Quadratic inequalities

Question

This is what I tried. I tried finding limits of y and then equating them with the given limits, but I could not simplify it further. The given options for this question are:

1. a+b=23
2. a^2+b^2=277
3. a+b=17
4. a^2=b^2=149

Answer 1

You want $-5 \le \dfrac{x^2+ax+b}{x^2+2x+3}\le 4$ for all real $x$.

Noting $x^2+2x+3 = (x+1)^2+2 > 0$, we can multiply with this throughout to get $$-5x^2-10x-15 \le x^2 + ax + b \le 4x^2+8x + 12$$

Now take the left inequality, and we have $6x^2+(a+10) x+(15+b) \ge 0$. As this needs to hold true for all real $x$, we can set a condition that the discriminant should never be positive. However if you also want equality to hold for some value of x, we want the discriminant to be zero. So $(a+10)^2 = 24(15+b)$ is one condition.

Similarly, looking at the right inequality, we have $3x^2 + (8-a)x+(12-b) \ge 0$, so the discriminant condition gives us $(8-a)^2 = 12(12-b)$.

Thus we have the two equations to solve for $a, b \in \mathbb{N}$. Solving these, we have $(a, b) = (14, 9)$.

Answer 2

First of all you need to be familiar with derivatives for this answer to make any sense. If you do not know how to differentiate a fraction like this one, then perhaps this is not the intended approach. If you're not more or less fluent in the use of derivatives, then I might suggest you try easier exercises before you start with this one. This is about as hairy as these problems get, ever.

The derivative of $y$ is another function, $y'$, and in this case it will be a fraction. More or less like $y$ itself, but with different numerator and denominator. You want to know for what $x$-values this fraction is equal to $0$ (those $x$-values are the same as the $x$-values for which $y$ itself is at its highest or lowest, more or less), so you solve the equation $y' = 0$. I assume you will get two solutions, both of them containing $a$ and $b$.

Once you have those, you put them in for $x$ in the original $y$-expression. You now have two expressions, both depending on (that means contains, in some form or another) $a$ and $b$. One of them should be equal to $-5$, the other to $4$. This is all you need to calculate $a$ and $b$ so that the maximum value of the function is $4$, and the minimum value is $-5$.