Problem 1 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find the mean rate of change of the following functions in the interval \([1,2]\) :- i. \(\quad f(x)=x^{2}\). \\{Ans. 3\\} ii. \(\quad f(x)=x^{3}\). \\{Ans. 7\(\\}\) iii. \(f(x)=\sqrt{x}\). \\{Ans. 0.414\\} iv. \(\quad f(x)=\frac{1}{x} .\) \\{ns. \(\left.-0.5\right\\}\) v. \(\quad f(x)=e^{x}\). \\{Ans. 4.67\\} vi. \(f(x)=\ln x .\\{\) Ans. \(0.693\\}\) vii. \(f(x)=\sin x .\\{\) Ans. \(0.068\\}\) viii. \(f(x)=\cos x\). \\{ns. \(-0.956\\}\)

Step-by-Step Solution

Answer

The mean rate of change for the given functions in the interval $[1,2]$ is as follows: 1. $f(x) = x^2$: \(3\) 2. $f(x) = x^3$: \(7\) 3. $f(x) = \sqrt{x}$: \(0.414\) 4. $f(x) = \frac{1}{x}$: \(-0.5\) 5. $f(x) = e^x$: \(4.67\) 6. $f(x) = \ln x$: \(0.693\) 7. $f(x) = \sin x$: \(0.068\) 8. $f(x) = \cos x$: \(-0.956\)

Step 1: 1. Mean rate of change of \(f(x) = x^{2}\)

First, we need to find f(1) and f(2). We have: \(f(1) = 1^{2} = 1\) \(f(2) = 2^{2}=4\) Now, apply the formula for the mean rate of change: Mean rate of change = \(\frac{4 - 1}{2 - 1} = \boxed{3}\)

Step 2: 2. Mean rate of change of \(f(x) = x^{3}\)

First, we need to find f(1) and f(2). We have: \(f(1) = 1^{3} = 1\) \(f(2) = 2^{3} = 8\) Now, apply the formula for the mean rate of change: Mean rate of change = \(\frac{8 - 1}{2 - 1} = \boxed{7}\)

Step 3: 3. Mean rate of change of \(f(x) = \sqrt{x}\)

First, we need to find f(1) and f(2). We have: \(f(1) = \sqrt{1} = 1\) \(f(2) = \sqrt{2} \approx 1.414\) Now, apply the formula for the mean rate of change: Mean rate of change = \(\frac{1.414 - 1}{2 - 1} \approx \boxed{0.414}\)

Step 4: 4. Mean rate of change of \(f(x) = \frac{1}{x}\)

First, we need to find f(1) and f(2). We have: \(f(1) = \frac{1}{1} = 1\) \(f(2) = \frac{1}{2} = 0.5\) Now, apply the formula for the mean rate of change: Mean rate of change = \(\frac{0.5 - 1}{2 - 1} = \boxed{-0.5}\)

Step 5: 5. Mean rate of change of \(f(x) = e^{x}\)

First, we need to find f(1) and f(2). We have: \(f(1) = e^{1} \approx 2.718\) \(f(2) = e^{2} \approx 7.389\) Now, apply the formula for the mean rate of change: Mean rate of change = \(\frac{7.389 - 2.718}{2 - 1} \approx \boxed{4.67}\)

Step 6: 6. Mean rate of change of \(f(x) = \ln x\)

First, we need to find f(1) and f(2). We have: \(f(1) = \ln 1 = 0\) \(f(2) = \ln 2 \approx 0.693\) Now, apply the formula for the mean rate of change: Mean rate of change = \(\frac{0.693 - 0}{2 - 1} = \boxed{0.693}\)

Step 7: 7. Mean rate of change of \(f(x) = \sin x\)

First, we need to find f(1) and f(2). We have: \(f(1) = \sin 1 \approx 0.841\) \(f(2) = \sin 2 \approx 0.909\) Now, apply the formula for the mean rate of change: Mean rate of change = \(\frac{0.909 - 0.841}{2 - 1} \approx \boxed{0.068}\)

Step 8: 8. Mean rate of change of \(f(x) = \cos x\)

First, we need to find f(1) and f(2). We have: \(f(1) = \cos 1 \approx 0.540\) \(f(2) = \cos 2 \approx -0.416\) Now, apply the formula for the mean rate of change: Mean rate of change = \(\frac{-0.416 - 0.540}{2 - 1} = \boxed{-0.956}\)

Key Concepts

Differential Calculus

Differential calculus is a branch of mathematics that deals with the study of rates at which quantities change. It is fundamentally concerned with the concept of the derivative, which measures the rate of change of a quantity. The mean rate of change is a concept that emerges directly from differential calculus and is used to describe the average rate at which a function changes over a certain interval.

For instance, consider the function \( f(x) = x^2 \). To find the mean rate of change over the interval [1,2], we evaluate the function at the endpoints of this interval, finding \( f(1) = 1^2 = 1 \) and \( f(2) = 2^2 = 4 \). The mean rate of change is the difference in these values, divided by the difference in the x-values: \( \frac{4 - 1}{2 - 1} = 3 \). This calculation represents the slope of the secant line that passes through the points \( (1, f(1)) \) and \( (2, f(2)) \) on the graph of the function

A positive mean rate of change indicates that the function is increasing, while a negative value suggests it is decreasing. Zero implies that the function's average rate of change is flat, or has no overall increase or decrease over the interval.

Function Analysis

Function analysis involves examining the properties of functions and understanding their behavior through various methods, including the calculation of rates of change. Examining a function's mean rate of change over an interval provides insight into its average behavior over that interval.

For more complex functions like \( f(x) = e^x \), the mean rate of change from x=1 to x=2 is computed in a similar fashion. After finding the values of the function at points 1 and 2, which are approximately 2.718 and 7.389 respectively, we find that the mean rate of change is roughly 4.67. Such a high rate indicates that the function is rapidly increasing in that region.

Understanding the mean rate of change requires evaluating the function at specific points and then applying the definition to arrive at the average change. This simple yet powerful concept is integral to both calculus and real-world applications, like calculating speeds and growth rates.

IIT JEE Mathematics

The Indian Institutes of Technology Joint Entrance Examination (IIT JEE) is a highly competitive exam used for admission to engineering programs in India. Mathematics is a critical component of the exam, and understanding concepts like the mean rate of change is essential for success. Problems involving the mean rate of change test a student's grasp of differential calculus and function analysis, which are integral topics in the IIT JEE mathematics syllabus.

Solving these problems effectively often requires a blend of conceptual understanding and problem-solving skills. For instance, questions may ask students to find the mean rate of change of a function over a specified interval, similar to the textbook examples provided. By mastering these types of problems, students develop their analytical abilities, which are necessary for tackling the rigorous questions present on the IIT JEE.

The mean rate of change, being a core concept in the IIT JEE mathematics section, showcases how fundamental calculus principles are to engineering and highlights the need for aspirants to be proficient in both constructing and interpreting mathematical arguments and solutions.