Problem 41 - A Complete Resource Book in Mathematics for JEE Main

Question

Which of the following expansions will have term containing \(x^{3} ?\) (A) \(\left(x^{-\frac{1}{5}}+2 x^{\frac{3}{5}}\right)^{25}\) (B) \(\left(x^{\frac{3}{5}}+2 x^{-\frac{1}{5}}\right)^{24}\) (C) \(\left(x^{\frac{3}{5}}-2 x^{-\frac{1}{5}}\right)^{23}\) (D) \(\left(x^{\frac{3}{5}}+2 x^{-\frac{1}{5}}\right)^{22}\)

Step-by-Step Solution

Answer

Option A contains the term with \(x^3\).

Step 1: Identify Binomial Expansion Formula

The general term in the binomial expansion of \((a + b)^n\) is given by \( T_k = \binom{n}{k} a^{n-k} b^k \) where \(k\) is the term number starting from 0.

Step 2: Apply Formula to Each Option

For option A, \( a = x^{-\frac{1}{5}} \), \( b = 2x^{\frac{3}{5}} \), and \( n = 25 \). The general term is: \[ T_k = \binom{25}{k} (x^{-\frac{1}{5}})^{25-k} (2x^{\frac{3}{5}})^k \]. Simplifying gives: \[ T_k = \binom{25}{k} 2^k x^{\frac{3k}{5} - \frac{25-k}{5}} = \binom{25}{k} 2^k x^{\frac{4k-25}{5}} \].Similarly, apply to options B, C, and D.

Step 3: Solve for Each Option to Find Term with \(x^3\)

For option A, set \( \frac{4k-25}{5}=3 \). This implies: \[ 4k - 25 = 15 \]\[ 4k = 40 \]\[ k = 10 \].For option B, set \( \frac{4k-12}{5}=3 \) and solve for \(k\).For option C, set \( \frac{4k-23}{5}=3 \) and solve for \(k\).For option D, set \( \frac{4k-22}{5}=3 \) and solve for \(k\).

Step 4: Verify which Option Contains \(x^3\)

After solving, you find:- Option A: \( k = 10 \)- Option B: \( 4k - 12 = 15 \), \( 4k = 27 \), not an integer- Option C: \( 4k - 23 = 15 \), \( k = 9.5 \), not an integer- Option D: \( 4k - 22 =15 \), \( 4k = 37 \), not an integerTherefore, only option A has a valid integer value of \(k\) that satisfies the equation.

Key Concepts

General Term Formula

The general term formula is a critical tool in the binomial expansion process. It helps us find any specific term without having to fully expand the expression. For any binomial raised to a power, such as \( (a + b)^n \), each term can be expressed using the formula:

• \( T_k = \binom{n}{k} a^{n-k} b^k \)

This formula allows you to calculate each term by varying the value of \( k \,\) where \( n \,\) is the power the binomial is raised to, and \( k \,\) is the specific term index starting from zero.
Knowing this formula greatly simplifies finding terms such as \( x^k \,\) as you don't need to expand everything manually. Just plug in your values and solve for \( k \,\) making it efficient especially for large powers.

Exponents

Exponents play a crucial role when dealing with polynomial equations or any expressions involving powers. In the context of binomial expansions, exponents determine how each part of the expression contributes to the terms.
For an expression like \( (x^{a} + b)^{n} \), the term involving \( x^{3} \) comes from carefully choosing values in the expansion that lead to the desired power, here 3 in \( x^{3} \).

• The key is to use the exponent rules: when multiplying powers with the same base, you add the exponents.
• In binomial expressions like \( (x^{-rac{1}{5}} + 2x^{\frac{3}{5}} )^{25} \,\) finding the correct exponent combination for terms is essential.

This involves identifying which \( k \,\) term will yield the power of three according to the binomial term formula: evaluating the exponents for each constituent of the term.

Polynomial Equations

Polynomial equations are expressions involving variables raised to various powers. Analyzing these powers, or exponents, helps us understand the structure and find specific terms within the equation.

• In binomial expansion, we treat it as a series of polynomial equations derived from the general term we compute.
• Each component of the expanded expression corresponds to a different polynomial term.

Finding a specific polynomial equation term, like one containing \( x^3 \), involves evaluating each option with the general term formula. Using the method demonstrated, solving \( \frac{4k - 25}{5} = 3 \) allows for solving polynomial contributions term by term. This systematic approach relies on understanding polynomial expressions as built-up from simpler base components, such as linear equations.