1. Understand the problem:
We need to analyze the nature of the point \( x = 2 \) for the function \( f(x) = x^3 - 6x^2 + 12x - 3 \).
2. Find the first derivative:
\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x - 3) = 3x^2 - 12x + 12 \]
3. Identify critical points:
Set \( f'(x) = 0 \):
\[ 3x^2 - 12x + 12 = 0 \implies x^2 - 4x + 4 = 0 \implies (x-2)^2 = 0 \implies x = 2 \]
Thus, \( x = 2 \) is a critical point.
4. Find the second derivative:
\[ f''(x) = \frac{d}{dx}(3x^2 - 12x + 12) = 6x - 12 \]
5. Evaluate the second derivative at \( x = 2 \):
\[ f''(2) = 6(2) - 12 = 0 \]
Since \( f''(2) = 0 \), the second derivative test is inconclusive.
6. Analyze higher-order derivatives or behavior:
Compute the third derivative:
\[ f'''(x) = 6 \neq 0 \]
Since the lowest-order non-zero derivative at \( x = 2 \) is odd, \( x = 2 \) is a point of inflection.
Correct Answer: (B) A point of inflexion
Find $f'(x)$ and $f''(x)$: \[ f'(x) = 3x^2 - 12x + 12, \quad f''(x) = 6x - 12. \] At $x = 2$: \[ f'(2) = 0, \quad f''(2) = 0. \] Check $f'''(x)$: \[ f'''(x) = 6 \implies f'''(2) = 6 \neq 0. \] Thus, $x = 2$ is a point of inflection.
Thus, the correct option is (B) A point of inflexion

A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:
(i)} Express the distance \( y \) between the wall and foot of the ladder in terms of \( h \) and height \( x \) on the wall at a certain instant. Also, write an expression in terms of \( h \) and \( x \) for the area \( A \) of the right triangle, as seen from the side by an observer.
Find the Derivative \( \frac{dy}{dx} \)
Given:\[ y = \cos(x^2) + \cos(2x) + \cos^2(x^2) + \cos(x^x) \]