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

Concept: Continuity at \( x=0 \): \[ \lim_{x \to 0} f(x) = f(0) \]

Step 1: Use small angle approximation.
\[ \tan ax \approx ax, \quad \tan x \approx x \]

Step 2: Substitute.
\[ \lim_{x\to 0} \frac{ax + (b+1)x}{x} = a + b + 1 \]

Step 3: Apply continuity.
\[ a + b + 1 = 5 \Rightarrow a + b = 4 \]