Question

When $ x $ is so small that its square and its higher powers may be neglected, then the value of $ \frac{(1 + \frac{3}{4}x)^{-4} \sqrt{(3 + x)}}{\sqrt{(3 - x)^3}} $ is approximately equal to

• A. $ \frac{1}{3} - \frac{7x}{9} $
• B. $ \frac{1}{3} + \frac{7x}{9} $
• C. $ \frac{1}{3} + \frac{11x}{18} $
• D. $ \frac{1}{3} - \frac{11x}{18} $

Hint:

Use binomial expansion $ (1+y)^n \approx 1+ny $ for small $ y $. Factor out constants before applying the expansion. Neglect higher order terms of $ x $.

Answer 1

The Correct Option is A

Step 1: Rewrite the expression in a form suitable for binomial expansion.
The given expression is $ \frac{(1 + \frac{3}{4}x)^{-4} \sqrt{(3 + x)}}{\sqrt{(3 - x)^3}} $.
Rewriting the terms: $ \frac{(1 + \frac{3}{4}x)^{-4} \sqrt{3} (1 + \frac{x}{3})^{\frac{1}{2}}}{3^{\frac{3}{2}} (1 - \frac{x}{3})^{\frac{3}{2}}} = \frac{1}{3} (1 + \frac{3}{4}x)^{-4} (1 + \frac{x}{3})^{\frac{1}{2}} (1 - \frac{x}{3})^{-\frac{3}{2}} $. 
Step 2: Apply the binomial expansion for small $ x $.
$ (1 + \frac{3}{4}x)^{-4} \approx 1 - 3x $
$ (1 + \frac{x}{3})^{\frac{1}{2}} \approx 1 + \frac{x}{6} $
$ (1 - \frac{x}{3})^{-\frac{3}{2}} \approx 1 + \frac{x}{2} $
Step 3: Substitute the approximations back into the expression.
$ \frac{1}{3} (1 - 3x) (1 + \frac{x}{6}) (1 + \frac{x}{2}) $ 
Step 4: Multiply the terms, neglecting terms with $ x^2 $ and higher powers.
$ \frac{1}{3} (1 - 3x) (1 + \frac{2}{3}x) = \frac{1}{3} (1 + \frac{2}{3}x - 3x) = \frac{1}{3} (1 - \frac{7}{3}x) = \frac{1}{3} - \frac{7x}{9} $. 
Step 5: Conclusion.
The approximate value is $ \frac{1}{3} - \frac{7x}{9} $.