Question

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Let \(f\) be a function defined by \[ f(x) = \begin{cases} \dfrac{\tan x}{x}, & x \neq 0 1, & x = 0 \end{cases} \] Assertion (A): \(x = 0\) is point of minima of \(f\).
Reason (R): \(f'(0) = 0\). In the light of the above statements, choose the most appropriate answer from the options given below:

• A. Both (A) and (R) are correct and (R) is the correct explanation of (A)
• B. Both (A) and (R) are correct but (R) is not the correct explanation of (A)
• C. (A) is correct but (R) is not correct
• D. (A) is not correct but (R) is correct

Hint:

Use series expansion near \(x=0\) to easily determine behavior of functions involving trigonometric limits.

Answer 1

The Correct Option is A

Concept: To determine whether a point is a minimum, we check:
• Continuity at the point
• First derivative equals zero
• Behaviour around the point

Step 1: Check continuity at \(x=0\). \[ \lim_{x \to 0} \frac{\tan x}{x} = 1 = f(0) \] Thus, function is continuous at \(x=0\).

Step 2: Find derivative at \(x=0\). Using expansion: \[ \tan x = x + \frac{x^3}{3} + \cdots \] \[ \frac{\tan x}{x} = 1 + \frac{x^2}{3} + \cdots \] Thus: \[ f(x) = 1 + \frac{x^2}{3} + \cdots \] \[ f'(x) = \frac{2x}{3} + \cdots \] So: \[ f'(0) = 0 \]

Step 3: Check nature of point. Since: \[ f(x) = 1 + \frac{x^2}{3} > 1 \text{ for } x \neq 0 \] Thus \(x=0\) gives minimum value.

Step 4: Conclusion.
• Assertion is true
• Reason is true and explains why minimum occurs \[ \boxed{(1)} \]