Mixing of concepts!

Calculus Level 5

A function is defined as f ( x ) = a x 4 + b x 3 + c x 2 + d x + λ f(x)=ax^4+bx^3+cx^2+dx+\lambda . This function has a special property: f ( 1 ) 1 = f ( 2 ) 2 = f ( 3 ) 3 \dfrac{f(1)}{1}=\dfrac{f(2)}{2}=\dfrac{f(3)}{3} . If lim ⁡ x → ∞ f ( x ) x 4 = 1 \displaystyle \lim _{ x\to \infty }{ \dfrac { f( x ) }{ { x }^{ 4 } } } =1 , then find f ( i ) + f ( − i ) f(i)+f(-i) where i = − 1 i=\sqrt{-1} .


The answer is -20.

Aditya Kumar by Aditya Kumar , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

展豪 張
May 16, 2016

Nice mix!
lim ⁡ x → ∞ f ( x ) x 4 = 1 ⇒ a = 1 \displaystyle \lim_{x\to \infty}\frac{f(x)}{x^4}=1 \Rightarrow a=1
f ( 1 ) 1 = f ( 2 ) 2 = f ( 3 ) 3 \dfrac{f(1)}1=\dfrac{f(2)}2=\dfrac{f(3)}3
⇒ 1 + b + c + d + λ = 8 + 4 b + 2 c + d + λ 2 = 27 + 9 b + 3 c + d + λ 3 \Rightarrow 1+b+c+d+\lambda=8+4b+2c+d+\dfrac\lambda 2=27+9b+3c+d+\dfrac\lambda 3
⇒ c = 11 + λ \Rightarrow c=11+\lambda
f ( i ) + f ( − i ) = ( a − b i − c + d i + λ ) + ( a + b i − c − d i + λ ) = 2 a − 2 c + 2 λ = 2 ( 1 ) − 2 ( 11 + λ ) + 2 λ = − 20 f(i)+f(-i)=(a-bi-c+di+\lambda)+(a+bi-c-di+\lambda)=2a-2c+2\lambda=2(1)-2(11+\lambda)+2\lambda=-20


8 Helpful 0 Interesting 1 Brilliant 0 Confused

As intended! Thanks for posting a solution!

Aditya Kumar - 5 years, 1 month ago

Exactly !!

Aakash Khandelwal - 5 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...