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
Solution
$f(x)=2 x^{n}+\lambda, \lambda \in \mathbb{R}, n \in \mathbb{N}$
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$f(4)=2 \cdot 4^{n}+\lambda=133, f(5)=2 \cdot 5^{n}+\lambda=255$
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$f(5)-f(4)=2 \cdot\left(5^{n} \cdot 4^{n}\right)=122 \Rightarrow n=3$
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$\Rightarrow f(3)-f(2)=2 \cdot\left(3^{n} \cdot 2^{n}\right)=2 \cdot\left(3^{3}-2^{3}\right)=2 \times 19$
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Required sum $=1+2+19+38=60$
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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