Problem 32 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The equation \(\sin ^{4} x+\cos ^{4} x=a\) has a solution for (A) all of values of \(a\) (B) \(a=1\) (C) \(a=\frac{1}{2}\) (D) \(\frac{1}{2}

Step-by-Step Solution

Answer

The correct answer is (D) \(\frac{1}{2} < a < 1\).

Step 1: Express Using Pythagorean Identity

First, recognize that \(\sin^4 x + \cos^4 x\) can be rewritten using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). Then, \(\sin^4 x + \cos^4 x\) can be expressed as \((\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2 \sin^2 x \cos^2 x\).

Step 2: Simplify Using Double Angle Formula

Simplify \(\sin^4 x + \cos^4 x\) using the double angle formula: \(\sin^2 x\cos^2 x = \frac{1}{4}\sin^2(2x)\). Thus, \(\sin^4 x + \cos^4 x = 1 - \frac{1}{2}\sin^2(2x)\).

Step 3: Determine Range of the Expression

The expression becomes \(a = 1 - \frac{1}{2}\sin^2(2x)\). The range of \(\sin^2(2x)\) is from \(0\) to \(1\). Therefore, \(1 - \frac{1}{2} \times 1 \leq a \leq 1\), or \(\frac{1}{2} \leq a \leq 1\).

Step 4: Identify Correct Answer

Recognize that the values \(\frac{1}{2} < a < 1\) are the values that make the equation solvable. Therefore, the correct answer is option (D).

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a crucial concept in trigonometry and states that for any angle \( x \), the equation \( \sin^2 x + \cos^2 x = 1 \) holds true. This identity is fundamental because it connects the squares of sine and cosine functions of an angle, providing a constant sum of 1.
This relation is often the starting point for simplifying complex trigonometric expressions. In the given exercise, we employ the Pythagorean Identity to reformulate \( \sin^4 x + \cos^4 x \) into an expression involving squares of trigonometric functions. This transformation helps set the stage for applying other trigonometric identities to solve the equation more effectively.

Double Angle Formula

The Double Angle Formula is a powerful tool used in trigonometry to express trigonometric functions of double angles. It is particularly helpful in simplifying expressions and solving equations. Specifically, the double angle formula for sine is \( \sin(2x) = 2\sin x \cos x \).
Given \( \sin^2 x \cos^2 x \) appears in the original expression, by using this formula, we can express \( \sin^2 x \cos^2 x = \frac{1}{4}\sin^2(2x) \). Incorporating the double angle formula allows us to simplify \( \sin^4 x + \cos^4 x \) to \( 1 - \frac{1}{2}\sin^2(2x) \). This simplification is crucial for determining the solution range of the original equation.

Solution Range

The solution range refers to the permissible values that the variable or expression can take while satisfying the equation. In this exercise, it involves determining the range for \( a \) in the expression \( a = 1 - \frac{1}{2}\sin^2(2x) \).
Knowing that \( \sin^2(2x) \) can vary from 0 to 1, we substitute these boundary values into the expression for \( a \):

• When \( \sin^2(2x) = 0 \), \( a = 1 \)
• When \( \sin^2(2x) = 1 \), \( a = \frac{1}{2} \)

Thus, the range for \( a \) is \( \frac{1}{2} \leq a \leq 1 \). However, since \( a \) cannot be exactly \( \frac{1}{2} \) or 1 in the choice provided, the appropriate solution is in the range \( \frac{1}{2} < a < 1 \), as highlighted in the correct answer.

JEE Main Mathematics

JEE Main Mathematics is a segment of the Joint Entrance Examination which tests an aspirant's understanding of various mathematical concepts at a high-school level. Trigonometry is a significant part of this exam, with questions often requiring a combination of identities and formulas for solutions.
In the context of this exercise, concepts such as the Pythagorean Identity, the Double Angle Formula, and determining solution ranges are integral. These concepts form the foundation for tackling more complex trigonometric problems, as seen in the solution to the given equation. Mastery in these areas not only aids in solving specific questions but also develops a robust mathematical intuition necessary for engineering entrance exams.