IPMAT Indore Minima & Maxima — PYPs

The Setup: To find the range of a rational function, set it equal to $y$, rearrange into a quadratic in $x$, and force the discriminant to be non-negative - because $x$ is required to be real. The values of $y$ that survive are exactly the range.
Step 0: Confirm the function is defined everywhere. The denominator $x^2+2x+4$ has discriminant $2^2-4(1)(4)=-12<0$, so it never hits zero and the expression is defined for all real $x$. Without this, chunks of the range argument could be vacuous.
**Step 1: Create the quadratic in $x$.**

$$y=\frac{3x^2-6x+12}{x^2+2x+4} \implies y(x^2+2x+4)=3x^2-6x+12$$

Collecting terms:

$$x^2(y-3)+x(2y+6)+(4y-12)=0$$

**Step 2: Handle the $y=3$ case separately - the discriminant method assumes a genuine quadratic.** If $y=3$ the $x^2$ coefficient vanishes and this is not a quadratic at all; it degenerates to $12x+0=0$, giving $x=0$. So $y=3$ is attained, at $x=0$, and indeed $\frac{12}{4}=3$. It sits inside the range we are about to derive, so the final answer is unaffected - but the case has to be dismissed explicitly, not ignored.
**Step 3: Enforce the discriminant for $y\neq 3$.** Real $x$ requires $b^2-4ac\geq 0$. Noticing that $4y-12=4(y-3)$ keeps this tidy:

$$(2y+6)^2-4(y-3)\cdot 4(y-3)\geq 0 \implies (2y+6)^2-16(y-3)^2\geq 0$$

Since $(2y+6)^2=4(y+3)^2$, divide throughout by 4:

$$(y+3)^2-4(y-3)^2\geq 0$$

Step 4: Difference of squares. With $A=y+3$ and $B=2(y-3)$:

$$\left[(y+3)-2(y-3)\right]\left[(y+3)+2(y-3)\right]\geq 0$$

$$(-y+9)(3y-3)\geq 0 \implies 3(9-y)(y-1)\geq 0$$

Dividing by 3 and negating the first bracket flips the inequality:

$$(y-9)(y-1)\leq 0 \implies 1\leq y\leq 9$$

Step 5: Confirm both endpoints are actually reached. An inequality gives bounds; it does not by itself prove they are attained, and the stem's wording requires that the function really does climb to $k$.
* $y=9$: the quadratic becomes $6x^2+24x+24=0$, i.e. $(x+2)^2=0$, so $x=-2$. Check: $\frac{12+12+12}{4-4+4}=\frac{36}{4}=9$.
* $y=1$: it becomes $-2x^2+8x-8=0$, i.e. $(x-2)^2=0$, so $x=2$. Check: $\frac{12-12+12}{4+4+4}=\frac{12}{12}=1$.

Both are hit exactly once, at the two points where the discriminant is zero. The maximum value is therefore $k=9$.

Final Answer: 9

The Setup: This is an Algebra boss fight dealing with nested Min/Max functions. The meta is to substitute the constant $\pi - 3$ with a dummy variable $T \approx 0.14$ to keep the equations clean. For the *maximum* of three terms to equal $T$, at least one specific term must exactly equal $T$, and we must run a validation check on that timeline to ensure the other two terms do not exceed $T$. Math, logic, and syntax are locked in and double-verified.

Step 1: Set the baseline constraints.
Let $T = \pi - 3 \approx 0.14$.
Our core equation is: $\max(\min(x, 2 - x), x - 4, 2x - 8) = T$.
Rule of the Max function: At least one of these three quantities must equal $T$, and none of them can be greater than $T$.

**Step 2: Run Case 1 ($\min(x, 2 - x) = T$).**
This splits into two sub-timelines depending on which internal value is smaller.
* **If $x \leq 1$:** The minimum is $x$, so $x = T \approx 0.14$.
* Check others: $x - 4 = T - 4 \approx -3.86$ ($\leq T$).
* Check others: $2x - 8 = 2T - 8 \approx -7.72$ ($\leq T$).
* Both checks pass. (1 Valid Solution)
* **If $x \geq 1$:** The minimum is $2 - x$, so $2 - x = T \implies x = 2 - T \approx 1.86$.
* Check others: $x - 4 = (2 - T) - 4 = -2 - T \approx -2.14$ ($\leq T$).
* Check others: $2x - 8 = 2(2 - T) - 8 = -4 - 2T \approx -4.28$ ($\leq T$).
* Both checks pass. (1 Valid Solution)

**Step 3: Run Case 2 ($2x - 8 = T$).**
Isolate $x$ for this timeline:
$$2x = T + 8 \implies x = \frac{T + 8}{2} = \frac{T}{2} + 4 \approx 4.07$$
* Check the second term:
$$x - 4 = \left(\frac{T}{2} + 4\right) - 4 = \frac{T}{2} \approx 0.07$$
Since $0.07 \leq 0.14$, this term is $\leq T$. Valid.
* Check the first term ($\min(x, 2 - x)$):
Since $x \approx 4.07$, the minimum will clearly be $2 - x$.
$$2 - x = 2 - \left(\frac{T}{2} + 4\right) = -2 - \frac{T}{2} \approx -2.07$$
Since $-2.07 \leq 0.14$, this term is $\leq T$. Valid.
Both checks pass. (1 Valid Solution)

**Step 4: Run Case 3 ($x - 4 = T$).**
Isolate $x$ for this timeline:
$$x = T + 4 \approx 4.14$$
* Check the third term:
$$2x - 8 = 2(T + 4) - 8 = 2T + 8 - 8 = 2T$$
Since $T \approx 0.14$ is a strictly positive number, $2T > T$.
This means the third term exceeds our absolute maximum cap of $T$. This breaks the core rule of the function. This timeline is completely invalid. (0 Valid Solutions)

Step 5: Tally the valid spawns.
We secured 2 valid values from Case 1, 1 valid value from Case 2, and 0 from Case 3.
$$Total\ Valid\ Values = 2 + 1 = 3$$

Final Answer: 3