JAMIA 2021 — Mathematics PYQ

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Question

JAMIA 2021 — Mathematics PYQ

JAMIA | Mathematics | 2021

Minimum value of $a^{a^x}+\frac a{a^{a^x}} \mathfrak{a}>\mathcal{O};\mathfrak{a},\mathfrak{x}\in\mathbb{R})$

Choose the correct answer:

Explanation

Solution:

Let the two positive terms be:

$A = a^{a^x}$

$B = \frac{a}{a^{a^x}}$

According to the AM-GM Inequality for positive numbers $A$ and $B$ :

\frac{A + B}{2} \geq \sqrt{A \cdot B}

A + B \geq 2\sqrt{A \cdot B}

Substituting the values of $A$ and $B$ :

a^{a^x} + \frac{a}{a^{a^x}} \geq 2\sqrt{a^{a^x} \cdot \frac{a}{a^{a^x}}}

The term $a^{a^x}$ cancels out inside the square root:

a^{a^x} + \frac{a}{a^{a^x}} \geq 2\sqrt{a}

Condition for Equality:

The minimum value occurs when $A = B$ :

a^{a^x} = \frac{a}{a^{a^x}}

(a^{a^x})^2 = a

a^{2a^x} = a^1

2a^x = 1 \implies a^x = \frac{1}{2}

x = \log_a\left(\frac{1}{2}\right)

(This $x$ exists for any $a > 0, a \neq 1$ . If $a=1$ , the expression is constant at $2$ ).

Minimum Value:

\text{Minimum value} = 2\sqrt{a}

Choose the correct answer:

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