Problem 24 - A Complete Resource Book in Mathematics for JEE Main 2017

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) \(-\frac{3}{2} \leq \alpha \leq \frac{1}{3}\) (D) none of these

Step-by-Step Solution

Answer

Option (C) \(-\frac{3}{2} \leq \alpha \leq \frac{1}{3}\).

Step 1: Simplify the Expression

Start by simplifying the expression \(\sin^4 x + \cos^4 x + \sin 2x\) to make calculations easier. Recall that \(\sin^4 x + \cos^4 x\) can be rewritten using the identity \(a^2 + b^2 = (a+b)^2 - 2ab\). Write it as \((\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - 2 \sin^2 x \cos^2 x\), since \(\sin^2 x + \cos^2 x = 1\). Use the double angle identity for sine, \(\sin 2x = 2 \sin x \cos x\), to get \(1 - 2\sin^2 x \cos^2 x + \sin 2x\).

Step 2: Express the Equation in Terms of a Single Variable

Let \(\sin x = a\) and \(\cos x = b\). Since \(a^2 + b^2 = 1\), the term \(2 \sin^2 x \cos^2 x\) becomes \(2a^2b^2 = 2(1 - (a^2+b^2)) = 2(1 - ab^2)\). Therefore, the expression becomes \(1 - 2(1-a^2b^2) + 2ab\) or \(-2a^2b^2 + 2ab + 1\).

Step 3: Simplify the Equation with the Condition Given

The original equation is \(-2a^2b^2 + 2ab + 1 + \alpha = 0\). Solve for \(\alpha\): \(\alpha = -1 + 2a^2b^2 - 2ab\). To find the range of \(\alpha\), consider the problem's constraints on \(x\). Because \(a = \sin x\) and \(b = \cos x\) are between -1 and 1, evaluate the potential minima and maxima of \(-2a^2b^2 + 2ab + 1\) when \(x\) is within these bounds.

Step 4: Finding Maxima and Minima

Since \(a^2b^2\) varies between 0 and 1/4 (when \(a = b = \pm\sqrt{1/2}\)), rewrite \(\alpha\) in terms of the extremum points. At \(a = b = 0\), \(\alpha = -1\), and at \(a = \pm 1/\sqrt{2}, b = 1/\sqrt{2}\) or vice versa, verify that \(-2a^2b^2 + 2ab\) fits within \([-3/2, 1/3]\).

Step 5: Conclusion from Calculations

From the calculations, the expression ranges from \(-\frac{3}{2}\) to \(\frac{1}{3}\). Therefore, the correct value of \(\alpha\) is \([-\frac{3}{2}, \frac{1}{3}]\), which matches option (C).

Key Concepts

Trigonometric Identities

Trigonometric identities are equations that hold true for all angles and are crucial for simplifying expressions and solving trigonometric equations. One of the most fundamental identities is the Pythagorean identity:

• \( \sin^2 x + \cos^2 x = 1 \)

This identity allows us to express one trigonometric function in terms of another, thereby simplifying equations.

Another commonly used identity is the sum and difference identities, which help in expressing compound angles. For instance:

• \( \sin(a \pm b) = \sin a \cos b \pm \cos a \sin b \)
• \( \cos(a \pm b) = \cos a \cos b \mp \sin a \sin b \)

In this particular exercise, we utilize the identity \( a^2 + b^2 = (a+b)^2 - 2ab \) to simplify the expression \( \sin^4 x + \cos^4 x \) into a more manageable form. This transformation helps in breaking down the problem into simpler components, enabling easier manipulation and solution finding.

Knowing and applying these identities becomes an integral part of working with trigonometric equations, expediting the process of simplification and yielding quicker results.

Double Angle Formulas

Double angle formulas are another set of trigonometric identities that specifically deal with expressions where angles are doubled. These formulas are particularly useful in solving equations involving double angles. The most common double angle formulas are:

• \(\sin 2x = 2 \sin x \cos x\)
• \(\cos 2x = \cos^2 x - \sin^2 x\)
• \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\)

In the original exercise, the formula \(\sin 2x = 2 \sin x \cos x\) is used to transform the term \(\sin 2x\) into a product of simpler trigonometric functions, \(2ab\), where \(a = \sin x\) and \(b = \cos x\).

This approach simplifies the equation, making it easier to handle and solve. Understanding and applying double angle formulas can significantly streamline the solving process by reducing the complexity of trigonometric expressions involving double angles.

Inequality Solutions

Solving inequalities in trigonometric equations involves finding the range of values that satisfy the given parameters. In this case, the problem is to solve for all possible values of \(\alpha\) in the equation \(-2a^2b^2 + 2ab + 1 + \alpha = 0\).

To determine the solutions, it requires analyzing the behavior of the expression \(-2a^2b^2 + 2ab + 1\) over the domain of possible values for \(a\) and \(b\), which, as sine and cosine, range from -1 to 1.

• Evaluate conditions that affect \(a^2b^2\), such as its maximum value \(1/4\) when \(a = b = \pm\sqrt{1/2}\).
• Check endpoints where \(a\) or \(b\) equals zero, impacting the outcome for \(\alpha\).

The entire expression thereby ranges from \(-\frac{3}{2}\) to \(\frac{1}{3}\), so effectively, the allowed values for \(\alpha\) are confined to those bounds. Understanding how to evaluate the minimum and maximum impact of component variables on these inequalities is key to mastering inequality-based trigonometric equations.