Question

A function \( f(x) \) is defined as: \[ f(x) = \begin{cases} ax^2 + bx + c, & x \leq -1
2x^2 + 4x + 1, & -1<x<1
cx^2 + bx + a, & x \geq 1 \end{cases} \] If \( f(x) \) is continuous on \( \mathbb{R} \) and \( \lim_{x \to -\frac{3}{2}} f(x) = 14 \), then \( \lim_{x \to 2} f(x) = \):

• A. \( 6 \)
• B. \( -8 \)
• C. \( 5 \)
• D. \( 1 \)

Hint:

Use continuity and substitution at transition points to build equations and solve the unknowns.

Answer 1

The Correct Option is B

Given continuity and function definitions, use matching at boundaries: - At \( x = -1 \), set: \[ \lim_{x \to -1^-} (ax^2 + bx + c) = \lim_{x \to -1^+} (2x^2 + 4x + 1) \Rightarrow a + (-b) + c = 2 + (-4) + 1 = -1 \Rightarrow a - b + c = -1 \quad \text{(1)} \] - At \( x = 1 \), match: \[ 2x^2 + 4x + 1 = cx^2 + bx + a \Rightarrow 2 + 4 + 1 = c + b + a = 7 \Rightarrow a + b + c = 7 \quad \text{(2)} \] Also given \( \lim_{x \to -3/2} f(x) = 14 \): \[ x = -\frac{3}{2} \in x \leq -1 \Rightarrow f(x) = a(\frac{9}{4}) - \frac{3}{2}b + c = 14 \Rightarrow \frac{9a}{4} - \frac{3b}{2} + c = 14 \quad \text{(3)} \] Solve the system (1), (2), (3) to find \( a, b, c \), then substitute into: \[ f(2) = c(4) + b(2) + a \Rightarrow \boxed{-8} \]