Problem 75 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f_{k}(x)=\frac{1}{k}\left(\sin ^{k} x+\cos ^{k} x\right)\) where \(x \in R\) and \(k \geq 1\) then, the value of \(f_{4}(x)-f_{6}(x)\) equals (A) \(\frac{1}{6}\) (B) \(\frac{1}{3}\) (C) \(\frac{1}{4}\) (D) \(\frac{1}{12}\)

Step-by-Step Solution

Answer

The value of \( f_{4}(x) - f_{6}(x) \) is \( \frac{1}{12} \), which is answer (D).

Step 1: Calculate f₄(x)

To calculate \( f_{4}(x) \), substitute \( k = 4 \) into the function: \[ f_{4}(x) = \frac{1}{4} \left( \sin^{4} x + \cos^{4} x \right) \]

Step 2: Simplify f₄(x)

Using the identity \( \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cdot \cos^2 x = 1 - 2\sin^2 x \cdot \cos^2 x \), substitute into the expression for \( f_{4}(x) \):\[ f_{4}(x) = \frac{1}{4} (1 - 2\sin^2 x \cdot \cos^2 x) \]

Step 3: Calculate f₆(x)

To calculate \( f_{6}(x) \), substitute \( k = 6 \) into the function: \[ f_{6}(x) = \frac{1}{6} \left( \sin^{6} x + \cos^{6} x \right) \]

Step 4: Simplify f₆(x)

Using the identity: \( \sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^3 - 3\sin^2 x\cos^2 x(\sin^2 x + \cos^2 x) = 1 - 3\sin^2 x \cdot \cos^2 x \),substitute into \( f_{6}(x) \):\[ f_{6}(x) = \frac{1}{6} (1 - 3\sin^2 x \cdot \cos^2 x) \]

Step 5: Calculate f₄(x) - f₆(x)

Subtract \( f_{6}(x) \) from \( f_{4}(x) \): \[ f_{4}(x) - f_{6}(x) = \left( \frac{1}{4}(1 - 2\sin^2 x \cdot \cos^2 x) \right) - \left( \frac{1}{6}(1 - 3\sin^2 x \cdot \cos^2 x) \right) \]

Step 6: Simplify the expression

Simplify the expression obtained in Step 5: \[ = \frac{1}{4} - \frac{2}{4}\sin^2 x \cos^2 x - \left( \frac{1}{6} - \frac{3}{6}\sin^2 x \cos^2 x \right) \] \[ = \frac{1}{4} - \frac{1}{6} + \left( \frac{3}{6} - \frac{2}{4} \right) \sin^2 x \cos^2 x \] \[ = \frac{3}{12} - \frac{2}{12} = \frac{1}{12} \]

Step 7: Conclude with the correct answer

The value of \( f_{4}(x) - f_{6}(x) \) is \( \frac{1}{12} \). Therefore, the correct answer is option (D).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools used in mathematics, particularly in calculus and trigonometry. They allow us to express complex trigonometric functions in simpler forms. For instance, one of the most commonly used identities is the Pythagorean identity, which tells us that \( \sin^2 x + \cos^2 x = 1 \).
This identity is incredibly useful because it helps in breaking down or combining terms that involve sine and cosine functions. It serves as a building block for more complicated identities such as \( \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cdot \cos^2 x \).

• These identities are used to simplify expressions and solve equations involving trigonometric functions.
• Understanding and applying these identities can greatly reduce the complexity of problems.
• They help transform trigonometric expressions into algebraic ones, facilitating easier manipulation.

Understanding and mastering these identities is key to successfully solving trigonometric problems in JEE mathematics.

Function Simplification

Function simplification involves reducing a mathematical expression to its simplest form. This process can make calculations easier and provide insights into the properties of the function. For example, in the given problem, we have to simplify \( f_{4}(x) = \frac{1}{4} \left( \sin^4 x + \cos^4 x \right) \) and \( f_{6}(x) = \frac{1}{6} \left( \sin^6 x + \cos^6 x \right) \).
By utilizing trigonometric identities, these equations are simplified further:

• For \( f_{4}(x) \), use the identity \( \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cdot \cos^2 x \) which simplifies the expression to \( \frac{1}{4} (1 - 2\sin^2 x \cdot \cos^2 x) \).
• Similarly, for \( f_{6}(x) \), employing the identity \( \sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cdot \cos^2 x \) simplifies it to \( \frac{1}{6} (1 - 3\sin^2 x \cdot \cos^2 x) \).

Simplifying functions in this manner ensures that we can handle them more systematically and efficiently when performing algebraic manipulations.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and rewriting algebraic expressions for ease of calculation or to derive solutions. In trigonometric expressions, algebraic manipulation often involves combining like terms, factoring, and expanding expressions.
In the exercise solution above, algebraic manipulation is used to find the value of \( f_{4}(x) - f_{6}(x) \):

• Start by substituting the simplified forms of the expressions for \( f_{4}(x) \) and \( f_{6}(x) \) into the difference equation.
• This involves subtracting \( \frac{1}{6}(1 - 3\sin^2 x \cdot \cos^2 x) \) from \( \frac{1}{4}(1 - 2\sin^2 x \cdot \cos^2 x) \).
• The terms are then rearranged to combine like terms and further reduce the equation to its simplest form.
• Ultimately, the algebraic simplification yields a neat answer: \( \frac{1}{12} \).

Mastering algebraic manipulation is essential in mathematics, allowing students to handle a wide range of problems, including those that appear on exams like the JEE.