Let $$\mathbb{R}$$ denote the set of all real numbers. Let $$f: \mathbb{R} \to \mathbb{R}$$ be defined by

$$f(x) = \begin{cases} \frac{6x + \sin x}{2x + \sin x} & \text{if } x \neq 0, \\ \frac{7}{3} & \text{if } x = 0. \end{cases}$$

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

The function is given by
$$f(x)=\dfrac{6x+\sin x}{2x+\sin x}\;,\;x\neq 0,\qquad f(0)=\dfrac73.$$

1. Behaviour at the origin

First-order Taylor expansion about $$x=0$$ gives

$$\sin x=x-\dfrac{x^{3}}6+O(x^{5}),\qquad \cos x=1-\dfrac{x^{2}}2+O(x^{4}).$$

Hence

$$6x+\sin x=6x+\left(x-\dfrac{x^{3}}6\right)=7x-\dfrac{x^{3}}6+O(x^{5}),$$ $$2x+\sin x=2x+\left(x-\dfrac{x^{3}}6\right)=3x-\dfrac{x^{3}}6+O(x^{5}).$$

Therefore

$$f(x)=\dfrac{7x-\dfrac{x^{3}}6+O(x^{5})}{3x-\dfrac{x^{3}}6+O(x^{5})} =\dfrac73\!\left[\,1+\dfrac{2x^{2}}{63}+O(x^{4})\right] =\dfrac73+\dfrac{2x^{2}}{27}+O(x^{4}).$$

Near the origin we thus have

$$f(x)=\dfrac73+\dfrac{2x^{2}}{27}+O(x^{4}).$$

This gives $$f'(0)=0$$ and $$f''(0)=\dfrac{4}{27}\;(\gt 0).$$ Hence $$x=0$$ is a point of local minimum. Option B is therefore correct, while Option A is not.

2. Critical points for $$x\neq 0$$

For $$x\neq 0$$ write $$N=6x+\sin x,\;D=2x+\sin x,$$ so that $$f(x)=\dfrac{N}{D}.$$ Using the quotient rule,

$$f'(x)=\dfrac{N' D-N D'}{D^{2}} =\dfrac{(6+\cos x)(2x+\sin x)-(6x+\sin x)(2+\cos x)}{(2x+\sin x)^{2}} =\dfrac{4\bigl(\sin x-x\cos x\bigr)}{(2x+\sin x)^{2}}.$$

Since $$(2x+\sin x)^{2}\gt 0\quad\text{for all }x\gt 0,$$ the sign of $$f'(x)$$ is governed solely by

$$g(x)=\sin x-x\cos x.$$

3. Location of critical points

Setting $$f'(x)=0$$ gives $$g(x)=0\iff \sin x-x\cos x=0 \iff\tan x = x.$$ The well-known equation $$\tan x=x$$ has one solution in every interval $$\bigl(n\pi,\;n\pi+\tfrac{\pi}{2}\bigr), \;n=1,2,3,\ldots$$ apart from $$x=0.$$ Denote these positive roots by $$x_1,x_2,x_3,\dots$$ in increasing order:

$$x_1\approx4.493,\;x_2\approx7.725,\;x_3\approx10.904,\; x_4\approx14.066,\;x_5\approx17.279,\;\ldots$$

4. Nature of each critical point

Differentiate $$g(x)$$:

$$g'(x)=\cos x-\cos x+x\sin x = x\sin x.$$

Thus, at a root $$x_n,$$ the sign of $$g'(x_n)=x_n\sin x_n$$ is the sign of $$\sin x_n.$$ Inside $$\bigl(n\pi,\;n\pi+\tfrac{\pi}{2}\bigr)$$ we have $$\sin x\;\begin{cases} \gt 0,& n\; \text{even},\\[4pt] \lt 0,& n\; \text{odd}. \end{cases}$$ Therefore

• If $$n$$ is odd ($$1,3,5,\dots$$) then $$g'(x_n)\lt 0:$$ $$g(x)$$ crosses the $$x$$-axis from positive to negative, so $$f'(x)$$ changes $$+\to-$$ and $$x_n$$ is a local maximum.
• If $$n$$ is even ($$2,4,6,\dots$$) then $$g'(x_n)\gt 0:$$ $$g(x)$$ crosses from negative to positive, so $$f'(x)$$ changes $$-\to+$$ and $$x_n$$ is a local minimum.

5. Counting extrema in the required intervals

(i) Interval $$[\pi,6\pi]\;( \approx[3.142,18.850] ):$$ roots inside this range are $$x_1,x_2,x_3,x_4,x_5.$$ Among these, maxima occur at the odd-indexed roots $$x_1,x_3,x_5.$$ Hence the number of local maxima is $$3.$$ Option C is correct.

(ii) Interval $$[2\pi,4\pi]\;( \approx[6.283,12.566] ):$$ only the root $$x_2\approx7.725$$ lies in this interval, and it is a local minimum (even index). Therefore there is exactly $$1$$ local minimum here. Option D is correct.

6. Final verdict

The true statements are:
Option B, Option C, and Option D.