Question

The greatest, the least values of the function, $f\left(x\right)=2-\sqrt{1+2x+x^2},x\in [-2,1]$ are respectively:

• A. $2,1$
• B. $2,-1$
• C. $2,0$
• D. $-2,3$

Answer 1

Answer: (C)

$2,0$

$f\left(x\right)=2-\sqrt{1+2x+x^2}Rangeof{x}^2+2x+1D=4-4=0\therefore Minimumvalue=\frac{-D}{4a}=0x^2+2x+1\ge 0letf\left(x\right)=x^2+2x+1f^{{}^{\prime }}\left(x\right)=2x+2\therefore f\left(x\right)isincreasingfunctionin[-2,1]x^2+2x+1=1+2+1=4\therefore 0\le \sqrt{1+2x+x^2}\le \sqrt{4}0\le \sqrt{x^2+2x+1}\le 20\ge -\sqrt{x^2+2x+1}\ge -2Add2onallsides2\ge 2-\sqrt{x^2+2x+1}\ge 0maximumvalue=2,Leastvalue=0Ans=(2,0)$