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 + 1
• B. a + $\frac{1}{a}$
• C. 2a
• D. 2$\sqrt{a}$

Hint:

When you encounter a function of the form $f(t) + \frac{k}{f(t)}$ where $f(t)$ is positive, immediately think of the AM-GM inequality. The minimum value will be $2\sqrt{k}$ and occurs when $f(t) = \sqrt{k}$.

Answer 1

The Correct Option is C

From the given equation, \[ \alpha^2=6\alpha+2,\qquad \beta^2=6\beta+2 \] Multiplying by \(\alpha^{n-2}\) and \(\beta^{n-2}\) respectively: \[ \alpha^n=6\alpha^{n-1}+2\alpha^{n-2} \] \[ \beta^n=6\beta^{n-1}+2\beta^{n-2} \] Subtracting, \[ a_n=6a_{n-1}+2a_{n-2} \] For \(n=10\), \[ a_{10}=6a_9+2a_8 \Rightarrow a_{10}-2a_8=6a_9 \] \[ \frac{a_{10}-2a_8}{3a_9} =\frac{6a_9}{3a_9}=2 \] Correct option: (C)