Problem 24 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Given \(\begin{aligned} f(x) &=\left(\frac{(1+x)^{\frac{1}{x}}}{e}\right)^{\frac{1}{x}}, \quad x<0 \\\ &=\frac{1}{\sqrt{e}}, \quad x=0 \\ &=(1+\ln (\cos (\sin x)))^{\frac{1}{x}}, \quad x>0 . \end{aligned}\)

Step-by-Step Solution

Answer

For the given piecewise function \(f(x)\), we analyze three cases: 1. For \(x<0\) (e.g., x = -2), \(f(x)=\left(\frac{(1+x)^{\frac{1}{x}}}{e}\right)^{\frac{1}{x}}\), and we find \(f(-2)=\sqrt{e}\). 2. For \(x=0\), \(f(x)=\frac{1}{\sqrt{e}}\), so \(f(0)=\frac{1}{\sqrt{e}}\). 3. For \(x>0\) (e.g., x = 1), \(f(x)=(1+\ln (\cos (\sin x)))^{\frac{1}{x}}\), and we find \(f(1)=0.5945\).

Step 1: Case 1: x < 0 (Negative values of x)

Let's find \(f(x)\) for a specific value of x in this case. We will utilize the given formula for negative x values: \(f(x)=\left(\frac{(1+x)^{\frac{1}{x}}}{e}\right)^{\frac{1}{x}}\) Let x = -2, so we will find \(f(-2)\): \(f(-2) = \left(\frac{(1+(-2))^{\frac{1}{-2}}}{e}\right)^{\frac{1}{-2}}\)

Step 1: Simplifying the innermost parentheses

Simplify \((1+(-2))\): \(f(-2) = \left(\frac{(-1)^{\frac{1}{-2}}}{e}\right)^{\frac{1}{-2}}\)

Step 2: Simplifying the exponent

Simplify the exponent \((-1)^{\frac{1}{-2}}\): \(f(-2) = \left(\frac{1}{e}\right)^{\frac{1}{-2}}\)

Step 3: Simplifying the final exponentiation

Simplify the final exponentiation \(\left(\frac{1}{e}\right)^{\frac{1}{-2}}\): \(f(-2) = \sqrt{e}\) Therefore, \(f(-2) = \sqrt{e}\).

Step 5: Case 2: x = 0

In this case, we are given the value of \(f(0)\): \(f(0) = \frac{1}{\sqrt{e}}\) There is no need to simplify any further. Thus, \(f(0) = \frac{1}{\sqrt{e}}\).

Step 6: Case 3: x > 0 (Positive values of x)

Let's find \(f(x)\) for a specific value of x in this case. We will use the given formula for positive x values: \(f(x) = (1+\ln (\cos (\sin x)))^{\frac{1}{x}}\) Let x = 1, so we will find \(f(1)\): \(f(1) = (1+\ln (\cos (\sin 1)))^{\frac{1}{1}}\)

Step 1: Evaluating the trigonometric functions

Evaluate \(\sin(1)\), then \(\cos(\sin(1))\): \(f(1) = (1+\ln (\cos (0.8415)))^{\frac{1}{1}}\) \(f(1) = (1+\ln (0.6666))^{\frac{1}{1}}\)

Step 2: Simplifying the logarithm

Evaluate \(\ln(0.6666)\): \(f(1) = (1 - 0.4055)^{\frac{1}{1}}\)

Step 3: Simplifying the final expression

Simplify the expression inside the parentheses, and remember that \((a)^{\frac{1}{1}} = a\): \(f(1) = 0.5945\) Therefore, \(f(1) = 0.5945\).

Key Concepts

Functional Analysis

Functional analysis is a branch of mathematics dealing with functions, their spaces, and transformations. In this exercise, the function \(f(x)\) behaves differently depending on the value of \(x\) (negative, zero, or positive). Understanding how a function changes with domain helps analyze behaviors such as continuity and differentiability.
For example, for negative \(x\), we transform the function using specific algebraic manipulations to find values like \(f(-2)\). Analyzing these changes helps students see patterns and properties in functions.

• Case Analysis: Involves checking different value ranges and adapting solutions accordingly.
• Continuity: Determines if functions maintain their form over changes in input.

Functional analysis is foundational for understanding complex systems in mathematics and physics.

Exponentiation

Exponentiation is raising a number to a power. This concept looks at repeated multiplication and extends to fractions or negative numbers. In the exercise, exponentiation is at the core of the function, particularly with expressions like \((1+x)^{\frac{1}{x}}\).
With negative bases, results might be non-real; hence, we simplify within the scope of real numbers (as seen with \((-1)^{\frac{1}{-2}} = \frac{1}{\sqrt{e}}\)).

• Power Laws: Understand the basic laws like \(a^m \cdot a^n = a^{m+n}\).
• Fractional Exponents: These represent roots and are crucial when \(x\) is not an integer.

Exponentiation in calculus helps us model exponential growth, decay, and more.

Logarithms

Logarithms are the inverses of exponentiation. They ask what power a base must be raised to in order to reach a given number. In the solution, logarithms appear in \(\ln(\cos(\sin x))\) when \(x > 0\).
The natural logarithm \(\ln\) has a base \(e\), and it is commonly used in calculus for solving problems involving growth and decay.

• Inverse Processes: Understand that \(\ln(e^x) = x\).
• Logarithmic Properties: Such as \(\ln(ab) = \ln a + \ln b\). They simplify complex expressions.

Logarithms provide insight into the multiplicative scaling of functions and simplify the computation of products and powers.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent relate angles to ratios of sides in right triangles. In our solution, we use \(\sin x\) and \(\cos(\sin x)\) to evaluate parts of \(f(x)\) for positive \(x\).
These functions oscillate between -1 and 1, making them useful for modeling periodic phenomena such as waves.

• Trigonometric Identities: Formulas that express relationships, such as \(\sin^2(x) + \cos^2(x) = 1\).
• Periodic Nature: Helps in modeling repeating cycles, crucial for understanding behaviors over an interval.

Grasping the use of trigonometric functions is essential for solving equations and understanding geometric interpretations in calculus.