If `f : IR rarr IR` be defined by
`f(x)={(2x,xgt3),(x^(2),1ltxle3."thenf(-1)+f(2)+f(5)=),(3x,xle1):}`

Options

A. 10
B. 11
C. 12
D. 14

Explanation

To solve the problem, we need to evaluate the function \( f \) at three different points: \( f(-1) \), \( f(2) \), and \( f(5) \). The function \( f \) is defined piecewise as follows:

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

Now, let's calculate each term step by step.

### Step 1: Calculate \( f(-1) \)
Since \(-1 < 1\), we use the first case of the function:
\[ f(-1) = 3 \times (-1) = -3 \]

### Step 2: Calculate \( f(2) \)
Since \(2\) lies between \(1\) and \(3\), we use the second case of the function:
\[ f(2) = 2^2 = 4 \]

### Step 3: Calculate \( f(5) \)
Since \(5 > 3\), we use the third case of the function:
\[ f(5) = 2 \times 5 = 10 \]

### Step 4: Combine the results
Now we sum the results from the three evaluations:
\[ f(-1) + f(2) + f(5) = -3 + 4 + 10 \]

Calculating this gives:
\[ -3 + 4 = 1 \\ 1 + 10 = 11 \]

### Final Result
Thus, the final result is:
\[ f(-1) + f(2) + f(5) = 11 \]