Problem 125 - Calculus for JEE ADVANCED

Question

$$ \begin{aligned} &\log _{4}\left(\frac{2 f(x)}{1-f(x)}\right)=x \\ &\Rightarrow \quad \frac{2 f(x)}{1-f(x)}=4^{x} \end{aligned} $$ Replacing \(x\) by \(1-x\), we get $$ \frac{2 f(1-x)}{1-f(1-x)}=4^{1-x} $$ Multiplying (1) and (2), we get $$ \begin{aligned} & \frac{2 f(x)}{1-f(x)}\left(\frac{2 f(1-x)}{1-f(1-x)}\right)=4^{1-x} \\ \Rightarrow & f(x) f(1-x)=1-f(x)-f(1-x)+f(x) f(1-x) \\ \Rightarrow & f(x)+f(1-x)=1 \end{aligned} $$ Putting \(x=2010\), we get $$ f(2010)+f(-2009)=1 $$

Step-by-Step Solution

Answer

The relationship is \( f(x) + f(1-x) = 1 \), so \( f(2010) + f(-2009) = 1 \).

Step 1: Understand the Given Equations

We are given the equation \( \log_{4}\left(\frac{2 f(x)}{1-f(x)}\right)=x \) which implies \( \frac{2 f(x)}{1-f(x)}=4^{x} \). The goal is to find the relationship between \( f(x) \) and \( f(1-x) \).

Step 2: Replace x by 1 - x

We substitute \(1 - x\) for \(x\) in the equation \( \frac{2 f(x)}{1-f(x)}=4^{x} \). This gives us \( \frac{2 f(1-x)}{1-f(1-x)}=4^{1-x} \).

Step 3: Multiply the Equations

We multiply the original equation \( \frac{2 f(x)}{1-f(x)}=4^{x} \) by the equation \( \frac{2 f(1-x)}{1-f(1-x)}=4^{1-x} \) resulting in \( \frac{2 f(x)}{1-f(x)} \cdot \frac{2 f(1-x)}{1-f(1-x)} = 4^{x} \cdot 4^{1-x} \).

Step 4: Simplify the Result

The right side simplifies to 4. The left side simplifies using the equation \(a \cdot b = c \cdot d\) to eliminate the fractions, giving us \( f(x) f(1-x) = 1-f(x)-f(1-x)+f(x) f(1-x) \).

Step 5: Solve for f(x) + f(1-x)

Rearrange the simplified equation to find that \( f(x) + f(1-x) = 1 \). This shows the relationship between \( f(x) \) and \( f(1-x) \).

Step 6: Apply the Relationship

Using the relationship \( f(x) + f(1-x) = 1 \), substitute \( x = 2010 \) to find \( f(2010) + f(-2009) = 1 \).

Key Concepts

Logarithmic Functions

Logarithmic functions are mathematical functions that are the inverses of exponential functions. A basic logarithmic function can be written as \( y = \log_b(x) \), where \( b \) is the base and it must be greater than zero and not equal to one. The logarithm answers the question: 'To what power must the base \( b \) be raised, to produce \( x \)?'In our problem, we have a logarithmic function with base 4: \( \log_{4}( \cdot ) \). The equation \( \log_{4}\left(\frac{2 f(x)}{1-f(x)}\right)=x \) sets up a relationship between a logarithm and a variable expression. Converting this from logarithmic to exponential form helps in solving for \( f(x) \) easily: \( \frac{2 f(x)}{1-f(x)}=4^{x} \). Understanding this conversion from logarithmic to exponential is crucial to solving functional equations involving logarithms.

Symmetry in Functions

Symmetry in functions refers to the property where a function maintains certain elements of invariance under transformations. The exercise reveals symmetry in the function \( f(x) \). By replacing \( x \) with \( 1-x \) and observing the equations \( \frac{2 f(x)}{1-f(x)} = 4^x \) and \( \frac{2 f(1-x)}{1-f(1-x)} = 4^{1-x} \), symmetry emerges.The key result is obtaining a relationship like \( f(x) + f(1-x) = 1 \). This unique property suggests a sort of equilibrium or balance between values at \( x \) and \( 1-x \), demonstrating how functional symmetry can provide significant insights into mathematical relationships.

Substitution Method

The substitution method involves replacing a variable with another expression to simplify an equation or solve it more easily. Here, substituting \( x \) with \( 1-x \) in the logarithmic function \( \log_{4}\left(\frac{2 f(x)}{1-f(x)}\right)=x \), gives a complementary expression \( \log_{4}\left(\frac{2 f(1-x)}{1-f(1-x)}\right)=1-x \).This manipulation allows us to tackle both expressions simultaneously. When the equations are multiplied, they simplify to reveal hidden relationships like \( f(x) + f(1-x) = 1 \), showcasing the power of substitution in solving complex equations efficiently, especially when symmetry is present.

Factorization Techniques

Factorization techniques help break down complex algebraic expressions into simpler, more manageable parts, often revealing crucial insights into the relationships between variables.In our problem, after multiplying and simplifying the expressions obtained from substituting \( x \) and \( 1-x \), the equation \( f(x) f(1-x) = 1 - f(x) - f(1-x) + f(x) f(1-x) \) is derived. Through simplification, this can be rewritten to yield \( f(x) + f(1-x) = 1 \).The use of factorization is vital here as it disentangles complex fractions and expressions into a simple form that showcases underlying relationships like symmetry. It also highlights how mathematical operations can demystify seemingly complicated equations by breaking them down into basic elements.