# CAT 2019 Quant Question with Solution 25

Question:
The number of the real roots of the equation $2 \cos (x(x+1))=2^{x}+2^{-x}$ is

1. 2
2. 1
3. infinite
4. 0

For any real value of x, the expression $2 \cos (x(x+1))=2^{x}+2^{-x}$ would always be positive.

Lets find the maximum value of $2 \cos (x(x+1))=2^{x}+2^{-x}$.

Applying AM-GM inequality we have

$\frac{{{2}^{x}}+{{2}^{-x}}}{2}\ge \sqrt{{{2}^{x}}\times {{2}^{-x}}}$

$\Rightarrow {{2}^{x}}+{{2}^{-x}}\ge 2\sqrt{{{2}^{0}}}$

$\Rightarrow {{2}^{x}}+{{2}^{-x}}\ge 2$

Therefore, $2\cos \left( x\left( x+1 \right) \right)\ge 2$

It is known that $-1\le \cos \theta \le 1$

$\Rightarrow 2\cos \left( x\left( x+1 \right) \right)=2$

Hence, the expression is valid only if ${{2}^{x}}+{{2}^{-x}}=2$, which is true for only one value of x i.e. 0.

Therefore, the expression has only one real solution.

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