CAT 2020 Set-1 | Question: 57

Question

The number of real$-$valued of the equation  $2^{x}+2^{-x}=2-(x-2)^{2}$ is

• A. infinite
• B. $1$
• C. $0$
• D. $2$

Answer 1

Given that, $2^{x} + 2^{-x} = 2 – (x – 2)^{2}$

$\Rightarrow 2^{x} + \frac{1}{2^{x}} = 2 – (x – 2)^{2} \quad \longrightarrow (1)$

We know that, $\text{AM} \geq \text{GM}$

$\Rightarrow \frac{2^{x} + \frac{1}{2^{x}}}{2} = \sqrt{ 2^{x} \cdot \frac{1}{2^{x}}}$

$\Rightarrow \boxed{ 2^{x} + \frac{1}{2^{x}} \geq 2}$

The minimum value of $2^{x} + \frac{1}{2^{x}}$ is $2,$ when $x=0.$

So, it follows $\text{LHS} \geq 2.$

And, $2 – (x-2)^{2} \leq 2$

$\Rightarrow – (x-2)^{2} \leq 0$

$\Rightarrow (x-2)^{2} \geq 0$

$\Rightarrow \boxed{x \geq 2}$

The maximum value of $2 – (x-2)^{2}$ is $2,$ when $x=2.$

Hence, there is no value of $x.$

Correct Answer$: \text{C}$