Problem 65 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the coefficients of \(\mathrm{x}^{2}\) and \(\mathrm{x}^{3}\) are both zero, in the expansion of the expression \(\left(1+a x+b x^{2}\right)(1-3 x)^{15}\) in powers of \(x\), then the ordered pair (a, b) is equal to: \(\quad\) (a) \((28,861)\) (b) \((-54,315)\) (c) \((28,315)\) (d) \((-21,714)\)

Step-by-Step Solution

Answer

The ordered pair (a, b) is \((28, 315)\).

Step 1: Understand the Problem

We need to find values of \(a\) and \(b\) such that the coefficients of \(x^2\) and \(x^3\) are zero in the expansion of \(\left(1+ax+bx^2\right)(1-3x)^{15}\). This involves a binomial expansion and combining like terms.

Step 2: Expand \((1 - 3x)^{15}\)

Use the binomial theorem to expand \((1 - 3x)^{15}\). According to the binomial theorem, the expansion is given by \((1 - 3x)^{15} = \sum_{k=0}^{15} \binom{15}{k} (-3x)^k\). Specifically, we'll need terms involving \(x^2\) and \(x^3\).

Step 3: Identify Terms for \(x^2\) and \(x^3\)

In the binomial expansion, the term involving \(x^k\) is \(\binom{15}{k} (-3)^k x^k\). For \(x^2\), the term is \(\binom{15}{2} (-3)^2 x^2\). For \(x^3\), it is \(\binom{15}{3} (-3)^3 x^3\).

Step 4: Combine Like Terms for \(x^2\)

The terms contributing to \(x^2\) are:1. From \((1 - 3x)^{15}\), it's \(\binom{15}{2} (-3)^2 x^2 = 105 \cdot 9 = 945x^2\).2. \(ax\cdot \text{term of } x = ax \cdot (-45x) = -45ax^2\).3. \(bx^2\cdot \text{constant term} = bx^2\).Thus, the coefficient of \(x^2\) is \(945 - 45a + b = 0\).

Step 5: Combine Like Terms for \(x^3\)

The terms contributing to \(x^3\) are:1. From \((1 - 3x)^{15}\), it's \(\binom{15}{3} (-3)^3 x^3 = 455 \cdot (-27) = -12285x^3\).2. \(ax \cdot \text{term of } x^2 = ax \cdot 945x^2 = 945ax^3\).3. \(bx^2 \cdot \text{term of } x = bx^2 \cdot (-45x) = -45bx^3\).Thus, the coefficient of \(x^3\) is \(-12285 + 945a - 45b = 0\).

Step 6: Solve the System of Equations

We need to solve the system of equations:1. \(945 - 45a + b = 0\)2. \(-12285 + 945a - 45b = 0\)First, simplify equation 1: \(b = 45a - 945\).Substitute into equation 2: \(-12285 + 945a - 45(45a - 945) = 0\).This simplifies to: \(-12285 + 945a - 2025a + 42525 = 0\).Combine like terms: \(-1080a + 30240 = 0\), solve for \(a\), \(a = \frac{30240}{1080} = 28\).Place \(a = 28\) into the equation for \(b\): \(b = 45 \times 28 - 945 = 1260 - 945 = 315\).

Step 7: Identify the Ordered Pair

The solution to the system is \(a = 28\) and \(b = 315\). Therefore, the ordered pair \((a, b)\) is \((28, 315)\).

Key Concepts

Coefficient of Polynomial

In polynomial expressions, coefficients are the numerical factors of the terms. When we expand or manipulate expressions, we often focus on these numbers. For example, in the expression \(ax^2 + bx + c\), \(a\) is the coefficient of \(x^2\), \(b\) is the coefficient of \(x\), and \(c\) is the constant term.

To find specific coefficients when dealing with expanded expressions, you need to carefully identify each term's contribution to relevant powers of \(x\). This process involves combining like terms. In our problem, we expanded the expression \((1 + ax + bx^2)(1 - 3x)^{15}\) and focused especially on terms that contribute to \(x^2\) and \(x^3\).

By setting the coefficients of \(x^2\) and \(x^3\) to zero, we formed a system of equations. Solving these equations gave us the values of \(a\) and \(b\), ensuring the specific coefficients vanished, as required by the problem.

System of Equations

A system of equations is a collection of multiple equations with the same set of unknowns. Solving systems of equations is a common task in algebra, as it allows us to find precise values for variables. In the given problem, we used a system of two equations derived from the requirements that the coefficients of \(x^2\) and \(x^3\) be zero.

The first equation was derived from the \(x^2\) terms: \(945 - 45a + b = 0\). The second equation came from the \(x^3\) terms: \(-12285 + 945a - 45b = 0\). By solving this system, we used substitution to express \(b\) in terms of \(a\), then substituted this relation into the second equation to find numerical values for \(a\) and \(b\).

This method of solving systems of equations—typically substitution or elimination—helps us isolate variables and find specific numerical values.

Expansion of Expressions

Expanding expressions involves multiplying expressions out fully, breaking them into simpler components. It's especially common when using the binomial theorem, a handy tool in algebra that helps expand expressions raised to a power.

The binomial theorem states that \((1 + x)^n\) can be expanded into the sum \(\sum_{k=0}^{n} \binom{n}{k} x^k\). In practice, this means expressing a binomial raised to a power as a series of terms, each based on combinations of \(x\) raised to a power.

In the original exercise, we used the binomial theorem to expand \((1 - 3x)^{15}\). This expansion helped us identify terms related to \(x^2\) and \(x^3\), which were crucial for solving the problem. By applying this systematic approach, we can break even complex algebraic expressions into manageable pieces.