Key Concept: Binomial Approximation for Small $x$

When $x$ is very small (approaching 0), we can use the binomial expansion to approximate expressions of the form $(1+y)^n$. The general formula is: $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!}y^2 + \frac{n(n-1)(n-2)}{3!}y^3 + \dots$ If $x$ is so small that $x^3$ and higher powers of $x$ can be neglected, we only need to keep terms up to $x^2$: $(1+y)^n \approx 1 + ny + \frac{n(n-1)}{2!}y^2$ This approximation is valid because as $x \to 0$, $x^3, x^4, \dots$ become negligibly small compared to $1, x, x^2$.

Step-by-Step Solution

We need to approximate the expression: $E = \frac{{{{\left( {1 + x} \right)}^{{3 \over 2}}} - {{\left( {1 + {1 \over 2}x} \right)}^3}}}{{{{\left( {1 - x} \right)}^{{1 \over 2}}}}}$

Our strategy is to expand each term in the numerator and the denominator separately using the binomial approximation up to $x^2$ (or as needed), and then combine them.

A. Expanding the Numerator Terms

1. Expand $(1+x)^{3/2}$: Here, $y=x$ and $n=3/2$. $(1+x)^{3/2} \approx 1 + \left(\frac{3}{2}\right)x + \frac{\frac{3}{2}\left(\frac{3}{2}-1\right)}{2!}x^2$ $= 1 + \frac{3}{2}x + \frac{\frac{3}{2} \cdot \frac{1}{2}}{2}x^2$ $= 1 + \frac{3}{2}x + \frac{3}{8}x^2 \quad \text{(Equation 1)}$ Explanation: We apply the binomial approximation formula directly, keeping terms up to $x^2$ as specified in the problem.

2. Expand $(1 + \frac{1}{2}x)^3$: Here, $y=\frac{1}{2}x$ and $n=3$. Since the power is an integer, we can use the direct binomial expansion $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. $(1 + \frac{1}{2}x)^3 = 1^3 + 3(1)^2\left(\frac{1}{2}x\right) + 3(1)\left(\frac{1}{2}x\right)^2 + \left(\frac{1}{2}x\right)^3$ $= 1 + \frac{3}{2}x + 3\left(\frac{1}{4}x^2\right) + \frac{1}{8}x^3$ $\approx 1 + \frac{3}{2}x + \frac{3}{4}x^2 \quad \text{(Equation 2)}$ Explanation: We expand $(1 + \frac{1}{2}x)^3$ completely. Then, we neglect the $x^3$ term (and any higher powers if they existed) as per the problem statement.

3. Subtract the expanded terms for the numerator: Numerator $= (1+x)^{3/2} - (1 + \frac{1}{2}x)^3$ Substitute from (Equation 1) and (Equation 2): $= \left(1 + \frac{3}{2}x + \frac{3}{8}x^2\right) - \left(1 + \frac{3}{2}x + \frac{3}{4}x^2\right)$ $= 1 + \frac{3}{2}x + \frac{3}{8}x^2 - 1 - \frac{3}{2}x - \frac{3}{4}x^2$ $= \left(1-1\right) + \left(\frac{3}{2}x - \frac{3}{2}x\right) + \left(\frac{3}{8}x^2 - \frac{3}{4}x^2\right)$ $= 0 + 0 + \left(\frac{3}{8} - \frac{6}{8}\right)x^2$ $= -\frac{3}{8}x^2 \quad \text{(Equation 3)}$ Explanation: We combine like terms. Notice that the constant terms and the $x$ terms cancel out, leaving only an $x^2$ term.

B. Expanding the Denominator Term

The denominator is $(1-x)^{1/2}$. To simplify the division, we can bring it to the numerator by changing the power to negative: $(1-x)^{-1/2}$. Here, $y=-x$ and $n=-1/2$. $(1-x)^{-1/2} \approx 1 + \left(-\frac{1}{2}\right)(-x) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{2!}(-x)^2$ $= 1 + \frac{1}{2}x + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}x^2$ $= 1 + \frac{1}{2}x + \frac{3}{8}x^2 \quad \text{(Equation 4)}$ Explanation: We apply the binomial approximation. When the denominator term is moved to the numerator, its exponent changes sign. We expand it up to the $x^2$ term.

C. Combining Numerator and Denominator Approximations

Now, substitute (Equation 3) and (Equation 4) back into the original expression for $E$: $E \approx \left(-\frac{3}{8}x^2\right) \left(1 + \frac{1}{2}x + \frac{3}{8}x^2\right)$ We need to multiply these two expressions and keep only terms up to $x^2$. $E \approx -\frac{3}{8}x^2 \cdot 1 \quad \text{(neglecting } x^3 \text{ and higher powers)}$ $E \approx -\frac{3}{8}x^2$ Explanation: Since the numerator is already an $x^2$ term ($-\frac{3}{8}x^2$), when we multiply it by the expanded form of $(1-x)^{-1/2}$ which is $(1 + \frac{1}{2}x + \frac{3}{8}x^2)$, any term from the denominator's expansion that is $x$ or higher ($x^1, x^2, \dots$) will result in $x^3$ or higher powers when multiplied by $x^2$. For example, $(-\frac{3}{8}x^2) \cdot (\frac{1}{2}x) = -\frac{3}{16}x^3$, which is neglected. Therefore, we only need to multiply the $x^2$ term from the numerator by the constant term (1) from the denominator's expansion.

The approximation for the given expression is $- {3 \over 8}{x^2}$.

Tip: Be very careful about how many terms to retain from each expansion. If the numerator simplifies to an $x^k$ term, and you are approximating up to $x^m$, you only need terms from the denominator's expansion $(1+ax+bx^2+\dots)$ such that when multiplied by $x^k$, the resulting power is less than or equal to $x^m$. In this case, $k=2$ and $m=2$, so we only need the constant term from the denominator's expansion ($1$) to avoid generating $x^3$ or higher powers.

Final Check: My derived answer is $- \frac{3}{8}x^2$, which corresponds to Option (C). However, the provided correct answer is (A) $1 - \frac{3}{8}x^2$. Given the precise mathematical steps, it appears there might be an issue with the question or the provided correct option. Based on the standard binomial approximations and the problem statement, the steps above lead unequivocally to $-\frac{3}{8}x^2$. If the intended answer was $1 - \frac{3}{8}x^2$, the original expression would likely need to be different (e.g., involve a leading constant term or a different numerator structure that doesn't fully cancel the constant terms).

Summary This problem demonstrates the application of the binomial theorem for approximation when $x$ is small. The key steps involve expanding each part of the expression up to the required power of $x$ (in this case, $x^2$), carefully handling signs and coefficients, and then multiplying/combining the approximated parts while neglecting terms of $x^3$ and higher powers. Careful calculation of each term is crucial to avoid errors.