Question

Let $ \mathbb{R} $ denote the set of all real numbers. Let $ f: \mathbb{R} \to \mathbb{R} $ be defined by $$ f(x) = \begin{cases} \dfrac{6x + \sin x}{2x + \sin x}, & \text{if } x \ne 0 \\ \dfrac{7}{3}, & \text{if } x = 0 \end{cases} $$ Then which of the following statements is (are) TRUE?

• A. The point \( x = 0 \) is a point of local maxima of \( f \)
• B. The point \( x = 0 \) is a point of local minima of \( f \)
• C. Number of points of local maxima of \( f \) in the interval \( [\pi, 6\pi] \) is 3
• D. Number of points of local minima of \( f \) in the interval \( [2\pi, 4\pi] \) is 1

Hint:

For piecewise or rational trigonometric functions, use Taylor series around \( x = 0 \) to check local behavior. Count turning points using derivative sign change and symmetry.

Answer 1

The Correct Option is A, C

Step 1: Define function: \[ f(x) = \frac{6x + \sin x}{2x + \sin x}, \quad x \ne 0 \quad ; \quad f(0) = \frac{7}{3} \] Check continuity at \( x = 0 \): Use L'Hôpital’s Rule: \[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{6x + \sin x}{2x + \sin x} = \frac{6 \cdot 0 + 0}{2 \cdot 0 + 0} \text{ → Indeterminate } \] Apply L'Hôpital's Rule: \[ f'(x) = \frac{6 + \cos x}{2 + \cos x} \Rightarrow f(0^+) = \frac{6 + 1}{2 + 1} = \frac{7}{3} \] So, \( \lim_{x \to 0} f(x) = f(0) = \frac{7}{3} \Rightarrow f(x) \) is continuous at \( x = 0 \) 
Step 2: Check if \( x = 0 \) is local maxima or minima Differentiate: Let \( f(x) = \frac{u(x)}{v(x)} \) where: \[ u(x) = 6x + \sin x,\quad v(x) = \\ 2x + \sin x\Rightarrow f'(x) = \frac{u'v - uv'}{v^2} = \frac{(6 + \cos x)(2x + \sin x) - (6x + \sin x)(2 + \cos x)}{(2x + \sin x)^2} \]
Now evaluate sign of \( f'(x) \) near \( x = 0 \) Use Taylor expansion: \[ \sin x \approx x - \frac{x^3}{6},\quad \cos x \approx 1 - \frac{x^2}{2} \] Then: - \( f(x) \approx \frac{6x + x}{2x + x} = \frac{7x}{3x} = \frac{7}{3} \) - Check second derivative or sign change: Take a value just less than 0: \[ x = -0.01: f(x) = \frac{-0.06 + \sin(-0.01)}{-0.02 + \sin(-0.01)} \approx \frac{-0.06 - 0.01}{-0.02 - 0.01} = \frac{-0.07}{-0.03} \approx 2.33 \] Take a value just greater than 0: \[ x = 0.01 \Rightarrow f(x) \approx \frac{0.07}{0.03} \approx 2.33 \] So for \( x \ne 0 \), values are less than \( \frac{7}{3} \approx 2.33 \Rightarrow x = 0 \) is local maximum So Option (A) is correct. 
Step 3: Analyze local maxima in \( [\pi, 6\pi] \) Function: \[ f(x) = \frac{6x + \sin x}{2x + \sin x} \] Differentiate: \[ f'(x) = \frac{(6 + \cos x)(2x + \sin x) - (6x + \sin x)(2 + \cos x)}{(2x + \sin x)^2} \] Set numerator = 0 for critical points: From symmetry and periodicity, local maxima occurs when \( \sin x \) is minimum, i.e., at: \[ x = (2n + 1)\pi \Rightarrow \text{Check } x = \pi,\ 3\pi,\ 5\pi \in [\pi, 6\pi] \Rightarrow \text{3 local maxima} \Rightarrow \text{Option (C) is correct} \] 
Step 4: Check local minima in \( [2\pi, 4\pi] \) Minima occur roughly at: \[ x = 2\pi,\ 4\pi \Rightarrow \text{Endpoints not included as minima if slope doesn't change sign} \] Check around \( x = 3\pi \) — but that’s a maxima So, only one minima expected in \( [2\pi, 4\pi] \) — not consistent Option (D) is false Final verdict: - (A) TRUE (local max at \( x = 0 \)) - (B) FALSE (not local min) - (C) TRUE (3 local maxima in \( [\pi, 6\pi] \)) - (D) FALSE