Problem 44 - A Complete Resource Book in Mathematics for JEE Main

Question

If the product of the roots of the equation \(x^{2}-3 k x+\) \(2 e^{2 \ln k}-1=0\) is 7 , then for real roots the value of \(k\) is equal to (A) 1 (B) 2 (C) 3 (D) 4

Step-by-Step Solution

Answer

The value of \( k \) is 2.

Step 1: Identify Formula for Product of Roots

For a quadratic equation of the form \[ ax^2 + bx + c = 0 \]the product of the roots is given by:\[ \frac{c}{a} \]

Step 2: Apply Given Information

In the quadratic equation \[ x^2 - 3kx + 2e^{2 \ln k} - 1 = 0 \]we can identify \( a = 1 \), \( b = -3k \), and \( c = 2e^{2 \ln k} - 1 \). The product of the roots is 7, so we set up the equation:\[ 2e^{2 \ln k} - 1 = 7 \]

Step 3: Solve for Expression Involving \(k\)

Simplify and solve the equation:\[ 2e^{2 \ln k} - 1 = 7 \]Add 1 to both sides to get:\[ 2e^{2 \ln k} = 8 \]Then divide both sides by 2:\[ e^{2 \ln k} = 4 \]

Step 4: Convert Exponential to Base \(k\)

Recall the property of logarithms and exponents, \( e^{\ln(x)} = x \):\[ e^{2 \ln k} = (e^{\ln k})^2 = k^2 \]This implies that the equation \( k^2 = 4 \) holds.

Step 5: Solve for \(k\)

Take the square root of both sides of the equation \( k^2 = 4 \):\[ k = \sqrt{4} \]The solutions for \( k \) are \( k = 2 \) and \( k = -2 \), but only positive \( k = 2 \) makes sense for real valued roots in this context.

Key Concepts

Roots of Quadratic Equation

Understanding the roots of a quadratic equation is crucial in solving many mathematical problems. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \). The solutions to this equation, known as the equation's roots, are the values of \( x \) that satisfy it. These roots can be found using the quadratic formula, completing the square, or factoring.
The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]- **Real Roots**: For the roots to be real numbers, the discriminant \( b^2 - 4ac \) must be greater than or equal to zero. If it equals zero, there is one real root, meaning the quadratic has a repeated root. If it is positive, there are two distinct real roots.- **Complex Roots**: If \( b^2 - 4ac \) is less than zero, the roots are complex numbers.In this particular exercise, we know the product of the roots and use it to find the value of \( k \), reinforcing the connection between the roots and coefficients of a quadratic equation.

Product of Roots

The product of the roots of a quadratic equation is another important aspect to understand fully. For any quadratic equation in the standard form \( ax^2 + bx + c = 0 \), the product of its roots \( \alpha \) and \( \beta \) can be derived from Viète's formulas. It is simply:\[ \alpha \cdot \beta = \frac{c}{a} \]This relation arises from the properties of polynomial roots and their coefficients.
In the exercise provided, this concept is applied as follows:- **Analyzing the Given Equation**: For the equation \( x^2 - 3kx + 2e^{2 \ln k} - 1 = 0 \), \( a = 1 \) and \( c = 2e^{2 \ln k} - 1 \).- **Equating Product to Given Value**: Using the product of roots given part as 7: \[ \frac{2e^{2 \ln k} - 1}{1} = 7 \]This equation is then solved to determine the expression \( k \), using exponential properties.This highlights how these mathematical relationships help solve quadratic equations, emphasizing how understanding the basic properties can simplify complex problems.

Exponential Functions

Exponential functions often appear in mathematical equations due to their unique growth properties. These functions are typically represented as \( f(x) = a \cdot e^{bx} \) or similar forms, where \( e \) is the base of the natural logarithm, an essential constant in mathematics approximated at 2.718.
In the original exercise, the expression \( e^{2 \ln k} \) is central to finding the solution. Let's break down this expression:- **Properties of Logarithms**: The property \( e^{\ln(x)} = x \) simplifies exponential expressions, allowing transformations using logarithms.- **Transforming Using Exponents**: Knowing \( \ln \) is the logarithm base \( e \), \( e^{2 \ln k} \) can be rewritten using properties as: \[ (e^{\ln k})^2 = k^2 \]This transformation helps us solve for \( k \) by focusing on simpler arithmetic expressions like \( k^2 = 4 \).This shows how understanding exponential and logarithmic properties is crucial for evaluating equations involving exponential functions. It demonstrates how these concepts facilitate complex calculations easily by transforming intricate expressions into understandable components.