Let R denote the set of all real numbers. Let f : R  R be defined by

$f\left(x\right)=\left\{\begin{array}{ll}\frac{6x+\sin x}{2x+\sin x}& \text{if }x\ne 0\\ \frac{7}{3}& \text{if }x=0\end{array}\right.$

Then which of the following statements is (are) TRUE?

To determine which of the statements are TRUE, let us analyze the function $f\left(x\right)$ given by:

$f\left(x\right)=\left\{\begin{array}{}\frac{6x+\sin x}{2x+\sin x}& \text{if}x\ne 0\\ \frac{7}{3}& \text{if}x=0\end{array}\right.$

First, let us check the continuity of $f\left(x\right)$ at $x=0$ by computing the limit:
$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{6x+\sin x}{2x+\sin x}$
Dividing the numerator and denominator by $x$ (since $x\ne 0$):
$\underset{x\to 0}{lim}\frac{6+\frac{\sin x}{x}}{2+\frac{\sin x}{x}}=\frac{6+1}{2+1}=\frac{7}{3}$
Since $\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)=\frac{7}{3}$, the function $f\left(x\right)$ is continuous at $x=0$.

Now, let us analyze the behavior of the derivative of $f\left(x\right)$ for $x\ne 0$. We can rewrite the expression for $f\left(x\right)$ as:
$f\left(x\right)=\frac{3(2x+\sin x)-2\sin x}{2x+\sin x}=3-\frac{2\sin x}{2x+\sin x}$
Differentiating $f\left(x\right)$ with respect to $x$ using the quotient rule:
$f^{\text{'}}\left(x\right)=-2·\frac{\cos x(2x+\sin x)-\sin x(2+\cos x)}{{(2x+\sin x)}^2}$
Simplifying the numerator:
$\cos x(2x+\sin x)-\sin x(2+\cos x)=2x\cos x+\sin x\cos x-2\sin x-\sin x\cos x=2(x\cos x-\sin x)$
Substituting this back into the derivative:
$f^{\text{'}}\left(x\right)=\frac{4(\sin x-x\cos x)}{{(2x+\sin x)}^2}$
Let $g\left(x\right)=\sin x-x\cos x$. The sign of $f^{\text{'}}\left(x\right)$ is determined solely by the sign of $g\left(x\right)$ since the denominator is always positive for $x\ne 0$ (note that $2x+\sin x=0$ only at $x=0$).

Let us analyze the behavior of $g\left(x\right)$ around $x=0$:
$g^{\text{'}}\left(x\right)=\cos x-(\cos x-x\sin x)=x\sin x$
For a small neighborhood around $x=0$ (excluding 0):
- If $x>0$ (with $x$ small), both $x>0$ and $\sin x>0$, so $g^{\text{'}}\left(x\right)>0$. Since $g\left(0\right)=0$, this implies $g\left(x\right)>0$ for $x>0$ near $0$.
- If $x<0$ (with $x$ small), both $x<0$ and $\sin x<0$, so $g^{\text{'}}\left(x\right)>0$. Since $g\left(0\right)=0$, this implies $g\left(x\right)<0$ for $x<0$ near $0$.
Consequently:
- For $x<0$ near $0$, $f^{\text{'}}\left(x\right)<0$ (function is decreasing).
- For $x>0$ near $0$, $f^{\text{'}}\left(x\right)>0$ (function is increasing).
Therefore, $x=0$ is a point of **local minima** of $f$.

Next, let us analyze the critical points where $f^{\text{'}}\left(x\right)=0$ for $x>0$:
$g\left(x\right)=0\Rightarrow \sin x=x\cos x\Rightarrow \tan x=x$
The points of local extrema correspond to the intersections of the curves $y=\tan x$ and $y=x$.

Let the consecutive positive roots of $\tan x=x$ be denoted by $x_1,x_2,x_3,...$ where:
- $x_1\in \left(\pi ,\frac{3\pi }{2}\right)$
- $x_2\in \left(2\pi ,\frac{5\pi }{2}\right)$
- $x_3\in \left(3\pi ,\frac{7\pi }{2}\right)$
- $x_4\in \left(4\pi ,\frac{9\pi }{2}\right)$
- $x_5\in \left(5\pi ,\frac{11\pi }{2}\right)$
- and so on.

Let us determine the nature of these critical points by observing the derivative of $g\left(x\right)$: $g^{\text{'}}\left(x\right)=x\sin x$.
- At $x_1\in \left(\pi ,\frac{3\pi }{2}\right)$, we have $\sin x_1<0$, so $g^{\text{'}}\left(x_1\right)<0$. This means $g\left(x\right)$ is changing from positive to negative, which corresponds to a **local maximum** for $f\left(x\right)$.
- At $x_2\in \left(2\pi ,\frac{5\pi }{2}\right)$, we have $\sin x_2>0$, so $g^{\text{'}}\left(x_2\right)>0$. This corresponds to a **local minimum** for $f\left(x\right)$.
- By extension, local maxima occur at odd-indexed roots $x_1,x_3,x_5,...$ and local minima occur at even-indexed roots $x_2,x_4,...$.

Let us check the number of local maxima in $[\pi ,6\pi ]$:
The critical points in this interval are $x_1,x_2,x_3,x_4,x_5$.
The local maxima are $x_1$, $x_3$, and $x_5$, which gives a total of **3** local maxima. Hence, the third statement is TRUE.

Let us check the number of local minima in $[2\pi ,4\pi ]$:
The critical points in this interval are $x_2$ and $x_3$.
Among these, the only local minimum is $x_2$, which gives a total of **1** local minimum. Hence, the fourth statement is TRUE.