Question

\[ \frac{1}{(D-2)^2}x^2= \]

• A. \(\dfrac{1}{4}\left(x^2+2x+\dfrac{3}{2}\right)\)
• B. \(\dfrac{1}{4}\left(x^2+\dfrac{3}{2}\right)\)
• C. \(-\dfrac{1}{4}\left(x^2+\dfrac{3}{2}\right)\)
• D. \(\dfrac{1}{4}\left(x^2+x+\dfrac{3}{4}\right)\)

Hint:

For inverse differential operators acting on polynomials, expand the operator and stop when higher derivatives become zero.

Answer 1

The Correct Option is A

Concept:
We need to evaluate \[ \frac{1}{(D-2)^2}x^2 \] where \[ D=\frac{d}{dx} \] For polynomial functions, we can expand the inverse operator.

Step 1: Rewrite the operator. \[ (D-2)^2=4\left(1-\frac{D}{2}\right)^2 \] Therefore, \[ \frac{1}{(D-2)^2} = \frac{1}{4}\left(1-\frac{D}{2}\right)^{-2} \]

Step 2: Use binomial expansion. We know that \[ (1-z)^{-2}=1+2z+3z^2+\cdots \] Here, \[ z=\frac{D}{2} \] So, \[ \left(1-\frac{D}{2}\right)^{-2} = 1+2\left(\frac{D}{2}\right)+3\left(\frac{D}{2}\right)^2+\cdots \] \[ =1+D+\frac{3D^2}{4}+\cdots \]

Step 3: Apply to \(x^2\). \[ \frac{1}{(D-2)^2}x^2 = \frac{1}{4}\left(1+D+\frac{3D^2}{4}\right)x^2 \] Higher derivatives after \(D^2\) vanish because \(x^2\) is a polynomial of degree \(2\). Now, \[ D(x^2)=2x \] and \[ D^2(x^2)=2 \] Therefore, \[ \frac{1}{(D-2)^2}x^2 = \frac{1}{4}\left(x^2+2x+\frac{3}{4}\cdot 2\right) \] \[ = \frac{1}{4}\left(x^2+2x+\frac{3}{2}\right) \]

Step 4: Final answer. \[ \boxed{\frac{1}{4}\left(x^2+2x+\frac{3}{2}\right)} \]