Question:
The number of the real roots of the equation $2 \cos (x(x+1))=2^{x}+2^{-x}$ is
- 2
- 1
- infinite
- 0
Correct Answer: Option: 2
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.
CAT Online Course
Also Check: 841+ CAT Quant Questions with Solutions