On which of the following intervals is the fu | vedclass

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

Question

(D) To determine where the function $f(x) = x^{100} + \sin x - 1$ is strictly increasing,we find its derivative:$f'(x) = 100x^{99} + \cos x$For the in

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

(D) To determine where the function $f(x) = x^{100} + \sin x - 1$ is strictly increasing,we find its derivative:$f'(x) = 100x^{99} + \cos x$For the interval $(0, 1)$:Since $x \in (0, 1)$,$x^{99} > 0$ and $\cos x > 0$. Thus,$f'(x) = 100x^{99} + \cos x > 0$.For the interval $(0, \pi / 2)$:Since $x \in (0, \pi / 2)$,$x^{99} > 0$ and $\cos x > 0$. Thus,$f'(x) = 100x^{99} + \cos x > 0$.For the interval $(\pi / 2, \pi )$:Here,$x > \pi / 2 \approx 1.57$,so $x^{99}$ is very large. $100x^{99} > 100(1.57)^{99}$,while $\cos x$ lies between $-1$ and $0$. Therefore,$100x^{99} + \cos x > 0$.Since the derivative is positive in all the given intervals,the function is strictly increasing in all of them.Therefore,the correct option is $D$.

On which of the following intervals is the function $f(x) = x^{100} + \sin x - 1$ strictly increasing?

Similar Questions

If $f(x) = kx^3 - 9x^2 + 9x + 3$ is monotonically increasing in each interval,then

Let $f(x) = \int\limits_1^x {\left( {t\ln(t) - \frac{{\ln(t)}}{t}} \right)dt}$ for $x > 1$. Then:

If $2f(x) + 3f\left(\frac{1}{x}\right) = x^2 + 1, x \neq 0$ and $y = 5x^2 f(x)$,then $y$ is strictly increasing in

If $5 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2, \forall x \neq 0$ and $y=9 x^2 f(x)$,then $y$ is strictly increasing in :

Find the intervals in which the function $f(x) = 10 - 6x - 2x^2$ is strictly increasing or strictly decreasing.

Vedclass Test Series

Exam Paper Generator

Online Exam Module