Question

If $f\left(x\right)=x^2+2x-2$ and if $f(s-1)=1$, find the smallest possible value of $s$.

• A. $-3$
• B. $-2$
• C. $-1$
• D. $1$
• E. $2$

Answer 1

The correct option is B $-2$

Given, $f\left(x\right)=x^2+2x-2$

$f(s-1)={(s-1)}^2+2(s-1)-2=s^2-2s+1+2s-2-2=s^2-3$
It should be equal to $1$, so we get $s^2-3=1$, which implies $s^2=4$
Therefore $s=\pm 2$ , the least possible value of $s$ is $-2$.