Problem 118 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f\) is an even function defined on the interval \([-5,5]\), then the real values of \(x\) satisfying the equation $$ f(x)=f\left(\frac{x+1}{x+2}\right) \text { are } $$ (A) \(\frac{-1 \pm \sqrt{5}}{2}\) (B) \(\frac{-3 \pm \sqrt{5}}{2}\) (C) \(\frac{-2 \pm \sqrt{5}}{2}\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is (A) \(\frac{-1 \pm \sqrt{5}}{2}\).

Step 1: Recognizing Properties of Even Functions

An even function is symmetric about the y-axis, which means for any real number \(x\), \(f(x) = f(-x)\). This property will be used when analyzing the problem.

Step 2: Understand the Equation

We are given \(f(x) = f\left(\frac{x+1}{x+2}\right)\). For\( f(x) = f(y) \) where \( y = \frac{x+1}{x+2} \), \(x\) and \(y\) must satisfy the property of the even function or else \(f(x)\) must be constant.

Step 3: Set up the Symmetric Condition

Let's consider the symmetry condition \(f(x) = f\left(\frac{x+1}{x+2}\right) = f\left(-\frac{x+1}{x+2}\right) \). Rearrange to \( \frac{x+1}{x+2} = -\frac{x+1}{x+2}\). Further simplification is not necessary by symmetry.

Step 4: Solve the Given Function Pair

Substitute \( \frac{x+1}{x+2} = -x \) and \(x = -x \), leading to the simplest roots possible from even function traits. This effectively shows changing signs which are often hidden. Simplify condition: \( \frac{x+1}{x+2} = x \) needs real solution adaptation.

Step 5: Solve for Roots

Rearrange \( \frac{x+1}{x+2} = x \) to \( (x+1) = x(x+2) \),\(\Rightarrow\), \((x+1) = x^2 + 2x\). Simplify to \(x^2 + x - 1 = 0\). Use quadratic formula to solve: \(x = \frac{-1 \pm \sqrt{5}}{2}\).

Step 6: Verify Solutions

Check each solution in \([-5, 5]\) and analyze they satisfy the equation in the symmetric property with the functional form constants. Each yielded value supports pair transitions can remain valid and verifiable for choice A.

Key Concepts

Symmetric Functions

Symmetric functions are fascinating because they have the unique property of being mirrored around a specific axis or point. In the case of an even function, this symmetry revolves around the y-axis. So, if a function is even, it means that the value of the function at any point is the same as its value at the mirror point—directly across the y-axis. Mathematically, this is expressed as:

• For any real number \(x\), an even function \(f(x)\) satisfies \(f(x) = f(-x)\).
• This property is crucial when solving equations involving even functions, as it allows for simplifications by setting equations with symmetric terms equal to one another.

Understanding this symmetry can often lead to elegant solutions without solving complex equations, simply by recognizing that similar values occur at equidistant points from the origin.
It also helps in graphing such functions, ensuring that one side of the graph is a mirror image of the other. This property is widely applicable across various mathematical problems and simplifies determining function behavior.

Quadratic Equation Solving

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). Solving them requires finding the values of \(x\) that satisfy this equation. In the given problem, after simplification, a quadratic equation \(x^2 + x - 1 = 0\) was obtained. Solving this involves using the quadratic formula, which is given by:

• \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Here, \(a = 1\), \(b = 1\), and \(c = -1\). Substituting these into the formula provides solutions to the equation. These solutions are real numbers when the discriminant \(b^2 - 4ac\) is non-negative.
The discriminant here is

• \(1^2 - 4 \times 1 \times (-1) = 5\), which is positive, indicating two distinct and real roots.

By substituting the values into the quadratic formula, we get:

• \[x = \frac{-1 \pm \sqrt{5}}{2}\]

This technique not only helps in finding the roots but also provides insights into the function's shape in graphical terms—where it crosses the x-axis, known as the roots or zeroes of the quadratic expression.

Functional Equations

Functional equations involve unknown functions and their properties instead of simple numeric relationships. Solving these equations requires understanding how functions interact with each other and often involves exploiting properties like symmetry or periodicity.
In the given problem, the functional equation was \(f(x) = f\left(\frac{x+1}{x+2}\right)\). With the knowledge that \(f\) is an even function, you can make crucial deductions:

• Since \(f(x)\) equals \(f\left(\frac{x+1}{x+2}\right)\), it means this mapping has a symmetrical quality inherent to even functions.
• The solution simplifies when considering both \(f(x)\) and \(f(-x)\) share the same value, allowing a transformation in the arguments without altering the equality.

Such equations provide an important link when different functional expressions equate to each other, as understanding the underlying properties often expose both trivial and non-trivial solutions. Sometimes, substituting back into the original form after simplification or transformation highlights the extent of these properties, ultimately leading to more efficient and insightful problem-solving.