JAMIA 2021 — Mathematics PYQ

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Question

JAMIA 2021 — Mathematics PYQ

JAMIA | Mathematics | 2021

The minimum value of $4^{x} + 4^{, 1 - x}$ , $x \in R$ , is

Choose the correct answer:

Explanation

1. Identify the Terms

Let:

$a = 4^x$
$b = 4^{1-x}$

Both terms are always positive for any $x \in \mathbb{R}$ .

2. Apply AM-GM Inequality

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

3. Simplify the Geometric Mean

Inside the square root, use the property $a^m \cdot a^n = a^{m+n}$ :

\sqrt{4^{x + 1 - x}} = \sqrt{4^1}

\sqrt{4} = 2

4. Solve for the Minimum Value

Substitute back into the inequality:

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

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

4^x + 4^{1-x} \ge 4

Final Result

The minimum value of $4^x + 4^{1-x}$ is 4.

Choose the correct answer:

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