Let f ( x )=2 x n +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

Question

Let f ( x )=2 x n +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 (A) 59 (B) 60 (C) 61 (D) 58

Tardigrade Admin · Accepted Answer

f( x )=2 x n +λ f(4)=133 f(5)=255 133=2 × 4 n +λ ...(1) 255=2 × 5 n +λ....(2) (2)-(1) 122=2(5 n -4 n ) ⇒ 5 n -4 n =61 ∴ n =3 λ=5 Now, f(3)-f(2)=2(33-23)=38 Number of Divisors is 1, 2, 19, 38 ; their sum is 60

Q. 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

59

60

61

58

f(x)=2xn+λ f(4)=133 f(5)=255 133=2×4n+λ...(1) 255=2×5n+λ....(2) (2)−(1) 122=2(5n−4n) ⇒5n−4n=61 ∴n=3&λ=5 Now, f(3)−f(2)=2(33−23)=38 Number of Divisors is 1,2,19,38; & their sum is 60

Q. Let $f (x) = 2 x^{n} + λ, λ \in R, n \in N$ , and $f (4) = 133$ , $f (5) = 255$ . Then the sum of all the positive integer divisors of $(f (3) - f (2))$ is