Answer is: option2

\( (0,0) \) and \( (1,-1) \)

Solution:

Finding Points of Inflection

A point of inflection occurs where the second derivative changes sign.

First Derivative:

\[ y' = \frac{d}{dx} (x^4 - 2x^3) = 4x^3 - 6x^2 \]

Second Derivative:

\[ y'' = \frac{d}{dx} (4x^3 - 6x^2) = 12x^2 - 12x \]

Setting \( y'' = 0 \):

\[ 12x^2 - 12x = 0 \]

Factoring out \( 12x \):

\[ 12x(x - 1) = 0 \]

Thus, \( x = 0 \) or \( x = 1 \).

Checking the Sign Change of \( y'' \):

• For \( x < 0 \) (e.g., \( x = -1 \)):

\[ y''(-1) = 12(-1)^2 - 12(-1) = 12 + 12 = 24 \quad \text{(positive)} \]

• For \( 0 < x < 1 \) (e.g., \( x = 0.5 \)):

\[ y''(0.5) = 12(0.5)^2 - 12(0.5) = 12(0.25) - 6 = 3 - 6 = -3 \quad \text{(negative)} \]

• For \( x > 1 \) (e.g., \( x = 2 \)):

\[ y''(2) = 12(2)^2 - 12(2) = 48 - 24 = 24 \quad \text{(positive)} \]

Since \( y'' \) changes sign at both \( x = 0 \) and \( x = 1 \), these are points of inflection.

Finding \( y(1) \):

\[ y(1) = 1^4 - 2(1)^3 = 1 - 2 = -1 \]

Thus, the points of inflection are:

\( (0,0) \) and \( (1,-1) \).

Final Answer: (B) (0,0) and (1,-1)