Question

f : R → R defined by f(x) = \(\left\{ \begin{aligned}     & 2x;x>3 \\     &x^2;1<x≤3\\     & 3x;x≤1 \end{aligned} \right.\) then f(-2) + f(3) + f(4) is

• A. 14
• B. 9
• C. 5
• D. 11

Answer 1

The Correct Option is D

We are given the piecewise function \( f(x) \) defined as:
\[ f(x) = \begin{cases} 2x & \text{if } x > 3 \\ x^2 & \text{if } 1 \leq x \leq 3 \\ 3x & \text{if } x \leq 1 \end{cases} \]
We are asked to find:
\[ f(-2) + f(3) + f(4) \]
Step 1: Find \( f(-2) \)
Since \( -2 \leq 1 \), we use the third case of the piecewise function, \( f(x) = 3x \):
\[ f(-2) = 3(-2) = -6 \]
Step 2: Find \( f(3) \)
Since \( 1 \leq 3 \leq 3 \), we use the second case of the piecewise function, \( f(x) = x^2 \):
\[ f(3) = 3^2 = 9 \]
Step 3: Find \( f(4) \)
Since \( 4 > 3 \), we use the first case of the piecewise function, \( f(x) = 2x \):
\[ f(4) = 2(4) = 8 \]
Step 4: Add the values
\[ f(-2) + f(3) + f(4) = -6 + 9 + 8 = 11 \]
Therefore, the correct answer is:
So, the correct answer is (D): 11

Answer 2

We are given the function \(f(x)\) defined as:

\(f(x) = \begin{cases} 2x & \text{if } x > 3 \\ x^2 & \text{if } 1 < x \leq 3 \\ 3x & \text{if } x \leq 1 \end{cases}\)

We need to find \(f(-2) + f(3) + f(4)\).

For \(f(-2)\), since -2 ≤ 1, we use the third case: \(f(-2) = 3(-2) = -6\).

For \(f(3)\), since 1 < 3 ≤ 3, we use the second case: \(f(3) = (3)^2 = 9\).

For \(f(4)\), since 4 > 3, we use the first case: \(f(4) = 2(4) = 8\).

Therefore, \(f(-2) + f(3) + f(4) = -6 + 9 + 8 = 3 + 8 = 11\).

The value of \(f(-2) + f(3) + f(4)\) is 11.