If f(x) = sin x - cos x, the function is decr | vedclass

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

Question

(D) Given $f(x) = \sin x - \cos x.$First,find the derivative $f'(x)$:$f'(x) = \cos x + \sin x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Given $f(x) = \sin x - \cos x.$First,find the derivative $f'(x)$:$f'(x) = \cos x + \sin x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x ight) = \sqrt{2} \sin(x + \pi/4).$For the function to be decreasing,we require $f'(x) $\sqrt{2} \sin(x + \pi/4) $\sin(x + \pi/4) In the interval $0 \le x \le 2\pi$,the argument $	heta = x + \pi/4$ ranges from $\pi/4$ to $9\pi/4$.The sine function is negative in the interval $(\pi, 2\pi)$.Therefore,$\pi Subtracting $\pi/4$ from all sides:$3\pi/4 Comparing this with the given options,none of the intervals provided are subsets of $(3\pi/4, 7\pi/4)$.Thus,the correct option is $(d)$.

If $f(x) = \sin x - \cos x,$ the function is decreasing in the interval $0 \le x \le 2\pi$ for which of the following?

Similar Questions

Let $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$,for $x \in (-\infty, \infty)$. For which interval is $f(x)$ an increasing function?

Prove that the function given by $f(x) = \cos x$ is increasing in $(\pi, 2\pi)$.

For the function $f(x) = \cos x - x + 1, x \in R$,consider the following two statements:
$(S1)$ $f(x) = 0$ for only one value of $x$ in $[0, \pi]$.
$(S2)$ $f(x)$ is decreasing in $[0, \frac{\pi}{2}]$ and increasing in $[\frac{\pi}{2}, \pi]$.

For what value of $\lambda$ is the function $f(x) = \lambda x + \cos x$ strictly increasing?

If $f(x)=e^{x}(x-2)^{2}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module