Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=x^2-(2k-1)x+k^2-5$. For exactly one root to lie in the interval $(-1,2)$, the values of $f(-1)$ and $f(2)$ must have opposite signs. That is, $f(-1)f(2)<0$. This condition automatically ensures distinct real roots. First, calculate $f(-1)$: $f(-1)=(-1{)}^2-(2k-1)(-1)+k^2-5=1+2k-1+k^2-5=k^2+2k-5$. Next, calculate $f(2)$: $f(2)=(2{)}^2-(2k-1)(2)+k^2-5=4-4k+2+k^2-5=k^2-4k+1$. We need $(k^2+2k-5)(k^2-4k+1)<0$. The roots of $k^2+2k-5=0$ are $k=\frac{-2\pm \sqrt{4-4(1)(-5)}}{2}=\frac{-2\pm \sqrt{24}}{2}=-1\pm \sqrt{6}$. The roots of $k^2-4k+1=0$ are $k=\frac{4\pm \sqrt{16-4(1)(1)}}{2}=\frac{4\pm \sqrt{12}}{2}=2\pm \sqrt{3}$. Arranging these roots in increasing order: $-1-\sqrt{6}$, $2-\sqrt{3}$, $-1+\sqrt{6}$, $2+\sqrt{3}$. The inequality $(k^2+2k-5)(k^2-4k+1)<0$ holds when $k$ is in the intervals where the product is negative. Using the sign scheme for a product of two quadratics, the solution is $k\in (-1-\sqrt{6},2-\sqrt{3})\cup (-1+\sqrt{6},2+\sqrt{3})$. Thus, option A is correct.