The minimum value of $4^x + 4^{1-x}$, $x \in R$, is (A) $2$ (B) $4$ (C) $1$ (D) $0$

The minimum value of 4x + 41-x, x ∈ R, is (A) 2 (B) 4 (C) 1 (D) 0

NCERT Exemplar · Q23

4^x + 4^{1-x}

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Concept understanding — Inequality Of Means

Inequality of Means

Take any two positive numbers, say 4 and 16. Add them and halve it — you get their arithmetic mean: $(4+16)/2 = 10$ . Multiply them and take the square root — you get their geometric mean: $\sqrt{4 \times 16} = 8$ . Notice something? $10 \ge 8$ . Try it with any other pair of positive numbers you like — the arithmetic mean is never smaller than the geometric mean. That simple, always-true observation is the Inequality of Means, usually written AM ≥ GM.

The precise statement

For two positive real numbers $a$ and $b$ :

\text{AM} = \frac{a+b}{2}, \qquad \text{GM} = \sqrt{ab}

\frac{a+b}{2} \ge \sqrt{ab}

with equality if and only if $a = b$ . If $a \ne b$ , the inequality is strict.

Why it is always true

Start from a fact that can never fail: the square of any real number is non-negative.

(\sqrt{a} - \sqrt{b})^2 \ge 0

Expand the left side:

a - 2\sqrt{a}\sqrt{b} + b \ge 0

a + b \ge 2\sqrt{ab}

Divide both sides by $2$ :

\frac{a+b}{2} \ge \sqrt{ab}

That's the whole proof — no assumptions beyond $a, b > 0$ (so that $\sqrt{a}, \sqrt{b}$ are real numbers). Since $(\sqrt a - \sqrt b)^2 = 0$ exactly when $\sqrt a = \sqrt b$ , equality holds exactly when $a = b$ .

Note

The inequality needs $a, b \ge 0$ . For negative numbers, $\sqrt{ab}$ may not even be real, so the "GM" isn't defined there.

Worked example

Find the AM and GM of $9$ and $25$ , and verify the inequality.

Step 1: $\text{AM} = \dfrac{9+25}{2} = 17$

Step 2: $\text{GM} = \sqrt{9 \times 25} = \sqrt{225} = 15$

Step 3: Check: $17 \ge 15$ ✓ — and since $9 \ne 25$ , the inequality is strict, exactly as the rule predicts.

A useful consequence: inserting a mean between two numbers

If $a$ and $b$ are two positive numbers and $G$ is inserted between them so that $a, G, b$ form a Geometric Progression, then $G = \sqrt{ab}$ — precisely the geometric mean. Comparing this $G$ against the arithmetic mean $A = \dfrac{a+b}{2}$ (the number that would sit between $a$ and $b$ in an Arithmetic Progression) is exactly an application of this inequality: $A \ge G$ always, so the AM-inserted term never sits below the GM-inserted term.

Watch out

A common slip is writing $\sqrt{ab}$ when $a$ or $b$ is negative, or applying the two-number formula directly to more than two numbers. For $n$ positive numbers $a_1, a_2, \dots, a_n$ , the generalised inequality is

$\frac{a_1+a_2+\cdots+a_n}{n} \ge \sqrt[n]{a_1 a_2 \cdots a_n}$ …

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Using the AM–GM inequality, the sum $4^x + 4^{1-x}$ is minimized when $4^x = 4^{1-x}$ , giving $x = \frac12$ and a minimum value of $4$ .

Concept first.

When you see a sum of two positive terms where one is the reciprocal (or near-reciprocal) of the other, the AM–GM inequality is often the fastest route. Here $4^x$ and $4^{1-x}$ are both positive for all real $x$ , and their product is constant:

$4^x \cdot 4^{1-x} = 4^{x + 1 - x} = 4^1 = 4$ .

That constant product is the key — it means the sum has a fixed lower bound.

Why AM–GM works here.

For any two non‑negative numbers $a$ and $b$ , the arithmetic mean is at least the geometric mean:

\frac{a+b}{2} \ge \sqrt{ab}.

Equality holds exactly when $a = b$ . So if we set $a = 4^x$ and $b = 4^{1-x}$ , we get a direct bound on the sum.

Step‑by‑step solution

Apply AM–GM Let $a = 4^x$ and $b = 4^{1-x}$ . Then

\frac{4^x + 4^{1-x}}{2} \ge \sqrt{4^x \cdot 4^{1-x}}.

Simplify the product

4^x \cdot 4^{1-x} = 4^{x + 1 - x} = 4^1 = 4.

So the right‑hand side becomes $\sqrt{4} = 2$ .

Obtain the inequality

\frac{4^x + 4^{1-x}}{2} \ge 2 \quad\Rightarrow\quad 4^x + 4^{1-x} \ge 4.

Find when equality occurs AM–GM gives equality when $a = b$ , i.e.

4^x = 4^{1-x}.

Since the base $4$ is positive and not $1$ , we equate exponents:

x = 1 - x \quad\Rightarrow\quad 2x = 1 \quad\Rightarrow\quad x = \frac12.

Verify the minimum …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2025Set 2025-A1 markMCQ
$A$
›Reveal solutionSolution
The problem asks for the minimum of $5^x + 5^{1-x}$ . Using the AM–GM inequality, the minimum occurs when $5^x = 5^{1-x}$ , giving $x = 1/2$ and the value $2\sqrt{5}$ . So the correct option is (A).

Concept & Intuition

We have two positive numbers: $5^x$ and $5^{1-x}$ . Their sum is what we want to minimize. For positive numbers, the arithmetic mean is always at least the geometric mean, with equality when the numbers are equal. That gives a lower bound on the sum — and that bound is actually achievable, so it’s the minimum.

Set up the AM–GM inequality For any positive $a$ and $b$ ,

$\frac{a+b}{2} \ge \sqrt{ab}.$
Here $a = 5^x$ , $b = 5^{1-x}$ . So
$\frac{5^x + 5^{1-x}}{2} \ge \sqrt{5^x \cdot 5^{1-x}}.$

Simplify the geometric mean

$5^x \cdot 5^{1-x} = 5^{x + (1-x)} = 5^1 = 5.$
Hence
$\sqrt{5^x \cdot 5^{1-x}} = \sqrt{5}.$

Apply the inequality

$\frac{5^x + 5^{1-x}}{2} \ge \sqrt{5} \quad\Rightarrow\quad 5^x + 5^{1-x} \ge 2\sqrt{5}.$

Check when equality occurs AM = GM when $a = b$ , i.e. $5^x = 5^{1-x} \quad\Rightarrow\quad x = 1-x \quad\Rightarrow\quad x = \frac12.$ …
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COMEDK 2025Set 2025-A1 markMCQ
$x, \cos \theta=x+\frac{1}{x}$
›Reveal solutionSolution
The key idea is that for real $x$ , the expression $x + \frac{1}{x}$ is either $\ge 2$ or $\le -2$ , but $\cos\theta$ must lie in $[-1, 1]$ . Since these ranges do not overlap, no real $\theta$ satisfies the equation. The correct option is (B).

We start with the given equation:
$\cos\theta = x + \frac{1}{x}$
for real values of $x$ . The question asks what we can conclude about $\theta$ .

Concept and Intuition

The expression $x + \frac{1}{x}$ is famous for having a restricted range when $x$ is real. Why? Because if $x > 0$ , by AM–GM inequality, $x + \frac{1}{x} \ge 2$ , with equality only at $x = 1$ . If $x < 0$ , then let $x = -t$ with $t > 0$ , so $x + \frac{1}{x} = -t - \frac{1}{t} = -\left(t + \frac{1}{t}\right) \le -2$ . So the sum is always either $\ge 2$ or $\le -2$ .

Meanwhile, $\cos\theta$ for any real $\theta$ is always between $-1$ and $1$ inclusive. So we are equating a number that lives outside $[-1, 1]$ (except possibly at the boundaries) to a number that must live inside $[-1, 1]$ . The only chance would be if $x + \frac{1}{x}$ could equal something in $[-1, 1]$ , but it cannot — the gap is clear.

Thus, no real $\theta$ can satisfy the equation for any real $x$ .

Step-by-step reasoning

Recall the range of $x + \frac{1}{x}$ for real $x$ For $x > 0$ , by AM–GM:

$x + \frac{1}{x} \ge 2\sqrt{x \cdot \frac{1}{x}} = 2$
For $x < 0$ , set $x = -t$ with $t > 0$ :
$x + \frac{1}{x} = -t - \frac{1}{t} = -\left(t + \frac{1}{t}\right) \le -2$
For $x = 0$ , the expression is undefined. So the possible values are $(-\infty, -2] \cup [2, \infty)$ .

Recall the range of $\cos\theta$ for real $\theta$ The cosine function always satisfies:

$-1 \le \cos\theta \le 1$

Compare the two ranges …
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COMEDK 2024Set 2024-E1 markMCQ
$3+2 \sqrt{2}: 3-2 \sqrt{2}$
›Reveal solutionSolution
The key idea is to set the two numbers as $a = (3+2\sqrt{2})k$ and $b = (3-2\sqrt{2})k$ , compute their arithmetic mean (AM) and geometric mean (GM), then simplify the ratio. The final ratio is $3:1$ , corresponding to option (D).

We are given two positive numbers in the ratio
$3+2\sqrt{2} : 3-2\sqrt{2}.$
We need the ratio of their arithmetic mean (AM) to their geometric mean (GM).

Concept and intuition:

When two numbers are in a given ratio, we can represent them as multiples of that ratio. The AM and GM are symmetric functions of the numbers, so the ratio of AM to GM will be independent of the scaling factor. The trick is to notice that $(3+2\sqrt{2})$ and $(3-2\sqrt{2})$ are conjugates — their product is a perfect square, which simplifies the GM nicely.

Step-by-step solution:

Set up the numbers. Let the two numbers be

$a = (3+2\sqrt{2})k, \quad b = (3-2\sqrt{2})k,$
where $k > 0$ is a common factor. This preserves the given ratio.

Compute the arithmetic mean (AM).

$\text{AM} = \frac{a+b}{2} = \frac{(3+2\sqrt{2})k + (3-2\sqrt{2})k}{2}.$
The $2\sqrt{2}$ terms cancel:
$a+b = 6k \quad \Rightarrow \quad \text{AM} = \frac{6k}{2} = 3k.$

Compute the geometric mean (GM).

$\text{GM} = \sqrt{ab} = \sqrt{(3+2\sqrt{2})k \cdot (3-2\sqrt{2})k}.$
The product inside is
$(3+2\sqrt{2})(3-2\sqrt{2}) \cdot k^2.$
Using the difference of squares:

$(3)^2 - (2\sqrt{2})^2 = 9 - 8 = 1.$ …
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Q.The minimum value of 4x+41−x4^x + 4^{1-x}4x+41−x, x∈Rx \in Rx∈R, is (A) 222 (B) 444 (C) 111 (D) 000

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Q.The minimum value of $4^x + 4^{1-x}$ , $x \in R$ , is
(A) $2$
(B) $4$
(C) $1$
(D) $0$

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