Let [t] represent the greatest integer not mo | vedclass

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

Question

(B) Let $g(x) = x^{1/x}$. We analyze the behavior of $g(x)$ for $x \in (0, \infty)$.Taking the natural logarithm,$\ln(g(x)) = \frac{\ln(x)}{x}$.Let $h

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let $g(x) = x^{1/x}$. We analyze the behavior of $g(x)$ for $x \in (0, \infty)$.Taking the natural logarithm,$\ln(g(x)) = \frac{\ln(x)}{x}$.Let $h(x) = \frac{\ln(x)}{x}$. Then $h'(x) = \frac{1 - \ln(x)}{x^2}$.The function $h(x)$ has a maximum at $x = e$,where $h(e) = \frac{1}{e} \approx 0.367$.Thus,$g(x) = e^{h(x)}$ has a maximum at $x = e$,where $g(e) = e^{1/e} \approx e^{0.367} \approx 1.44$.As $x 	o 0^+$,$g(x) 	o 0$,and as $x 	o \infty$,$g(x) 	o 1$.The range of $g(x)$ is $(0, e^{1/e}]$.The greatest integer function $[g(x)]$ is discontinuous whenever $g(x)$ takes an integer value.Since the range is $(0, 1.44]$,the only integer value $g(x)$ can take is $1$.$x^{1/x} = 1$ implies $x = 1$.At $x = 1$,$f(1) = [1^{1/1}] = [1] = 1$.For $x$ slightly less than $1$,$x^{1/x} For $x$ slightly greater than $1$,$x^{1/x} > 1$ (but less than $1.44$),so $[x^{1/x}] = 1$.Since the left-hand limit and right-hand limit are different at $x = 1$,the function is discontinuous at $x = 1$.Thus,there is only $1$ point of discontinuity.

Let $[t]$ represent the greatest integer not more than $t$. Then the number of discontinuous points of $f(x) = [x^{1/x}]$ in $(0, \infty)$ is

Similar Questions

If $f(x) = \begin{cases} x \frac{e^{(1/x)} - e^{(-1/x)}}{e^{(1/x)} + e^{(-1/x)}}, & x \ne 0 \\ 0, & x = 0 \end{cases}$ then which of the following is true?

Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx^2, & \text{if } x \le 2 \\ 3, & \text{if } x > 2 \end{cases}$ at $x=2$.

If $f(x) = \begin{cases} \frac{1 - \sin x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ \lambda, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then the value of $\lambda$ is:

$f(x) = \begin{cases} \frac{(2x^2 - ax + 1) - (ax^2 + 3bx + 2)}{x + 1} & ; x \neq -1 \\ k & ; x = -1 \end{cases}$ is a real-valued function. If $a, b, k \in R$ and $f$ is continuous on $R$,then $k =$

Let $a$ be a positive real number. If a real valued function $f(x) = \begin{cases} \frac{6^x-3^x-2^x+1}{1-\cos \left(\frac{x}{a}\right)} & \text{if } x \neq 0 \\ \log 3 \log 4 & \text{if } x=0 \end{cases}$ is continuous at $x=0$,then $a=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module