Which one of the following functions is not c | vedclass

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

Question

(D) We analyze the continuity of each function on the interval $(0, \pi )$: $1$. $f(x) = \cot x = \frac{\cos x}{\sin x}$. This function is discontinuo

Vedclass · Accepted Answer

$l(x) = \begin{cases} x \sin x & 0 < x \le \frac{\pi}{2} \ \frac{\pi}{2} \sin(x + \pi) & \frac{\pi}{2} < x < \pi \end{cases}$

Vedclass · Answer

(D) We analyze the continuity of each function on the interval $(0, \pi )$: $1$. $f(x) = \cot x = \frac{\cos x}{\sin x}$. This function is discontinuous at $x = \pi /2$ is incorrect,it is discontinuous at $x = n\pi$. Within $(0, \pi )$,$\sin x$ is never $0$,so $f(x)$ is continuous. $2$. $g(x) = \int_{0}^{x} t \sin \frac{1}{t} \, dt$. Since the integrand $t \sin(1/t)$ is bounded and continuous on $(0, \pi )$,the integral function $g(x)$ is continuous on $(0, \pi )$. $3$. $h(x)$ is defined piecewise. At $x = 3\pi /4$,the left-hand limit is $1$. The right-hand limit is $2 \sin(\frac{2}{9} \cdot \frac{3\pi}{4}) = 2 \sin(\frac{\pi}{6}) = 2(1/2) = 1$. Since the limits match,$h(x)$ is continuous. $4$. $l(x)$ is defined piecewise. At $x = \pi /2$,the left-hand limit is $\frac{\pi}{2} \sin(\frac{\pi}{2}) = \frac{\pi}{2}$. The right-hand limit is $\frac{\pi}{2} \sin(\frac{\pi}{2} + \pi) = \frac{\pi}{2} \sin(\frac{3\pi}{2}) = \frac{\pi}{2}(-1) = -\frac{\pi}{2}$. Since $\frac{\pi}{2} 
eq -\frac{\pi}{2}$,$l(x)$ is discontinuous at $x = \pi /2$.

Which one of the following functions is not continuous on $(0, \pi )$?

Similar Questions

For real $x$ with $-10 \leq x \leq 10$,define $f(x) = \int_{-10}^x 2^{[t]} dt$,where for a real number $r$,we denote by $[r]$ the greatest integer less than or equal to $r$. The number of points of discontinuity of $f$ in the interval $(-10, 10)$ is

If $f(x) = \begin{cases} k, & \text{for } x = 1 \\ \frac{(9x-1)(\sqrt{x}-1)}{3x^2+2x-5}, & \text{for } x \neq 1 \end{cases}$ is continuous on $[0, \infty)$, then $k =$

If $f(x) = \begin{cases} \frac{(8-2x)^{\frac{1}{3}} - 2}{3 - (243+5x)^{\frac{1}{5}}}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$, then $k =$

If the function $f$ defined as $f(x) = \frac{1}{x} - \frac{k - 1}{e^{2x} - 1}$,$x \neq 0$,is continuous at $x = 0$,then the ordered pair $(k, f(0))$ is equal to?

Let $f: R \rightarrow R$ be a continuous function satisfying $f(0)=1$ and $f(2x)-f(x)=x$ for all $x \in R$. If $\lim_{n \rightarrow \infty} \{f(x)-f(\frac{x}{2^n})\} = G(x)$,then $\sum_{r=1}^{10} G(r^2)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module