The minimum value of f(x) = a^{a^{x}} + a^{1- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the function $f(x) = a^{a^{x}} + a^{1-a^{x}}$.We can rewrite the second term as $a^{1-a^{x}} = \frac{a}{a^{a^{x}}}$.So,$f(x) = a^{a^{x}} + \

Vedclass · Accepted Answer

$2\sqrt{a}$

Vedclass · Answer

(B) Given the function $f(x) = a^{a^{x}} + a^{1-a^{x}}$.We can rewrite the second term as $a^{1-a^{x}} = \frac{a}{a^{a^{x}}}$.So,$f(x) = a^{a^{x}} + \frac{a}{a^{a^{x}}}$.Since $a > 0$,both $a^{a^{x}}$ and $\frac{a}{a^{a^{x}}}$ are positive real numbers.By the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality,for any two positive numbers $u$ and $v$,$u + v \geq 2\sqrt{uv}$.Let $u = a^{a^{x}}$ and $v = \frac{a}{a^{a^{x}}}$.Then $f(x) = u + v \geq 2\sqrt{u \cdot v} = 2\sqrt{a^{a^{x}} \cdot \frac{a}{a^{a^{x}}}} = 2\sqrt{a}$.Thus,the minimum value is $2\sqrt{a}$.

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

Similar Questions

For $x \in \mathbb{R}$,the range of $3 \cos (4x - 5) + 4$ lies in the interval:

If $A + B + C = \pi ,$ then $\frac{\cos A}{\sin B \sin C} + \frac{\cos B}{\sin C \sin A} + \frac{\cos C}{\sin A \sin B} = $

The maximum value of the function $f(x) = 3 \sin x + 4 \cos x$ is

The maximum value of the function $f(x) = a \sin x + b \cos x$ is:

The number of integral values of $k$,for which the equation $7 \cos x + 5 \sin x = 2k + 1$ has a solution,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module