Problem 181 - A Complete Resource Book in Mathematics for JEE Main

Question

Let the real values \(a, b\) be such that the function \(\mathrm{f}\) given by \(f(x)=\ln |x|+b x^{2}+a x, x \neq 0\) has extreme values at \(x=-1\) and \(x=2 .\) Statement \(1: f\) has local maximum at \(x=-1\) and at \(x=2\) Statement 2: \(a=\frac{1}{2}\) and \(b=\frac{-1}{4}\) (A) Statement 1 is false, statement 2 is true (B) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 (C) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1 (D) Statement 1 is true, statement 2 is false

Step-by-Step Solution

Answer

(B) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Step 1: Find the Derivative

Given the function \( f(x)=\ln |x|+b x^{2}+a x \). To find the extreme values, we need to first find the derivative of this function: \[ f'(x) = \frac{d}{dx}(\ln |x|) + \frac{d}{dx}(b x^2) + \frac{d}{dx}(a x) \]We know that \(\frac{d}{dx}(\ln |x|) = \frac{1}{x}\), \(\frac{d}{dx}(b x^2) = 2bx\), and \(\frac{d}{dx}(a x) = a\). Therefore, \[ f'(x) = \frac{1}{x} + 2bx + a \]

Step 2: Set the Derivative to Zero

For \(f(x)\) to have extreme values at \(x = -1\) and \(x = 2\), the derivative needs to equal zero at these points. Thus, \[ \frac{1}{x} + 2bx + a = 0 \]must hold true for \(x = -1\) and \(x = 2\).

Step 3: Solve the Equation for x = -1

Substitute \(x = -1\) into the derivative equation:\[ \frac{1}{-1} + 2b(-1) + a = 0 \]which simplifies to \[ -1 - 2b + a = 0 \] so\[ a = 1 + 2b \].

Step 4: Solve the Equation for x = 2

Substitute \(x = 2\) into the derivative equation:\[ \frac{1}{2} + 2b(2) + a = 0 \]which simplifies to \[ \frac{1}{2} + 4b + a = 0 \].

Step 5: Solve Simultaneous Equations

You have two equations: 1. \(a = 1 + 2b\)2. \(\frac{1}{2} + 4b + a = 0\)Substitute the equation from Step 3 into the second equation:\[ \frac{1}{2} + 4b + (1 + 2b) = 0 \]Simplify and solve \[ \frac{1}{2} + 4b + 1 + 2b = 0 \]\[ \frac{3}{2} + 6b = 0 \]\[ 6b = -\frac{3}{2} \]\[ b = -\frac{1}{4} \]Substitute \(b = -\frac{1}{4}\) back into \(a = 1 + 2b \)\[ a = 1 + 2(-\frac{1}{4}) \]\[ a = 1 - \frac{1}{2} \]\[ a = \frac{1}{2} \].

Step 6: Analyze the Statements

From the solution, we found \(a = \frac{1}{2}\) and \(b = -\frac{1}{4}\). Therefore, Statement 2 is true.Now, we need to check if the function has local maxima or minima at these points.Substitute \(x = -1\) and \(x = 2\) in the second derivative: \[ f''(x) = -\frac{1}{x^2} + 2b \]At \(x = -1\), \[ f''(-1) = -1 + 2(-\frac{1}{4}) = -1 - \frac{1}{2} = -\frac{3}{2} < 0 \] which means a local maxima.At \(x = 2\), \[ f''(2) = -\frac{1}{4} - \frac{1}{2} + 1 = -\frac{1}{4} < 0 \] which means a local maxima. This confirms that Statement 1 is true.

Key Concepts

Extreme Values

Extreme values refer to the maximum or minimum values a function can attain, either locally or globally. These values occur where the derivative of the function equals zero or is undefined, often at critical points. For such points, the function experiences a change in direction, suggesting a potential peak or trough.

To find these extreme values, one must determine the derivative of the function and set it equal to zero. For example, for the function \(f(x) = \ln |x| + b x^2 + a x \), finding extreme values involves setting \(f'(x) = \frac{1}{x} + 2bx + a = 0\). Solving this provides critical points, which can then be analyzed further to determine the nature of the extreme values.

Derivatives

Derivatives play a critical role in calculus, allowing us to understand the rate at which a function changes. For a function like \(f(x) = \ln |x| + b x^2 + a x \), the derivative \(f'(x)\) is computed by taking the derivative of each term separately. Understanding how to find the derivative is essential for identifying where a function increases or decreases.

The derivative \(f'(x)\) gives insights into the behavior of the function. It helps locate points where the rate of change switches from positive to negative or vice versa, indicating possible extreme values. Calculating derivations accurately is foundational to solving for roots and analyzing the function's overall curve. It's like a mathematical tool that reveals the underlying narrative of the function's motion.

Local Maxima

Local maxima occur at points where the function reaches a peak, surrounded by lower values in its immediate vicinity. To identify these points, after finding the derivative and setting it to zero, you must examine the second derivative \(f''(x)\). If \(f''(x) < 0\) at a critical point, it is a local maximum.

In this context, for \(f(x) = \ln |x| + b x^2 + a x\), we check the second derivative at \(x = -1\) and \(x = 2\). Both give a negative result, signalling local maxima. Hence, confirming that these points are peaks on the graph of the function, where the slope of the tangent is zero, and the curve turns downwards around these points.