Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=(m-1)x^2-2(m+1)x+m+3$. For exactly one root to lie in $(0,2)$, we must have $f(0)f(2)<0$. We calculate $f(0)=m+3$ and $f(2)=4(m-1)-4(m+1)+(m+3)=4m-4-4m-4+m+3=m-5$. So, $(m+3)(m-5)<0$, which implies $m\in (-3,5)$. We also consider the case when $m-1=0$, i.e., $m=1$. If $m=1$, the equation becomes $-4x+4=0$, so $x=1$. Since $x=1$ is in $(0,2)$, $m=1$ is a valid value. The interval $(-3,5)$ includes $m=1$. Thus, the solution is $m\in (-3,5)$.