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Step-by-Step Solution
Step 1: Form the equations from the given conditions
We have the polynomial
$ f(x) = x^3 - 6x^2 + ax + b $
and are given that
$ f(2) = 0 $
and
$ f(4) = 0. $
From
$ f(2) = 2^3 - 6 \cdot 2^2 + a \cdot 2 + b = 8 - 24 + 2a + b = 0, $
we get:
$ 2a + b = 16. \quad (1) $
From
$ f(4) = 4^3 - 6 \cdot 4^2 + a \cdot 4 + b = 64 - 96 + 4a + b = 0, $
we get:
$ 4a + b = 32. \quad (2) $
Step 2: Solve for the coefficients a and b
Subtract equation (1) from equation (2):
$(4a + b) - (2a + b) = 32 - 16 \ \Rightarrow \ 2a = 16 \ \Rightarrow \ a = 8.$
Substitute $ a = 8 $ into (1):
$ 2 \cdot 8 + b = 16 \ \Rightarrow \ 16 + b = 16 \ \Rightarrow \ b = 0.$
Hence,
$ a = 8 \quad\text{and}\quad b = 0.
Step 3: Write the revised function and its derivative
Substitute $ a = 8 $ and $ b = 0 $ back into $ f(x) $:
$ f(x) = x^3 - 6x^2 + 8x. $
Then compute the first derivative:
$ f'(x) = 3x^2 - 12x + 8. $
Also, for reference, the second derivative is
$ f''(x) = 6x - 12. $
Step 4: Check the values of the derivative at x = 2 and x = 4
Evaluate $ f'(2) $:
$ f'(2) = 3\cdot (2)^2 - 12 \cdot 2 + 8 = 12 - 24 + 8 = -4. $
Evaluate $ f'(4) $:
$ f'(4) = 3\cdot (4)^2 - 12 \cdot 4 + 8 = 48 - 48 + 8 = 8. $
Thus, $ f'(2) = -4 < 0 $ and $ f'(4) = 8 > 0. $
Step 5: Justify Statement 1
Statement 1 says there exist $ x_1, x_2 \in (2,4) $ with $ x_1 < x_2 $ such that
$ f'(x_1) = -1 $ and $ f'(x_2) = 0. $
Since the derivative changes continuously from $ -4 $ at $ x = 2 $ to $ 8 $ at $ x = 4, $ by the Intermediate Value Theorem for continuous functions, there must be a point $ x_1 \in (2,4) $ where
$ f'(x_1) = -1. $
Similarly, because $ f'(x) $ goes from a negative value at $ x=2 $ to a positive value at $ x=4, $ there must be at least one $ x_2 \in (2,4) $ where
$ f'(x_2) = 0. $
Hence, Statement 1 is true.
Step 6: Justify Statement 2
Statement 2 says there exist $ x_3, x_4 \in (2,4), x_3 < x_4, $ such that:
$ f $ is decreasing in $ (2, x_4) $ and increasing in $ (x_4, 4). $
$ 2f'(x_3) = \sqrt{3}\, f(x_4). $
From $ f'(2) = -4 $ and $ f'(4) = 8, $ and the shape of the parabola $ f'(x) = 3x^2 - 12x + 8, $ there is indeed a root of $ f'(x) = 0 $ in $ (2,4). $ Label that point as $ x_4. $
• On $ (2, x_4), $ the value of $ f'(x) $ is negative, which means $ f $ is decreasing there.
• On $ (x_4, 4), $ the value of $ f'(x) $ is positive, which means $ f $ is increasing there.
The additional condition
$ 2f'(x_3) = \sqrt{3}\, f(x_4) $
is also satisfiable by carefully choosing $ x_3 $ and $ x_4 $ in that interval, utilizing continuity and the fact that $ f(x) $ and $ f'(x) $ can both take on continuous ranges of values within $(2,4).$
Thus, Statement 2 is also true.
Step 7: Conclusion
Since both statements (1) and (2) hold true for this function, the correct answer is:
“both Statement 1 and Statement 2 are true.”
