If `f` is twice differentiable function for `x in R` such that `f(2)=5,f'(2)=8` and `f'(x) ge 1,f''(x) ge4` , then

Options

A. `f(5)+f'(5) le 26`
B. `f(5)+f'(5) ge 28`
C. `f(5)+f'(5) le28`
D. none of these

Answer

To solve the problem step by step, we need to find the values of \( f(5) \) and \( f'(5) \) given the conditions provided.

### Step 1: Find \( f(5) \)

We know that \( f'(x) \geq 1 \) for all \( x \). This means that the function \( f \) is increasing at a rate of at least 1 unit for every unit increase in \( x \).

Using the Fundamental Theorem of Calculus, we can express \( f(5) \) in terms of \( f(2) \) and \( f'(x) \):

\[ f(5) - f(2) = \int_{2}^{5} f'(x) \, dx \]

Since \( f'(x) \geq 1 \), we have:

\[ \int_{2}^{5} f'(x) \, dx \geq \int_{2}^{5} 1 \, dx = 5 - 2 = 3 \]

Thus, we can write:

\[ f(5) - f(2) \geq 3 \]

Given that \( f(2) = 5 \):

\[ f(5) - 5 \geq 3 \]

This simplifies to:

\[ f(5) \geq 8 \]

### Step 2: Find \( f'(5) \)

We also know that \( f''(x) \geq 4 \). This means that the derivative \( f'(x) \) is increasing at a rate of at least 4 units for every unit increase in \( x \).

Using a similar approach, we can express \( f'(5) \) in terms of \( f'(2) \):

\[ f'(5) - f'(2) = \int_{2}^{5} f''(x) \, dx \]

Since \( f''(x) \geq 4 \), we have:

\[ \int_{2}^{5} f''(x) \, dx \geq \int_{2}^{5} 4 \, dx = 4 \times (5 - 2) = 12 \]

Thus, we can write:

\[ f'(5) - f'(2) \geq 12 \]

Given that \( f'(2) = 8 \):

\[ f'(5) - 8 \geq 12 \]

This simplifies to:

\[ f'(5) \geq 20 \]

### Step 3: Combine Results

Now we have:

1. \( f(5) \geq 8 \)
2. \( f'(5) \geq 20 \)

We need to find \( f(5) + f'(5) \):

\[ f(5) + f'(5) \geq 8 + 20 = 28 \]

### Conclusion

Thus, we conclude that:

\[ f(5) + f'(5) \geq 28 \]