Problem 5 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

For what value of \(x\) in the interval \(\left(0, \frac{\pi}{2}\right)\), the maximum value of \(\sin \left(x+\frac{\pi}{6}\right)+\cos \left(x+\frac{\pi}{6}\right)\) is attained?

Step-by-Step Solution

Answer

The maximum value is attained for \(x = \frac{\pi}{6}\).

Step 1: Find the function in terms of \(x\)

From the problem, the function is \(f(x) = \sin \left(x+\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{6}\right)\). This function is a combination of sine and cosine functions.

Step 2: Differentiate the function

Differentiate \(f(x)\) with respect to \(x\) to get \(f'(x) = \cos \left(x+\frac{\pi}{6}\right) - \sin\left(x+\frac{\pi}{6}\right)\). We differentiate to find where the function's slope equals zero, because these points are the potential maxima and minima.

Step 3: Find critical points

Setting \(f'(x) = 0\), we find the critical points. We solve \(\cos \left(x+\frac{\pi}{6}\right) - \sin\left(x+\frac{\pi}{6}\right) = 0\). This gives \(x = \frac{\pi}{6}, \frac{5\pi}{6}\) but we only consider \(x = \frac{\pi}{6}\) as the other solution lies outside the given interval (0, \(\frac{\pi}{2}\)).

Step 4: Check for maximum

From step 3, we only find one critical point, hence that must be the point where the maximum value is achieved. However, further confirmation can be obtained using the Second Derivative Test, which in this case is not necessary since we only have one critical point within the given interval and the function is continuous.

Key Concepts

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental components in mathematics. They are used to describe periods and patterns found in waves. The exercise given involves combining a sine and a cosine function. The function is given by

• \[ f(x) = \sin \left(x+\frac{\pi}{6}\right) + \cos \left(x+\frac{\pi}{6}\right) \]

This is simply a transformation of the basic sine and cosine functions, where each argument is shifted by \(\frac{\pi}{6}\). Such transformations affect the period and phase of the wave, resulting in a unique pattern or orientation. Understanding these transformations allows us to grasp how a wave behaves under different shifts and compositions.

Critical Points

The critical points of a function are values of \(x\) where the first derivative is zero or undefined. These are points where the graph has a potential peak or trough. To find critical points, we set the first derivative equal to zero:

• \( f'(x) = \cos \left(x+\frac{\pi}{6}\right) - \sin\left(x+\frac{\pi}{6}\right) = 0 \)

By solving this equation, we identify the x-values that could represent key changes in the function's behavior, such as maximums or minimums. In this specific example, the solution gives us critical points at \(x = \frac{\pi}{6}, \frac{5\pi}{6}\), but only \(x = \frac{\pi}{6}\) is valid within the interval \((0, \frac{\pi}{2})\). Identifying critical points is crucial as they guide us to where extremities occur on the function's graph.

Maxima and Minima

In calculus, maxima are points where a function reaches its highest value, while minima are points where it reaches its lowest. These are found by examining critical points. After identifying a critical point within the specified interval, we determine if it is a maximum or minimum.

• For this function, at \(x = \frac{\pi}{6}\), we are interested in checking whether it's a maximum since only one critical point lies in the range.

If more critical points were present, further analysis using additional tests (like the Second Derivative Test) would be required. Nonetheless, understanding the concepts of maxima and minima helps to describe peak values, vital in optimizing real-world problems.

Differentiation

Differentiation is the process of finding the derivative of a function. It represents the rate of change or slope of the function at any point. In our problem, the differentiation of

• \[ f(x)= \sin \left(x+\frac{\pi}{6}\right) + \cos \left(x+\frac{\pi}{6}\right) \]

gives us

• \[ f'(x) = \cos \left(x+\frac{\pi}{6}\right) - \sin\left(x+\frac{\pi}{6}\right) \]

This derivative helps us locate critical points, which are essential in finding where the function's slope changes direction, indicating potential extrema. Mastery of differentiation techniques is crucial for exploring and understanding calculus problems effectively.

Second Derivative Test

The Second Derivative Test helps in determining whether a critical point is a maximum or minimum. This is done by taking the second derivative of the function and evaluating it at the critical points.

• If the second derivative is positive at a critical point, the function has a local minimum there.
• If it is negative, the function has a local maximum.

In this exercise, since we only had one critical point within the valid interval, we bypassed confirming with the Second Derivative Test. However, being familiar with this test is important as it provides a handy tool in thoroughly analyzing the nature of critical points. It's a definitive way to classify extrema when understanding complex functions.