The minimum value of 2^(sin x) + 2^(cos x) is: (A) 2^(-1+√2) (B) 2^(1-√2) (C) 2^(1-1/√2) (D) 2^(-1+1

Q: Wait, why can't I just plug in $x=0$ or $x=\pi/2$ to find the minimum?

That is a classic trap! Plugging in specific values like $0$ or $\pi/2$ gives you $2^0 + 2^1 = 3$ or $2^1 + 2^0 = 3$. However, the minimum might occur at a point where $\sin x$ and $\cos x$ are balanced differently. Always look for the global minimum using inequalities or calculus rather than guessing points.

Q: Why does AM-GM work here? Isn't it only for constants?

AM-GM is a powerful tool that holds for any two positive real numbers $a$ and $b$. Since $2^{\sin x}$ and $2^{\cos x}$ are strictly positive for all real $x$, they qualify as valid inputs for the inequality at every single point in the domain.

Q: I got stuck at $2^{1 + \frac{\sin x + \cos x}{2}}$. How do I know if I should minimize or maximize the exponent?

Great observation! Since the base of our exponent is $2$ (which is greater than $1$), the function $g(u) = 2^u$ is an increasing function. Therefore, to make the whole expression as small as possible, we must make the exponent $u$ as small as possible.

Q: Is the range of $\sin x + \cos x$ always $[-\sqrt{2}, \sqrt{2}]$?

Exactly! By using the harmonic addition theorem, $\sin x + \cos x = \sqrt{2} (\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x) = \sqrt{2} \sin(x + \pi/4)$. Since the sine function oscillates between $-1$ and $1$, the entire expression oscillates between $-\sqrt{2}$ and $\sqrt{2}$.

Q: Why did we simplify $\frac{\sqrt{2}}{2}$ to $\frac{1}{\sqrt{2}}$?

It is purely for matching the format of the given options. In competitive exams like JEE, always keep an eye on the options to see how they want the final answer expressed. Simplifying radicals is a standard practice to ensure your result aligns with the provided choices.

Question

The minimum value of $2^{\sin x} + 2^{\cos x}$ is:

Options:

(A) $2^{-1+\sqrt{2}}$

(B) $2^{1-\sqrt{2}}$

(C) $2^{1-\frac{1}{\sqrt{2}}}$

(D) $2^{-1+\frac{1}{\sqrt{2}}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that at first glance looks like a simple trigonometric expression, but beneath the surface, it hides a beautiful interplay between the Arithmetic Mean-Geometric Mean (AM-GM) inequality and the harmonic nature of waves.

We are tasked with finding the minimum value of $f(x) = 2^{\sin x} + 2^{\cos x}$.

### The Power of AM-GM

When you see a sum of two positive terms, your mathematical intuition should immediately scream, "AM-GM!" The AM-GM inequality, which states that $\frac{a+b}{2} \ge \sqrt{ab}$ for any positive $a$ and $b$, is the most elegant way to bridge the gap between a sum and a product.

Here, our terms are $a = 2^{\sin x}$ and $b = 2^{\cos x}$. Since $2^t$ is always positive for any real $t$, we are perfectly safe to proceed. Applying the inequality, we get:

$$ \frac{2^{\sin x} + 2^{\cos x}}{2} \ge \sqrt{2^{\sin x} \cdot 2^{\cos x}} $$

Look at the beauty of the right-hand side. By the laws of exponents, $a^m \cdot a^n = a^{m+n}$, so the product inside the square root simplifies to $2^{\sin x + \cos x}$. Now, recalling that $\sqrt{k} = k^{1/2}$, our inequality transforms into:

$$ \frac{2^{\sin x} + 2^{\cos x}}{2} \ge (2^{\sin x + \cos x})^{1/2} = 2^{\frac{\sin x + \cos x}{2}} $$

### Taming the Trigonometric Beast

We are now left with $2^{\sin x} + 2^{\cos x} \ge 2 \cdot 2^{\frac{\sin x + \cos x}{2}}$. Using the rule $a^1 \cdot a^n = a^{1+n}$, we can write this as:

$$ 2^{\sin x} + 2^{\cos x} \ge 2^{1 + \frac{\sin x + \cos x}{2}} $$

This is the moment where many students freeze. We have successfully reduced the problem to minimizing the exponent. Since the base $2$ is greater than $1$, the function $2^u$ is strictly increasing.

To minimize the whole expression, we simply need to find the minimum value of the exponent $u = 1 + \frac{\sin x + \cos x}{2}$. This boils down to finding the minimum value of $\sin x + \cos x$.

### The Harmonic Shift

Think of $\sin x + \cos x$ as a single wave. We can rewrite it using the harmonic addition theorem:

$$ \sin x + \cos x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x \right) = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) $$

The range of this expression is clearly $[-\sqrt{2}, \sqrt{2}]$. Therefore, the minimum value of $\sin x + \cos x$ is $-\sqrt{2}$.

### The Final Flourish

Now, we substitute this minimum value back into our inequality:

$$ f(x)_{min} = 2^{1 + \frac{-\sqrt{2}}{2}} $$

To match the options provided, we simplify the exponent. Since $\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{\sqrt{2}}$, our final expression becomes:

$$ f(x)_{min} = 2^{1 - \frac{1}{\sqrt{2}}} $$

And there it is! We have navigated the inequality, tamed the trigonometric wave, and arrived at the **final answer** of $2^{1 - \frac{1}{\sqrt{2}}}$. Remember, in JEE, it is not just about the calculation; it is about recognizing the tools that make the problem collapse into simplicity.

Animated Solution for Mathematics - Sequence and Series: The minimum value of $2^{s i n x} + 2^{c o s x}$ is:

Select Answer:

Visualized Solution

Defining the Function $f (x)$

The AM-GM Inequality Tool

Setting up the Inequality

Simplifying the Product

Handling the Square Root

Combining the Exponents

Minimizing the Trig Sum

Substituting the Minimum Value

Final Simplification

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

The Power of AM-GM

Taming the Trigonometric Beast

The Harmonic Shift

The Final Flourish

Similar Questions

The minimum value of $f (x) = a^{a^{x}} + a^{1 - a^{x}}$ , where $a, x \in R$ and $a > 0$ , is equal to:

If $α \in (0, π /2)$ , then $x^{2} + x + \frac{t a n ^{2} α}{x ^{2} + x}$ is always greater than or equal to

If $a_{1}, a_{2}, \dots, a_{n}$ are positive real numbers whose product is a fixed number $c$ , then the minimum value of $a_{1} + a_{2} + \dots + a_{n - 1} + 2 a_{n}$ is

Let $x, y > 0$ . If $x^{3} y^{2} = 2^{15}$ , then the least value of $3 x + 2 y$ is

Let $f : R \to R$ be such that for all $x \in R$ , $(2^{1 + x} + 2^{1 - x})$ , $f (x)$ and $(3^{x} + 3^{- x})$ are in A.P., then the minimum value of $f (x)$ is :

Let $f : R \to R$ be such that for all $x \in R (2^{1 + x} + 2^{1 - x}), f (x)$ and $(3^{x} + 3^{- x})$ are in A.P., then the minimum value of $f (x)$ is :

If the minimum value of $f (x) = \frac{5 x ^{2}}{2} + \frac{α}{x ^{5}}, x > 0$ , is 14, then the value of $α$ is equal to :

The minimum value of the sum of real numbers $a^{- 5}, a^{- 4}, 3 a^{- 3}, 1, a^{8}$ and $a^{10}$ where $a > 0$ is .........

Let $x, y$ be positive real numbers and $m, n$ positive integers. The maximum value of the expression $\frac{x ^{m} y ^{n}}{( 1 + x ^{2 m} ) ( 1 + y ^{2 n} )}$ is:

For $x \geq 0$ , the least value of $K$ , for which $4^{1 + x} + 4^{1 - x}, \frac{K}{2}, 1 6^{x} + 1 6^{- x}$ are three consecutive terms of an A.P., is equal to :

Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The minimum value of 2sinx+2cosx is: