Let f(x) = 2xⁿ + λ ,λ ∈ R,n ∈ N, and f(4) = 133,f(5) = 255. Then the sum of all the positive integer divisors of (f(3) - f(2)) is

Solution

Maths sir · 4 steps

Subtract the two given values first: λ cancels and n falls out at once.

1. Remove lambda. f(5) - f(4) = 2(5ⁿ - 4ⁿ) = 255 - 133 = 122, so 5ⁿ - 4ⁿ = 61.
2. Find n. Try small values: n = 1 gives 1, n = 2 gives 9, n = 3 gives 125 - 64 = 61. So n = 3, and λ = 133 - 2 × 64 = 5, although you do not need it.
3. Compute the difference. f(3) - f(2) = 2(3³ - 2³) = 2(27 - 8) = 38. Here λ cancels again.

   38 = 2 × 19 ⇒ σ(38) = (1+2)(1+19) = 60

4. So. The positive divisors are 1, 2, 19 and 38, and their sum is 60, option (a).

The idea to keep. When a constant is added to a function, take differences of its values and the constant disappears. For the sum of divisors, factorise into primes and multiply (1 + p + … + pᵏ) for each prime.

Answer from the NTA final answer key. Solution by PrepWiser.