Let f ( x )=2 x ^{ n }+λ, λ ∈ R , n ∈ N , and f (4)=133 , f(5)=255 . Then sum of all positive

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1.Relation and Function

hard

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

A

$61$

B

$60$

C

$58$

D

$59$

(JEE MAIN-2023)

Solution

$f(x)=2 x^{ n }+\lambda$

$f(4)=133$

$f(5)=255$

$133=2 \times 4^{ n }+\lambda……(1)$

$255=2 \times 5^{ n }+\lambda……(2)$

$(2) -(1)$

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

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

$\therefore n =3\, and\, \lambda=5$

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

Number of Divisors is $1,2,19,38$; and their sum is $60$.

Standard 12

Mathematics

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