f(x) is differentiable on mathbb{R} and f^{pr | vedclass

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

Question

(C) Given the limit expression: $\lim _{x \rightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$.We can rewrite the numerator by adding and subt

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given the limit expression: $\lim _{x ightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$.We can rewrite the numerator by adding and subtracting $m f(m)$:$\lim _{x ightarrow m} \frac{x f(m)-m f(m)+m f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$$\lim _{x ightarrow m} \left[ f(m) \frac{x-m}{x-m} - m \frac{f(x)-f(m)}{x-m} ight] + f^{\prime}(m) = f(m)$$f(m) - m f^{\prime}(m) + f^{\prime}(m) = f(m)$$-m f^{\prime}(m) + f^{\prime}(m) = 0$$f^{\prime}(m)(1-m) = 0$Since it is given that $f^{\prime}(m) 
eq 0$,we must have $1-m = 0$,which implies $m = 1$.

$f(x)$ is differentiable on $\mathbb{R}$ and $f^{\prime}(m) \neq 0, \,m \in \mathbb{R}$. If $\lim _{x \rightarrow m} \frac{x f(m)-m f(x)}{x-m}+f^{\prime}(m)=f(m)$,then $m=$

Similar Questions

Differentiate with respect to $x$,the following function:
$e^{\sec ^{2} x}+3 \cos ^{-1} x$

Let $f: R \rightarrow R$ be a function such that $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6, x \in R$,then $f(2)$ equals

If $y=\log \left(\frac{1+x}{1-x}\right)^{1 / 4}-\frac{1}{2} \tan ^{-1}(x)$,then $\frac{d y}{d x}$ at $x=\frac{1}{\sqrt{2}}$ equals

If the function $f(x)$ is defined by $f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1$,then $f'(0) = $

If $y = x \left[ \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right) \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right) + \sin x \right] + \frac{1}{2\sqrt{x}}$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module