cat 2019 Question 19

Question

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

Accepted Answer

$2 \cos (x(x + 1)) = 2^x + 2^{-x}$

The maximum value of LHS is 2 when $\cos (x(x + 1))$ is 1 and the minimum value of RHS is 2 using AM $\geq$ GM

Hence LHS and RHS can only be equal when both sides are 2. For LHS, cosx(x+1)=1 => x(x+1)=0 => x=0,-1

For RHS minimum value, x=0

Hence only one solution x=0

Question 19.