Question

\[ \text{If the function } f(x) = \begin{cases} 1 + \cos x, & x \leq 0 \\ a - x, & 0 < x \leq 2 \\ x^2 - b^2, & x > 2 \end{cases} \text{ is continuous everywhere, then } a^2 + b^2 =\ ? \]

• A. \( 4 \)
• B. \( 8 \)
• C. \( 6 \)
• D. \( 12 \)

Hint:

For piecewise functions to be continuous everywhere, the left-hand and right-hand limits must match the function value at the joining points.

Answer 1

The Correct Option is B

Step 1: Ensure continuity at \( x = 0 \) From the left: \[ \lim_{x \to 0^-} f(x) = 1 + \cos 0 = 1 + 1 = 2 \] From the right: \[ \lim_{x \to 0^+} f(x) = a - 0 = a \] For continuity at \( x = 0 \): \[ a = 2 \] Step 2: Ensure continuity at \( x = 2 \) From the left: \[ \lim_{x \to 2^-} f(x) = a - 2 = 2 - 2 = 0 \] From the right: \[ \lim_{x \to 2^+} f(x) = 2^2 - b^2 = 4 - b^2 \] For continuity at \( x = 2 \): \[ 4 - b^2 = 0 \Rightarrow b^2 = 4 \Rightarrow b = \pm 2 \] Step 3: Compute \( a^2 + b^2 \) \[ a = 2, b^2 = 4 \Rightarrow a^2 + b^2 = 4 + 4 = 8 \] % Final Answer \[ \boxed{8} \]