10:I[6034,["878","static/chunks/9da6db1e-321469e27dc51aa4.js","954","static/chunks/d3ac728e-23c45ec216700eea.js","972","static/chunks/972-cf0c19383a04ee27.js","605","static/chunks/605-e6d88733e2ed50de.js","718","static/chunks/718-16004f17ca4d7da3.js","957","static/chunks/957-15c5f8c3ab68f8ee.js","34","static/chunks/34-c897dff191c58929.js","801","static/chunks/app/questions/%5Bslug%5D/page-d9aea32ec2d4c76b.js"],"default"] 11:I[8845,["878","static/chunks/9da6db1e-321469e27dc51aa4.js","954","static/chunks/d3ac728e-23c45ec216700eea.js","972","static/chunks/972-cf0c19383a04ee27.js","605","static/chunks/605-e6d88733e2ed50de.js","718","static/chunks/718-16004f17ca4d7da3.js","957","static/chunks/957-15c5f8c3ab68f8ee.js","34","static/chunks/34-c897dff191c58929.js","801","static/chunks/app/questions/%5Bslug%5D/page-d9aea32ec2d4c76b.js"],"default"] 12:T707,We want $\frac{x^2 - kx + 9}{x^2 + 1} < 2$. Since $x^2 + 1 > 0$, multiply by the denominator: $x^2 - kx + 9 < 2x^2 + 2$, which simplifies to $x^2 + kx - 7 > 0$. For this quadratic to be strictly positive for all $x$, its discriminant $D = k^2 - 4(1)(-7)$ must be less than $0$. Thus $k^2 + 28 < 0$, which is impossible. Re-evaluating the inequality: $x^2 + kx - 7 > 0$ is NOT possible for all $x$ because the parabola opens upward. Wait, checking the original expression: $x^2 - kx + 9 < 2x^2 + 2 \implies x^2 + kx - 7 > 0$. Actually, the condition $x^2 - kx + 9 < 2(x^2 + 1)$ leads to $0 < x^2 + kx - 7$. For this to hold for all $x$, the quadratic must have no real roots and be positive. But the leading coefficient is 1 (positive), so it will eventually be positive, but it crosses the axis if $D > 0$. To never be less than 2, the minimum of $f(x)$ must be $\geq 2$ or the maximum must be $< 2$. Actually, checking $f(x) < 2$: $x^2 - kx + 9 < 2x^2 + 2 \iff x^2 + kx - 7 > 0$. This inequality cannot hold for all $x$ because as $x \to -\infty$, the expression goes to $+\infty$, but for $x=0$, it is $-7 < 0$. The question implies $f(x) < 2$ is impossible for any $k$ if the domain is a

Given the function f(x)=x2−kx+9x2+1f(x) = \frac{x^2 - kx + 9}{x^2 + 1}f(x)=x2+1x2−kx+9​, find all real values of kkk such that f(x)<2f(x) < 2f(x)<2 for all x∈Rx \in \mathbb{R}x∈R.

Given the function $f(x) = \frac{x^2 - kx + 9}{x^2 + 1}$ , find all real values of $k$ such that $f(x) < 2$ for all $x \in \mathbb{R}$ .