Functions Question 7
Question 7 - 25 January - Shift 2
Let $f(x)=2 x^{n}+\lambda, \lambda \in \mathbb{R}, n \in \mathbb{N}$, and $f(4)=133$, $f(5)=255$. Then the sum of all the positive integer divisors of ( $f(3)-f(2))$ is
(1) 61
(2) 60
(3) 58
(4) 59
Show Answer
Answer: (2)
Solution:
Formula: Operations on functions, Multinomial theorem
$ \begin{aligned} & f(x)=2 x^{n}+\lambda \\ & f(4)=133 \\ & f(5)=255 \\ & 133=2 \times 4^{n}+\lambda \ldots \ldots (1)\\ & 255=2 \times 5^{n}+\lambda \ldots \ldots (2) \\ & (2)-(1) \\ & 122=2(5^{n}-4^{n}) \\ & \Rightarrow 5^{n}-4^{n}=61 \\ & \therefore n=3 \& \lambda=5 \end{aligned} $
Now, $f(3)-f(2)=2(3^{3}-2^{3})=38$
Number of Divisors is 1, 2, 19, 38 ;
Their sum is 60
