Problem 20 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

If \(f(x)=\cos x, 0 \leq x \leq \frac{\pi}{2}\), then the real num- ber ' \(\mathrm{c}\) ' for the mean value theorem is [MPPET-1994] (a) \(\pi / 6\) (b) \(\pi / 4\) (c) \(\sin ^{-1}(2 / \pi)\) (d) \(\cos ^{-1}(2 / \pi)\)

Step-by-Step Solution

Answer

Option (c) \(\sin^{-1}(2/\pi)\) is correct.

Step 1: Understanding the Mean Value Theorem

The Mean Value Theorem (MVT) for derivatives states that for a continuous function on the interval \([a, b]\) and differentiable on \((a, b)\), there exists at least one real number \(c\) in \((a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). For this problem, \(f(x) = \cos x\), \(a = 0\), and \(b = \frac{\pi}{2}\).

Step 2: Calculate \(f(b) - f(a)\)

First, evaluate \(f(x)\) at the endpoints of the interval: \(f(\frac{\pi}{2}) = \cos\left(\frac{\pi}{2}\right) = 0\) and \(f(0) = \cos(0) = 1\). Thus, \(f(b) - f(a) = 0 - 1 = -1\).

Step 3: Calculate \(b - a\)

Subtract the endpoints of the interval: \(b - a = \frac{\pi}{2} - 0 = \frac{\pi}{2}\).

Step 4: Use MVT to find \(f'(c)\)

According to the MVT, \(f'(c) = \frac{f(b) - f(a)}{b - a} = \frac{-1}{\frac{\pi}{2}} = -\frac{2}{\pi}\). Now, find \(c\) for which this holds.

Step 5: Differentiate \(f(x)\) and solve for \(c\)

Differentiate \(f(x) = \cos x\), giving \(f'(x) = -\sin x\). We set \(-\sin c = -\frac{2}{\pi}\), which implies \(\sin c = \frac{2}{\pi}\).

Step 6: Solve for \(c\)

To find \(c\), use the inverse sine function: \(c = \sin^{-1}\left(\frac{2}{\pi}\right)\). This calculation gives us the value of \(c\) in the interval \((0, \frac{\pi}{2})\).

Key Concepts

Continuous Function

Continuous functions play a critical role in calculus. A function is said to be continuous over an interval if you can draw it without lifting your pencil from the paper. This means there are no abrupt jumps, breaks, or holes in the graph of the function on that interval.
For example, consider the function \(f(x) = \cos x\) over the interval \[0, \frac{\pi}{2}\]. This function is continuous because you can trace the cosine wave from \(x = 0\) to \(x = \frac{\pi}{2}\) smoothly.

• Continuous functions are essential for applying the Mean Value Theorem (MVT).
• They ensure that a change in input produces a predictable and gradual output change.
• For continuity, a function must be defined at every point in the interval.

Understanding continuity is key to solving problems that involve changing functions over specific intervals.

Differentiable Function

A differentiable function means a function that has a derivative at every point within its domain. This concept goes hand in hand with continuity but adds the requirement of smoothness in the change.
For the Mean Value Theorem, the function must be not only continuous but also differentiable on the open interval between the endpoints.
A great example of differentiability is again \(f(x) = \cos x\). On the interval \((0, \frac{\pi}{2})\), the function is differentiable because its derivative, \(f'(x) = -\sin x\), exists at every point.

• Differentiability indicates that the function's graph does not have sharp corners or cusps.
• This property allows us to discuss exactly how the function is changing at each point, not just over broader intervals.
• If a function is differentiable on an interval, it is also continuous on that interval. The converse might not always be true.

Knowing if a function is differentiable helps in calculating rates of change and is crucial for more advanced calculus concepts.

Interval Notation

Interval notation is a mathematical concept used to denote the set of real numbers between two endpoints. It is a concise way of expressing a range of values.
In this context, the function \(f(x) = \cos x\) is considered on the interval \[0, \frac{\pi}{2}\]. The bracket \[\]\ indicates that the endpoints are included in the interval, meaning the function is continuous at these points.

• Brackets [ ] indicate closed intervals, meaning the endpoints are part of the set.
• Parentheses ( ) indicate open intervals, meaning the endpoints are not included.
• Interval notation helps to clearly specify which portions of a graph or function are under consideration.

Understanding interval notation is vital for analyzing and visualizing functions within specific bounds.

Derivative Calculation

Calculating the derivative is a core skill in calculus used to determine the rate of change of a function. When you differentiate a function, you are finding a new function that describes this rate of change.
For the function \(f(x) = \cos x\), its derivative is \(f'(x) = -\sin x\). This tells us how the function \(\cos x\) is changing at each point on its interval.

• The derivative provides the slope of the tangent line to the graph of the function at any given point.
• It is used in the Mean Value Theorem to find specific values where the average rate of change matches the instantaneous rate of change.
• Using derivatives, we solve various real-world problems involving motion, optimization, and change.

Mastering derivative calculation is key for success in calculus and for a deeper understanding of how functions behave.