Problem 10 - A Complete Resource Book in Mathematics for JEE Main

Question

The number of values of \(x\), where the function \(f(x)=\) \(\cos x+\cos (\sqrt{2} x)\) attains its maximum, is (A) 0 (B) 1 (C) 2 (D) infinite

Step-by-Step Solution

Answer

(A) 0

Step 1: Understand the Function

We are given the function \(f(x) = \cos x + \cos (\sqrt{2}x)\). This function involves two cosine terms, which oscillate between -1 and 1 for all real numbers \(x\). We need to find the values of \(x\) where this function reaches its maximum value.

Step 2: Determine the Maximum Possible Value

Each cosine value oscillates between -1 and 1. So, the maximum value of \(\cos x\) is 1 and \(\cos (\sqrt{2}x)\) is also 1. Thus, the maximum value of \(f(x) = \cos x + \cos (\sqrt{2}x)\) is 2, occurring when both terms are equal to 1.

Step 3: Find Conditions for Maximum Value

For \(f(x)\) to equal 2, both \(\cos x = 1\) and \(\cos (\sqrt{2}x) = 1\) must simultaneously hold. \(\cos x = 1\) when \(x = 2k\pi\) for \(k \in \mathbb{Z}\), and \(\cos (\sqrt{2}x) = 1\) when \(\sqrt{2}x = 2m\pi\), or equivalently, \(x = \frac{2m\pi}{\sqrt{2}} = \sqrt{2}m\pi\) for \(m \in \mathbb{Z}\).

Step 4: Solving the Equations Simultaneously

Both conditions imply \(x = 2k\pi = \sqrt{2}m\pi\). Equating these gives \(2k = \sqrt{2}m\), which becomes \(\frac{m}{k} = \sqrt{2}\). This equation implies \(m\) and \(k\) must be chosen such that their ratio is \(\sqrt{2}\), an irrational number, which cannot happen for integer \(m\) and \(k\).

Step 5: Conclusion

Since \(m\) and \(k\) cannot satisfy the condition \(\frac{m}{k} = \sqrt{2}\) for integers, there are no values of \(x\) where both conditions are satisfied simultaneously.

Key Concepts

Cosine Function

The cosine function, denoted as \(\cos(x)\), is one of the fundamental functions in trigonometry. It describes the horizontal component of a unit circle as an angle \(x\) changes. This function has a distinctive wave-like pattern known as a sinusoidal shape. Here are some key properties of the cosine function:

• Periodicity: The cosine function repeats its values in regular intervals of \(2\pi\). This is known as the period of \(\cos(x)\).
• Range: The values of the cosine function vary between -1 and 1, making it bounded.
• Symmetry: Cosine is an even function, which means \(\cos(x) = \cos(-x)\).

In the given exercise, the function \(f(x) = \cos x + \cos (\sqrt{2}x)\) features two cosine components. Each component reaches its peak value of 1, contributing to a comprehensive maximum potential of 2 for the function. This behavior reflects the oscillatory nature of the cosine function, which oscillates as \(x\) changes over time.

Maximum Value

In trigonometric functions, the concept of maximum value refers to the peak amplitude a function can reach. For the provided function \(f(x) = \cos x + \cos (\sqrt{2}x)\), identifying this maximum value involves understanding when both cosine components simultaneously hit their peak values.

• Individual Maximum: Each cosine function, \(\cos(x)\) and \(\cos(\sqrt{2}x)\), independently achieves its maximum value of 1.
• Combined Maximum: When both do so together, we attain the maximum value of the sum: 1 + 1 = 2.

However, reaching this combined maximum requires both functions to equal 1 at the same \(x\), which can be calculated by setting \(\cos x = 1\) and \(\cos (\sqrt{2}x) = 1\) simultaneously. By exploring scenarios where these conditions hold, you can conclude that no real \(x\) leads to this situation under integer constraints, hence, no maximum value for real \(x\) is conclusively found.

Simultaneous Equations

Simultaneous equations are a set of equations containing multiple variables that are solved together since they share common solutions. To investigate when \(f(x) = \cos x + \cos(\sqrt{2}x)\) reaches its maximum of 2, we need to work with two conditions:

• \(\cos x = 1\), which occurs when \(x = 2k\pi\) where \(k\) is an integer.
• \(\cos(\sqrt{2}x) = 1\), equivalent to \(x = \sqrt{2}m\pi\), with \(m\) as an integer.

Solving these conditions requires setting \(2k\pi = \sqrt{2}m\pi\). Simplifying gives \(2k = \sqrt{2}m\) and finally \(\frac{m}{k} = \sqrt{2}\). This condition creates a challenging scenario because\(\sqrt{2}\) is irrational, making a rational integer ratio unachievable. Thus, these simultaneous equations reveal there are no integer solutions that satisfy both, confirming no \(x\) leads to the desired maximum.

Rational and Irrational Numbers

Rational and irrational numbers are fundamental in understanding number theory and its applications. A rational number can be expressed as a fraction \(\frac{a}{b}\), where both \(a\) and \(b\) are integers and \(b eq 0\). In contrast, an irrational number cannot be represented as a simple fraction. It has infinite, non-repeating decimal places.

• Examples: Rational numbers include fractions like \(\frac{1}{2}\), whereas \(\pi\) and \(\sqrt{2}\) are irrational.
• Properties: The irrationality of a number indicates that no two integers can provide its exact ratio.

In the exercise, when seeking to solve \(2k = \sqrt{2}m\), finding \(\frac{m}{k} = \sqrt{2}\) highlights the core issue. No such integer \(m\) and \(k\) will satisfy this equation because \(\sqrt{2}\) lacks a precise rational representation, hence confirming the non-existence of an \(x\) that maximizes \(f(x)\). Understanding this distinction helps in analyzing problems involving trigonometric functions and simultaneous equations.