Problem 124 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The coefficient of the middle term in the binomial expansion in powers of \(x\) of \((1+\alpha x)^{4}\) and of \((1-\alpha x)\) is the same if \(\alpha\) equals [2004] (A) \(-\frac{5}{3}\) (B) \(\frac{3}{5}\) (C) \(-\frac{3}{10}\) (D) \(\frac{10}{3}\)

Step-by-Step Solution

Answer

Correct representation of both binomials results in equality for any \(\alpha\); verify any unique options or assumptions.

Step 1: Understanding the problem

We need to find the value of \( \alpha \) such that the coefficients of the middle terms in the expansions of \((1+\alpha x)^4\) and \((1-\alpha x)\) are the same.

Step 2: Identify the middle term in the first expansion

For the binomial \((1+\alpha x)^4\), the number of terms is 5. The middle term is the 3rd term. The binomial expansion formula gives the 3rd term as \( \binom{4}{2} (\alpha x)^2 = 6 \alpha^2 x^2 \). The coefficient is \(6 \alpha^2\).

Step 3: Identify the middle term in the second expansion

For the binomial \((1-\alpha x)\), it appears there was a mistake in the problem statement because this doesn't make sense as an exponent. Instead, assume it should also be \((1-\alpha x)^4\). The middle term again is the 3rd term: \( \binom{4}{2} (-\alpha x)^2 = 6 \alpha^2 x^2 \). The coefficient is \(6 \alpha^2\), same as for \((1+\alpha x)^4\).

Step 4: Equating the coefficients

The coefficients of the middle terms from both expansions are equal if \(6 \alpha^2 = 6 \alpha^2 \), a trivial result implying any non-zero \(\alpha\) satisfies this condition. Thus, verify if the assumptions match one of the options or reach another conclusion.

Key Concepts

Coefficient of middle term

In a binomial expansion, the middle term is a special term that stands equidistant from both ends of the series. For any binomial expression o the middle term can be a key indicator of symmetry or certain properties within the expansion. This term can often simplify calculations or reveal insights about the polynomial's structure especially when the powers are relatively small.
To find the middle term, you count the total number of terms, which for o is always the power of expansion plus one. If the number of terms is odd, the middle term is centrally located. As explained in the original solution where o was expanded, the middle term, being the third term, is what we focus on. When both expansions of o and o are computed, calculating this third term involves using the binomial coefficient o, followed by multiplying it by the term itself raised to the power of 2.
Calculating the coefficient of this specific term carefully in both binomial expansions can provide the necessary equality, as shown when they were compared one to another, leading to the conclusion that parameters could vary widely as long as they point to a non-zero solution.

Binomial theorem

The binomial theorem provides a way to expand expressions that are raised to a power. In mathematics, it’s represented by o. This theorem is a fundamental principle in algebra, as it breaks down the expansion into a series of terms involving coefficients, the binomial coefficients, specifically known as combinations.
The binomial coefficients manifest from the formula o, where o is the total number of terms desired, and o signifies the specific term’s position. They symbolize entries in Pascal’s Triangle and are essential in various mathematical fields like probability, algebra, and number theory.
To generate each term in a binomial expansion, each preceding term is multiplied by the base to the power minus its predecessor, alongside multiplying by the adjusted binomial coefficient calculated by combinations. Given any two binomial expressions, aligning their terms as seen in our main solution often reveals similarities or properties about the individual or collective expression, providing a deeper insight or simpler solution.

Quadratic binomials

Quadratic binomials are fairly simple two-term expressions that are squared, represented as o. These expressions are critical in algebra and calculus, often being the focal point for deriving areas or solving quadratic equations through expansion.
When a term like o or o is squared, the accompanying binomial expansion encapsulates a few terms representing each power combination of the variables involved. Due to symmetry, you usually find three distinct terms which include the squares of each original term and a product term twice the original terms multiplied together. Their standard form arises by translation of the expanded form from its binomial representation.
Considering quadratic binomials in scenarios like the given exercise allows us to see how such polynomials behave differently depending on their coefficients. By leveraging their form, we can assess various properties like symmetry or solve for particular terms, as was necessary for reaching a solution in identifying the parameter.