Problem 31 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Suppose a differentiable function \(f(x)\) satisfies the identity \(f(x+y)=f(x)+f(y)+x y^{2}+x^{2} y\), for all real \(x\) and \(y\). If \(\lim _{x \rightarrow 0} \frac{f(x)}{x}=1\), then \(f^{\prime}(3)\) is equal to

Step-by-Step Solution

Answer

10

Step 1: Understand the Problem

We are given a differentiable function \(f(x)\) which satisfies a specific functional equation \(f(x+y)=f(x)+f(y)+xy^2+x^2y\) and a limit condition \(\lim_{x \rightarrow 0} \frac{f(x)}{x}=1\). We need to find \(f'(3)\).

Step 2: Investigate the Functional Equation by Setting y = 0

Set \(y = 0\) in the equation \(f(x+y)=f(x)+f(y)+xy^2+x^2y\). This gives us \(f(x) = f(x) + f(0) + 0 + 0\), which simplifies to \(f(0) = 0\).

Step 3: Use Limit Condition to Determine the Derivative at x=0

The limit suggests we find \(f'(0)\). From \(\lim_{x \rightarrow 0} \frac{f(x)}{x}=1\), this implies \(f'(0) = 1\).

Step 4: Differentiate the Given Functional Equation

Differentiate both sides of \(f(x+y)=f(x)+f(y)+xy^2+x^2y\) with respect to \(y\). This gives \(\frac{d}{dy}[f(x+y)] = f'(x+y)\). For the right hand side, the derivative is \(0 + f'(y) + x(2y) + x^2\). Therefore, we have:\[f'(x+y) = f'(y) + 2xy + x^2\].

Step 5: Apply x = 0 to the Derivative Equation

Substitute \(x = 0\) into the differentiated equation. This gives \(f'(y) = f'(y) + 0 + 0\), which is consistent and doesn't help directly. We need another substitution.

Step 6: Apply y = 0 to the Derivative Equation

Substitute \(y = 0\) into the differentiated equation. This gives \(f'(x) = f'(0) + 0 + x^2\). Since \(f'(0) = 1\), we have:\[f'(x) = 1 + x^2\].

Step 7: Calculate f'(3)

Using the expression derived \(f'(x) = 1 + x^2\), substitute \(x = 3\):\[f'(3) = 1 + 3^2 = 1 + 9 = 10\].

Key Concepts

Functional Equation

A functional equation is an equation that specifies a function in implicit form. It describes how the function behaves and relates values of the function at different points. In the given problem, the functional equation is \[ f(x+y) = f(x) + f(y) + xy^2 + x^2y \]This means that for any real numbers \( x \) and \( y \), the value of \( f(x+y) \) is determined by its values at \( x \) and \( y \), along with some polynomial terms related to \( x \) and \( y \).

• Set \( y = 0 \): This simplifies our equation to \( f(x) = f(x) + f(0) \). It tells us \( f(0) = 0 \), providing a base point.
• Solve for specific \( x \) or \( y \): helps uncover the function's properties and behavior further.

Understanding the functional equation provides a way to describe the behavior of \( f \) and famously supports solving differential and difference equations.

Limit Evaluation

Limit evaluation is crucial in calculus for finding the behavior of a function around a point, especially when directly substituting the point leads to forms like \( \frac{0}{0} \). In this exercise, \[ \lim _{x \rightarrow 0} \frac{f(x)}{x} = 1 \]tells us something important about the derivative at zero.Limits help us determine differentiability and continuity of functions. Here:

• The limit equals 1, indicating \( f'(0) = 1 \). It's derived from the definition of the derivative.
• A limit value like this frequently suggests the function's slope or rate of change at that exact point.

Evaluating limits correctly is essential for calculating derivatives and understanding function behavior at particular points.

Derivative Calculation

Derivative calculation relies on understanding the rate of change of a function. It gives us a way to analyze how a function behaves locally. In the problem provided, \[ f'(x) = rac{d}{dx}[f(x)] \]gives us the slope of the tangent line to the curve of the function at any point \( x \). For the function described by the problem:

• Start with differentiating the given functional equation concerning y and x.
• Finding the general form, like \( f'(x) = 1 + x^2 \), helps compute specific derivatives by substituting values.

For \( x=3 \), the derivative is easily calculated by substitution:
\[ f'(3) = 1 + 3^2 = 10 \]
Understanding derivatives is pivotal in calculus as they show us the changing dynamics of functions.

Calculus Problem Solving

Solving calculus problems, such as this, often involves multiple techniques like handling functional equations, limits, and derivatives simultaneously. Solving this specific problem involves:

• Identifying the Type of Problem: Recognizing it's a functional equation accompanied by a limit condition.
• Analyzing Function Properties: Investigating behaviors using simplification tricks like \( y=0 \) and considering limits.
• Solve Using Multiple Facets: Here, using differentiation maintains both the algebraic and calculus aspects of problem-solving.

Approaching calculus problems involves breaking them into manageable parts and applying different calculus rules creatively:

-Balance algebraic manipulation with calculus approaches, leading to solutions that leverage both differentiation and limit evaluation.
Each step builds on the previous, offering a structured pathway to a solution and deepening the understanding needed to tackle similar problems in the future.