The function f(x) = x + cos x is | vedclass

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

Question

(A) Given the function $f(x) = x + \cos x$. To determine the nature of the function,we find its derivative with respect to $x$: $f'(x) = \frac{d}{dx}(

Vedclass · Accepted Answer

Always increasing

Vedclass · Answer

(A) Given the function $f(x) = x + \cos x$. To determine the nature of the function,we find its derivative with respect to $x$: $f'(x) = \frac{d}{dx}(x + \cos x) = 1 - \sin x$. We know that for any real number $x$,the range of $\sin x$ is $[-1, 1]$. Therefore,$1 - \sin x \ge 0$ for all $x \in \mathbb{R}$. Specifically,$f'(x) = 0$ only at points where $\sin x = 1$ (i.e.,$x = 2n\pi + \frac{\pi}{2}$),and $f'(x) > 0$ elsewhere. Since $f'(x) \ge 0$ for all $x$ and does not vanish on any interval,the function $f(x)$ is strictly increasing on $\mathbb{R}$.

The function $f(x) = x + \cos x$ is

Similar Questions

The function $f(x) = x^x$ is increasing when:

If $f(x) = x \cdot e^{x(1-x)}$,then $f(x)$ is

In which interval is the function $f(x) = \frac{x}{\ln x}$ increasing?

For which value$(s)$ of $a$,$f(x) = -x^3 + 4ax^2 + 2x - 5$ is decreasing for every $x$?

Prove that $y = \frac{4 \sin \theta}{2 + \cos \theta} - \theta$ is an increasing function of $\theta$ in the interval $\left[0, \frac{\pi}{2}\right].$

Vedclass Test Series

Exam Paper Generator

Online Exam Module