$f(x) = \cos x - 1 + \frac{x^2}{2!}, x \in R$. Then $f(x)$ is
- Adecreasing function
- Bincreasing function
- Cneither increasing nor decreasing
- Dconstant $\forall x > 0$
Explore More
Similar Questions
The value of $K$ such that $f(x) = \sin x - \cos x - Kx + 5$ decreases for all positive real values of $x$ is given by
MediumWBJEE 2017
View SolutionLet $f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}+x^{3}+x+19$. Then,$f(x)=0$ has
The interval in which the function $f(x) = 2x^2 - \log x$,for $x > 0$ decreases is
The function $f(x) = \frac{x}{x^2-6x-16}$,where $x \in \mathbb{R} - \{-2, 8\}$,
DifficultJEE MAIN 2024
View SolutionThe function $f(x)=\log (1+x)-\frac{2 x}{2+x}$ is increasing on
MediumKCET 2022
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo