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Question

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A. $\dfrac{\pi }{3}$

B. $\dfrac{\pi }{2}$

C. $\dfrac{3\pi }{2}$

D. $\pi $

Answer

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Hint: Take $f(x)=3\sin x-4{{\sin }^{2}}x$ and differentiate it that is find ${{f}^{'}}(x)$. So if the function is increasing, put${{f}^{'}}(x)\ge 0$. You will get the answer.

Complete step-by-step answer:

Regarding intervals of increasing or decreasing on a graph, it is a popular convention to use only "open" interval notation. However, it is considered correct to use either "open" or "closed" notation when describing intervals of increasing or decreasing. References to $\pm $ infinity, however, are always "open" notation.

A function is increasing, if as x increases (reading from left to right), y also increases. In plain English, as you look at the graph, from left to right, the graph goes up-hill. The graph has a positive slope.

A function is strictly increasing on an interval, if when ${{x}_{1}}<{{x}_{2}}$, then $f({{x}_{1}})

If the function notation is bothering you, this definition can also be thought of as stating ${{x}_{1}}<{{x}_{2}}$implies ${{y}_{1}}<{{y}_{2}}$. As the $x$ get larger, then $y$ get larger.

A function is decreasing, if as $c$ increases (reading from left to right), $y$ decreases. In plain English, as you look at the graph, from left to right, the graph goes down-hill. The graph has a negative slope.

A function is strictly decreasing on an interval, if when ${{x}_{1}}<{{x}_{2}}$,

then $f({{x}_{1}})>f({{x}_{2}})$

If the function notation is bothering you, this definition can also be thought of as stating ${{x}_{1}}<{{x}_{2}}$implies ${{y}_{1}}>{{y}_{2}}$.

As the $x$ get larger the$y$ get smaller.

So now in question, we have given that, $f(x)=3\sin x-4{{\sin }^{2}}x$.

So we know the identity$3\sin x-4{{\sin }^{2}}x=\sin 3x$.

So we get$f(x)=\sin 3x$

So now let us find${{f}^{'}}(x)$.

${{f}^{'}}(x)=3\cos 3x$

So $f(x)$it is increasing.

So now putting${{f}^{'}}(x)\ge 0$.

So $3\cos 3x\ge 0$.

$3\cos 3x\ge \cos \dfrac{\pi }{2}$ or $3\cos 3x\ge \cos \dfrac{3\pi }{2}$

So we can see that in the interval $[0{}^\circ ,360{}^\circ ),$ $3x$ should be in the first or fourth quadrant.

$\Rightarrow 3x\in [0,90{}^\circ )\cup (270{}^\circ ,360{}^\circ )$

So $x\in [0{}^\circ ,30{}^\circ )\cup [90{}^\circ ,150{}^\circ )\cup [210{}^\circ ,270{}^\circ )\cup [330{}^\circ ,360{}^\circ ).$

So in the above, it can be seen that the length of the longest interval for which the function is increasing is $60{}^\circ $=$\dfrac{\pi }{3}$radians.

So the final answer we get the longest term for $\sin 3x$ is $\dfrac{\pi }{3}$.

So option(A) is correct.

Note: Read the question in a careful manner. So you should be familiar with the concepts of increasing and decreasing functions. You should also know when open brackets are used and when closed brackets are used. You should also know the identities such as$3\sin x-4{{\sin }^{2}}x=\sin 3x$. Most of the mistakes occur at$3\cos 3x\ge 0$ that is to put$\le or\ge $so don’t make mistakes.

Complete step-by-step answer:

Regarding intervals of increasing or decreasing on a graph, it is a popular convention to use only "open" interval notation. However, it is considered correct to use either "open" or "closed" notation when describing intervals of increasing or decreasing. References to $\pm $ infinity, however, are always "open" notation.

A function is increasing, if as x increases (reading from left to right), y also increases. In plain English, as you look at the graph, from left to right, the graph goes up-hill. The graph has a positive slope.

A function is strictly increasing on an interval, if when ${{x}_{1}}<{{x}_{2}}$, then $f({{x}_{1}})

If the function notation is bothering you, this definition can also be thought of as stating ${{x}_{1}}<{{x}_{2}}$implies ${{y}_{1}}<{{y}_{2}}$. As the $x$ get larger, then $y$ get larger.

A function is decreasing, if as $c$ increases (reading from left to right), $y$ decreases. In plain English, as you look at the graph, from left to right, the graph goes down-hill. The graph has a negative slope.

A function is strictly decreasing on an interval, if when ${{x}_{1}}<{{x}_{2}}$,

then $f({{x}_{1}})>f({{x}_{2}})$

If the function notation is bothering you, this definition can also be thought of as stating ${{x}_{1}}<{{x}_{2}}$implies ${{y}_{1}}>{{y}_{2}}$.

As the $x$ get larger the$y$ get smaller.

So now in question, we have given that, $f(x)=3\sin x-4{{\sin }^{2}}x$.

So we know the identity$3\sin x-4{{\sin }^{2}}x=\sin 3x$.

So we get$f(x)=\sin 3x$

So now let us find${{f}^{'}}(x)$.

${{f}^{'}}(x)=3\cos 3x$

So $f(x)$it is increasing.

So now putting${{f}^{'}}(x)\ge 0$.

So $3\cos 3x\ge 0$.

$3\cos 3x\ge \cos \dfrac{\pi }{2}$ or $3\cos 3x\ge \cos \dfrac{3\pi }{2}$

So we can see that in the interval $[0{}^\circ ,360{}^\circ ),$ $3x$ should be in the first or fourth quadrant.

$\Rightarrow 3x\in [0,90{}^\circ )\cup (270{}^\circ ,360{}^\circ )$

So $x\in [0{}^\circ ,30{}^\circ )\cup [90{}^\circ ,150{}^\circ )\cup [210{}^\circ ,270{}^\circ )\cup [330{}^\circ ,360{}^\circ ).$

So in the above, it can be seen that the length of the longest interval for which the function is increasing is $60{}^\circ $=$\dfrac{\pi }{3}$radians.

So the final answer we get the longest term for $\sin 3x$ is $\dfrac{\pi }{3}$.

So option(A) is correct.

Note: Read the question in a careful manner. So you should be familiar with the concepts of increasing and decreasing functions. You should also know when open brackets are used and when closed brackets are used. You should also know the identities such as$3\sin x-4{{\sin }^{2}}x=\sin 3x$. Most of the mistakes occur at$3\cos 3x\ge 0$ that is to put$\le or\ge $so don’t make mistakes.

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