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AM,GM,HM Inequality

The minimum v...

Question

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

A

a + \frac{1}{a}

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B

a + 1

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C

2 a

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D

2 \sqrt{a}

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Solution

The correct option is D $2 \sqrt{a}$
Using $A M \geq G M$ inequality, we get
$\frac{a^{a^{x}} + \frac{a}{a^{a^{x}}}}{2} \geq {(a^{a^{x}} \cdot \frac{a}{a^{a^{x}}})}^{\frac{1}{2}} \Rightarrow a^{a^{x}} + a^{1 - a^{x}} \geq 2 \sqrt{a}$

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Similar questions

Q. The minimum value of

f (x) = a^{a^{x}} + a^{1 - a^{x}}

a, x \in R

a > 0

Q. The minimum value of

f (x) = a^{a^{x}} + a^{1 - a^{x}}

a, x \in R

a > 0

A = {(x, y) \in R \times R | 2 x^{2} + 2 y^{2} - 2 x - 2 y = 1},

B = {(x, y) \in R \times R | 4 x^{2} + 4 y^{2} - 16 y + 7 = 0}

C = {(x, y) \in R \times R | x^{2} + y^{2} - 4 x - 2 y + 5 \leq r^{2}}

.
Then the minimum value of

| r |

A \cup B \subseteq C

f (x) = e^{a x} + e^{b x},

a \neq b

f^{''} (x) - 2 f^{'} (x) - 15 f (x) = 0

x \in R

. Then the product

a b

f : R \to R

be a function defined by

f (x) = {\begin{matrix} [x], x \leq 2 0, x > 2 \end{matrix},

[x]

is the greatest integer less than or equal to

x .

I = 2 \int - 1 \frac{x f (x^{2})}{2 + f (x + 1)} d x,

then the value of

(4 I - 1)

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AM,GM,HM Inequality

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