Problem 78 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(x\) is so small that \(x^{3}\) and higher powers of \(x\) may be neglected, then \(\frac{(1+x)^{\frac{3}{2}}-\left(1+\frac{1}{2} x\right)^{3}}{(1-x)^{\frac{1}{2}}}\) may be approximated as (a) \(1-\frac{3}{8} x^{2}\) (b) \(3 x+\frac{3}{8} x^{2}\) (c) \(-\frac{3}{8} x^{2}\) (d) \(\frac{x}{2}-\frac{3}{8} x^{2}\)

Step-by-Step Solution

Answer

Option (c) \(-\frac{3}{8} x^{2}\) is correct as higher power terms are negligible and the numerator becomes zero.

Step 1: Approximate \((1+x)^{\frac{3}{2}}\)

For small values of \(x\), use the binomial expansion and neglect terms higher than \(x^2\): \[(1+x)^{\frac{3}{2}} = 1 + \frac{3}{2}x + \frac{3}{8}x^2.\]

Step 2: Approximate \((1+\frac{1}{2}x)^{3}\)

Similarly, expand \((1+\frac{1}{2}x)^{3}\) and neglect terms higher than \(x^2\):\[ (1+\frac{1}{2}x)^{3} = 1 + \frac{3}{2}x + \frac{3}{4}(\frac{1}{2}x)^2 = 1 + \frac{3}{2}x + \frac{3}{8}x^2.\]

Step 3: Calculate the numerator difference

Subtract the expressions from Step 1 and Step 2 to find the difference: \[\left((1+x)^{\frac{3}{2}} - \left(1+\frac{1}{2}x\right)^3\right) = \left(1 + \frac{3}{2}x + \frac{3}{8}x^2\right) - \left(1 + \frac{3}{2}x + \frac{3}{8}x^2\right) = 0.\]

Step 4: Simplify \((1-x)^{\frac{1}{2}}\) denominator

Use the binomial expansion to approximate \((1-x)^{\frac{1}{2}}\):\[(1-x)^{\frac{1}{2}} = 1 - \frac{1}{2}x.\]

Step 5: Final simplification

Since the numerator is 0, the expression simplifies entirely to:\[\frac{0}{1 - \frac{1}{2}x} = 0.\] The higher power terms in \(x\) don't affect the final zero result.

Key Concepts

Approximation

Approximation is a powerful tool in mathematics, especially when dealing with complex functions. It allows us to simplify a problem by considering only the most significant terms. In the context of binomial expansions, approximation becomes particularly handy. For small values of a variable, like \(x\), we might disregard terms that have negligible impact on the final result.

The approximation technique leverages the binomial theorem. This theorem can expand expressions of the form \((1+x)^n\) into a series. For small \(x\), we only consider the first few terms of this series, ignoring higher-order terms because they contribute insignificantly. In our problem, when \((1+x)^{\frac{3}{2}}\) is expanded, higher powers like \(x^3\) and above are left out to approximate the function as:

• \((1+x)^{\frac{3}{2}} = 1 + \frac{3}{2}x + \frac{3}{8}x^2\)

Doing so simplifies calculations, making it easier to manage and solve the problem effectively.

Small x Assumption

The small \(x\) assumption simplifies many problems, especially in calculus and algebra. It is based on the idea that if \(x\) is very small, its higher powers become extremely small, and their influence can be considered negligible.

This assumption works well with binomial expansions used in our exercise. By assuming \(x\) is small, we skip expanding terms beyond \(x^2\). For example:

• \((1+\frac{1}{2}x)^{3}\)
• Expands roughly to \(1 + \frac{3}{2}x + \frac{3}{8}x^2\)

Ignoring higher-order terms does not significantly alter the outcome, as they contribute very minute values. Consequently, this approach provides a quick and straightforward solution without unnecessary complexity.

Numerator and Denominator Simplification

Simplifying the numerator and the denominator is crucial for evaluating expressions effectively. In the exercise, after applying binomial expansion, the expression's numerator simplifies to zero.

Numerator simplification involves subtracting equivalent parts derived from the binomial expansions:

• \((1+x)^{\frac{3}{2}}\) and \((1+\frac{1}{2}x)^3\)
• Both ultimately yield \(1 + \frac{3}{2}x + \frac{3/8}x^2\), leading to zero difference.

For the denominator, a similar expansion is applied:

• \((1-x)^{\frac{1}{2}} = 1 - \frac{1}{2}x\)

These simplifications underline the efficiency of approximations in algebraic problems. When the numerator becomes zero, the overall expression simplifies directly to zero, bypassing further evaluation for any terms involving \(x\). This exercise beautifully illustrates how elementary algebraic operations align with binomial expansion techniques for streamlined problem-solving.