Problem 56 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=\cos ^{-1}\left(\frac{x^{-1}-x}{x^{-1}+x}\right)\), then \(f^{\prime}(x)\) is (A) odd (B) even (C) periodic (D) None of these

Step-by-Step Solution

Answer

The derivative \( f'(x) \) is odd.

Step 1: Simplify the Function

The function given is \( f(x) = \cos^{-1}\left( \frac{x^{-1} - x}{x^{-1} + x} \right) \). Simplify the expression inside the inverse cosine function. Let \( y = x^{-1} = \frac{1}{x} \). Thus, \( f(x) = \cos^{-1}\left( \frac{y - x}{y + x} \right) \). This further simplifies to \( f(x) = \cos^{-1}\left( \frac{\frac{1}{x} - x}{\frac{1}{x} + x} \right) \). By rationalizing the denominator by multiplying numerator and denominator by \( x \), we get \( f(x) = \cos^{-1}\left( \frac{1-x^2}{1+x^2} \right) \).

Step 2: Determine Even/Odd Nature of the Function

To determine whether the function \( f(x) \) is even, odd, or neither, substitute \( -x \) into the function and simplify: \( f(-x) = \cos^{-1}\left( \frac{1-(-x)^2}{1+(-x)^2} \right) \). This simplifies to the same expression \( f(x) = \cos^{-1}\left( \frac{1-x^2}{1+x^2} \right) \), hence \( f(-x) = f(x) \), meaning \( f(x) \) is even.

Step 3: Verify the Derivative for Even/Odd Property

To confirm the behavior in derivatives, find \( f'(x) \). The derivative of \( \cos^{-1}(u) \) is \( -\frac{1}{\sqrt{1-u^2}} \cdot u' \). Here, \( u = \frac{1-x^2}{1+x^2} \), so \( f'(x) = -\frac{1}{\sqrt{1-u^2}} \cdot \left( \frac{-4x}{(1+x^2)^2} \right) \). Compute \( u' \) by differentiating the expression for \( u \). After simplification, since \( u'(-x) = -u'(x) \), the derivative is odd.

Key Concepts

Even and Odd Functions

First things first, let’s understand what even and odd functions are! An even function is one in which the equation is symmetric along the y-axis. This means that for any input value \( x \), the output value at \( f(x) \) will be the same as that at \( f(-x) \). A classic example of an even function is \( f(x) = x^2 \), because \( (-x)^2 = x^2 \), so \( f(x) = f(-x) \) for all \( x \).

On the flip side, an odd function is symmetric around the origin. For an odd function, the equation \( f(-x) = -f(x) \) holds true. A typical odd function example is \( f(x) = x^3 \), as \((-x)^3 = -x^3\), resulting in \( f(-x) = -f(x) \).

• For the function \( f(x) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \), substituting \(-x\) returns the same expression. Therefore, this function is even.

Recognizing these properties is essential in calculus because they can simplify integration and differentiation tasks. By noting symmetry, mathematical calculations can become less complicated.

Inverse Trigonometric Functions

Inverse trigonometric functions, often called arc functions, are essentially the reverse functions of the usual trigonometric functions like sine, cosine, and tangent. They are used to determine the angle whose trigonometric function value is known.

There are six primary inverse trigonometric functions: \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), \( \tan^{-1}(x) \), \( \cot^{-1}(x) \), \( \sec^{-1}(x) \), and \( \csc^{-1}(x) \). The function \( \cos^{-1}(x) \), for instance, gives the angle whose cosine is \( x \).

• In our example, \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \) depicts the angle whose cosine equals the simplified form \( \frac{1-x^2}{1+x^2} \).
• This practice helps in analyzing problems where angles—and not just their sine, cosine, or other values—are pivotal.

Such functions are crucial when dealing with equations involving trigonometric functions, helping to translate them into algebraic terms. When studying calculus, knowing how to manage these functions enables tackling otherwise complex problems.

Derivative of Functions

Derivatives are fundamental in calculus. They tell us everything we need to know about rates of change! If a function illustrates how one quantity depends on another, the derivative provides the rate at which this dependency changes.

For any given function \( f(x) \), its derivative \( f'(x) \) represents the slope of the tangent to the function's graph at any point. This allows us to understand how the function behaves around that area which can be incredibly useful for optimization and modeling real-world scenarios.

• In our exercise, we determined \( f'(x) \) using the derivative of the \( \cos^{-1}(u) \), simplifying it through the chain rule.
• The general form used, \( -\frac{1}{\sqrt{1-u^2}} \cdot u' \), highlights how derivatives can be systematically simplified, reflecting function symmetry.

Calculating derivatives of inverse trigonometric functions often includes understanding how changes in input values translate to changes in function outputs. This essential calculus skill allows students to describe motion, growth, and changes in scientific and engineering contexts.