Problem 62 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The term independent of \(x\) in the expansion of \(\left(\frac{1}{60}-\frac{x^{8}}{81}\right) \cdot\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}\) is equal to: (a) \(-72\) (b) 36 (c) \(-36\) (d) \(-108\)

Step-by-Step Solution

Answer

The term independent of \(x\) is \(-72\), option (a).

Step 1: Understand the Problem

We need to find the term in the given expression that does not include the variable \(x\). This term is often called the constant term or the term independent of \(x\).

Step 2: Expand the Binomial Expression

Consider the binomial expression \( \left(2x^2 - \frac{3}{x^2}\right)^6 \). Using the binomial theorem, the general term \(T_r\) in the expansion is given by: \[ T_r = \binom{6}{r} (2x^2)^{6-r} \left(-\frac{3}{x^2}\right)^r \]Simplifying, we get:\[ T_r = \binom{6}{r} \cdot 2^{6-r} \cdot (-3)^r \cdot x^{2(6-r) - 2r} \]\[ T_r = \binom{6}{r} \cdot 2^{6-r} \cdot (-3)^r \cdot x^{12-4r} \]

Step 3: Find the Term Independent of x

The term independent of \(x\) in the expansion of \( \left(2x^2 - \frac{3}{x^2}\right)^6 \) occurs when the power of \(x\) is zero. Set the exponent of \(x\) in \(T_r\) to zero:\[ 12 - 4r = 0 \]Solving for \(r\), we find that \(r = 3\).

Step 4: Compute the Independent Term in Binomial Expansion

Use \(r = 3\) to find the term:\[ T_3 = \binom{6}{3} \cdot 2^{6-3} \cdot (-3)^3 \cdot x^0 \]\[ T_3 = \binom{6}{3} \cdot 2^3 \cdot (-3)^3 \]\[ T_3 = 20 \cdot 8 \cdot (-27) \]\[ T_3 = -4320 \]

Step 5: Multiply with the Remaining Expression

Now, we need to multiply the constant term found with \( \frac{1}{60} \):\[ -4320 \times \frac{1}{60} = -72 \]

Step 6: Conclusion

Thus, the term independent of \(x\) in the given expression is \(-72\). This matches option (a).

Key Concepts

Binomial Theorem

The Binomial Theorem offers a systematic way of expanding expressions raised to a power, such as \((a + b)^n\). This theorem is incredibly useful in algebra, as it enables us to express complex polynomials in a structured manner. Each term in the expansion is formed based on its position and can be directly calculated using a formula. Here’s how it works in more detail:

• For any non-negative integer \(n\), the expansion of \((a + b)^n\) is given as a series of terms \(a^{n-k}b^k\) where \(k\) ranges from 0 to \(n\).
• The coefficient of each term \(a^{n-k}b^k\) is determined by the binomial coefficient, represented by \(\binom{n}{k}\). This coefficient is calculated as \(\frac{n!}{k!(n-k)!}\).
• The Binomial Theorem allows for quick calculation of any specific term in the expansion without expanding the entire expression.

In our exercise, we applied the Binomial Theorem to expand the expression \((2x^2 - \frac{3}{x^2})^6\), generating a series of terms from which we identified the constant term.

Constant Term

The constant term in a polynomial is the part that does not change with the variable, or in other words, does not include any variables at all. When dealing with expressions, especially those involving powers, identifying the constant term requires some strategic thinking.

• To find the constant term, you need to determine the terms where the variable results in a neutral or zero exponent. This is because any variable raised to the power of zero equals one, essentially nullifying its effect.
• In polynomial expansions like the one we are considering, the power of \(x\) in each term is manipulated to achieve a scenario where it cancels out or results in zero, leaving a constant coefficient as the remainder.
• It's crucial to set the power of \(x\) to zero in the formula derived from the Binomial Theorem to isolate the constant term. Mathematically, this involves solving equations to determine the exact term required.

In the given expression \((2x^2 - \frac{3}{x^2})^6\), the term independent of \(x\) occurred when the power of \(x\) was zero, leading to a direct and efficient way to identify and compute its value.

Polynomial Expansion

Polynomial Expansion involves expressing a polynomial power as a sum of simpler terms. This process is often necessary in algebra to simplify calculations, find specific terms, or solve complex equations involving multiple variables and higher powers.

• Polynomial expansion transforms a compact expression like \((a + b)^n\) into a more explicit sum of terms, allowing better analysis and computation of specific terms.
• Using the Binomial Theorem, each term of the expansion becomes structured, with coefficients determined by binomial coefficients, reflecting its place in the series.
• This method simplifies problems, especially when searching for specific properties like the coefficient, degree, or constant terms within the expression.

In our exercise, polynomial expansion allowed us to systematically unravel \((2x^2 - \frac{3}{x^2})^6\) into its constituent parts, making it simpler to single out and compute the term independent of \(x\). This approach enhances the efficiency and accuracy of tackling algebraic problems.