Let f(x) = cos left(frac{pi}{x}right), x neq | vedclass

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

Question

(A) Given $f(x) = \cos \left(\frac{\pi}{x}\right)$.Taking the derivative with respect to $x$:$f'(x) = -\sin \left(\frac{\pi}{x}\right) \cdot \left(-\f

Vedclass · Accepted Answer

$f(x)$ increases in the interval $\left(\frac{1}{2k+1}, \frac{1}{2k}\right)$

Vedclass · Answer

(A) Given $f(x) = \cos \left(\frac{\pi}{x}ight)$.Taking the derivative with respect to $x$:$f'(x) = -\sin \left(\frac{\pi}{x}ight) \cdot \left(-\frac{\pi}{x^2}ight) = \frac{\pi}{x^2} \sin \left(\frac{\pi}{x}ight)$.For $f(x)$ to be increasing, $f'(x) > 0$.Since $\frac{\pi}{x^2} > 0$ for all $x 
eq 0$, we need $\sin \left(\frac{\pi}{x}ight) > 0$.This occurs when $2k\pi Dividing by $\pi$, we get $2k Taking the reciprocal, the inequality reverses: $\frac{1}{2k+1} Thus, $f(x)$ increases in the interval $\left(\frac{1}{2k+1}, \frac{1}{2k}ight)$.For $f(x)$ to be decreasing, $f'(x) This occurs when $(2k+1)\pi Dividing by $\pi$, we get $2k+1 Taking the reciprocal, we get $\frac{1}{2k+2} Thus, $f(x)$ decreases in the interval $\left(\frac{1}{2k+2}, \frac{1}{2k+1}ight)$.

Let $f(x) = \cos \left(\frac{\pi}{x}\right), x \neq 0$. Assuming $k$ is an integer, which of the following is true?

Similar Questions

The set of all real values of $a$ such that the real valued function $f(x) = x^3 + 2ax^2 + 3(a+1)x + 5$ is strictly increasing in its entire domain is

If $f(x) = \int_{x^2}^{x^2 + 1} e^{-t^2} dt$,then $f(x)$ increases in

Define $f(x) = \frac{1}{2}[|\sin x| + \sin x]$, $0 < x \leq 2\pi$. Then, $f$ is

The interval in which the function $f(x) = {x^2}{e^{ - x}}$ is non-decreasing is:

If $f(x) = x^{3/2}(3x - 10)$,$x \geq 0$,then in which interval is $f(x)$ an increasing function?

Vedclass Test Series

Exam Paper Generator

Online Exam Module