Question

If \( f(x) = \begin{cases} 1 + 6x - 3x^2, & x \leq 1 \\ x + \log_2(b^2 + 7), & x > 1 \end{cases} \) is continuous at all real \( x \), then \( b \) is:

• A. \( \pm 1 \)
• B. \( 0 \)
• C. \( \pm 5 \)
• D. \( \pm 2 \)

Hint:

When solving for continuity, make sure to equate the left-hand and right-hand limits at the point of interest. Don't forget to simplify logarithmic equations correctly.

Answer 1

The Correct Option is A

We are given a piecewise function: \[ f(x) = \begin{cases} 1 + 6x - 3x^2, & x \leq 1 \\x + \log_2(b^2 + 7), & x>1 \end{cases} \] To ensure that the function is continuous at \( x = 1 \), the left-hand limit as \( x \to 1^- \) must equal the right-hand limit as \( x \to 1^+ \), i.e., \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) \] From the left-hand side: \[ \lim_{x \to 1^-} f(x) = 1 + 6(1) - 3(1)^2 = 1 + 6 - 3 = 4 \] From the right-hand side: \[ \lim_{x \to 1^+} f(x) = 1 + \log_2(b^2 + 7) \] Setting these equal: \[ 4 = 1 + \log_2(b^2 + 7) \] Solving for \( b \): \[ 3 = \log_2(b^2 + 7) \] Exponentiating both sides: \[ 2^3 = b^2 + 7 \] \[ 8 = b^2 + 7 \] \[ b^2 = 1 \] \[ b = \pm 1 \]