We want to find the coefficient of \(x\) in the expansion of
\[ \frac{(1 - 4x)^2 (1 - 2x^2)^{1/2}}{(4 - x)^{3/2}}. \]
We can write
\[ \frac{(1 - 4x)^2 (1 - 2x^2)^{1/2}}{(4 - x)^{3/2}} = (1 - 4x)^2 (1 - 2x^2)^{1/2} \cdot 4^{-3/2} \left( 1 - \frac{x}{4} \right)^{-3/2} \] \[ = \frac{1}{8} (1 - 8x + 16x^2) (1 - x^2 + \ldots) \left( 1 + \frac{3}{8} x + \ldots \right). \]
We are only interested in the coefficient of \(x\), so we can ignore terms of degree 2 or higher. Then
\[ \frac{(1 - 4x)^2 (1 - 2x^2)^{1/2}}{(4 - x)^{3/2}} = \frac{1}{8} (1 - 8x) \left( 1 + \frac{3}{8} x \right) + O(x^2) \] \[ = \frac{1}{8} \left( 1 - 8x + \frac{3}{8} x - 3x^2 \right) + O(x^2) \] \[ = \frac{1}{8} \left( 1 - \frac{61}{8} x \right) + O(x^2) \] \[ = \frac{1}{8} - \frac{61}{64} x + O(x^2). \]
Therefore, the coefficient of \(x\) is \(-\frac{61}{64}\).
To solve this problem, we first perform the partial fraction decomposition of the given expression:
\[ \frac{4x^2 + 5}{(x - 2)^4} = \frac{A}{(x - 2)} + \frac{B}{(x - 2)^2} + \frac{C}{(x - 2)^3} + \frac{D}{(x - 2)^4}. \]
Step 1: Clear the denominator
Multiply both sides by \( (x - 2)^4 \) to eliminate the denominators:
\[ 4x^2 + 5 = A(x - 2)^3 + B(x - 2)^2 + C(x - 2) + D. \]
Step 2: Expand the right-hand side
\[ A(x - 2)^3 = A(x^3 - 6x^2 + 12x - 8), \] \[ B(x - 2)^2 = B(x^2 - 4x + 4), \] \[ C(x - 2) = Cx - 2C, \] \[ D = D. \]
Combine these terms:
\[ A(x^3 - 6x^2 + 12x - 8) + B(x^2 - 4x + 4) + Cx - 2C + D. \]
Step 3: Equate coefficients
Compare the expanded form with the left-hand side \( 4x^2 + 5 \).
Step 4: Solve for \( A, B, C, D \)
\[ A = 0, \quad B = 4, \quad C = 16, \quad D = 21. \]
Step 5: Compute the expression
\[ \sqrt{\frac{A}{C} + \frac{B}{C} + \frac{D}{C}} = \frac{5}{4}. \]
Final Answer: \(\frac{5}{4}\)
What are X and Y respectively in the following set of reactions?
What are X and Y respectively in the following reactions?
Observe the following reactions:
The correct answer is: