Problem 258 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } f(x)=\cos (\log x) \text { , then evaluate } f(x) f(y)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right] \text { . } $$

Step-by-Step Solution

Answer

The final evaluated value of the given exercise is \( \cos (\log x) \cos (\log y) -\frac{1}{2}\left[\cos (\log \frac{x}{y}) + \cos (\log xy)\right] \).

Step 1: Understand the given function.

The function given is \( f(x)=\cos (\log x) \). So, Same function for y will be \( f(y)=\cos (\log y) \).

Step 2: Substitute x/y and xy into the function.

For \( f\left(\frac{x}{y}\right) \), replace \( x \) with \( \frac{x}{y} \) in the function to get \( f\left(\frac{x}{y}\right) = \cos (\log \frac{x}{y}) \). Similarly, for \( f(xy) \), replace \( x \) with \( xy \) in function to get \( f(xy) = \cos (\log xy) \).

Step 3: Consolidate all the terms.

Substitute all the evaluated function values in the equation given, \( f(x)f(y)-\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f(x y)\right] \) is equal to \( \cos (\log x) \cos (\log y) -\frac{1}{2}\left[\cos (\log \frac{x}{y}) + \cos (\log xy)\right] \). This is the evaluation of the given exercise.

Key Concepts

Cosine Function

The cosine function, denoted as \( \cos(x) \), is a fundamental trigonometric function, often used in the study of angles and the representation of periodic phenomena.

• The cosine of an angle in a right triangle is the ratio of the adjacent side to the hypotenuse.
• In the unit circle, \( \cos(\theta) \) represents the x-coordinate of a point on the circle's circumference.
• It is periodic with a period of \( 2\pi \), which means every \( 2\pi \) units, the function repeats its values.
• The function oscillates between -1 and 1, with a peak at 0 and a trough at \( \pi \).

The cosine function's properties make it integral in analyzing cyclic patterns, such as waves or harmonic motions. In calculi, the derivation and integration of the cosine function play crucial roles in solving complex problems involving oscillations.

Logarithmic Function

Logarithmic functions, represented as \( \log(x) \), are the inverses of exponential functions. They arise naturally in various contexts where exponential growth or decay processes are present.

• A logarithm answers the question: "To what power must a given base be raised, to yield a certain number?"
• In base \( 10 \), known as the common logarithm, \( \log_{10}(x) \) is often written simply as \( \log(x) \). For the base \( e \) (approximately 2.718), it's written as \( \ln(x) \).
• Properties include: \( \log(ab) = \log a + \log b \) and \( \log(a^b) = b \log a \).

The logarithmic function is essential in transforming multiplicative relationships into additive ones, simplifying calculus operations. In the function \( f(x) = \cos(\log x) \), the logarithm helps adjust the scale of \( x \), modifying the input for the cosine function and creating a composite function.

Function Substitution

Substitution is a fundamental technique in calculus used to simplify complex functions or problems by replacing one variable or expression with another.

• This method helps transform integrals or differentials into more manageable forms.
• It involves identifying parts of a function that can be rewritten using different variables or expressions.
• In our context, \( f\left( \frac{x}{y} \right) = \cos(\log \frac{x}{y}) \) and \( f(xy) = \cos(\log xy) \), these substitutions help evaluate expressions without directly handling complex initial forms.
• It's particularly effective in revealing underlying relationships or symmetries within an equation.

Using substitution within \( f(x) = \cos(\log x) \) aids in systematically breaking down and solving the original problem. It's an invaluable tool for tackling problems that seem daunting at first glance.