Problem 2 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The minimum value of \(2^{\sin x}+2^{\cos x}\) is: \([\) Sep. 04, \(\mathbf{2 0 2 0}\) (II)](a) \(\begin{array}{llll}2^{-1+\frac{1}{\sqrt{2}}} & \text { (b) } 2^{-1+\sqrt{2}} & \text { (c) } 2^{1-\sqrt{2}} & \text { (d) } 2^{1-\frac{1}{\sqrt{2}}}\end{array}\)

Step-by-Step Solution

Answer

The minimum value of \(2^{\sin x}+2^{\cos x}\) is \(2^{1 - \frac{1}{\sqrt{2}}}\), option (d).

Step 1: Analyze the Expression

We are given the expression \( 2^{\sin x} + 2^{\cos x} \). We need to find the minimum value of this expression. Let's denote \( a = 2^{\sin x} \) and \( b = 2^{\cos x} \), so that the problem reduces to finding the minimum value of \( a + b \) where \( a = 2^{\sin x} \) and \( b = 2^{\cos x} \).

Step 2: Use Arithmetic Mean - Geometric Mean Inequality (AM-GM)

According to the AM-GM inequality, for non-negative numbers \( a \) and \( b \), \( \frac{a + b}{2} \geq \sqrt{ab} \). Applying this to our terms gives \( \frac{2^{\sin x} + 2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}} \).

Step 3: Simplify the inequality

The product inside the square root can be expressed as \( \sqrt{2^{\sin x + \cos x}} = 2^{\frac{1}{2}(\sin x + \cos x)} \). Thus, the AM-GM inequality becomes \( 2^{\sin x} + 2^{\cos x} \geq 2 \cdot 2^{\frac{1}{2}(\sin x + \cos x)} \).

Step 4: Analyze the Range of \( \sin x + \cos x \)

Since \( \sin x + \cos x \leq \sqrt{2} \), the maximum value of \( \frac{1}{2}(\sin x + \cos x) \) is \( \frac{1}{2} \sqrt{2} \). Substitute this into the inequality, \( 2^{\frac{1}{2}\sqrt{2}} \), which gives the lower bound of \( 2^{1 - \frac{1}{\sqrt{2}}} \) for \( 2^{\sin x} + 2^{\cos x} \).

Step 5: Find the Minimum Value

From our analysis, the minimum value is \( 2^{1 - \frac{1}{\sqrt{2}}} \) which corresponds to option (d).

Key Concepts

Trigonometric Functions

Trigonometric functions, like sine and cosine, are fundamental in mathematics, especially in the analysis of periodic phenomena. These functions relate the angles of a triangle to the lengths of its sides and are essential for describing rotation and oscillations.
In our problem, we make use of the sine function, denoted as \(\sin x\), and the cosine function, \(\cos x\). These functions can take any real number \(x\) and map it to a value between -1 and 1, which is crucial because this range impacts the calculations involving these functions.

• Sine Function (\(\sin x\)): This represents the vertical component of an angle in the unit circle.
• Cosine Function (\(\cos x\)): This shows the horizontal component of an angle in the unit circle.

For solving problems, understanding the behavior of these functions and their periodic nature is critical. They have peaks at specific points, such as \(\pi/2\) and \(3\pi/2\) for sine, and 0 and \(\pi\) for cosine, which influence where functions like \(2^{\sin x}\) and \(2^{\cos x}\) attain minimum values.

AM-GM Inequality

The Arithmetic Mean-Geometric Mean (AM-GM) inequality is a basic but powerful tool in mathematics, especially in optimization and problem solving. It states that for any non-negative numbers, the arithmetic mean (average) is greater than or equal to the geometric mean. This is written as, \(\frac{a+b}{2} \geq \sqrt{ab}\) for any non-negative \(a\) and \(b\).

• Use in Problems: The AM-GM inequality helps in finding minimum or maximum values easily by providing a straightforward comparison between averages and products.
• Why It Works: The inequality is based on fundamental properties of numbers that ensure the sum or average tends to flatten compared to products, which can vary widely.

In our context, applying the AM-GM inequality to \(2^{\sin x}\) and \(2^{\cos x}\) lets us find bounds for their sum, which aids in determining the minimum possible value of the expression. By multiplying under the square root, we take advantage of how exponential forms simplify under multiplication, aiding in solutions.

Minimization Problem

Minimization problems involve finding the smallest value that a function can attain. This requires a good understanding of calculus and various inequalities, like AM-GM, to find such values without having to compute derivatives directly.

• Steps to Solve:
  • Identify the function to minimize, here \(2^{\sin x} + 2^{\cos x}\).
  • Apply known mathematical inequalities to find possible bounds.
  • Determine the domain and range of the variables involved, \(\sin x\) and \(\cos x\), to narrow down the possibilities.
  • Find extreme values using these bounds.

For the given exercise, knowing that \(\sin x + \cos x\) reaches a maximum of \(\sqrt{2}\) helps in setting up and solving inequality. The challenge is in simplifying the exponential terms and recognizing valid transformation rules to make logical deductions about the expression's minimum value, here resolved as \(2^{1 - \frac{1}{\sqrt{2}}}\).