $\lim _{x \rightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$,then the value of $k$ is:

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}{x} = $

The value of $\lim _{x \rightarrow 0} \frac{1}{x^3} \int_0^x \frac{t \ln (1+t)}{t^4+4} dt$ is

The value of $\mathop {\lim }\limits_{x \to \infty } \frac{{\log x}}{{{x^n}}}, \; n > 0$ is

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^{ - 1}}x}}{x}$ is

Let $[x]$ denote the greatest integer less than or equal to $x$ and $k \geq 2$ be an integer. Then $\lim_{x \rightarrow k} \frac{\sin \left(2 \pi\left([x]-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module