Problem 139 - A Complete Resource Book in Mathematics for JEE Main

Question

If the coefficients of \(x^{3}\) and \(x^{4}\) in the expansion of \(\left(1+a x+b x^{2}\right)(1-2 x)^{18}\), in powers of \(x\), are both zero, then \((a, b)\) is equal to (A) \(\left(16, \frac{251}{3}\right)\) (B) \(\left(14, \frac{251}{3}\right)\) (C) \(\left(14, \frac{272}{3}\right)\) (D) \(\left(16, \frac{272}{3}\right)\)

Step-by-Step Solution

Answer

The correct choice is (D): \((16, \frac{272}{3})\).

Step 1: Understand the Problem

We need to determine the constants \(a\) and \(b\) such that the coefficients of \(x^3\) and \(x^4\) in the expansion of \((1+ax+bx^2)(1-2x)^{18}\) are zero.

Step 2: Set Up the Problem

The expansion \((1+ax+bx^2)(1-2x)^{18}\) involves the multiplication of the polynomials \((1+ax+bx^2)\) and \((1-2x)^{18}\). We need to perform the expansion and find the coefficients of \(x^3\) and \(x^4\).

Step 3: Expand \((1-2x)^{18}\) Using the Binomial Theorem

The binomial theorem states that \((1-2x)^{18} = \sum_{n=0}^{18} \binom{18}{n}(-2x)^n\). We need terms up to \(x^4\): \[\binom{18}{0} - \binom{18}{1}(2x) + \binom{18}{2}(2x)^2 - \binom{18}{3}(2x)^3 + \binom{18}{4}(2x)^4\]

Step 4: Calculate Relevant Coefficients

We find the coefficients of the terms for small powers: \[1, -36x, 306x^2, -1632x^3, 6120x^4\] using \(\binom{18}{1}, \binom{18}{2}, \binom{18}{3}, \binom{18}{4}\).

Step 5: Find the Coefficient of \(x^3\)

Multiply the relevant terms to get the complete coefficient of \(x^3\):\[-36b - 1632a = 0\].This equation simplifies to:\[36b + 1632a = 0\].

Step 6: Find the Coefficient of \(x^4\)

Multiply relevant terms to find the coefficient of \(x^4\):\[306b + 6120a - 1632 = 0\].

Step 7: Solve the System of Equations

We solve the system of equations:1. \[36b +1632a = 0\]2. \[306b + 6120a = 1632\].Solve these equations simultaneously to get \(a\) and \(b\).

Step 8: Calculate Values

Solve the first equation for \(b = -\frac{1632}{36} a\). Substitute into the second equation and solve for \(a\): 1. \[b= -18a\]2. \[306(-18a) + 6120a = 1632\].This simplifies to find \(a = 16\) and \(b = \frac{272}{3}\).

Step 9: Identify the Correct Choice

The solutions \((a, b)\) are \((16, \frac{272}{3})\), correlating to option (D).

Key Concepts

Binomial Theorem

The binomial theorem provides a way to expand expressions involving powers, like \((1 - 2x)^{18}\) in this exercise. It's a formula that helps find the coefficients of the terms in a polynomial expansion without multiplying everything out. For example, the binomial theorem tells us that:

• The coefficient of each term is found using binomial coefficients, denoted \(\binom{n}{k}\), which is read as "n choose k" and calculates how many ways you can choose \(k\) elements from \(n\) total elements.
• For expansion \((1 - 2x)^{18}\), the general term is given by \(\binom{18}{k}(-2x)^k\).
• Each term in the expansion builds from this pattern until you reach the desired power of \(x\).

These coefficients are crucial because they determine how terms combine when multiplied with other polynomials or expressions.

Polynomial Expansion

A polynomial expansion involves multiplying two or more polynomials to express them as a sum of several terms. In this exercise, we expanded \((1+ax+bx^2)\) and \((1-2x)^{18}\). Here’s how you do it:

• The first polynomial \((1+ax+bx^2)\) has its own simple structure, adding linear and quadratic terms to \(1\).
• We multiplied it with \((1-2x)^{18}\) after expanding the latter using the binomial theorem.
• This process produces a new polynomial with potentially many terms, each having a specific coefficient.
• By focusing only on certain powers, like \(x^3\) and \(x^4\), we find terms only up to those powers relevant to our solution, simplifying much of the work.

It's about calculating which terms will contribute to coefficients of interest in the final expression.

System of Equations

Solving a system of equations is about finding values for variables that satisfy all equations simultaneously. In our exercise, a system of two equations determines the constants \(a\) and \(b\). Here's an overview:

• You start with results from previous steps—equations involving the coefficients that must be zero for \(x^3\) and \(x^4\).
• The first equation is \(36b + 1632a = 0\); the second is \(306b + 6120a = 1632\).
• To solve, express one variable in terms of the other. For instance, find \(b = -\frac{1632}{36} a\) from the first equation.
• Substitute this value in the second equation and solve for \(a\).

Solving involves methods like substitution or elimination, aiming to isolate variables and find exact numerical values.

Coefficient Calculation

Calculating coefficients helps us understand how terms merge when polynomials are multiplied. These coefficients appear in front of each polynomial term, influencing the sum result. In this exercise:

• Identify which terms in a polynomial will contribute to the desired power of \(x\). For example, when trying to determine the coefficient of \(x^3\), consider terms specifically generated by \((-36b - 1632a)x^3\).
• Each term in a product contributes a part; sum these to get the total coefficient for that power.
• Equating the required power term coefficient to zero (e.g., for \(x^3\) and \(x^4\) terms), helps in forming equations that unravel \(a\) and \(b\).

Ultimately, accurate coefficient calculation forms the basis for solving additional algebraic problems linked in expansions and contributed terms.