Question

If \(x\) is numerically so small so that \(x^2\) and higher powers of \(x\) can be neglected, then
\[ \left(1+\frac{2x}{3}\right)^{3/2}\left(32+5x\right)^{-1/5} \] is approximately equal to

• A. \(\frac{32+31x}{64}\)
• B. \(\frac{31+32x}{64}\)
• C. \(\frac{31-32x}{64}\)
• D. \(\frac{1-2x}{64}\)

Hint:

For small \(x\), use \((1+x)^n \approx 1+nx\) and ignore \(x^2\) terms while multiplying expansions.

Answer 1

The Correct Option is A

Step 1: Expand first term using binomial approximation.
\[ \left(1+\frac{2x}{3}\right)^{3/2} \approx 1+\frac{3}{2}\cdot\frac{2x}{3} =1+x \]
Step 2: Rewrite second term.
\[ (32+5x)^{-1/5} =32^{-1/5}\left(1+\frac{5x}{32}\right)^{-1/5} \]
Now \(32=2^5\), so:
\[ 32^{-1/5}=2^{-1}=\frac{1}{2} \]
Step 3: Expand \(\left(1+\frac{5x}{32}\right)^{-1/5\).}
\[ (1+u)^n \approx 1+nu \]
Here \(u=\frac{5x}{32}\), \(n=-\frac{1}{5}\).
\[ \left(1+\frac{5x}{32}\right)^{-1/5} \approx 1-\frac{1}{5}\cdot\frac{5x}{32} =1-\frac{x}{32} \]
So:
\[ (32+5x)^{-1/5}\approx \frac{1}{2}\left(1-\frac{x}{32}\right) \]
Step 4: Multiply both approximations.
\[ (1+x)\cdot \frac{1}{2}\left(1-\frac{x}{32}\right) \approx \frac{1}{2}\left(1+x-\frac{x}{32}\right) \]
\[ = \frac{1}{2}\left(1+\frac{31x}{32}\right) = \frac{1}{2}+\frac{31x}{64} \]
\[ = \frac{32}{64}+\frac{31x}{64} = \frac{32+31x}{64} \]
Final Answer:
\[ \boxed{\frac{32+31x}{64}} \]