Let `f(x) =int_-1^x e^(t^2) dt and h(x)=f(1+g(x)),` where `g (x)` is defined for all `x, g'(x)` exists for all `x, and g(x) < 0 for x > 0.` If `h'(1)=e and g'(1)= 1,` then the possible values which `g(1)` can take

Options

A. 0
B. `-1`
C. `-2`
D. `-4`

Answer

To solve the problem step by step, let's break it down:

**Step 1: Define the functions**
We are given:
- \( f(x) = \int_{-1}^{x} e^{t^2} dt \)
- \( h(x) = f(1 + g(x)) \)

**Step 2: Differentiate \( h(x) \)**
Using the chain rule, we differentiate \( h(x) \):
\[ h'(x) = f'(1 + g(x)) \cdot g'(x) \]

**Step 3: Evaluate at \( x = 1 \)**
We need to find \( h'(1) \):
\[ h'(1) = f'(1 + g(1)) \cdot g'(1) \]
We know \( h'(1) = e \) and \( g'(1) = 1 \), so we can substitute these values:
\[ e = f'(1 + g(1)) \cdot 1 \quad \Rightarrow \quad f'(1 + g(1)) = e \]

**Step 4: Differentiate \( f(x) \)**
Now, we differentiate \( f(x) \):
\[ f'(x) = e^{x^2} \]
Thus, at \( x = 1 + g(1) \):
\[ f'(1 + g(1)) = e^{(1 + g(1))^2} \]

**Step 5: Set up the equation**
From the previous steps, we have:
\[ e^{(1 + g(1))^2} = e \]

**Step 6: Solve for \( g(1) \)**
Taking the natural logarithm of both sides:
\[ (1 + g(1))^2 = 1 \]
This gives us two cases:
1. \( 1 + g(1) = 1 \) which leads to \( g(1) = 0 \)
2. \( 1 + g(1) = -1 \) which leads to \( g(1) = -2 \)

**Step 7: Check the conditions**
We know that \( g(x) < 0 \) for \( x > 0 \). Therefore, \( g(1) = 0 \) is not valid since it does not satisfy \( g(1) < 0 \). Hence, the only valid solution is:
\[ g(1) = -2 \]

**Final Answer:**
The possible value that \( g(1) \) can take is \( -2 \).

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