JEE Advanced Maths: Logarithms and Infinite Series questions with solutions

Q1. The value of 6 + log_(3/2)( (1 / (3*sqrt(2))) * sqrt(4 - (1 / (3*sqrt(2))) * sqrt(4 - (1 / (3*sqrt(2))) * sqrt(4 -... (to infinity))))) is

1. 2
2. 4
3. 6
4. 8

Answer: 4

Setting A equal to the infinite nested expression and solving the resulting quadratic gives A = 4/9. Then log_(3/2)(4/9) = log_(3/2)((2/3)²) = 2 * log_(3/2)(2/3) = 2 * (-1) = -2. So the answer is 6 + (-2) = 4.