JEE Advanced Maths: Logarithms / Basic Mathematics questions with solutions

Q1. Find the greatest value of the expression f(x) = 4*log₁₀(x) - logₓ(0.0001) for 0 < x < 1.

1. 4
2. -4
3. 8
4. -8

Answer: -4

Let t = log₁₀(x) < 0 (since 0 < x < 1). Then logₓ(0.0001) = -4/t. So f = 4t - (-4/t) = 4t + 4/t. For t < 0, both terms are negative; by AM-GM on |t| and 1/|t|, the maximum (least negative) value is -4*2 = -8... but we need to re-examine: f = 4t + 4/t = 4(t + 1/t). For t < 0 let t = -s (s > 0): f = 4(-s - 1/s) = -4(s + 1/s). By AM-GM, s + 1/s >= 2, so f <= -8. As x -> 1⁻, t -> 0⁻ and f -> -infinity; as x -> 0⁺, f -> -infinity. The function reaches its maximum at s = 1, i.e., t = -1, x = 0.1, giving f = -8. Wait — rechecking: the greatest value is -8. But option '-4' is also listed; let me recheck the problem expression: 4*log₁₀(x) - logₓ(0.0001). logₓ(0.0001) = log₁₀(0.0001)/log₁₀(x) = (-4)/t. So f = 4t - (-4/t) = 4t + 4/t. For t = -1: f = -4 + (-4) = -8. Maximum is -8.

Q2. If logₖ(A) * log₅(k) = 3, then A is equal to (given k > 0, k not equal to 1).

1. 5k³
2. k³
3. 125
4. 243

Answer: 125

By the chain rule: logₖ(A) * log₅(k) = log₅(A). Setting this equal to 3 gives log₅(A) = 3, so A = 5³ = 125.