Problem 139 - A Complete Resource Book in Mathematics for JEE Main 2017

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 [2014] (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 value of \((a, b)\) is \(\left(16, \frac{272}{3}\right)\), so the correct choice is (D).

Step 1: Expand Each Term Individually

First, break down the expression into two parts: \((1 + ax + bx^2)\) and \((1-2x)^{18}\). For the term \((1-2x)^{18}\), we'll be using the binomial expansion formula: \( (1-2x)^{18} = \sum_{k=0}^{18} \binom{18}{k} (-2x)^k = \sum_{k=0}^{18} \binom{18}{k} (-2)^k x^k \).

Step 2: Determine Coefficient of x^3

To find the coefficient of \(x^3\) in the product, we consider contributions from the different parts of the expansion. We have: (i) coefficient from \( (1)(1-2x)^{18} \) is \( \binom{18}{3}(-2)^3 = -816 \). (ii) coefficient from \( ax(1-2x)^{18} \) is \( a\times\binom{18}{2}(-2)^2 = 306a \) since only the \(x^2\) term from \((1-2x)^{18}\) contributes to \(x^3\) when multiplied by \(ax\). (iii) coefficient from \( bx^2(1-2x)^{18} \) is \( b \times \binom{18}{1} (-2)^1 = -36b \). The sum of these must be zero: \(-816 + 306a - 36b = 0\).

Step 3: Simplify Constraint for x^3

From \(-816 + 306a - 36b = 0\), we simplify this to \(306a - 36b = 816\). Divide the entire equation by 6 to simplify: \(51a - 6b = 136\).

Step 4: Determine Coefficient of x^4

Next, calculate the contribution to \(x^4\): (i) term \((1)\) with \(x^4\) term from \((1-2x)^{18}\) gives \(\binom{18}{4}(-2)^4 = 3060 \). (ii) term \((ax)\) with \(x^3\) term from \((1-2x)^{18}\) gives \( a\times\binom{18}{3}(-2)^3 = -816a \). (iii) term \((bx^2)\) with \(x^2\) from \((1-2x)^{18}\) gives \( b\times\binom{18}{2}(-2)^2 = 153b \). Again, sum of these must be zero: \(3060 - 816a + 153b = 0\).

Step 5: Simplify Constraint for x^4

From \(3060 - 816a + 153b = 0\), simplify to \(-816a + 153b = -3060\). Divide the entire equation by 3: \(-272a + 51b = -1020\).

Step 6: Solve System of Equations

We now have two equations: (1) \(51a - 6b = 136\) and (2) \(-272a + 51b = -1020\). Solve this system using substitution or elimination. Multiplying equation (1) by 51 and equation (2) by 6 and adding gives: \(2601a - 306b + -1632a + 306b = 816 + 6120\). Solve to find \(a = 16\) and substituting in equation (1) gives \(b = 272/3\).

Step 7: Confirm Answer

Now that we have \(a = 16\) and \(b = 272/3\), check the choices. \((a, b) = (16, 272/3)\) corresponds to choice (D).

Key Concepts

Polynomial Coefficients

The concept of polynomial coefficients is crucial when dealing with expressions involving powers, like binomial expansions. In such expansions, a polynomial's degree tells us how many solutions or roots it may have, and the coefficients provide the specific values for these solutions. For instance, in the expression \[(1+a x+b x^{2})(1-2 x)^{18}\],consider expanding to determine which coefficients (of terms like \(x^3\) and \(x^4\)) affect the calculation.
These coefficients serve as multipliers for each term in the expansion, and they play a significant role in solving equations derived from such expressions.
If the coefficients of certain powers are set to zero, like in the exercise above, it can lead directly to finding specific parameters (\(a\) and \(b\) in this case) that zero out those terms. Identifying which coefficients zero out certain terms helps simplify the polynomial significantly, saving time and effort and focusing solves on key parameters.

System of Equations

A system of equations arises when multiple equations are set up to find a common solution set, often involving several variables. In the exercise, once we expand the polynomials, we deduce two equations based on the coefficients of \(x^3\) and \(x^4\), namely:

• \(51a - 6b = 136\)
• \(-272a + 51b = -1020\)

These equations are derived by setting the conditions that their corresponding polynomial coefficients should be zero, leading to a direct problem-solving approach. Solving a system like this usually involves either substitution or elimination methods characterized by:

• Substitution: Solving for one variable in one equation and substituting into the other.
• Elimination: Canceling out a variable by algebraically adding or subtracting the equations.

For the given system, multiplying and aligning equations to eliminate a variable, followed by solving, must yield the values for \(a\) and \(b\). This involves ensuring accurate arithmetic handling to achieve precise solutions, central to algebraic problem solving.

Exponentials

Exponentials in polynomial expressions refer to terms that involve powers raised to a base number. They are prominently seen in binomial expansions, such as \[(1-2x)^{18}\]. This term represents an exponential because \(18\) is the power to which the base \((1-2x)\) is raised.
The binomial theorem essentially helps expand these exponential terms into a sum of terms with decreasing powers. By using the binomial expansion formula, \(C_n^k\ ((base)^n)\), we break this down into simpler terms, each contributing a polynomial coefficient based on its index:

• \((1-2x)^{18} = \sum_{k=0}^{18} \binom{18}{k} (-2x)^k\)

Understanding how these exponential terms contribute to the polynomial's overall form is important when solving for specific coefficients as required in problems asking for zero coefficients of certain powers. It identifies which terms mix when multiplied with factors from another polynomial, like \((1+ax+bx^2)\). Each exponential term influences if it produces zero or non-zero contributions to the product.