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

$f\left(x\right)$ is given as

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

$\because f\left(4\right)=133$

$\Rightarrow 133=2\times 4^{\mathrm{n}}+\lambda ...\left(1\right)$

$\because f\left(5\right)=255$

$\Rightarrow 255=2\times 5^{\mathrm{n}}+\lambda ...\left(2\right)$

Now, $\left(2\right)-\left(1\right)$

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

$\Rightarrow 5^{\mathrm{n}}-4^{\mathrm{n}}=61$

$\therefore \mathrm{n}=3 \& \lambda =5$

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

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