Problem 7 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If the fourth term in the Binomial expansion of \(\left(\frac{2}{x}+x^{\log _{8} x}\right)^{6}(x>0)\) is \(20 \times 8^{7}\), then a value of \(x\) is: (a) \(8^{3}\) (b) \(8^{2}\) (c) 8 (d) \(8^{-2}\)

Step-by-Step Solution

Answer

The correct answer is (a) \(x = 8^3\).

Step 1: Understanding the Formula

In a binomial expansion \((a+b)^n\), the general term \(T_k\) is given by \(T_k = \binom{n}{k-1} a^{n-k+1} b^{k-1}\). For the expression \(\left(\frac{2}{x} + x^{\log_{8} x}\right)^6\), identify \(a = \frac{2}{x}\) and \(b = x^{\log_8 x}\) with \(n = 6\).

Step 2: Finding the Fourth Term

The fourth term is \(T_4\), where \(k = 4\). Use the general term formula: \(T_4 = \binom{6}{3} \left(\frac{2}{x}\right)^{6-3} \left(x^{\log_8 x}\right)^{3}\). Evaluate \(\binom{6}{3} = 20\).

Step 3: Substitute into the General Term

Plug values into \(T_4 = 20 \left(\frac{2}{x}\right)^3 \left(x^{\log_8 x}\right)^3\). This simplifies to \(20 \frac{8}{x^3} x^{3\log_8 x}\) or \(20 \times 8 x^{3 \log_8 x - 3}\).

Step 4: Solving for x

Since \(T_4 = 20 \times 8^7\), set \(20 \times 8 x^{3 \log_8 x - 3} = 20 \times 8^7\). Cancel 20 from both sides: \(8 x^{3 \log_8 x - 3} = 8^7\). Further simplifies to \(x^{3 \log_8 x - 3} = 8^6\).

Step 5: Equating Exponents

Recognize that \(8 = 2^3\), so \(8^6 = (2^3)^6 = 2^{18}\). Also, express \(x^{3 \log_8 x - 3}\) as \((2^3)^{3 \log_8 x - 3}\). Equating exponents gives \(3(3 \log_8 x - 3) = 18\).

Step 6: Simplify and Find x

Solve the equation \(9 \log_8 x - 9 = 18\). Add 9 to both sides: \(9 \log_8 x = 27\). Dividing by 9 gives \(\log_8 x = 3\). Converting to exponential form, \(x = 8^3\).

Step 7: Choice Validation

Checking options, \(x = 8^3\) is option (a).

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool for expanding expressions raised to a power. It explains how to expand expressions like \((a+b)^n\) into a sum of terms of the form\(\binom{n}{k}a^{n-k}b^k\). This tool is essential when working with polynomial expansions.
In our problem, we use the binomial expression \(\left(\frac{2}{x} + x^{\log_8 x}\right)^6\). According to the Binomial Theorem, we can express each term using

• \(\binom{n}{k}\): the binomial coefficient that determines each term's weight in the expansion.
• \(a^{n-k}\): the component of the first term raised to a power decreasing from \(n\).
• \(b^k\): the component of the second term raised to a power increasing from 0 to \(n\).

For our fourth term, the key is using these steps to identify corresponding values and find the power composition in the expression.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions. They answer the question: "to what exponent must we raise the base to obtain a given number?"
In our exercise, the function \(\log_8 x\) tells us the power to which 8 must be raised to produce \(x\). It's helpful in manipulating exponents and understanding the growth of a function.

• \(\log_b a = c\) means \(b^c = a\). For instance, in our problem, if \(\log_8 x = 3\), then \(x\) is \(8^3\).
• The change of base formula can rewrite logs with different bases: \(\log_b a = \frac{\log_c a}{\log_c b}\).

Recognizing how the logarithm \(\log_8 x\) is used, allows simplification of exponents, contributing to identifying the correct value of \(x\).

Exponents

Exponents are used in mathematics to denote repeated multiplication of a base number. In exponential notation, \(b^n\) indicates \(b\) multiplied by itself \(n\) times.
Exponents follow specific rules which are crucial in many mathematical operations, including simplifying expressions, solving equations, and more.

• Product of Powers: \(a^m \times a^n = a^{m+n}\).
• Power of a Power: \((a^m)^n = a^{m \times n}\).
• Negative Exponent: \(a^{-n} = \frac{1}{a^n}\).
• Zero Exponent: \(a^0 = 1\) for \(a eq 0\).

In the problem, understanding exponents is vital because it helps simplify expressions, especially when manipulating logarithmic functions that express numbers as powers, like recognizing that \(8^6 = (2^3)^6 = 2^{18}\). Such insights lead to solving for \(x\) effectively.