Functions Question 427
Question: The domain of the function $ f(x)=\sqrt{x^{14}-x^{11}+x^{6}-x^{3}+x^{2}+1} $ is
Options:
A) $ (-\infty ,\infty ) $
B) $ [0,\infty ) $
C) $ (-\infty ,0] $
D) $ R\backslash [0,1] $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given $ f(x)=\sqrt{x^{14}-x^{11}+x^{6}-x^{3}+x^{2}+1} $ For $ f(x) $ to be defined, $ x^{14}-x^{11}+x^{6}-x^{3}+x^{2}+1\ge 0 $ Case 1: $ x\ge 1 $ $ x^{14}-x^{11}+x^{6}+x^{3}+x^{2}+1 $ $ =(x^{14}-x^{11})+(x^{6}-x^{3})+(x^{2}+1)>0 $ Case 2: $ 0\le x\le 1 $ $ x^{14}-x^{11}+x^{6}-x^{3}+x^{2}+1 $ $ =x^{14}{(x^{11}-x^{11})+(x^{3}-x^{2})+1>0 $ $ {\because ,x^{11}-x^{6}\le 0,x^{3}-x^{2}\le 0} $ Case 3: $ x<0 $ $ x^{14}-x^{11}+x^{6}-x^{3}+x^{2}+1>0 $ $ (\because x^{11}<0,x^{3}<0,x^{14},x^{6},x^{2}>0) $ Thus for all area, $ x,x^{14}-x^{11}+x^{6}-x^{3}+x^{2}+1\ge 0 $ Hence the domain of $ f(x)=R=(-\infty ,\infty ) $
BETA
