CAT 2019

Slot 1

QA

Question & Solution

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

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