Problem 159 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The function \(f(x)=\cot ^{-1} x+x\) increases in the interval: (A) \((1, \infty)\) (B) \((-1, \infty)\) (C) \((-\infty, \infty)\) (D) \((0, \infty)\)

Step-by-Step Solution

Answer

The function is increasing in the interval \((- fty, fty)\).

Step 1: Find the derivative of f(x)

To determine where the function \(f(x) = \cot^{-1} x + x\) is increasing, we first find the derivative. The derivative of \(\cot^{-1} x\) is \(-\frac{1}{1+x^2}\), and the derivative of \(x\) is \(1\). Thus, the derivative \(f'(x)\) is: \[ f'(x) = -\frac{1}{1+x^2} + 1 = 1 - \frac{1}{1+x^2} \]

Step 2: Simplify the derivative

Rewrite the expression for \(f'(x)\) with a common denominator: \[ f'(x) = \frac{1+x^2}{1+x^2} - \frac{1}{1+x^2} = \frac{1+x^2 - 1}{1+x^2} = \frac{x^2}{1+x^2} \]

Step 3: Determine where f'(x) is positive

The function is increasing where \(f'(x) > 0\). From the expression \(f'(x) = \frac{x^2}{1+x^2}\), note that \(x^2\) is always non-negative. The expression will be positive wherever the denominator \(1+x^2\) is positive, which is for all \(x\). Thus, \(f(x)\) is increasing for all \(x\) since \(x^2 > 0\) whenever \(x eq 0\).

Step 4: Considering all real numbers for x

Since \(x^2\) is positive for any \(xeq 0\), and the derivative \(f'(x) = \frac{x^2}{1+x^2}\) is positive for all real numbers, the function \(f(x)\) is increasing on the interval \((-\infty, \infty)\).

Key Concepts

Function Derivatives

Understanding derivatives is essential in calculus, especially when analyzing the behavior of functions. The derivative of a function represents the rate at which the function's value changes as the input changes. For trigonometric functions and their inverses, like \(f(x) = \cot^{-1} x + x\), it's crucial to know how to derive each component.To find the derivative of \(f(x)\), identify the derivatives of its parts:

• The derivative of \(\cot^{-1} x\) is \(-\frac{1}{1+x^2}\). This is a standard formula for the derivative of inverse trigonometric functions.
• The derivative of \(x\) is simply 1, as the change in \(x\) over itself is steady.

Putting them together, \(f'(x) = 1 - \frac{1}{1+x^2}\). This derivative can then be simplified to show how the function behaves over its domain.

Increasing and Decreasing Functions

A foundational concept in understanding function behavior is determining where a function increases or decreases. A function is said to be increasing on an interval if its derivative is positive throughout that interval.For the function \((f'(x) = \frac{x^2}{1+x^2})\), it's positive whenever the numerator is non-negative, which occurs when \(x^2 > 0\) for \(x eq 0\). Furthermore, the denominator \(1+x^2\) is always positive for all real numbers, ensuring that the derivative remains positive.This positivity indicates that the function \(f(x)\) increases on the entire real line, except at \(x = 0\). Therefore, \(f(x)\) is increasing over \((-\infty, \infty)\). This condition helps us understand the nature of the function without detailed pointwise evaluation.

Calculus

Calculus is a branch of mathematics that allows us to study change and motion through derivatives and integrals. Derivatives, as seen earlier, are fundamental in understanding how functions change.Specifically for \(f(x) = \cot^{-1} x + x\), calculus provides us tools to determine where and how quickly it increases or decreases. Through differentiation, we obtained \(f'(x) = \frac{x^2}{1+x^2}\), which was key to concluding on the nature of \(f(x)\).Utilizing calculus concepts like differentiation not only helps in exploring functions like inverse trigonometric ones but also offers broader applications in physics, engineering, and other sciences. By mastering these concepts, students can better interpret real-world phenomena and solve complex problems more effectively.