Problem 23 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The solution of the equation \(1+\sin ^{2} a x=\cos x\), where \(a\) is irrational, is (A) \(x=0\) (B) \(x=\frac{n \pi}{a}\) (C) \(x=2 n \pi\) (D) none of these

Step-by-Step Solution

Answer

The solution is (D) none of these.

Step 1: Understand the Equation

The given equation is \(1 + \sin^2(ax) = \cos(x)\), where \(a\) is irrational. Our goal is to determine for which values of \(x\) this equation holds.

Step 2: Analyze the Range of the Functions

Let's consider the ranges of the functions involved. The function \(\sin^2(ax)\) ranges from 0 to 1 since \(\sin(ax)\) ranges from -1 to 1. Therefore, \(1 + \sin^2(ax)\) ranges from 1 to 2. On the other hand, \(\cos(x)\) ranges from -1 to 1.

Step 3: Compare the Ranges of Each Side

The function \(1 + \sin^2(ax)\) has a minimum at 1, while \(\cos(x)\) can go as low as -1. To satisfy the equation \(1 + \sin^2(ax) = \cos(x)\), both sides must equal, meaning both the minimum and maximum values should be identical. However, \(\cos(x)\) cannot reach 2, and \(1 + \sin^2(ax)\) cannot go below 1.

Step 4: Conclusion

Since the ranges of \(1 + \sin^2(ax)\) and \(\cos(x)\) do not overlap, there can be no solution for any real \(x\). This implies that neither side will ever equal the other for any \(x\), making the equation unsolvable.

Key Concepts

Irrational Numbers

Understanding irrational numbers is crucial when approaching trigonometric equations like the one in our exercise. An irrational number cannot be expressed as a simple fraction, meaning it has a non-repeating and non-terminating decimal representation. For example, the number \( \sqrt{2} \) is irrational. In our equation, \( a \), an irrational number, plays a significant role, particularly in affecting the periodicity of the function \( \sin(ax) \).

Periodicity typically refers to how often a function repeats itself. When \( a \) is irrational, the usual repetitive patterns we might expect from the sine and cosine functions become disrupted. This disruption makes it challenging to find solutions where the patterns align. Hence, equations involving trigonometric functions and irrational numbers are often unsolvable, as seen in this exercise.

Function Ranges

When solving the trigonometric equation \( 1 + \sin^2(ax) = \cos(x) \), understanding function ranges is fundamental. Each trigonometric function has a specific range, which tells us the set of values the function can take.

The range of \( \sin^2(ax) \) is from 0 to 1 because \( \sin(ax) \) ranges between -1 and 1, and squaring non-negative results. Consequently, \( 1 + \sin^2(ax) \) has a range from 1 to 2.
In contrast, \( \cos(x) \) ranges from -1 to 1.

The key point is that the ranges must overlap for the equation to have possible solutions. In this case, there's no overlap between the ranges of \( 1 + \sin^2(ax) \) and \( \cos(x) \), which is why no solutions exist for this equation.

Equation Solving

Solving the equation \( 1 + \sin^2(ax) = \cos(x) \) involves analyzing and comparing the ranges of both sides. Equation solving in trigonometry often requires setting each side to be equal across a possible range of values.

Since \( 1 + \sin^2(ax) \) never dips below 1, while \( \cos(x) \) never rises above 1, there's no common value \( x \) that can satisfy the equation at all points in their ranges.

• No common values mean no solutions.
• Considering additional factors like periodicity or phase shifts does not alter this fact, due to the irrational nature of \( a \).

Thus, exhaustive consideration of the trigonometric properties and inequalities suggests the equation is unsolvable.

Comparing Trigonometric Functions

Comparing trigonometric functions is a staple in solving such equations. It involves evaluating where two or more trigonometric expressions can possibly be equal to one another. Here, comparing \( 1 + \sin^2(ax) \) and \( \cos(x) \) reveals the strict range limitations preventing them from meeting at any point.

This comparison's insight is:

• \( 1 + \sin^2(ax) \) is always greater than or equal to 1.
• \( \cos(x) \) is always less than or equal to 1.

The lack of overlapping values due to these limits confirms the impossibility of the equation being satisfied. Understanding these comparisons illuminates broader methods in trigonometry for evaluating function equality.