JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 17)
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
60
58
61
59
Explanation
$f(x)=2 x^{n}+\lambda, \lambda \in \mathbb{R}, n \in \mathbb{N}$
$f(4)=2 \cdot 4^{n}+\lambda=133, f(5)=2 \cdot 5^{n}+\lambda=255$
$f(5)-f(4)=2 \cdot\left(5^{n} \cdot 4^{n}\right)=122 \Rightarrow n=3$
$\Rightarrow f(3)-f(2)=2 \cdot\left(3^{n} \cdot 2^{n}\right)=2 \cdot\left(3^{3}-2^{3}\right)=2 \times 19$
Required sum $=1+2+19+38=60$
$f(4)=2 \cdot 4^{n}+\lambda=133, f(5)=2 \cdot 5^{n}+\lambda=255$
$f(5)-f(4)=2 \cdot\left(5^{n} \cdot 4^{n}\right)=122 \Rightarrow n=3$
$\Rightarrow f(3)-f(2)=2 \cdot\left(3^{n} \cdot 2^{n}\right)=2 \cdot\left(3^{3}-2^{3}\right)=2 \times 19$
Required sum $=1+2+19+38=60$
Comments (0)
