If x is real , the numbers $ 5^{1+x} + 5^{1-x} $ , a/2 , $ 25^x + 25^{-x} $ form an A.P. then to which interval must ' a ' belong to?

Question

A) [ 1 , 5 ]

B) [ 2 , 5 ]

C) [ 5 , 12 ]

D) [ 12 , $ \infty $ ]

My progress so far :

Let $ 5^{x} = y $

Since they are in AP so ,

=> $ 5y $ + $ 5\over y $ + $ y^2 $ + $ 1\over (y^2) $ = a

=> $ {5y^3 + 5y + y^4 + 1 }\over (y)^2 $ = a

As you can see , It becomes a mess . How do I solve it with a better approach .

Answer 1

For questions like this I try to find useful values of $x$. The three that immediately suggest themselves to me here are $x=0, x\to +\infty,x \to -\infty$ If $x=0$, we need $10, a/2, 2$ to form an AP, which works with $a=12$. That rules out two of your choices. Now consider what happens when $x$ is large and positive and again when $x$ is large and negative.