### Related Formula The function min\f(x), g(x)\$\min\{f(x), g(x)\}$ chooses the lower vertical path between the two curves at any coordinate x$x$. ### Core Logic Analyze the conditions for the \right-hand function min\|x-3|, |x+2|\$\min\{|x-3|, |x+2|\}$: - The intersection of |x-3| = |x+2|$|x-3| = |x+2|$ happens at x - 3 = -(x + 2) Rightarrow 2x = 1 Rightarrow x = 0.5$x - 3 = -(x + 2) \Rightarrow 2x = 1 \Rightarrow x = 0.5$. - For x le 0.5$x \le 0.5$, |x+2| le |x-3| Rightarrow min = |x+2|$|x+2| \le |x-3| \Rightarrow \min = |x+2|$. - For x > 0.5$x > 0.5$, |x-3| le |x+2| Rightarrow min = |x-3|$|x-3| \le |x+2| \Rightarrow \min = |x-3|$.

### Step 1: Check Interval x le -2$x \le -2$ Here, |x+2| = -(x+2) = -x-2$|x+2| = -(x+2) = -x-2$: x^2 + 3x + 2 = -x - 2 Rightarrow x^2 + 4x + 4 = 0$x^2 + 3x + 2 = -x - 2 \Rightarrow x^2 + 4x + 4 = 0$ (x+2)^2 = 0 Rightarrow x = -2$(x+2)^2 = 0 \Rightarrow x = -2$ This is a valid solution as it lies precisely within the interval condition boundary. ### Step 2: Check Interval -2 < x le 0.5$-2 < x \le 0.5$ Here, |x+2| = x+2$|x+2| = x+2$: x^2 + 3x + 2 = x + 2 Rightarrow x^2 + 2x = 0$x^2 + 3x + 2 = x + 2 \Rightarrow x^2 + 2x = 0$ x(x+2) = 0 Rightarrow x = 0 quad textor quad x = -2$x(x+2) = 0 \Rightarrow x = 0 \quad \text{or} \quad x = -2$ Only x = 0$x = 0$ fits inside this interval. ### Step 3: Check Interval x > 0.5$x > 0.5$ Here, min = |x-3| = 3-x$\min = |x-3| = 3-x$: x^2 + 3x + 2 = 3 - x Rightarrow x^2 + 4x - 1 = 0$x^2 + 3x + 2 = 3 - x \Rightarrow x^2 + 4x - 1 = 0$ x = frac-4 pm sqrt16 - 4(1)(-1)2 = -2 pm sqrt5$x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = -2 \pm \sqrt{5}$ Evaluating values: -2 + sqrt5 approx 0.236$-2 + \sqrt{5} \approx 0.236$, which does not satisfy x > 0.5$x > 0.5$. Thus, no real roots occur in this span. Combining valid points, we find exactly 2 distinct real solutions (x = -2, 0$x = -2, 0$). ### Pattern Recognition Sketching a rough visualization showing the parabola crossing below the sharp wedge of the combined absolute values makes it visually clear that there are exactly two crossing points, confirming the algebraic count. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Quadratic Equations Class 12 Mathematics: Limits, Continuity and Differentiability