Problem 12 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Investigate the following functions for extremum at \(x=0:-\) i. \(\quad f(x)=\sin x-x\). ii. \(\quad f(x)=\sin x-x+\frac{x^{3}}{3 !}\). iii. \(f(x)=\sin x-x+\frac{x^{3}}{3 !}-\frac{x^{4}}{4 !}\). iv. \(\quad f(x)=e^{\frac{1}{x}}, \quad x \neq 0\) \(=0, \quad x=0 .\) v. \(f(x)=\cosh x+\cos x\). vi. \(f(x)=\cos x-1+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}\). vii. \(f(x)=\cos x-1+\frac{x^{2}}{2}\). viii. \(f(x)=x+x^{\frac{2}{3}}\). ix. \(f(x)=x^{2}+x^{\frac{1}{3}}\). x. \(f(x)=x^{\frac{4}{3}}+2 .\) xi. \(f(x)=x^{\frac{5}{3}}-3 .\)

Step-by-Step Solution

Answer

Answer: i. No extremum at x=0 ii. No extremum at x=0 iii. No extremum at x=0 iv. Indeterminate v. Extremum at x=0 vi. Extremum at x=0 vii. Extremum at x=0 viii. No extremum at x=0 ix. Indeterminate x. No extremum at x=0 xi. No extremum at x=0

Step 1: i. f(x) = sin(x) - x

: 1. Find first derivative: \(f'(x) = \cos(x) - 1\) 2. Set first derivative to zero: \(\cos(x) - 1 = 0 \Rightarrow \cos(x) = 1\) There is no solution for x=0. Conclusion: No extremum at x=0.

Step 2: ii. f(x) = sin(x) - x + (x^3)/6

: 1. Find first derivative: \(f'(x) = \cos(x) - 1 + x^2\) 2. Set first derivative to zero: \(\cos(x) - 1 + x^2 = 0\) At x=0, \(\cos(0)-1+0^2 = -1\) Conclusion: No extremum at x=0.

Step 3: iii. f(x) = sin(x) - x + (x^3)/6 - (x^5)/120

: 1. Find first derivative: \(f'(x) = \cos(x) - 1 + x^2 - \frac{x^4}{24}\) 2. Set first derivative to zero: \(\cos(x) - 1 + x^2 - \frac{x^4}{24} = 0\) At x=0, \(\cos(0)-1+0^2 - \frac{0^4}{24} = -1\) Conclusion: No extremum at x=0.

Step 4: iv. f(x) = e^(1/x) for x ≠ 0, and f(x) = 0 for x=0

: 1. Find first derivative: \(f'(x) = -\frac{1}{x^2} e^\frac{1}{x}\) for x ≠ 0, and undefined for x=0. We can't set the derivative to zero, so we can't find an extremum by following the standard procedure. Conclusion: Indeterminate.

Step 5: v. f(x) = cosh(x) + cos(x)

: 1. Find first derivative: \(f'(x) = \sinh(x) - \sin(x)\) 2. Set first derivative to zero: \(\sinh(x) - \sin(x) = 0\) At x=0, \(\sinh(0) - \sin(0) = 0\) Since \(\sinh(x)\) is an odd function, and its limit as x approaches zero is equal to zero, while \(-\sin(x)\) is also an odd function and its limit as x approaches zero is also equal to zero, we can conclude that there is an extremum at x=0.

Step 6: vi. f(x) = cos(x) - 1 + (x^2)/2 - (x^3)/6

: 1. Find first derivative: \(f'(x) = -\sin(x) + x - \frac{x^2}{2}\) 2. Set first derivative to zero: \(-\sin(x) + x - \frac{x^2}{2} = 0\) At x=0, \(-\sin(0) + 0 - \frac{0^2}{2} = 0\) Conclusion: There is an extremum at x=0.

Step 7: vii. f(x) = cos(x) - 1 + x^2/2

: 1. Find first derivative: \(f'(x) = -\sin(x) + x\) 2. Set first derivative to zero: \(-\sin(x) + x = 0\) At x=0, \(-\sin(0) + 0 = 0\) Conclusion: There is an extremum at x=0.

Step 8: viii. f(x) = x + x^(2/3)

: 1. Find first derivative: \(f'(x) = 1 + \frac{2}{3}x^{-\frac{1}{3}}\) 2. Set first derivative to zero: \(1 + \frac{2}{3}x^{-\frac{1}{3}} = 0\) There is no solution for x=0. Hence, no extremum at x=0.

Step 9: ix. f(x) = x^2 + x^(1/3)

: 1. Find first derivative: \(f'(x) = 2x + \frac{1}{3}x^{-\frac{2}{3}}\) 2. Set first derivative to zero: \(2x + \frac{1}{3}x^{-\frac{2}{3}} = 0\) At x=0, the derivative is undefined. Conclusion: Indeterminate.

Step 10: x. f(x) = x^(4/3) + 2

: 1. Find first derivative: \(f'(x) = \frac{4}{3}x^{\frac{1}{3}}\) 2. Set first derivative to zero: \(\frac{4}{3}x^{\frac{1}{3}} = 0\) There is no solution for x=0. Conclusion: No extremum at x=0.

Step 11: xi. f(x) = x^(5/3) - 3

: 1. Find first derivative: \(f'(x) = \frac{5}{3}x^{\frac{2}{3}}\) 2. Set first derivative to zero: \(\frac{5}{3}x^{\frac{2}{3}} = 0\) There is no solution for x=0. Conclusion: No extremum at x=0.

Key Concepts

Calculus

Calculus is a branch of mathematics that studies how things change. It is built around two main concepts: differentiation and integration. Differentiation allows us to find rates of change, like speed or the slope of a curve. Integration is about finding the total amount or the area under curves. In our context, we use differentiation to determine the extremum of functions, which are points where the function reaches a local maximum or minimum.
The extremum of a function is found by calculating its derivative and setting it to zero. This process identifies critical points where a function may have an extremum. The nature of change at these points (increasing or decreasing) helps us confirm whether these points are indeed maxima or minima.

First Derivative Test

The first derivative test is a tool in calculus used to find and confirm the existence of local maxima and minima of a function. Once you find the derivative of a function, you look for where this derivative equals zero or is undefined.

• If a function changes from increasing to decreasing at a point, the function has a local maximum.
• Conversely, if it changes from decreasing to increasing, it has a local minimum.

For example, if you are evaluating an extremum test and find that the derivative equals zero at a point, you then check the function's behavior immediately around this point to conclude if the point is an extremum.
In the provided solutions, steps were taken to find critical points for given functions. Some functions did not provide valid solutions at all, showing stability at those points but without extremum behavior.

Trigonometric Functions

Trigonometric functions, such as sine (\( \sin x\)) and cosine (\( \cos x \)), are foundational in mathematics, especially calculus. They arise in circular and periodic phenomena. In extremum exploration, they often appear when analyzing functions' periodic behavior.
These functions provide smooth curves with well-known derivatives that can show the rates at which these curves slope upwards or downwards. In our exercise, you noticed factors like \( \cos(x) \) - 1 and \( \cosh(x) \) as parts of expressions. Finding extremum involves looking for points where these trigonometric derivatives are zero, providing critical points where the function stops changing or "turns around."

Exponential Functions

Exponential functions, like \( e^{x} \), have the unique property of growing continuously and significantly. They appear in various fields, modeling growth processes such as populations or investments. In the exercise, the function \( e^{1/x} \) pushes us into complex behavior near the origin where standard derivative handling sometimes becomes difficult due to undefined regions.
These exponential components showcase how rapidly outputs can shift, especially near points like \( x = 0 \), which impacts fundamental calculus analysis for extremum testing. Understanding how these functions behave near extreme points aids in interpreting and solving complex mathematical problems involving growth and rates of change.

Hyperbolic Functions

Hyperbolic functions, such as \( \cosh(x) \) and \( \sinh(x) \), are analogs to trigonometric functions, yet they relate to hyperbolas instead of circles. These functions help model many real-world components such as hanging cables (i.e., catenary curves), where their slopes (derivatives) provide insight into how these curves behave.
In analyzing the extremum of functions in this exercise, the hyperbolic cosine function (\( \cosh x\)) played a role when mixed with \( \cos x \). Its derivative relation with the hyperbolic sine provides powerful tools to investigate local extremum. These allow us to explore function behavior deeply, crucial when combined function components demand more careful derivation checks around critical points.