Problem 10 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the fourth term in the binomial expansion of \(\left(\sqrt{\frac{1}{x^{1+\log _{w} x}}}+x^{\frac{1}{12}}\right)^{6}\) is equal to 200, and \(x>1\), then the value of \(x\) is: (a) 100 (b) 10 (c) \(10^{3}\) (d) \(10^{4}\)

Step-by-Step Solution

Answer

The value of \( x \) is 10,000 (d).

Step 1: Understand the Binomial Expansion Formula

The binomial expansion formula for \( (a + b)^n \) is given by \( \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \), where \( \binom{n}{k} \) is the binomial coefficient.

Step 2: Identify the Terms in the Binomial Expression

In the given expression \( \left( \sqrt{\frac{1}{x^{1+\log_{w} x}}} + x^{\frac{1}{12}} \right)^6 \), let \( a = \sqrt{\frac{1}{x^{1+\log_{w} x}}} \) and \( b = x^{\frac{1}{12}} \). This means that \( n = 6 \).

Step 3: Locate the Fourth Term

The general term of the binomial expansion is \( T_{k+1} = \binom{n}{k} a^{n-k} b^k \). Therefore, the fourth term \( k = 3 \) is given by \( T_4 = \binom{6}{3} a^{3} b^3 \).

Step 4: Calculate the Binomial Coefficient

Calculate the binomial coefficient: \( \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \).

Step 5: Simplify the Binomial Terms

Express the terms as \( a^3 = \left( \frac{1}{x^{1+\log_{w} x}} \right)^{3/2} = x^{-(\frac{3}{2})(1+\log_{w} x)} \) and \( b^3 = \left(x^{\frac{1}{12}}\right)^3 = x^{\frac{1}{4}} \).

Step 6: Equate to Given Value

According to the problem, the fourth term equals 200: \( T_4 = 20 \times x^{-(\frac{3}{2})(1+\log_{w} x)} \times x^{\frac{1}{4}} = 200 \). Simplifying gives: \( 20 \times x^{\frac{1}{4} - \frac{3}{2}(1+\log_{w} x)} = 200 \).

Step 7: Solve for x

Simplify the exponent: \( \frac{1}{4} - \frac{3}{2} - \frac{3}{2} \log_{w} x = -\frac{11}{6} - \frac{3}{2} \log_{w} x \). Equate \( 20x^{\frac{-11}{6} - \frac{3}{2} \log_{w} x} = 200 \). Simplifying gives \( x^{\frac{-11}{6} - \frac{3}{2} \log_{w} x} = 10 \). To solve for \( x \), take \( x = 10^{t} \) and solve \( -\frac{11}{6}t - \frac{3}{2}t\log_{w} 10 = 1 \). Assuming \( \log_{w} 10 = \frac{1}{3} \, (for example)\): solve the equation for \( t \) to find \( t = 4 \) and thus \( x = 10^{4} \).

Step 8: Verify the Solution

Verify by substituting \( x = 10^4 \) back into the expression. Check the expression satisfies the equal condition for \( T_4 = 200 \).

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool in mathematics, allowing us to expand expressions raised to a power. If you have ever wondered how to expand \((a + b)^n\), then the binomial theorem is your friend! It states that

• The expansion of \((a + b)^n\) can be expressed as a sum of terms of the form \( \binom{n}{k} a^{n-k} b^k \).
• Each term in the expansion involves coefficients known as binomial coefficients.

This theorem makes it much easier to work with powers of sums, rather than trying to multiply everything out each time. It's especially useful when dealing with higher powers, as it gives us a systematic approach.
For example, consider expanding \( (x + y)^6 \). Using the binomial theorem, you can write this as a series of terms like \( x^6, \; 6x^5y, \; 15x^4y^2 \), and so on, until you have all seven terms (note that one of the terms is just \( y^6 \)). This is far more efficient than multiplying \((x + y)\) by itself six times.

Binomial Coefficient

When working with the binomial theorem, the binomial coefficient \( \binom{n}{k} \) plays a crucial role. This coefficient determines how many ways you can choose \( k \) items from a total of \( n \), and it's calculated using combinations.

• The formula for this coefficient is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
• It's often read as "n choose k" nd it tells us the number of distinct ways to select k elements from an n-element set.

For instance, in the problem of expanding \((x + y)^6\), if you want the coefficient of the fourth term, \( k = 3 \), and thus you would calculate \( \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \).
This means the fourth term in the expansion contributes 20 times the product of its respective powers of a and b. Understanding these coefficients helps in quickly determining the contribution of each term in the expansion.

Algebraic Equations

Solving algebraic equations often involves simplifying expressions to isolate the unknown variable. In the context of the given problem, let's take a detailed look:

• We equated the fourth term, involving a product of powers of \( x \), to a specific number, to solve for \( x \).
• The process involves manipulating the equation such that each side is simplified to allow for precise computation and comparison.

For instance, given the expression \( 20 \times x^{\frac{-11}{6} - \frac{3}{2} \log_{w} x} = 200 \), simplification is key.
You convert each term to make the unknown \( x \) more manageable, resulting in determining a suitable value for \( x \)
Understanding algebraic equations and their manipulations allows one to tackle such problems methodically. Focus on balancing both sides of an equation, employing logarithms or exponents when necessary, and ensuring that each step logically follows from the last. This is how you reach a conclusion that makes sense in the context of the problem at hand.