A) 50, 50
B) 99, 1
C) 98, 2
D) None of these
Correct Answer: B
Solution :
Let one number is\[(100-x)\] and then another is x. Therefore \[f(x)=2(100-x)+{{x}^{2}}={{x}^{2}}-2x+200\] \[f'(x)=0\Rightarrow 2x-2=0\Rightarrow x=1\] Here \[{f}''\,(x)=2>0\] Therefore function is minimum at \[x=1\] So the numbers are 99 and 1.You need to login to perform this action.
You will be redirected in
3 sec