Question

Consider the polynomial \( f(x) = x^3 - 6x^2 + 11x - 6 \) on the domain \( S \) given by \( 1 \leq x \leq 3 \). The first and second derivatives are \( f'(x) \) and \( f''(x) \).
Consider the following statements:
I. The given polynomial is zero at the boundary points \( x = 1 \text{ and } x = 3. \)
II. There exists one local maxima of \( f(x) \text{ within the domain } S. \)
III. The second derivative \( f''(x)>0 \text{ throughout the domain } S. \)
IV. There exists one local minima of \( f(x) \text{ within the domain } S. \)
The correct option is:

• A. Only statements I, II and III are correct.
• B. Only statements I, II and IV are correct.
• C. Only statements I and IV are correct.
• D. Only statements II and IV are correct.

Hint:

For polynomials, always use the first and second derivatives to find critical points and determine the nature of those points (maxima or minima).

Answer 1

The Correct Option is B

Let’s evaluate the correctness of each statement:
Statement I: "The given polynomial is zero at the boundary points \( x = 1 \) and \( x = 3 \)."
We need to evaluate the polynomial at \( x = 1 \) and \( x = 3 \):
- For \( x = 1 \): \[ f(1) = 1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0. \] - For \( x = 3 \): \[ f(3) = 3^3 - 6(3)^2 + 11(3) - 6 = 27 - 54 + 33 - 6 = 0. \] Thus, the polynomial is zero at both \( x = 1 \) and \( x = 3 \). So, Statement I is True.
Statement II: "There exists one local maxima of \( f(x) \) within the domain \( S \)."
To find local maxima or minima, we first compute the first derivative \( f'(x) \) and set it equal to zero:
\[ f'(x) = 3x^2 - 12x + 11. \] Solving \( f'(x) = 0 \) for critical points:
\[ 3x^2 - 12x + 11 = 0. \] Using the quadratic formula: \[ x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(11)}}{2(3)} = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} = \frac{12 \pm 2\sqrt{3}}{6}. \] Thus, the critical points are \( x = \frac{6 + \sqrt{3}}{3} \) and \( x = \frac{6 - \sqrt{3}}{3} \), both of which lie within the domain \( 1 \leq x \leq 3 \).
We check the second derivative to confirm the nature of these points:
\[ f''(x) = 6x - 12. \] At the critical points, if \( f''(x)<0 \), we have a local maxima. Since \( f''(x) \) is negative for one of the critical points, there exists one local maxima within the domain. Hence, Statement II is True.
Statement III: "The second derivative f''(x) \(>\) 0 \text{ throughout the domain } S."
The second derivative is: \[ f''(x) = 6x - 12. \] This is positive for \( x>2 \) and negative for \( x<2 \). Therefore, \( f''(x) \) is not positive throughout the entire domain. Statement III is False.
Statement IV: "There exists one local minima of \( f(x) \) within the domain \( S \)."
Using the second derivative test: Since \( f''(x)>0 \) for \( x>2 \), there is one local minima within the domain. Hence, Statement IV is True.
Conclusion:
- Statement I is True,
- Statement II is True,
- Statement III is False,
- Statement IV is True.
Thus, the correct answer is (B).