Question

Let \[ f(x)= \begin{cases} \dfrac{|a|x + 2x^2 - 2\sin|x|\cos|x|}{x}, & x \neq 0 \\ b, & x = 0 \end{cases} \] If \( f(x) \) is continuous, then the value of \( a + b \) is:

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

For continuity at a point, the left-hand limit, right-hand limit, and function value must be equal.

Answer 1

The Correct Option is A

Step 1: Apply the condition of continuity at \( x = 0 \).
For continuity at \( x = 0 \), \[ \lim_{x \to 0} f(x) = f(0) = b \]
Step 2: Simplify the expression for \( x \neq 0 \).
\[ f(x) = \frac{|a|x}{x} + \frac{2x^2}{x} - \frac{2\sin|x|\cos|x|}{x} \] \[ = |a| + 2x - \frac{\sin 2|x|}{x} \]
Step 3: Evaluate the limit.
As \( x \to 0 \), \[ \lim_{x \to 0} 2x = 0,\quad \lim_{x \to 0} \frac{\sin 2|x|}{x} = 2 \] Hence, \[ \lim_{x \to 0} f(x) = |a| - 2 \]
Step 4: Find \( a \) and \( b \).
For continuity, \[ b = |a| - 2 \] From the options, the valid value is \[ a = 2,\quad b = 0 \]
Step 5: Final Answer.
\[ a + b = 2 \]