Meditaliano IMAT Prep

Example: Determine the condition of $x = 2$ relative to $x^2 = 4$.

Let $p: x = 2$ and $q: x^2 = 4$.

• - $p \implies q$: "If $x = 2 \implies x^2 = 4$" is **True**.
• - $q \implies p$: "If $x^2 = 4 \implies x = 2$" is **False** (since $x$ could also be $-2$).

Therefore, $x = 2$ is a **sufficient but not necessary** condition for $x^2 = 4$.

Example: Solve $|2x - 3| < 5$.

$$ -5 < 2x - 3 < 5 $$

Add 3 to all parts: $$ -2 < 2x < 8 $$

Divide by 2: $$ -1 < x < 4 $$

Example (Trig Inequality): Solve $2\cos^2 x + \sin x - 1 > 0$ for $0 \le x < 2\pi$.

Step 1: Convert to a single trig function.

Substitute $\cos^2 x = 1 - \sin^2 x$:

$$ 2(1 - \sin^2 x) + \sin x - 1 > 0 \implies -2\sin^2 x + \sin x + 1 > 0 $$

Multiply by $-1$ (reverse the inequality):

$$ 2\sin^2 x - \sin x - 1 < 0 $$

Step 2: Factorize.

$$ (2\sin x + 1)(\sin x - 1) < 0 \implies -\frac{1}{2} < \sin x < 1 $$

Step 3: Solve on the Unit Circle.

$\sin x = -1/2$ at $x = \frac{7\pi}{6}$ and $\frac{11\pi}{6}$. Since $\sin x < 1$, $x \ne \frac{\pi}{2}$.

The solution is: $$ 0 \le x < \frac{7\pi}{6} \quad \text{and} \quad \frac{11\pi}{6} < x < 2\pi \quad (\text{with } x \ne \frac{\pi}{2}) $$

Example (Log Inequality): Solve $\log_{0.5}(x-1) > \log_{0.5}(5-x)$.

Step 1: Determine the Domain.

$$ x-1 > 0 \implies x > 1 $$

$$ 5-x > 0 \implies x < 5 $$

Combined Domain: $1 < x < 5$.

Step 2: Solve the inequality (Base $0.5 < 1 \implies$ reverse the sign).

$$ x-1 < 5-x \implies 2x < 6 \implies x < 3 $$

Step 3: Combine with Domain.

Combining $x < 3$ with $1 < x < 5$ gives:

$$ 1 < x < 3 $$

Example (Inscribed Rectangle): A rectangle is inscribed under the parabola $y = 12 - x^2$ such that its base lies on the x-axis, and its top two corners touch the parabola. Find the dimensions of the rectangle that maximize its area.

Thought Process:

1. Translate: Let the top-right corner of the rectangle on the parabola be $P(x, y) = P(x, 12-x^2)$ where $x > 0$.
   Because of the parabola's symmetry across the y-axis, the top-left corner must be $(-x, 12-x^2)$.
2. Express Dimensions:
   - The width of the rectangle is the distance from $-x$ to $x$, which is $2x$.
   - The height of the rectangle is $y = 12 - x^2$.
3. Formulate Area Function: The area $A$ is:
   $$ A(x) = \text{width} \times \text{height} = 2x(12 - x^2) = 24x - 2x^3 $$
4. Optimize (Advanced): For those who know calculus, $A'(x) = 24 - 6x^2 = 0 \implies x^2 = 4 \implies x = 2$.
   Alternatively, you can test values or use inequalities to find that $x=2$ maximizes the cubic function in the positive domain.
5. Calculate Final Dimensions:
   - Width = $2x = 2(2) = 4$.
   - Height = $12 - x^2 = 12 - 4 = 8$.
   - Maximum Area = $4 \times 8 = 32$.

Example (Locus 1 - Parabola): Find the equation of the locus of points P(x, y) that are equidistant from point A(0, 2) and the line $L: y = -2$.

Thought Process:
1. Translate: The condition is $PA = PL$, where $PL$ is the perpendicular distance from P to the line $y=-2$.
2. Identify Tools: Distance Formula. The distance from $(x, y)$ to $y=-2$ is $|y - (-2)| = |y+2|$.
3. Connect: $PA^2 = PL^2$ (squaring removes the root).
$PA^2 = (x-0)^2 + (y-2)^2 = x^2 + y^2 - 4y + 4$.
$PL^2 = (y+2)^2 = y^2 + 4y + 4$.
4. Execute: $x^2 + y^2 - 4y + 4 = y^2 + 4y + 4$
$x^2 = 8y \implies y = \frac{1}{8}x^2$.
Solution: The locus is an upward-opening parabola with its vertex at (0,0). This is the definition of a parabola.

Example (Locus 2 - Apollonian Circle): Find the locus of points P(x, y) such that the ratio of distances from A(1, 0) to B(4, 0) is $PA:PB = 1:2$.

Thought Process:
1. Translate: $PA:PB = 1:2 \implies 2 \cdot PA = PB \implies 4 \cdot PA^2 = PB^2$.
2. Identify Tools: Distance Formula.
3. Execute: $4 \left( (x-1)^2 + (y-0)^2 \right) = (x-4)^2 + (y-0)^2$
$4(x^2 - 2x + 1 + y^2) = x^2 - 8x + 16 + y^2$
$4x^2 - 8x + 4 + 4y^2 = x^2 - 8x + 16 + y^2$
$3x^2 + 3y^2 = 12 \implies x^2 + y^2 = 4$.
Solution: The locus is a circle centered at (0, 0) with radius 2.

Example (Region): Find the area of the region defined by $x^2 + y^2 \le 4$ and $y \ge |x|$.

Thought Process:
1. Translate: $x^2 + y^2 \le 4$ is the interior of a circle (center (0,0), $r=2$). $y \ge |x|$ means $y \ge x$ AND $y \ge -x$. This is the V-shaped region above the lines $y=x$ and $y=-x$.
2. Diagram: Sketch the circle and the two lines. The region is a "slice of pizza" pointing upwards.
3. Identify Tools: Area of a circular sector: $A = \pi r^2 \times (\frac{\theta_{deg}}{360^\circ})$.
4. Connect: The line $y=x$ is at $45^\circ$. The line $y=-x$ is at $135^\circ$. The angle between them is $135^\circ - 45^\circ = 90^\circ$.
5. Execute: The area of the full circle is $A_{full} = \pi r^2 = \pi(2)^2 = 4\pi$. The region is exactly $90/360 = 1/4$ of the circle.
Solution: $A_{region} = \frac{1}{4} (4\pi) = \pi$.

Example (AM-GM): A rectangle is built in the first quadrant with one corner at (0,0) and the opposite corner on the line $x+2y=8$. Find the maximum area of the rectangle.

Thought Process:
1. Translate: Let the corner be $(x, y)$. The area is $A = xy$. We know $x > 0$, $y > 0$, and $x+2y=8$.
2. Identify Tools: We want to maximize a product ($xy$) given a sum ($x+2y$). This is a classic AM-GM setup.
3. Connect: Apply AM-GM to the terms $x$ and $2y$.
$\frac{x + 2y}{2} \ge \sqrt{x \cdot 2y} \implies \frac{8}{2} \ge \sqrt{2xy}$
$4 \ge \sqrt{2xy} \implies 16 \ge 2xy \implies 8 \ge xy$.
4. Execute: The maximum value of $xy$ (the area) is 8.
Solution: The maximum area is 8.

Example (d=r): Find the values of $k$ for which $y = x + k$ is tangent to $x^2 + y^2 = 8$.

Thought Process (Using $d=r$):
1. Translate: Circle center is $C(0, 0)$. Radius is $r = \sqrt{8} = 2\sqrt{2}$. The line is $x - y + k = 0$.
2. Identify Tool: Distance from point to line $d = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$.
3. Execute: $d = \frac{|1(0) - 1(0) + k|}{\sqrt{1^2 + (-1)^2}} = \frac{|k|}{\sqrt{2}}$.
4. Set $d=r$: $\frac{|k|}{\sqrt{2}} = 2\sqrt{2} \implies |k| = 2\sqrt{2} \cdot \sqrt{2} = 4$.
Solution: $k = 4$ or $k = -4$. (This is much faster than $D=0$).

Example (Two Circles): How many intersection points do $C_1: x^2+y^2=9$ and $C_2: (x-1)^2+y^2=16$ have?

Thought Process:
1. Identify Tools: $C_1$: Center (0,0), $r_1=3$. $C_2$: Center (1,0), $r_2=4$.
2. Calculate $d$: Distance between centers $d = \sqrt{(1-0)^2 + (0-0)^2} = 1$.
3. Calculate $r_1+r_2$ and $|r_1-r_2|$:
Sum: $r_1+r_2 = 3+4 = 7$.
Difference: $|r_1-r_2| = |3-4| = 1$.
4. Compare: We see that $d = 1$ and $|r_1-r_2| = 1$.
Solution: Since $d = |r_1-r_2|$, the circles are internally tangent (1 intersection point).

Example: Find $k$ such that $x^2 - 2kx + (k+2) = 0$ has two distinct *positive* roots.

Thought Process: You must check three separate conditions.
1. Two distinct roots: Discriminant $D > 0$.
$D = (-2k)^2 - 4(1)(k+2) = 4k^2 - 4k - 8 > 0$
$k^2 - k - 2 > 0 \implies (k-2)(k+1) > 0 \implies k < -1$ or $k > 2$.
2. Roots are positive (Sum): If both roots $x_1, x_2$ are positive, their sum must be positive. $x_1 + x_2 > 0$.
From Vieta's formulas, $x_1 + x_2 = -b/a = -(-2k)/1 = 2k$.
$2k > 0 \implies k > 0$.
3. Roots are positive (Product): If both roots are positive, their product must be positive. $x_1 \cdot x_2 > 0$.
From Vieta's formulas, $x_1 \cdot x_2 = c/a = (k+2)/1$.
$k+2 > 0 \implies k > -2$.
4. Synthesis: Find the intersection of all three conditions.
- (1) $k < -1$ or $k > 2$
- (2) $k > 0$
- (3) $k > -2$
Solution: The only range that satisfies all three is $k > 2$.