The length of the longest interval in which t | vedclass

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

Question

(A) Given function is $f(x) = 3\sin x - 4\sin^3 x$.
Using the trigonometric identity $\sin(3x) = 3\sin x - 4\sin^3 x$, we can write $f(x) = \sin(3x)$.

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(A) Given function is $f(x) = 3\sin x - 4\sin^3 x$. Using the trigonometric identity $\sin(3x) = 3\sin x - 4\sin^3 x$, we can write $f(x) = \sin(3x)$. For the function to be increasing, its derivative must be positive: $f'(x) > 0$. $f'(x) = \frac{d}{dx}(\sin(3x)) = 3\cos(3x)$. Setting $3\cos(3x) > 0$, we get $\cos(3x) > 0$. The cosine function is positive in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ for its argument. Thus, $-\frac{\pi}{2} < 3x < \frac{\pi}{2}$. Dividing by $3$, we get $-\frac{\pi}{6} < x < \frac{\pi}{6}$. The length of this interval is $\frac{\pi}{6} - (-\frac{\pi}{6}) = \frac{2\pi}{6} = \frac{\pi}{3}$.

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

Similar Questions

If $f(x) = \sin x - \cos x - ax + b$ is a decreasing function for all $x \in R$,then:

If $f(x) = x^3 - x^2 + 100x + 1001$,then:

Let $\phi(x) = f(x) + f(1-x)$ and $f^{\prime \prime}(x) < 0$ in $[0, 1]$, then

If $y = ax^3 + 3x^2 + (2a + 1)x + 1000$ is a strictly increasing function for all values of $x$,then:

The function $f(x) = \frac{\log(\pi + x)}{\log(e + x)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module