Finding the interval in which $a$ lies

Question

Question

If $x\in \mathbb{R}$, the numbers $5^{1+x}+5^{1-x},\frac{a}{2},25^{x}+25^{-x}$ form an Arithmetic Progression, then $a$ must lie in the interval:

(A) $[1,5]$

(B) $[2,5]$

(C) $[5,12]$

(D) $[12,\infty)$

My Approach

Since the numbers $5^{1+x}+5^{1-x},\frac{a}{2},25^{x}+25^{-x}$ form an Arithmetic Progression,

$\implies a=5^{1+x}+5^{1-x}+25^{x}+25^{-x}$

$\implies a=5^{1+x}+5^{1-x}+5^{2x}+5^{-2x}$

After this, I am unable to find the range of $a$ to determine in which interval it lies. Usually to determine range, I use to either draw the graph of the function or equate the function to $y$, i.e., $y=f(x)$ then express $x$ as a function of $y$. The range of the original function then would be the domain of the new function $x=g(y)$. Here I am unable to determine the range by these methods.

Could you please help how to proceed or is there any alternative approach for this problem?

Kindly clarify my doubt.

Answer 1

Hint:

For lower bound use $$x+y\geq 2\sqrt{xy}$$ if $x,y$ are positive. Notice that we get an equality iff $x=y$.

So $$5^{1+x}+ 5^{1-x} \geq 2\sqrt{5^{1+x} 5^{1-x} } = 10$$ with equality iff $5^{1+x}= 5^{1-x}$ i.e. $x=0$.

And similary for other two...

Since $\lim_{x\to \infty} 25^x =\infty$ we see $a$ has no upper bound.

Answer 2

Notice that $$a=10\cosh{(\ln{(5)}x)}+2\cosh{(2\ln{(5)}x)}$$ which has a minimum exactly when $x=0$ and both hyperbolic cosines are equal to $1$. Thus we have $$a\ge12\implies a\in[12,\infty)$$

Answer 3

$$2\cdot \dfrac a2=5\left(b^2+\dfrac1{b^2}\right)+b^4+\dfrac1{b^4}=5\left(b^2+\dfrac1{b^2}\right)+\left(b^2+\dfrac1{b^2}\right)^2-2$$ where $b^2=5^x>0,25^x=(5^2)^x=(5^x)^2=b^4$

$$a=\left(b^2+\dfrac1{b^2}+\dfrac52\right)^2-2-\left(\dfrac52\right)^2$$

Now $b^2+\dfrac1{b^2}=\left(b-\dfrac1b\right)^2+2\ge2$

$$\implies a\ge\left(2+\dfrac52\right)^2-2-\left(\dfrac52\right)^2=2+10$$