Question

The minimum possible value of \(\frac{x^2−6x+10}{3−x}\),for x<3,is

• A. \(\frac{1}{2}\)
• B. \(-\frac{1}{2}\)
• C. 2
• D. -2

Answer 1

The Correct Option is C

Step 1: Differentiate the Expression

Let: \[ f(x) = \frac{u}{v}, \quad \text{where } u = x^2 - 6x + 10, \quad v = 3 - x \] Using the quotient rule: \[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \] Compute derivatives: \[ u' = 2x - 6, \quad v' = -1 \] So, \[ f'(x) = \frac{(3 - x)(2x - 6) - (x^2 - 6x + 10)(-1)}{(3 - x)^2} \] Simplify numerator: \[ (3 - x)(2x - 6) + (x^2 - 6x + 10) = (6x - 18 - 2x^2 + 6x) + x^2 - 6x + 10 = -2x^2 + 6x + 6x + x^2 - 6x - 18 + 10 = -x^2 + 6x - 8 \] Wait — better to directly simplify the original expression properly: Rewriting numerator: \[ (3 - x)(2x - 6) - (x^2 - 6x + 10)(-1) = (6x - 18 - 2x^2 + 6x) + x^2 - 6x + 10 = -2x^2 + 12x + x^2 - 6x - 18 + 10 = -x^2 + 6x - 8 \] So, \[ f'(x) = \frac{-x^2 + 6x - 8}{(3 - x)^2} \] Set \( f'(x) = 0 \Rightarrow -x^2 + 6x - 8 = 0 \)

Step 2: Solve the Critical Points

\[ x^2 - 6x + 8 = 0 \Rightarrow x = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2} = 2 \quad \text{and} \quad 4 \] Only \( x = 2 \) satisfies the domain \( x < 3 \)

Step 3: Calculate Minimum Value at \( x = 2 \)

\[ f(2) = \frac{2^2 - 6 \cdot 2 + 10}{3 - 2} = \frac{4 - 12 + 10}{1} = \frac{2}{1} = \boxed{2} \]

Final Answer:

\[ \boxed{2} \quad \text{(Option C)} \]

Answer 2

Step 1: Let \( y = 3 - x \)

Since \( x < 3 \), we have: \[ y = 3 - x > 0 \] Now rewrite the numerator: \[ x^2 - 6x + 10 = (x - 3)^2 + 1 \] So the expression becomes: \[ \frac{(x - 3)^2 + 1}{3 - x} = \frac{(3 - x)^2 + 1}{3 - x} = \frac{y^2 + 1}{y} = y + \frac{1}{y} \]

Step 2: Minimize \( y + \frac{1}{y} \) for \( y > 0 \)

By the AM-GM inequality: \[ y + \frac{1}{y} \geq 2 \] Equality occurs when \( y = \frac{1}{y} \Rightarrow y = 1 \) So the minimum value of the expression is: \[ \boxed{2} \]

Final Answer:

\[ \boxed{\text{Option (C): } 2} \]