Question

A real valued function \( f(x)=\left|x^{2}-3x+2\right|+2x-3 \) is defined on \( [-2,1] \). If m and M are absolute minimum and absolute maximum values of \( f \) respectively then \( M-4m= \)

• A. 0
• B. 1
• C. 15
• D. 10

Hint:

Checking the sign of the quadratic inside the modulus is crucial. Since the interval ends at a root (x=1), the expression does not change sign within the interval \([-2, 1)\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

We first simplify the function by analyzing the sign of the expression inside the modulus over the interval \( [-2, 1] \). Then we find the critical points and evaluate the function at the critical points and endpoints to find the global maximum (M) and minimum (m).
Step 2: Key Formula or Approach:

1. \( |A| = A \) if \( A \ge 0 \). 2. Critical points occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined. 3. Absolute Extrema on \([a, b]\) are found by comparing \( f(a), f(b), \) and \( f(c) \) where \( c \) are critical points.
Step 3: Detailed Explanation:

The expression inside the modulus is \( g(x) = x^2 - 3x + 2 = (x-1)(x-2) \). Roots are at \( x=1 \) and \( x=2 \). For \( x \in [-2, 1] \): Test \( x=0 \): \( g(0) = 2 \textgreater 0 \). So \( x^2 - 3x + 2 \ge 0 \) on \( [-2, 1] \). Thus, \( f(x) = (x^2 - 3x + 2) + 2x - 3 = x^2 - x - 1 \). Find the derivative: \[ f'(x) = 2x - 1 \] Set \( f'(x) = 0 \implies x = 1/2 \). This critical point lies in \( [-2, 1] \). Evaluate \( f(x) \) at critical point and endpoints: 1. \( f(-2) = (-2)^2 - (-2) - 1 = 4 + 2 - 1 = 5 \). 2. \( f(1) = (1)^2 - 1 - 1 = -1 \). 3. \( f(1/2) = (1/2)^2 - (1/2) - 1 = 1/4 - 2/4 - 4/4 = -5/4 \). Comparing values: Maximum \( M = 5 \). Minimum \( m = -5/4 \). Calculate \( M - 4m \): \[ M - 4m = 5 - 4\left(-\frac{5}{4}\right) = 5 + 5 = 10 \]
Step 4: Final Answer:

The value is 10.