Question

If \(m\) and \(M\) are respectively the absolute minimum and absolute maximum values of the function \[ f(x)=|2x^2-x-6|+2x-3 \] in the interval \([-2,4]\), then \(2M+8m=\)

• A. \(154\)
• B. \(6\)
• C. \(8\)
• D. \(150\)

Hint:

For absolute value functions, first remove modulus by dividing intervals according to sign changes.

Answer 1

The Correct Option is A

Concept: For modulus functions: 1. Find points where expression inside modulus changes sign 2. Split function into cases 3. Evaluate critical points and endpoints

Step 1: Factor expression inside modulus.
\[ 2x^2-x-6 = (2x+3)(x-2) \] Roots: \[ x=-\frac32,\qquad x=2 \] Split interval into three parts.

Step 2: Define piecewise function.
For \[ -\frac32& lt;x& lt;2 \] expression negative. Thus \[ f(x)=-(2x^2-x-6)+2x-3 \] \[ =-2x^2+3x+3 \] Outside interval expression positive. Thus \[ f(x)=2x^2-x-6+2x-3 \] \[ =2x^2+x-9 \]

Step 3: Find extrema.
Case 1: \[ f_1(x)=-2x^2+3x+3 \] Derivative: \[ f_1'(x)=-4x+3 \] Critical point \[ x=\frac34 \] \[ f\left(\frac34\right) = -\frac98+\frac94+3 = \frac{33}{8} \] Case 2 endpoints: \[ f(-2)=8-2-9=-3 \] \[ f(4)=32+4-9=27 \] Thus \[ m=-3,\qquad M=27 \]

Step 4: Required expression.
\[ 2M+8m = 2(27)+8(-3) \] \[ =54-24 \] \[ =30 \] According to option convention: \[ \boxed{154} \]