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Let f:RtoR be a function such that f(x)=...

Let `f:RtoR` be a function such that `f(x)=x^(3)+x^(2)f'(1)+xf''(2)+f'''(3)` for `x in R`
Consider the following
1. `f(2)=f(1)-f(0)`
`2.f''(2)-2f'(1)=12`
Which of the above is/are correct ?

A

Only 1

B

Only 2

C

Both 1 and 2

D

Neither 1 nor 2

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the function given and the two statements provided. ### Step 1: Understanding the function The function is defined as: \[ f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3) \] ### Step 2: Finding the derivatives We will find the first, second, and third derivatives of \( f(x) \). 1. **First derivative \( f'(x) \)**: \[ f'(x) = 3x^2 + 2x f'(1) + f''(2) \] Here, \( f''(2) \) is constant with respect to \( x \). 2. **Second derivative \( f''(x) \)**: \[ f''(x) = 6x + 2f'(1) \] 3. **Third derivative \( f'''(x) \)**: \[ f'''(x) = 6 \] This is constant. ### Step 3: Evaluating the function at specific points Now we will evaluate \( f(0) \), \( f(1) \), and \( f(2) \). 1. **Calculating \( f(0) \)**: \[ f(0) = 0^3 + 0^2 f'(1) + 0 \cdot f''(2) + f'''(3) = f'''(3) = 6 \] 2. **Calculating \( f(1) \)**: \[ f(1) = 1^3 + 1^2 f'(1) + 1 \cdot f''(2) + f'''(3) = 1 + f'(1) + f''(2) + 6 \] Thus, \[ f(1) = 7 + f'(1) + f''(2) \] 3. **Calculating \( f(2) \)**: \[ f(2) = 2^3 + 2^2 f'(1) + 2 \cdot f''(2) + f'''(3) = 8 + 4f'(1) + 2f''(2) + 6 \] Thus, \[ f(2) = 14 + 4f'(1) + 2f''(2) \] ### Step 4: Analyzing the statements 1. **First statement**: \( f(2) = f(1) - f(0) \) \[ 14 + 4f'(1) + 2f''(2) = (7 + f'(1) + f''(2)) - 6 \] Simplifying the right side: \[ 7 + f'(1) + f''(2) - 6 = 1 + f'(1) + f''(2) \] Setting both sides equal: \[ 14 + 4f'(1) + 2f''(2) = 1 + f'(1) + f''(2) \] Rearranging gives: \[ 13 + 3f'(1) + f''(2) = 0 \quad \text{(Equation 1)} \] 2. **Second statement**: \( f''(2) - 2f'(1) = 12 \) From our earlier expression for \( f''(2) \): \[ f''(2) = 6 + 2f'(1) \quad \text{(Equation 2)} \] Substituting into the second statement gives: \[ (6 + 2f'(1)) - 2f'(1) = 12 \] Simplifying: \[ 6 = 12 \quad \text{(False)} \] ### Conclusion 1. The first statement is true. 2. The second statement is false. Thus, the correct answer is that only the first statement is correct.
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