Let $\mathbb{R}$ denote the set of all real numbers. Let $f: \mathbb{R} \to \mathbb{R}$ be defined by
$$f(x) = \begin{cases} \dfrac{6x + \sin x}{2x + \sin x} & \text{if } x \neq 0, \\ \dfrac{7}{3} & \text{if } x = 0. \end{cases}$$
Then which of the following statements is (are) TRUE?
The point $x = 0$ is a point of local maxima of $f$
The point $x = 0$ is a point of local minima of $f$
Number of points of local maxima of $f$ in the interval $[\pi, 6\pi]$ is 3
Number of points of local minima of $f$ in the interval $[2\pi, 4\pi]$ is 1
Step-by-Step Solution
Key Concept: Using Taylor series expansion of $\sin x$ to evaluate the limit at the boundary point, and analyzing the sign of $f'(x)$ using the graph of $ an x = x$.
Let's check $f(x) - f(0)$ for $x$ close to 0:
$$f(x) - \dfrac{7}{3} = \dfrac{6x + \sin x}{2x + \sin x} - \dfrac{7}{3} = \dfrac{4x - 4\sin x}{3(2x + \sin x)}$$
Using Taylor expansions: $\sin x \approx x - x^3/6$:
$$f(x) - \dfrac{7}{3} \approx \dfrac{4(x^3/6)}{3(3x)} = \dfrac{2}{27}x^2 > 0 \text{ for all } x \neq 0.$$
Thus, $x = 0$ is a point of local minima (Option B is True, A is False).
Differentiate $f(x)$ for $x \neq 0$ using the quotient rule:
$$f'(x) = \dfrac{4(\sin x - x\cos x)}{(2x+\sin x)^2} = \dfrac{4\cos x(\tan x - x)}{(2x+\sin x)^2}$$
The critical points occur when $f'(x) = 0 \implies \tan x = x$.
For $x_n \in (n\pi, n\pi + \pi/2)$, the sign of $f'(x)$ is determined by $\cos x (\tan x - x)$:
- When $n$ is odd, $\cos x < 0$. $f'(x)$ changes from positive to negative at $x_n \implies$ local maximum.
- When $n$ is even, $\cos x > 0$. $f'(x)$ changes from negative to positive at $x_n \implies$ local minimum.
Now check intervals:
- For $[\pi, 6\pi]$: The roots are $x_1$ (odd, max), $x_2$ (even, min), $x_3$ (odd, max), $x_4$ (even, min), $x_5$ (odd, max). Total local maxima = 3 ($x_1, x_3, x_5$) (Option C is True).
- For $[2\pi, 4\pi]$: The only critical point is $x_2 \in (2\pi, 5\pi/2)$ (even, min). $x_3$ is a max, and $x_4 > 4\pi$. Total local minima = 1 ($x_2$) (Option D is True).
Correct Answer: B, C, D