JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Quadratic Equations and Inequalities

Let $f(x)=2x^2-8x+k$. Its parabola opens upward, so for one root in $(1,2)$ and the other in $(2,3)$ the endpoint values must satisfy $f(1)>0$, $f(2)<0$ and $f(3)>0$. Computing, $f(1)=k-6$, $f(2)=k-8$ and $f(3)=k-6$. These give $k>6$, $k<8$ and $k>6$ respectively, so together $6<k<8$. The only integer in this range is $k=7$, for which the roots are $2\pm \frac{1}{\sqrt{2}}$, correctly placed in the two intervals. Hence the number of integral values of $k$ is $1$.