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Q. Let f(x)=2xn+λ,λR,nN, and f(4)=133, f(5)=255. Then the sum of all the positive integer divisors of (f(3)f(2)) is

JEE MainJEE Main 2023Relations and Functions

Solution:

f(x)=2xn+λ
f(4)=133
f(5)=255
133=2×4n+λ...(1)
255=2×5n+λ....(2)
(2)(1)
122=2(5n4n)
5n4n=61
n=3&λ=5
Now, f(3)f(2)=2(3323)=38
Number of Divisors is 1,2,19,38; & their sum is 60