Problem 7 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If in the expansion of \(\left(2^{x}+\frac{1}{4^{x}}\right)^{n}, \frac{T_{3}}{T_{2}}=7\) and the sum of the coefficients of 2 nd and 3 rd terms is 36 , then the value of \(x\) is (A) \(-\frac{1}{3}\) (B) \(-\frac{1}{2}\) (C) \(\frac{1}{3}\) (D) \(\frac{1}{2}\)

Step-by-Step Solution

Answer

The value of \(x\) is \(-\frac{1}{3}\).

Step 1: Identify the Binomial Expansion

In the given expression \(\left(2^{x}+\frac{1}{4^{x}}\right)^{n}\), we recognize it as a binomial expansion. Each term in the expansion is given by \(T_{r+1} = \binom{n}{r} (2^x)^{n-r} \left(\frac{1}{4^{x}}\right)^r\).

Step 2: Express T2 and T3

The second term \(T_2\) is given by \(T_2 = \binom{n}{1} (2^x)^{n-1} \left(\frac{1}{4^x}\right)^1 = n \cdot (2^x)^{n-1} \cdot \frac{1}{4^x}\). The third term \(T_3\) is given by \(T_3 = \binom{n}{2} (2^x)^{n-2} \left(\frac{1}{4^x}\right)^2 = \frac{n(n-1)}{2} \cdot (2^x)^{n-2} \cdot \frac{1}{16^x}\).

Step 3: Set Up the Equation for Fraction of Terms

We know from the problem that \(\frac{T_3}{T_2} = 7\). Substituting the expressions of \(T_2\) and \(T_3\) into this equation, we get:\[\frac{\frac{n(n-1)}{2} \cdot (2^x)^{n-2} \cdot \frac{1}{16^x}}{n \cdot (2^x)^{n-1} \cdot \frac{1}{4^x}} = 7\]Simplifying this equation yields:\[\frac{(n-1) \cdot \frac{1}{16^x}}{2 \cdot (2^x) \cdot \frac{1}{4^x}} = 7\]

Step 4: Simplify the Equation

Simplifying further:\[\frac{(n-1)}{2 \cdot 2^{3x}} = 7\]\(n-1 = 14 \cdot 2^{3x}\).

Step 5: Use the Sum of Coefficients Condition

The sum of the coefficients from the 2nd and 3rd terms is given as 36. That is, \(n + \frac{n(n-1)}{2} = 36\). This simplifies to:\[2n + n(n-1) = 72\]\[n^2 + n - 72 = 0\]

Step 6: Solve the Quadratic Equation

Solving the quadratic equation \(n^2 + n - 72 = 0\), we use the quadratic formula: \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=1\), \(c=-72\). This gives:\[n = \frac{-1 \pm \sqrt{1 + 288}}{2}\]\[n = \frac{-1 \pm 17}{2}\]This results in \(n = 8\) or \(n = -9\). Since \(n\) must be positive, \(n = 8\).

Step 7: Find the Value of x

Substituting \(n = 8\) into \(n-1 = 14 \cdot 2^{3x}\), we get \(7 = 14 \cdot 2^{3x}\). Thus, \(2^{3x} = \frac{1}{2}\), implying \(3x = -1\). Solving for \(x\) gives: \(x = -\frac{1}{3}\).

Key Concepts

Coefficient

In mathematics, a coefficient is a numerical or constant factor in front of variables in an algebraic expression. For example, in the term \(5x\), the coefficient is 5. Coefficients allow us to analyze and compute the size or magnitude of terms in an expression.

In the context of a binomial expansion, coefficients are crucial because they determine the weight contributed by each term in the expansion. When we expand \((a + b)^n\), each term takes a coefficient from the Binomial Theorem, written as \( \binom{n}{r} \), based on combinations. These coefficients help us calculate various properties such as the sum of specific terms in the expansion.

Understanding coefficients is essential because they influence the overall expression's outcome, particularly in equations derived from expansions, such as when solving for unknowns basic on terms like \( T_2 \) and \( T_3 \), which were mentioned in the exercise.

Exponential Equation

An exponential equation involves terms wherein the variable impacts the exponent, often seen as the form \(a^x\). Solving such equations typically requires manipulating the exponential terms to isolate the variable.

In the given problem, terms like \(2^x\) are indicative of exponential behaviors. This is crucial because an exponential function's key property is its growth rate, making exponential equations vital in expressing and solving practical scenarios where growth or decay is non-linear.

Manipulating exponential equations often involves utilizing logarithms to bring down the exponent for simpler solving. The exercise achieved this by equating and simplifying the terms involving exponential expressions, ultimately finding the value of \(x\). For example, reducing \(2^{3x} = \frac{1}{2}\) to \(x = -\frac{1}{3}\) required understanding how to invert and manipulate the exponential base.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two, generally in the form \(ax^2 + bx + c = 0\). Solving quadratics can be done using factorization, completing the square, or applying the quadratic formula. The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is particularly useful when factorization is not easily applicable.

In this exercise, a secondary quadratic equation \(n^2 + n - 72 = 0\) arose from working with the coefficients' summed terms. Solving it allowed us to determine a viable positive \(n\) to further solve for \(x\). This demonstrates how quadratic equations interface with other mathematical contexts beyond their basic structure, facilitating the solution of more complex multi-step problems.

Quadratic equations are ubiquitous in math due to their broad applicability in describing various phenomena and their direct utility in solving sums or products problems like this one.