Question

Define \( f(x) = \begin{cases} x^2 + bx + c, & x< 1 \\ x, & x \geq 1 \end{cases} \). If f(x) is differentiable at x=1, then b−c is equal to

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

Hint:

For piecewise functions, ensure that both continuity and differentiability are checked at the transition point.

Answer 1

The Correct Option is A

Step 1: Continuity at \( x = 1 \). For differentiability, the function must be continuous at \( x = 1 \). 

Thus, we require: \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x). \] Substitute: \[ 1^2 + b(1) + c = 1 \quad \Rightarrow \quad 1 + b + c = 1 \quad \Rightarrow \quad b + c = 0. \]

 Step 2: Differentiability at \( x = 1 \). The derivatives from the left and right must also be equal at \( x = 1 \): \[ \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x). \] \[ \lim_{x \to 1^-} (2x + b) = \lim_{x \to 1^+} 1. \] 

Substitute \( x = 1 \) into the equation: \[ 2(1) + b = 1 \quad \Rightarrow \quad b = -1. \] 

From the equation \( b + c = 0 \), we find: \[ c = 1. \] 

Step 3: Compute \( b - c \). Now, calculate: \[ b - c = -1 - 1 = -2. \] 

Final Answer: \[ \boxed{-2} \]