A twice differentiable function f(x)is defined for all real numbers and satisfies the following conditions `f(0) = 2; f'(0)--5 and f"(0) = 3`. The function `g(x)` is defined by `g(x) = e^(ax) + f (x) AA x in R`, where 'a' is any constant If `g'(0) + g"(0)=0`. Find the value(s) of 'a'

Options

A. 1
B. `-1`
C. 2
D. `-2`

Explanation

To solve the problem step by step, we need to find the values of 'a' such that \( g'(0) + g''(0) = 0 \) given the function \( g(x) = e^{ax} + f(x) \) and the conditions on \( f(x) \).

### Step 1: Find \( g'(x) \)

The function \( g(x) \) is defined as:
\[ g(x) = e^{ax} + f(x) \]

To find the first derivative \( g'(x) \):
\[ g'(x) = \frac{d}{dx}(e^{ax}) + \frac{d}{dx}(f(x)) = ae^{ax} + f'(x) \]

### Step 2: Evaluate \( g'(0) \)

Now, we need to evaluate \( g'(0) \):
\[ g'(0) = ae^{a \cdot 0} + f'(0) = a \cdot 1 + f'(0) = a + f'(0) \]
Given that \( f'(0) = -5 \):
\[ g'(0) = a - 5 \tag{1} \]

### Step 3: Find \( g''(x) \)

Next, we find the second derivative \( g''(x) \):
\[ g''(x) = \frac{d}{dx}(g'(x)) = \frac{d}{dx}(ae^{ax} + f'(x)) = a^2 e^{ax} + f''(x) \]

### Step 4: Evaluate \( g''(0) \)

Now, we evaluate \( g''(0) \):
\[ g''(0) = a^2 e^{a \cdot 0} + f''(0) = a^2 \cdot 1 + f''(0) = a^2 + f''(0) \]
Given that \( f''(0) = 3 \):
\[ g''(0) = a^2 + 3 \tag{2} \]

### Step 5: Set up the equation \( g'(0) + g''(0) = 0 \)

Now we can set up the equation:
\[ g'(0) + g''(0) = 0 \]
Substituting equations (1) and (2):
\[ (a - 5) + (a^2 + 3) = 0 \]
This simplifies to:
\[ a^2 + a - 2 = 0 \]

### Step 6: Solve the quadratic equation

Now we solve the quadratic equation:
\[ a^2 + a - 2 = 0 \]
Factoring:
\[ (a + 2)(a - 1) = 0 \]
Setting each factor to zero gives:
\[ a + 2 = 0 \quad \Rightarrow \quad a = -2 \]
\[ a - 1 = 0 \quad \Rightarrow \quad a = 1 \]

### Step 7: Conclusion

The values of \( a \) are:
\[ a = -2 \quad \text{and} \quad a = 1 \]