Question

The set $ \{{{x}^{3}}-12x:-3\le x\le 3\} $ is equal to

• A. $ \{x:-16\le x\le 16\} $
• B. $ \{x:-12\le x\le 12\} $
• C. $ \{x:-9\le x\le 9\} $
• D. $ \{x:0\le x\le 10\} $

Answer 1

Answer: (A)

Let $ y={{x}^{3}}-12x $
$ \frac{dy}{dx}=3{{x}^{2}}-12 $
Put $ \frac{dy}{dx}=0,\,\,3{{x}^{2}}-12=0 $
$ \Rightarrow $ $ x=\pm 2 $
At $ x=2,\,y={{2}^{3}}-12(2)=-16 $
At $ x=-2,\,y={{(-2)}^{3}}-12(-2)=16 $
Hence, option is correct.