The range of function $\frac{x+m}{x^2+1}$ contains interval [0,1] if m>3/k then k must be

Question

The range of function $\frac{x+m}{x^2+1}$ $(m\in R)$ contains interval [0,1] if $m>\frac{3}{k}$ then k must be.
I am not getting exact approach to solve this problem. T tried getting quadratic equation in x and then applying $D\geq 0 $ but it didn't work can anyone help thanks

Answer 1

The question is essentially asking you to pick the minimum value of $m$ such that the maximum value of the function $f(x) = \frac{x+m}{x^2+1}$ is greater than or equal to $ 1$

Now, let $f(x) = \frac{x+m}{x^2+1}$

For $f(x) \geq 1$, we must have a non empty solution set for

$$\frac{x+m}{x^2+1} \geq 1 $$

$$\implies x^2 - x + (1-m) \leq 0$$

Since the leading coefficient for the quadratic is positive, the solution can only exist if the discriminant of the quadratic is greater than zero

$$D = 1 - 4(1-m) > 0$$

$$\implies -3 + 4m > 0$$

$$\implies m > \frac{3}{4}$$