Problem 54 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=\frac{\cos ^{2} x+\sin ^{4} x}{\sin ^{2} x+\cos ^{4} x}\) for \(x \in R\), then \(f(2002)=\) (A) (B) 2 (C) 3 (D) 4

Step-by-Step Solution

Answer

1, because the function simplifies to a constant value.

Step 1: Simplify the Function

First, observe the expression for the function: \(f(x)=\frac{\cos ^{2} x+\sin ^{4} x}{\sin ^{2} x+\cos ^{4} x}\). Notice that both the numerator and the denominator contain trigonometric identities. Try to simplify these expressions using known identities like \(\sin^2 x + \cos^2 x = 1\).

Step 2: Express in Known Identity

Re-write the numerator: \(\cos^2 x + \sin^4 x = \cos^2 x + \sin^2 x \cdot \sin^2 x\). The denominator: \(\sin^2 x + \cos^4 x = \sin^2 x + \cos^2 x \cdot \cos^2 x\).

Step 3: Transform Using Trigonometric Identity

Substitute \(\sin^2 x + \cos^2 x = 1\) in both expressions. The numerator becomes \(\cos^2 x + (1-\cos^2 x) \cdot \sin^2 x\), and the denominator becomes \(\sin^2 x + (1-\sin^2 x) \cdot \cos^2 x\).

Step 4: Simplify the Expression Further

Perform simple algebra in both the numerator and the denominator: - Numerator: \(\cos^2 x + \sin^2 x - \sin^2 x \cdot \cos^2 x\)- Denominator: \(\sin^2 x + \cos^2 x - \sin^2 x \cdot \cos^2 x\).Both simplify to \(1 - \sin^2 x \cdot \cos^2 x\).

Step 5: Simplified Function

Now, the expression becomes \(f(x)=\frac{1 - \sin^2 x \cdot \cos^2 x}{1 - \sin^2 x \cdot \cos^2 x}\), which simplifies to 1 as long as the denominator \(1 - \sin^2 x \cdot \cos^2 x eq 0\).

Step 6: Evaluate at Specific Point

Evaluate at the given point: \(f(2002)\). Since \(f(x) = 1\), this implies \(f(2002) = 1\). Therefore, regardless of the value of \(x\), as long as \(1 - \sin^2 x \cdot \cos^2 x eq 0\), we have \(f(x) = 1\).

Key Concepts

Trigonometric Identities

Trigonometric identities are very useful in simplifying expressions or equations involving trigonometric functions like sine and cosine. One of the most essential identities you should remember is the Pythagorean identity:

• \(\sin^2 x + \cos^2 x = 1\)

This identity is often used to transform or simplify trigonometric expressions by allowing you to replace one component of the identity with the other to balance or solve equations. In the context of the given exercise, this identity helps us adjust the numerator and denominator expressions for the function to make them more manageable.
You'll notice that, during simplification, expressions like \(\sin^4 x\) were rewritten in terms of \(\sin^2 x\) and the known identity. By understanding and utilizing these identities, you can better navigate complex-looking trigonometric functions.

Simplification

Simplification is the process of making expressions easier to work with by reducing complexity. In the given function, there's a need to simplify both the numerator and the denominator. To achieve this, we carefully utilized the Pythagorean identity and basic algebraic manipulation.
First, we can transform expressions like \(\cos^2 x + \sin^4 x\) by rewriting as \(\cos^2 x + \sin^2 x \cdot \sin^2 x\). Similarly, transforming the denominator from \(\sin^2 x + \cos^4 x\) to \(\sin^2 x + \cos^2 x \cdot \cos^2 x\) allows us to simplify further using the identity that \(\sin^2 x + \cos^2 x = 1\).
After simplification, both numerator and denominator become \(1 - \sin^2 x \cdot \cos^2 x\). Simplification not only helps remove complexity but also reveals important features of the function that provide insights, like in this case, where both reduce to the same value.

Function Evaluation

Function evaluation involves determining the value of a function for a particular input. Once the function \(f(x)\) was simplified, it became clear that the expression was consistently 1 when \(1 - \sin^2 x \cdot \cos^2 x\) was not zero.
In this scenario, evaluating at a specific point such as \(x = 2002\) does not change the outcome, because the simplified form of the function is independent of \(x\), given it doesn't cause division by zero. To perform function evaluation:

• First, ensure the expression is in its simplest form, as we did with \(f(x)\) becoming 1.
• Then substitute the given value, checking for undefined conditions like zero in the denominator.
• If no issues arise, the function value remains consistent across different inputs.

Here, \(f(2002) = 1\) since the simplified expression assures this outcome across all valid \(x\). This demonstrates the power of simplification and identity use for easier evaluation of functions.