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Letfand g be differentiable on the inter...

Letfand g be differentiable on the interval I and let a, `b in I,a lt b ` . Then

A

If f(a) =0 =f(b), the equation f"(x) +f(x)g'(x) = 0 is solvable in (a, b)

B

If f(x)= 0 -f(b), the equation f'(x)+f(x)g'(x) =0 may not be solvable in (a, b).

C

If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 is solvable in (a, b), k `in`R.

D

If g(a) = 0 = g(b), the equation g'(x) + kg(x) = 0 more than may not be solvable in (a, b), k `in`R.

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To solve the given problem, we need to analyze the statements regarding the differentiable functions \( f \) and \( g \) on the interval \( I \) where \( a, b \in I \) and \( a < b \). ### Step-by-Step Solution: 1. **Understanding the Conditions**: We have two functions \( f \) and \( g \) that are differentiable on the interval \( I \). We are given specific conditions about the values of these functions at the endpoints \( a \) and \( b \). 2. **Analyzing Option 1**: - If \( f(a) = 0 \) and \( f(b) = 0 \), we can apply Rolle's Theorem. According to this theorem, since \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists at least one \( c \in (a, b) \) such that \( f'(c) = 0 \). - Now consider the equation \( f'(x) + f(x) g'(x) = 0 \). We can define a function \( h(x) = f'(x) + f(x) g'(x) \). - Since \( h(a) = f'(a) + f(a)g'(a) = f'(a) \) and \( h(b) = f'(b) + f(b)g'(b) = f'(b) \), and knowing that \( f(a) = 0 \) and \( f(b) = 0 \), we can conclude that \( h(a) \) and \( h(b) \) must have opposite signs (assuming \( f' \) does not change sign). - Thus, by the Intermediate Value Theorem, \( h(x) = 0 \) has at least one solution in \( (a, b) \). 3. **Analyzing Option 2**: - This option states that if \( f(x) = 0 \) at \( a \) and \( b \), the equation may not be solvable in \( (a, b) \). However, we have already established that under the conditions of the first option, it is indeed solvable. Hence, this option is incorrect. 4. **Analyzing Option 3**: - If \( g(a) = 0 \) and \( g(b) = 0 \), we can similarly apply Rolle's Theorem to \( g \). There exists at least one \( d \in (a, b) \) such that \( g'(d) = 0 \). - Now consider the equation \( g'(x) + k g(x) = 0 \). We can analyze this as well. Since \( g(a) = 0 \) and \( g(b) = 0 \), it follows that \( g'(a) \) and \( g'(b) \) must also have opposite signs. - Therefore, by the Intermediate Value Theorem, this equation also has at least one solution in \( (a, b) \). 5. **Analyzing Option 4**: - This option states that the equation may not be solvable in \( (a, b) \). However, we have shown that it is indeed solvable under the conditions given, making this option incorrect. ### Conclusion: The correct options based on the analysis are: - Option 1: If \( f(a) = 0 \) and \( f(b) = 0 \), the equation \( f'(x) + f(x) g'(x) = 0 \) is solvable in \( (a, b) \). - Option 3: If \( g(a) = 0 \) and \( g(b) = 0 \), the equation \( g'(x) + k g(x) = 0 \) is solvable in \( (a, b) \). Thus, the final answer is **Option 1 and Option 3**.
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