Question

Let \( f(x) = \begin{cases} \frac{\tan(ax) + (b+1)\tan(x)}{x}, & x \neq 0 \\ 5, & x = 0 \end{cases} \) be continuous at \( x = 0 \). Then the value of \( a + b \) is equal to

• A. \(2 \)
• B. \(3 \)
• C. \(4 \)
• D. \(5 \)
• E. \(6 \)

Hint:

For continuity at 0, replace functions using standard limits like \( \tan x \approx x \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a function to be continuous at a point \(x=c\), the limit of the function as x approaches c must exist and be equal to the function's value at that point. In this case, \(\lim_{x \to 0} f(x) = f(0)\).
Step 2: Key Formula or Approach:
1. We are given \(f(0) = 5\). 2. We must calculate the limit \(\lim_{x \to 0} \frac{\tan \alpha x + (\beta+1)\tan x}{x}\). 3. Set the result of the limit equal to 5 and solve for the required expression. 4. We will use the standard trigonometric limit \(\lim_{u \to 0} \frac{\tan u}{u} = 1\).
Step 3: Detailed Explanation:
For continuity at x=0, we must have: \[ \lim_{x \to 0} f(x) = f(0) \] \[ \lim_{x \to 0} \frac{\tan(\alpha x) + (\beta+1)\tan x}{x} = 5 \] We can split the fraction and the limit: \[ \lim_{x \to 0} \left( \frac{\tan(\alpha x)}{x} + \frac{(\beta+1)\tan x}{x} \right) = 5 \] \[ \lim_{x \to 0} \frac{\tan(\alpha x)}{x} + \lim_{x \to 0} (\beta+1)\frac{\tan x}{x} = 5 \] Now, we evaluate each limit separately. For the first limit, we use the substitution \(u = \alpha x\). As \(x \to 0\), \(u \to 0\). \[ \lim_{x \to 0} \frac{\tan(\alpha x)}{x} = \lim_{x \to 0} \alpha \cdot \frac{\tan(\alpha x)}{\alpha x} = \alpha \cdot \lim_{u \to 0} \frac{\tan u}{u} = \alpha \cdot 1 = \alpha \] For the second limit, we use the standard limit directly: \[ \lim_{x \to 0} (\beta+1)\frac{\tan x}{x} = (\beta+1) \lim_{x \to 0} \frac{\tan x}{x} = (\beta+1) \cdot 1 = \beta+1 \] Now, substitute these results back into the main equation: \[ \alpha + (\beta+1) = 5 \] \[ \alpha + \beta + 1 = 5 \] \[ \alpha + \beta = 4 \] Step 4: Final Answer:
The value of \(\alpha + \beta\) is 4.