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

If $R$ is the least value of $a$ such that the function $f(x) = x^{2} + ax + 1$ is increasing on $[1, 2]$ and $S$ is the greatest value of $a$ such that the function $f(x) = x^{2} + ax + 1$ is decreasing on $[1, 2]$,then the value of $|R - S|$ is ..... .

If $x > 0$,then $\frac{x}{1+x} - \log(1+x)$

The function $f(x)=x^{2}-2x$ is strictly decreasing in the interval

Let $f(x) = \frac{x}{\sqrt{a^2+x^2}} - \frac{d-x}{\sqrt{b^2+(d-x)^2}}$,$x \in R$,where $a, b, d$ are non-zero real constants. Then

Which of the following is true about $f(x) = 3 \sinh(x) - 2 \cosh(x)$ for all $x \in R$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module