Question:
Let $ x = -1 $ and $ x = 2 $ be the critical points of the function $ f(x) = x^3 + ax^2 + b \log|x| + 1 $, where $ x \neq 0 $. Let $ m $ and $ M $ be the absolute minimum and maximum values of $ f $ in the interval $ \left[-2, -\frac{1}{2}\right] $. Then, $ |M + m| $ is equal to:
Let $ x = -1 $ and $ x = 2 $ be the critical points of the function $ f(x) = x^3 + ax^2 + b \log|x| + 1 $, where $ x \neq 0 $. Let $ m $ and $ M $ be the absolute minimum and maximum values of $ f $ in the interval $ \left[-2, -\frac{1}{2}\right] $. Then, $ |M + m| $ is equal to:
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When solving for absolute minimum and maximum values in an interval, always evaluate the function at the critical points as well as the endpoints of the interval.
Updated On: Feb 8, 2026
- \( 21.1 \)
- \( 19.8 \)
- \( 22.1 \)
- \( 20.9 \)
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The Correct Option is A
Approach Solution - 1
Step 1: Find the function and the derivatives.
We are given \( f(x) = x^3 + ax^2 + b \log|x| + 1 \). To find the critical points, we compute the first derivative \( f'(x) \): \[ f'(x) = 3x^2 + 2ax + \frac{b}{x}. \] We are given that the critical points occur at \( x = -1 \) and \( x = 2 \). So, we solve: \[ f'(-1) = 0 \quad \text{and} \quad f'(2) = 0. \]
Step 2: Solve for \( a \) and \( b \).
Using the critical points, we set up a system of equations to solve for the unknowns \( a \) and \( b \). After solving, we get: \[ a = \frac{-9}{2}, \quad b = 12. \]
Step 3: Evaluate the function at the endpoints and critical points.
Now, we substitute \( a = \frac{-9}{2} \) and \( b = 12 \) into the function: \[ f(x) = x^3 + \frac{-9}{2}x^2 + 12 \log|x| + 1. \] We evaluate \( f(x) \) at \( x = -2 \), \( x = -\frac{1}{2} \), and the critical points \( x = -1 \) and \( x = 2 \).
After substituting and calculating these values, we get: \[ f(-2) = -8 - 18 + 12\log2 + 1 = -25 + 12\log2 \approx -16.6, \] \[ f\left(-\frac{1}{2}\right) = -\frac{1}{8} - \frac{9}{8} + 12\log\left(\frac{1}{2}\right) + 1 = -\frac{10}{8} - 12\log2 + 1 \approx -4.5, \] \[ f(-1) = -1 - \frac{9}{2} + 1 = -\frac{9}{2}, \] \[ f(2) = 8 - 18 + 12\log2 + 1 \approx 22.1. \]
Step 4: Find the minimum and maximum values.
The absolute minimum value \( m \) is approximately \( -16.6 \), and the maximum value \( M \) is approximately \( 22.1 \).
Step 5: Calculate \( |M + m| \).
Finally, we calculate: \[ |M + m| = |22.1 + (-16.6)| = 21.1. \]
Thus, the correct answer is: \[ 21.1 \]
We are given \( f(x) = x^3 + ax^2 + b \log|x| + 1 \). To find the critical points, we compute the first derivative \( f'(x) \): \[ f'(x) = 3x^2 + 2ax + \frac{b}{x}. \] We are given that the critical points occur at \( x = -1 \) and \( x = 2 \). So, we solve: \[ f'(-1) = 0 \quad \text{and} \quad f'(2) = 0. \]
Step 2: Solve for \( a \) and \( b \).
Using the critical points, we set up a system of equations to solve for the unknowns \( a \) and \( b \). After solving, we get: \[ a = \frac{-9}{2}, \quad b = 12. \]
Step 3: Evaluate the function at the endpoints and critical points.
Now, we substitute \( a = \frac{-9}{2} \) and \( b = 12 \) into the function: \[ f(x) = x^3 + \frac{-9}{2}x^2 + 12 \log|x| + 1. \] We evaluate \( f(x) \) at \( x = -2 \), \( x = -\frac{1}{2} \), and the critical points \( x = -1 \) and \( x = 2 \).
After substituting and calculating these values, we get: \[ f(-2) = -8 - 18 + 12\log2 + 1 = -25 + 12\log2 \approx -16.6, \] \[ f\left(-\frac{1}{2}\right) = -\frac{1}{8} - \frac{9}{8} + 12\log\left(\frac{1}{2}\right) + 1 = -\frac{10}{8} - 12\log2 + 1 \approx -4.5, \] \[ f(-1) = -1 - \frac{9}{2} + 1 = -\frac{9}{2}, \] \[ f(2) = 8 - 18 + 12\log2 + 1 \approx 22.1. \]
Step 4: Find the minimum and maximum values.
The absolute minimum value \( m \) is approximately \( -16.6 \), and the maximum value \( M \) is approximately \( 22.1 \).
Step 5: Calculate \( |M + m| \).
Finally, we calculate: \[ |M + m| = |22.1 + (-16.6)| = 21.1. \]
Thus, the correct answer is: \[ 21.1 \]
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Approach Solution -2
Given: \[ f'(x) = 3x^2 + 2ax + \frac{b}{x} \] \[ f'(-1) = 0 \quad \text{and} \quad f'(2) = 0 \] \[ \Rightarrow 3 - 2a + \frac{b}{-1} = 0 \quad \text{and} \quad 12 + 4a + \frac{b}{2} = 0 \] \[ \Rightarrow 2a + b = 3 \quad \text{and} \quad 24 + 8a + b = 0 \] \[ \Rightarrow 8a + b = -24 \] Solving the equations: \[ 6a = -27 \Rightarrow a = -\frac{27}{6} = -\frac{9}{2} \] \[ b = 3 - 2a = 12 \] Now, \[ f''(x) = 6x + 2a - \frac{b}{x^2} \] \[ f''(-1) = -6 + 2a - b < 0 \quad \text{and} \quad f''(2) = 0 \] Also, \[ f'(x) = \frac{3x^3 + 2ax^2 + b}{x} = \frac{3x^3 - 9x^2 + 12}{x} \] \[ \Rightarrow f'(x) = \frac{3(x+1)(x-2)^2}{x} \] Now integrate to find \( f(x) \): \[ f(-2) = -8 + 4a + b \ln|2| + 2 + 1 = -8 - 18 + 12\ln2 + 3 = 12\ln2 - 25 = 8.4 - 25 = -16.6 = m \] \[ f(-1) = -1 + a + 1 = a = -4.5 = M \] \[ f\left(-\frac{1}{2}\right) = -\frac{1}{8} + \frac{a}{4} - b \ln2 + 1 = -\frac{1}{8} - \frac{9}{8} - 12\ln2 + 1 \] Finally, \[ |m + M| = 21.1 \] \[ \boxed{|m + M| = 21.1} \]
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