Problem 63 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x+y)=f(x)+f(y)+2 x y-1 \forall x, y \in R\). If \(f(x)\) is differentiable and \(f^{\prime}(0)=\sin \theta\), then (A) \(f(x)>0 \forall x \in R\) (B) \(f(x)<0 \forall x \in R\) (C) \(f(x)=\sin \theta \forall x \in R\) (D) None of these

Step-by-Step Solution

Answer

(A) \( f(x) > 0 \forall x \in \mathbb{R} \).

Step 1: Define the base case

Start by substituting zero for both variables in the given functional equation: \( f(0+0) = f(0) + f(0) + 2 \cdot 0 \cdot 0 - 1 \). Simplifying gives \( f(0) = 2f(0) - 1 \), which results in \( f(0) = 1 \).

Step 2: Differentiate the functional equation

Since \( f(x) \) is differentiable, differentiate both sides of the equation \( f(x+y) = f(x) + f(y) + 2xy - 1 \) with respect to \( x \):\( \frac{d}{dx}f(x+y) = \frac{d}{dx}[f(x) + f(y) + 2xy - 1] \).Using the chain rule on the left side, we get \( f'(x+y) \). On the right side, the differentiated terms are: \( f'(x) \) and \( 2y \). Hence,\( f'(x+y) \cdot 1 = f'(x) + 0 + 2y \) because \( f(y) \) and \( -1 \) are constant on differentiating with respect to \( x \).

Step 3: Evaluate the derivative at x = 0

Put \( x = 0 \) in the derivative expression \( f'(x+y) = f'(x) + 2y \):This gives us \( f'(y) = f'(0) + 2y \). Since \( f'(0) = \sin \theta \), substitute this to find:\( f'(y) = \sin \theta + 2y \).

Step 4: Integrate the derivative to find f(x)

Integrate \( f'(y) = \sin \theta + 2y \) to find \( f(y) \):\[ f(y) = \int (\sin \theta + 2y) \, dy = \sin \theta y + y^2 + C \].Given \( f(0) = 1 \), substitute \( y = 0 \) into \( f(y) = 1 \): \( 1 = \sin \theta \cdot 0 + 0^2 + C \). Thus, \( C = 1 \).Therefore, \( f(y) = \sin \theta y + y^2 + 1 \).

Step 5: Check the options

Substitute \( f(y) = \sin \theta y + y^2 + 1 \) into each of the given options:- Option (A): \( f(x) > 0 \forall x \in \mathbb{R} \). Indeed, for all \( x \), \( y^2 + 1 > 0 \) because the square of a real number is always non-negative and \( +1 \) ensures positivity. - Option (B): \( f(x) < 0 \forall x \in \mathbb{R} \). Not possible since \( f(x) > 0 \).- Option (C): \( f(x) = \sin \theta \forall x \in \mathbb{R} \). Clearly false as \( f(x) \) depends on \( x \).Hence, the correct choice is (A).

Key Concepts

Functional Equations

Functional equations are mathematical expressions involving a function, its values at various points, and possibly other variables. In the given problem, we have the functional equation:

• \( f(x+y) = f(x) + f(y) + 2xy - 1 \) for all real numbers \( x \) and \( y \).

To solve for the unknown function \( f(x) \), we first substitute zero for both variables. This step simplifies the equation and helps uncover details about \( f(x) \). The results often reveal base cases such as \( f(0) = 1 \). This is crucial in building a foundation to understand more complex behavior of \( f \).
Functional equations require analyzing the behavior of \( f(x) \) under different operations, like addition in this case. This helps establish relationships between different points of the function and aids in further steps like differentiation.

Derivative

The derivative of a function is a key concept in calculus. It measures the rate at which the function changes as its input changes. Here,

• We differentiate the functional equation \( f(x+y) = f(x) + f(y) + 2xy - 1 \) with respect to \( x \).

Differentiation involves applying rules such as the chain rule. When differentiating the equation, we obtain \( f'(x+y) \), which simplifies the process by breaking down the function into more manageable parts.
A derivative, represented as \( f'(x) \), is used to analyze intricate details about how \( f(x) \) behaves across its domain, providing insights that allow finding solutions or verifying properties of the function.

Differentiation

Differentiation is the mathematical technique used to compute a derivative. It involves rules and formulas to systematically find how and at what rate functions change. Here,

• We differentiated the functional equation to find the expression \( f'(x+y) = f'(x) + 2y \).
• This step leads us to understand how \( f'(y) \) behaves when \( x = 0 \), giving \( f'(y) = \sin \theta + 2y \).

Differentiation helps break down complex functions into simpler, smaller pieces. This is essential for problems involving variable interaction, as it provides clarity and direction on how to proceed with finding the original function or solving equations.
Grasping differentiation is critical, as it allows you to predict or determine patterns about changes in functions, which is integral to solving calculus problems like this one.

Integration

Integration is the mathematical process of finding the original function from its derivative. It is, in a sense, the inverse operation of differentiation. From the derivative \( f'(y) = \sin \theta + 2y \),

• We integrate to obtain: \[ f(y) = \int (\sin \theta + 2y) \, dy = \sin \theta y + y^2 + C \].
• Using the base case \( f(0) = 1 \), we solve for the constant \( C \).

This results in \( f(y) = \sin \theta y + y^2 + 1 \).
Integration ties back into our functional equation and verifies that the solution meets the criteria from previous steps. It allows us to reconstruct \( f(y) \) and ensures the function aligns with the conditions set out by the original problem statement, such as being differentiable.