Problem 18 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f:(0, \infty) \rightarrow(0, \infty)\) be a differentiable function such that \(f(1)=e\) and \(\lim _{t \rightarrow x} \frac{t^{2} f^{2}(x)-x^{2} f^{2}(t)}{t-x}=0\). If \(f(x)=1\), then \(x\) is equal to: \(\quad\) [Sep. 04, 2020 (II)] (a) \(\frac{1}{e}\) (b) \(2 e\) (c) \(\frac{1}{2 e}\) (d) \(e\)

Step-by-Step Solution

Answer

The value of \(x\) is \(e\) (option d).

Step 1: Analyze the given limit

The problem involves the limit \( \lim_{t \rightarrow x} \frac{t^{2} f^{2}(x) - x^{2} f^{2}(t)}{t-x} = 0 \). This resembles the definition of the derivative of a function. Rewriting it gives us a hint:\[\frac{d}{dt} \left( t^2 f^2(t) \right) = 0\]This indicates that \(t^2 f^2(t)\) might be a constant with respect to \(t\).

Step 2: Assume a solution form and check

Assume \(t^2 f^2(t) = C\) for some constant \(C\). This gives:\[ f^2(t) = \frac{C}{t^2} \]Evaluate using the initial condition \(f(1) = e\): \[f^2(1) = C \implies e^2 = C \implies f^2(t) = \frac{e^2}{t^2}\].

Step 3: Solve for given condition \(f(x)=1\)

Now, substitute \(f(x) = 1\) into the derived function form:\[1^2 = \frac{e^2}{x^2} \]Simplifying gives:\[x^2 = e^2\]Taking the square root on both sides results in:\[x = e\]

Step 4: Confirm the value of \(x\)

The solution satisfies the condition \(f(x) = 1\), and all steps followed directly from the given problem and its constraints.Thus, the value of \(x\) that satisfies \(f(x) = 1\) is \(e\).

Key Concepts

Limit Definition

In calculus, the limit is a fundamental concept used to define various mathematical constructs, such as derivatives and continuity. Limits help us understand the behavior of functions as inputs approach a specific value. When we talk about the limit of a function as a variable approaches a certain point, we're interested in finding out what value the function tends to.

In the given exercise, the limit given is \[\lim_{t \rightarrow x} \frac{t^{2} f^{2}(x) - x^{2} f^{2}(t)}{t-x} = 0.\]This resembles the definition of a derivative. The expression inside the limit indicates that as \(t\) approaches \(x\), the numerator approaches zero at a rate balanced by the denominator. Such a relation suggests that there is a certain symmetry or constancy in the function \(f\). Here, the limit helps identify that \(t^2 f^2(t)\) behaves like a constant, leading us to further exploration in the problem.

Derivative

The concept of the derivative is central to calculus. It represents the rate of change or the slope of the tangent line to the function at any given point. This is formally defined by the limit:\[\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}.\]In the context of the exercise, the limit \[\lim_{t \rightarrow x} \frac{t^{2} f^{2}(x) - x^{2} f^{2}(t)}{t-x}\]is akin to a derivative because it looks like the change in a function's value over the change in its input.

• This indicates an element of differentiation, as this form suggests a derivative of some composite or transformed function.
• The outcome points to \(t^2 f^2(t)\) being a constant function of \(t\), as the derivative equaling zero suggests a lack of change.

Understanding derivatives helps us evaluate how functions evolve as their variables change, critical for solving differentiation-related problems.

Constant Function

A constant function is one that does not change, no matter what the input value is. Mathematically, if a function \(f(x)\) returns the same value for all \(x\), it is a constant function. In many calculus problems, identifying parts of expressions as constant can provide significant insights. In this problem, the speculation that \(t^2 f^2(t) = C\) stems from the derivative-like limit condition provided:

• When we solve \(f^2(t) = \frac{C}{t^2}\), and use the given initial condition \(f(1) = e\), it simplifies to a straightforward form with \(C = e^2\).
• This reveals that the expression behaves as a constant with respect to \(t\), linking back to the nature of the derivative presented earlier.

Recognizing constants in calculus simplifies the computation and understanding of many problems, as it dramatically reduces the variability being dealt with.

Initial Condition

Initial conditions are specific values provided for variables, often used to solve differential equations or specify a solution uniquely. They give context and constraints to a problem, guiding the path to the solution. The problem uses the initial condition \(f(1) = e\). This piece of information is crucial because:

• It allows us to determine the constant \(C\) in the expression \(t^2 f^2(t) = C\).
• By substituting \(t = 1\) into \(f^2(t) = \frac{C}{t^2}\), we find \(C = e^2\), simplifying the function to \(f^2(t) = \frac{e^2}{t^2}\).

Initial conditions offer a starting point, ensuring that the derived general solutions fit the specific cases presented in the problems. They are critical in tailoring any function or equation to a particular, given scenario.