If \(f(x) = \begin{cases} x^2-4x & \quad \text{if } x≥2 \text{}\\ x

1.	If \(f(x) = \begin{cases} x^2-4x & \quad \text{if } x≥2 \text{}\\ x+2 & \quad \text{if } x<2 \text{} \end{cases}\), then f(0) is (a) 2 (b) 5 (c) -1 (d) 0
Answer» (a) 2 \(f(x) = \begin{cases} x^2-4x & \quad \text{if } x≥0 \text{}\\ x+2 & \quad \text{if } x<2 \text{} \end{cases}\) f(0) = 0 + 2 = 2 [For x = 0 take f(x) = x + 2]

If \(f(x) = \begin{cases} x^2-4x & \quad \text{if } x≥2 \text{}\\ x+2 & \quad \text{if } x<2 \text{} \end{cases}\), then f(0) is (a) 2 (b) 5 (c) -1 (d) 0

Answer»

(a) 2

\(f(x) = \begin{cases} x^2-4x & \quad \text{if } x≥0 \text{}\\ x+2 & \quad \text{if } x<2 \text{} \end{cases}\)

f(0) = 0 + 2 = 2

[For x = 0 take f(x) = x + 2]