Problem 45 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the constant term in the binomial expansion of \(\left(\sqrt{x} \frac{k}{x^{2}}\right)^{10}\) is 405, then \(|k|\) equals: \(\quad\) (a) 9 (b) 1 (c) 3 (d) 2

Step-by-Step Solution

Answer

The value of |k| is 3.

Step 1: Understand the Binomial Expression

Given the expression \(\left(\sqrt{x} \frac{k}{x^{2}}\right)^{10}\), we need to rewrite it in a workable form. Each term in the binomial expansion can be expressed using the general term formula for a binomial expression.

Step 2: Simplify Binomial Terms

Rewrite the given expression. When expanded, the general term of \((a + b)^n\) is \(T_r = \binom{n}{r} a^{n-r} b^r\). For this exercise, the term will take the form \(T_r = \binom{10}{r} (\sqrt{x})^{10-r} \left(\frac{k}{x^2}\right)^r\).

Step 3: Simplify Powers

Substitute in the terms: \[(\sqrt{x})^{10-r} = x^{\frac{10-r}{2}}\] and \[\left(\frac{k}{x^2}\right)^r = k^r x^{-2r}\]. Thus, \(T_r = \binom{10}{r} \cdot x^{\frac{10-r}{2}} \cdot k^r \cdot x^{-2r}\).

Step 4: Find the Exponent of x for the Constant Term

The exponential term becomes \(x^{\frac{10-r}{2} - 2r} = x^{\frac{10 - r - 4r}{2}} = x^{\frac{10 - 5r}{2}}\). The constant term implies \(x^{0}\), so set \(\frac{10 - 5r}{2} = 0\).

Step 5: Solve for r

Solve \(\frac{10 - 5r}{2} = 0\). Multiply through by 2 to get \(10 - 5r = 0\). This simplifies to \(10 = 5r\) leading to \(r = 2\).

Step 6: Calculate |k| for Constant Term Value

Use \(r = 2\) in the binomial term. \(T_2 = \binom{10}{2} \cdot k^2\). Calculate \(\binom{10}{2} = \frac{10 \cdot 9}{2} = 45\). The term becomes \(45k^2\). This is equal to 405, so solve \(45k^2 = 405\).

Step 7: Solve for k

Solve for \(k^2\): \(k^2 = \frac{405}{45}\). Simplify to find \(k^2 = 9\) leading to \(|k| = \sqrt{9} = 3\).

Key Concepts

Constant Term

The constant term in a binomial expansion is a crucial aspect, especially when you're asked to find specific values like in our example. The constant term is the term in the expansion that does not contain the variable, meaning its exponent is zero.
In the exercise, we needed to find the term in the expansion of the expression \[\left(\sqrt{x} \frac{k}{x^{2}}\right)^{10}\]that resulted in a constant. Thinking about constants can be simplified by the fact that the variable part must completely "cancel out" for the exponent to become zero.

• First, recall that any value raised to the power of zero is one, which is why we set the power of \(x^{\frac{10 - 5r}{2}}\) to zero.
• The power equation \(\frac{10 - 5r}{2} = 0\) is derived from seeking this cancellation and helps us find the specific index \(r\) where the magic happens, making the term constant.

Binomial Theorem

The binomial theorem allows us to efficiently expand expressions with a power such as \((a + b)^n\). For the expression in question, we have to apply this theorem cleverly. Here's a breakdown.

• The binomial theorem states that \((a + b)^n\) can be expanded as the sum of terms \(\binom{n}{r} a^{n-r} b^r\).
• This involves binomial coefficients \(\binom{n}{r}\), calculated as \(\frac{n!}{r!(n-r)!}\).
• In our case, the binomial terms took the form \(\binom{10}{r} (\sqrt{x})^{10-r} \left(\frac{k}{x^2}\right)^r\).

In the exercise, we specifically focused on a term where the exponent of \(x\) became zero to isolate the constant term for further calculations on \(|k|\). The power management in these terms is a key application of the theorem, showing its versatility in not just expanding, but also solving complex problems.

Algebraic Expressions

Understanding algebraic expressions is foundational when dealing with binomial expansions. Algebraic expressions are combinations of terms that can involve constants, variables, and operations. In this problem, it involved simplifying and managing powers within the expression.

• Initially, the expression \(\left(\sqrt{x} \frac{k}{x^{2}}\right)^{10}\) had to be reformulated. Recognizing \(\sqrt{x}\) as \(x^{\frac{1}{2}}\) is an accelerator.
• The effective use of algebraic manipulations simplified the calculation of powers: \((\sqrt{x})^{10-r} = x^{\frac{10-r}{2}}\) and \(\left(\frac{k}{x^2}\right)^r = k^r x^{-2r}\). The key here was managing these powers correctly.
• Working through algebraic expressions is about seeing through the layers of terms, operations, and transformations to reach a solution. It involves reversing and reorganizing steps as needed for simplification.

This exercise illustrated these core algebraic skills, highlighting how subtle interpretations and manipulations can lead to solutions like finding \(|k|\).