Question

If \(|x|\) is so small that \(x^2\) and higher powers of \(x\) may be neglected, and hence \[ \frac{(8-3x)^{1/3}}{(2+3x)^3} \approx \frac{1}{a}(1+bx), \] then \(2ab=\) 

• A. \( -\frac{37}{2} \)
• B. \( -\frac{35}{2} \)
• C. \( -37 \)
• D. \( -35 \)

Hint:

Always ensure the constant term inside the parenthesis is \( 1 \) before applying the binomial approximation. If it is not, factor out the constant raised to the power first.

Answer 1

The Correct Option is C

Concept: When \( |x| \) is very small, we can use the binomial approximation for any rational index:
• \( (1+x)^n \approx 1 + nx \) for \( |x| \ll 1 \).
• Neglect terms containing \( x^2, x^3, \dots \) during expansion.

Step 1: Factor out the constants to form \( (1+z)^n \).
The given expression is \( \frac{(8-3x)^{1/3}}{(2+3x)^3} \). \[ = \frac{8^{1/3}\left(1 - \frac{3x}{8}\right)^{1/3}}{2^3\left(1 + \frac{3x}{2}\right)^3} = \frac{2\left(1 - \frac{3x}{8}\right)^{1/3}}{8\left(1 + \frac{3x}{2}\right)^3} = \frac{1}{4} \left(1 - \frac{3x}{8}\right)^{1/3} \left(1 + \frac{3x}{2}\right)^{-3} \]

Step 2: Apply binomial approximation.
Using \( (1+z)^n \approx 1+nz \): \[ \approx \frac{1}{4} \left[ 1 + \frac{1}{3}\left(-\frac{3x}{8}\right) \right] \left[ 1 + (-3)\left(\frac{3x}{2}\right) \right] \] \[ = \frac{1}{4} \left( 1 - \frac{x}{8} \right) \left( 1 - \frac{9x}{2} \right) \]

Step 3: Multiply and find \( a \) and \( b \).
Neglecting the \( x^2 \) term: \[ \approx \frac{1}{4} \left( 1 - \frac{9x}{2} - \frac{x}{8} \right) = \frac{1}{4} \left( 1 - \frac{36x + x}{8} \right) = \frac{1}{4} \left( 1 - \frac{37x}{8} \right) \] Comparing with \( \frac{1}{a}(1+bx) \), we get \( a = 4 \) and \( b = -\frac{37}{8} \). Therefore, \( 2ab = 2 \times 4 \times \left(-\frac{37}{8}\right) = -37 \).