If x in (0, 1),then what type of function is | vedclass

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

Question

(A) Given the function $f(x) = x^{100} + \sin x - 1$. To determine the nature of the function,we find its derivative $f'(x)$. $f'(x) = \frac{d}{dx}(x^

Vedclass · Accepted Answer

Monotonically increasing

Vedclass · Answer

(A) Given the function $f(x) = x^{100} + \sin x - 1$. To determine the nature of the function,we find its derivative $f'(x)$. $f'(x) = \frac{d}{dx}(x^{100} + \sin x - 1) = 100x^{99} + \cos x$. For $x \in (0, 1)$,we observe that $x^{99} > 0$,so $100x^{99} > 0$. Also,for $x \in (0, 1)$,$\cos x > 0$ because $x$ is in the first quadrant. Since both terms $100x^{99}$ and $\cos x$ are positive for all $x \in (0, 1)$,their sum $f'(x) = 100x^{99} + \cos x > 0$. Since $f'(x) > 0$ for all $x \in (0, 1)$,the function $f(x)$ is monotonically increasing on the interval $(0, 1)$.

If $x \in (0, 1)$,then what type of function is $f(x) = x^{100} + \sin x - 1$?

Similar Questions

For what values of $x$ is the function $f(x) = \tan^{-1}(\sin x + \cos x)$ strictly increasing?

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

Which of the following functions are strictly decreasing on $\left(0, \frac{\pi}{2}\right)?$

Assertion $(A)$: The function $f(x) = x - \log \left(\frac{1+x}{x}\right), x > 0$ has no maximum. Reason $(R)$: If a function $f(x)$ is strictly increasing in an interval $(a, b)$, then at any point in $(a, b)$, $f^{\prime}(x) \neq 0$. The correct option among the following is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module