Problem 24 - A Complete Resource Book in Mathematics for JEE Main

Question

The values of \(\alpha\) for which the equation \(\sin ^{4} x+\cos ^{4} x+\sin 2 x+\alpha=0\) may be valid, are (A) \(-\frac{3}{2} \leq \alpha \leq 1\) (B) \(0 \leq \alpha \leq \frac{1}{2}\) (C) \(-\frac{3}{2} \leq \alpha \leq \frac{1}{2}\) (D) none of these

Step-by-Step Solution

Answer

The answer is (D) none of these.

Step 1: Simplify the given equation

Start with the equation \( \sin^4 x + \cos^4 x + \sin 2x + \alpha = 0 \). We know that \( \sin^4 x + \cos^4 x \) can be rewritten using the identity \( (\sin^2 x + \cos^2 x)^2 = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x \). Therefore, \( \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x \).

Step 2: Use Trigonometric Identity

Recall that \( \sin 2x = 2\sin x \cos x \). Substitute back into the equation: \( 1 - 2\sin^2 x \cos^2 x + 2\sin x \cos x + \alpha = 0 \)

Step 3: Maximum and Minimum Values

Simplify the expression \( -2\sin^2 x \cos^2 x + 2\sin x \cos x \) to determine its range. This expression can take values from \(-1\) to \(1\) since both \(-2\sin^2 x \cos^2 x\) and \(2\sin x \cos x\) are bounded by \(-1\) and \(1\). Thus, \( 1 - 1 \leq 1 - 2\sin^2 x \cos^2 x + 2\sin x \cos x \leq 1 + 1 \), giving \( 0 \leq \sin^4 x + \cos^4 x + \sin 2x \leq 2 \).

Step 4: Solve for \(\alpha\)

Set the bounds for \(\sin^4 x + \cos^4 x + \sin 2x \) from Step 3 into the equation \( \sin^4 x + \cos^4 x + \sin 2x + \alpha = 0 \). Solving for \(\alpha\), we get:\( \alpha = - (\sin^4 x + \cos^4 x + \sin 2x) \). By substituting the bounds:\( \alpha = -2 \text{ when } 1 + 1 = 2 \text{, and } \alpha = 0 \text{ when } 1 + -1 = 0 \). Thus, \(-2 \leq \alpha \leq 0\).

Step 5: Choose the Correct Option

From the derived range \(-2 \leq \alpha \leq 0\), none of the options (A), (B), or (C) in the problem statement matches this range. Thus, the correct answer is option (D) none of these.

Key Concepts

Trigonometric identities

Trigonometric identities are useful because they allow us to simplify complex expressions and equations, transforming them into more manageable forms. In the problem given, we utilize two important identities:

• The Pythagorean identity:
  \[ \sin^2 x + \cos^2 x = 1 \]
• The double angle identity for sine:
  \[ \sin 2x = 2\sin x \cos x \]

Using these identities, especially the Pythagorean identity squared, helps in breaking down the term \(\sin^4 x + \cos^4 x\) into something more familiar:
\[ 1 - 2\sin^2 x\cos^2 x \]Additionally, incorporating the double angle identity allows us to further simplify the equation expressionally.
These steps show how recognized patterns in trigonometry open ways to substitute and simplify, which is crucial for solving equations.

Range of trigonometric expressions

Understanding the range of trigonometric expressions is fundamental when analyzing how trigonometric functions behave. These functions have set boundaries within which they oscillate.
For the particular problem, the expression \(-2\sin^2 x \cos^2 x + 2\sin x \cos x\) needs to be evaluated to determine its range of values. This range goes from

• \(-1\) for its minimum
• \(1\) for its maximum

This results because both sine and cosine independently take values between
-1 and 1. When compounded in forms such as \(2\sin x \cos x\), it reflects those boundaries in its range.Through understanding these limits, students can properly set the boundaries for their calculations and predict the possible outputs of an expression.

Solving inequalities

Solving inequalities is about placing expressions within a specified range. In trigonometry, this often means setting bounds for an equation and determining permissible values for unknowns.
In this exercise, once the equation \(\sin^4 x + \cos^4 x + \sin 2x\) is simplified, we determine its total output range:

• The sum cannot be less than 0
• It also cannot exceed 2

We then isolate \(\alpha\) by using the equation \(\alpha = - (\sin^4 x + \cos^4 x + \sin 2x) \).
Solving for \(\alpha\), given the established range above, determines that \(\alpha\) itself ranges from -2 to 0.
Recognizing this endpoint is essential, as it guides students in matching their calculated range with the correct choice from given options, illustrating how steps in solving inequalities distinctively narrow down possible solutions.