Answer is: option1

Solution:

Since \( f'(x) > 0 \), the graph is increasing.

Since \( f''(x) < 0 \), the graph is concave down (slopes are decreasing).

So the function increases but at a decreasing rate — like a curve that flattens as it rises.

The increase from \( x = 1 \) to \( x = 3 \) is:

\( f(3) - f(1) = 12 - 7 = 5 \)

Because \( f \) is concave down, the rate of increase slows down beyond \( x = 3 \). So from \( x = 3 \) to \( x = 5 \), the increase must be less than 5.

If we again had an increase of 5, we’d get \( f(5) = 12 + 5 = 17 \), but that would suggest a constant rate of increase, not a slowing one.

So:
\( f(5) < 17 \)

\( f(5) = 16 \) is a possible value