Let f: R rightarrow R be a differentiable fun | vedclass

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

Question

(A) For $g(x)$ to be continuous at $x=3$,we must have $K = \lim_{x 	o 3} g(x) = \lim_{x 	o 3} \int_3^{f(x)} \frac{3t^2}{x-3} dt$. Since the integral

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) For $g(x)$ to be continuous at $x=3$,we must have $K = \lim_{x 	o 3} g(x) = \lim_{x 	o 3} \int_3^{f(x)} \frac{3t^2}{x-3} dt$. Since the integral is of the form $\frac{0}{0}$ as $x 	o 3$ (because $f(3)=3$),we apply $L$'$H$ôpital's Rule and Leibniz's Rule for differentiation under the integral sign. Let $I(x) = \int_3^{f(x)} 3t^2 dt$. Then $g(x) = \frac{I(x)}{x-3}$. By $L$'$H$ôpital's Rule,$K = \lim_{x 	o 3} \frac{\frac{d}{dx} \int_3^{f(x)} 3t^2 dt}{\frac{d}{dx} (x-3)}$. Using Leibniz's Rule,$\frac{d}{dx} \int_3^{f(x)} 3t^2 dt = 3(f(x))^2 \cdot f^{\prime}(x)$. Thus,$K = \lim_{x 	o 3} \frac{3(f(x))^2 \cdot f^{\prime}(x)}{1} = 3(f(3))^2 \cdot f^{\prime}(3)$. Substituting the given values $f(3)=3$ and $f^{\prime}(3)=\frac{1}{27}$: $K = 3 \cdot (3)^2 \cdot \frac{1}{27} = 3 \cdot 9 \cdot \frac{1}{27} = 27 \cdot \frac{1}{27} = 1$.

Let $f: R \rightarrow R$ be a differentiable function such that $f(3)=3$ and $f^{\prime}(3)=\frac{1}{27}$. If $g(x)=\begin{cases} \int_3^{f(x)} \frac{3t^2}{x-3} dt & \text{if } x \neq 3 \\ K & \text{if } x=3 \end{cases}$ is continuous at $x=3$,then $K=$

Similar Questions

If $a$ and $b$ $(a > b)$ are points of discontinuity of the function $f(x) = \begin{cases} 3-2x^2, & \text{for } x \leq 0 \\ 2x+3, & \text{for } 0 < x \leq 1 \\ 2x^2-3x, & \text{for } 1 < x < 2 \\ 2x-3, & \text{for } 2 \leq x < 3 \\ |x|, & \text{for } x \geq 3 \end{cases}$,then $3a-b = $

Let $f:[0, \infty) \rightarrow [0, \infty)$ be defined as $f(x) = \int_{0}^{x} [y] \, dy$,where $[x]$ is the greatest integer less than or equal to $x$. Which of the following is true?

$f(x) = \begin{cases} [x] + [-x], & \text{when } x \neq 2 \\ \lambda, & \text{when } x = 2 \end{cases}$
If $f(x)$ is continuous at $x = 2$, the value of $\lambda$ will be

If the function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1}, & x \neq 2 \\ k, & x = 2 \end{cases}$ is right continuous at $x = 2$, then $k =$

If in the interval $[0,3]$,$f(x) = \begin{cases} x\{x\}^2, & x \notin I \\ x, & x \in I \end{cases}$,then which of the following statements is correct? (where $\{.\}$ denotes the fractional part function)

Vedclass Test Series

Exam Paper Generator

Online Exam Module