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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Functions

NDA 2020 · Mathematics · Functions

If f(x)=2 x-x^2 , then what is the value of f(x+2)+f(x-2) when x=0 ?

If f(x)=2x−x2f(x)=2 x-x^2, then what is the value of f(x+2)+f(x−2)f(x+2)+f(x-2) when x=0x=0 ?

Options

  1. A-8
  2. B-4
  3. C8
  4. D4

Correct answer

A. -8

Step-by-step solution

f(x)=2x−x2f(x+2)+f(x−2)\begin{aligned} & f(x)=2 x-x^2 \\ & f(x+2)+f(x-2) \end{aligned}=2(x+2)−(x+2)2+2(x−2)−(x−2)2=4+4x−x2−4−4x−x2−4+4x−4\begin{aligned} & =2(x+2)-(x+2)^2+2(x-2)-(x-2)^2 \\ & =4+4 x-x^2-4-4 x-x^2-4+4 x-4\end{aligned} =8 x-2 x^2-8WhenWhenx=0,then,, then,f(x+2)+f(x-2)=-8$
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