Let f ( x ) = 2 x ^ n + λ, λ ∈ ℝ, n ∈ ℕ , and f ( 4 ) = 133 , f ( 5...

$f ( x )$ is given as

$f ( x ) = 2{x}^{n} + \lambda, \lambda \in R, n \in N$

$∵ f ( 4 ) = 133$

$⇒ 133 = 2 \times {4}^{n} + \lambda . . . ( 1 )$

$∵ f ( 5 ) = 255$

$⇒ 255 = 2 \times {5}^{n} + \lambda . . . ( 2 )$

Now, $( 2 ) - ( 1 )$

$122 = 2\left({5}^{n} - {4}^{n}\right)$

$⇒ {5}^{n} - {4}^{n} = 61$

$∴ n = 3 \text{ and } \lambda = 5$

Now $f ( 3 ) - f ( 2 ) = 2\left({3}^{3} - {2}^{3}\right) = 38$

Number of divisors $= 1, 2, 19, 38$ and their sum is $60$