A) \[\{x:-16\le x\le 16\}\]
B) \[\{x:-12\le x\le 12\}\]
C) \[\{x:-9\le x\le 9\}\]
D) \[\{x:0\le x\le 10\}\]
Correct Answer: A
Solution :
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 [a] is correct.
You need to login to perform this action.
You will be redirected in
3 sec
