Problem 74 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(y=a \log |x|+b x^{2}+x\) has its extremum values at \(x=-1\) and \(x=2\), then (A) \(a=2, b=-1\) (B) \(a=2, b=-1 / 2\) (C) \(a=-2, b=1 / 2\) (D) None of these

Step-by-Step Solution

Answer

Option (B): \(a=2, b=-1/2\).

Step 1: Understand the Problem

The given function is \( y = a \log |x| + bx^2 + x \) and we need to find values of \(a\) and \(b\) such that the function has extremum values at \(x = -1\) and \(x = 2\). "Extremum values" means that the derivative of the function at these points is 0.

Step 2: Differentiate the Function

Differentiate the function with respect to \(x\). The derivative \(y'\) is given by: \( y' = \frac{d}{dx}(a \log |x| + bx^2 + x) = \frac{a}{x} + 2bx + 1 \).

Step 3: Set Derivative to Zero for Extremum Points

Set the derivative equal to zero at the given points. For \(x = -1\), \( \frac{a}{-1} + 2b(-1) + 1 = 0 \) results in \(-a - 2b + 1 = 0\). For \(x = 2\), \( \frac{a}{2} + 2b(2) + 1 = 0 \) results in \(\frac{a}{2} + 4b + 1 = 0\).

Step 4: Solve the System of Equations

Solve the two equations obtained from the previous step:1. \(-a - 2b + 1 = 0\)2. \(\frac{a}{2} + 4b + 1 = 0\)Multiply the second equation by 2 to clear the fraction: \(a + 8b + 2 = 0\). Now we have:\[\begin{align*}-a - 2b + 1 &= 0, \a + 8b + 2 &= 0.\end{align*}\]Adding these equations results in \(6b + 3 = 0\), giving \(b = -\frac{1}{2}\). Substituting \(b = -\frac{1}{2}\) into \(-a - 2b + 1 = 0\) results in \(a = 2\).

Step 5: Check Consistency with Options

The solution we found is \(a = 2\) and \(b = -\frac{1}{2}\). Compare this with the given options. Option (B) matches our solution: \(a=2, b=-1 / 2\).

Key Concepts

Derivative

In calculus, a derivative indicates how a function changes as its input changes. In our exercise, we explore the derivative of the function \( y = a \log |x| + bx^2 + x \), to find out where the function has "extremum values". Derivatives are calculated with respect to a variable, commonly denoted by \( x \).

• The derivative of \( \log |x| \) is \( \frac{1}{x} \), but since \( a \) is a constant multiplier, it becomes \( \frac{a}{x} \).
• The derivative of \( x^2 \) is \( 2x \), and thus \( bx^2 \) becomes \( 2bx \).
• The derivative of \( x \) is simply 1.

Bringing all these components together, the derivative of our function is \( y' = \frac{a}{x} + 2bx + 1 \). This expression is key in finding the extremum values through the next steps.

Extremum Values

Extremum values refer to the points at which a function reaches either a minimum or a maximum. To find these points for the function \( y = a \log |x| + bx^2 + x \), set its derivative to zero. This is because, at extremum points, the slope of the function—or its rate of change—is zero. To determine when our function's derivative equals zero, we calculate:

• \( \frac{a}{x} + 2bx + 1 = 0 \), for each specified \( x \), specifically, \( x = -1 \) and \( x = 2 \).

These points are hypothesized to be where the function changes direction from increasing to decreasing or vice versa, hence ensuring the presence of an extremum.

System of Equations

Once we have calculated the derivative and set it to zero at the extremum points \( x = -1 \) and \( x = 2 \), we end up with a system of equations. Solving this system is crucial to finding the values of \( a \) and \( b \).

• Equation 1: \( -a - 2b + 1 = 0 \) is derived from substituting \( x = -1 \) into the derivative.
• Equation 2: \( \frac{a}{2} + 4b + 1 = 0 \), modified into \( a + 8b + 2 = 0 \) after multiplying by 2, arises from substituting \( x = 2 \).

These equations are solved simultaneously. By adding them, we eliminate \( a \) and solve for \( b \), and subsequently, substitute \( b \) back to solve for \( a \). Solving systems of equations is a critical skill in calculus, often revealing values essential for the analysis of functions.

Logarithmic Differentiation

Logarithmic differentiation is a technique used for finding derivatives of functions that involve logarithms or variables raised to variable powers. In our given problem, we have a logarithmic term \( a \log |x| \). The absolute value within the log is considered to ensure the domain of the logarithm is non-negative.Here are the key points about logarithmic differentiation:

• When you have a logarithm in the form \( \log |x| \), the derivative becomes \( \frac{1}{x} \).
• A constant multiplier \( a \) in front of the log modifies the derivative to \( \frac{a}{x} \).

In this problem, logarithmic differentiation is combined with polynomial differentiation to handle the function's mixed terms effectively. Mastery of this concept allows handling complex functions with a mix of logarithmic and polynomial parts.