What is the length of the longest interval in | vedclass

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

Question

(A) Given the function $f(x) = 3\sin x - 4\sin^3 x$. Using the trigonometric identity,we know that $3\sin x - 4\sin^3 x = \sin(3x)$. So,$f(x) = \sin(3

Vedclass · Accepted Answer

$\pi /3$

Vedclass · Answer

(A) Given the function $f(x) = 3\sin x - 4\sin^3 x$. Using the trigonometric identity,we know that $3\sin x - 4\sin^3 x = \sin(3x)$. So,$f(x) = \sin(3x)$. For the function to be increasing,its derivative must be positive: $f'(x) > 0$. $f'(x) = \frac{d}{dx}(\sin(3x)) = 3\cos(3x)$. We require $3\cos(3x) > 0$,which means $\cos(3x) > 0$. The cosine function is positive in the interval $(-\pi/2, \pi/2)$. Therefore,$-\frac{\pi}{2} Dividing by $3$,we get $-\frac{\pi}{6} The length of this interval is $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.

What is the length of the longest interval in which the function $f(x) = 3\sin x - 4\sin^3 x$ is an increasing function?

Similar Questions

In which interval is the function $f(x) = x^2 - x + 1$ not monotonic?

The interval in which the function $f(x) = \frac{\log(7+x)}{\log(3+x)}$ for $x > 0$ decreases is:

Show that the function given by $f(x) = \sin x$ is decreasing in $\left(\frac{\pi}{2}, \pi\right)$.

The equation $x^3+x-1=0$ has

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module