Let $a, b \in \mathbb{R}$ be such that the function $f$ given by $f(x) = \ln |x| + bx^2 + ax, x \neq 0$ has extreme values at $x = -1$ and $x = 2$. \\ \textbf{Statement-1:} $f$ has local maximum at $x = -1$ and at $x = 2$. \\ \textbf{Statement-2:} $a = 1/2$ and $b = -1/4$. Options: (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.

### Analyzing the Setup We are given the function $f(x) = \ln |x| + bx^2 + ax$, where $x \neq 0$. This function combines logarithmic and quadratic behaviors, creating a unique landscape of extrema. Our goal is to determine the constants $a$ and $b$ and verify the nature of the extrema at $x = -1$ and $x = 2$. ### The Hunt for Critical Points In calculus, an extreme value (a peak or a valley) occurs where the tangent line is horizontal. This implies that the first derivative, $f'(x)$, must be zero at these points. We differentiate the function term by term: $$f'(x) = \frac{1}{x} + 2bx + a$$ This expression represents the slope of the curve at any valid point $x$. Since we know extrema occur at $x = -1$ and $x = 2$, we set $f'(x) = 0$ at these specific locations. ### The System of Equations At $x = -1$, the condition $f'(-1) = 0$ yields: $$\frac{1}{-1} + 2b(-1) + a = 0 \implies a - 2b = 1$$ At $x = 2$, the condition $f'(2) = 0$ yields: $$\frac{1}{2} + 2b(2) + a = 0 \implies a + 4b = -\frac{1}{2}$$ We now solve this system of two linear equations. Subtracting the first equation from the second eliminates $a$: $$(a + 4b) - (a - 2b) = -\frac{1}{2} - 1$$ $$6b = -\frac{3}{2} \implies b = -\frac{1}{4}$$ Substituting $b = -\frac{1}{4}$ back into the first equation: $$a - 2\left(-\frac{1}{4}\right) = 1 \implies a + \frac{1}{2} = 1 \implies a = \frac{1}{2}$$ ### The Second Derivative Test To confirm the nature of these extrema, we use the **Second Derivative Test**. We differentiate $f'(x)$ to find $f''(x)$: $$f''(x) = -\frac{1}{x^2} + 2b$$ Substituting our value $b = -\frac{1}{4}$ into the expression: $$f''(x) = -\frac{1}{x^2} - \frac{1}{2}$$ Since $x^2 > 0$ for all $x \neq 0$, the term $-\frac{1}{x^2}$ is always negative. Consequently, $f''(x) < 0$ for all $x$ in the domain. Because the second derivative is strictly negative, the function is concave down everywhere. This confirms that the points at $x = -1$ and $x = 2$ are indeed **local maxima**.

JEE Main 2012

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $a, b \in R$ be such that the function $f$ given by $f (x) = ln ∣ x ∣ + b x^{2} + a x, x \neq = 0$ has extreme values at $x = - 1$ and $x = 2$ . \\ \textbf{Statement-1:} $f$ has local maximum at $x = - 1$ and at $x = 2$ . \\ \textbf{Statement-2:} $a = 1/2$ and $b = - 1/4$ .

Q: Wait, why is the derivative of $\ln|x|$ just $1/x$? Shouldn't the absolute value change something?

That is a fantastic question! It often trips students up. Remember that for $x > 0$, $\ln|x| = \ln(x)$, and its derivative is $1/x$. For $x < 0$, $\ln|x| = \ln(-x)$. By the chain rule, the derivative is $\frac{1}{-x} \cdot (-1)$, which simplifies back to $1/x$. So, in both domains, the derivative is elegantly $1/x$! It is a beautiful consistency in calculus.

Q: Do I really need the second derivative test? Can't I just look at the function and guess?

Guessing is the enemy of precision in JEE Advanced! The second derivative test is your mathematical proof. It tells us the concavity of the curve. Without it, you are just hoping; with it, you are certain. In this problem, the second derivative test provides the rigorous proof required to confirm that both points are indeed local maxima.

Q: Why do we reject the case where the derivative is undefined at $x=0$?

The problem statement explicitly defines the domain as $x \neq 0$. Furthermore, the function $\ln|x|$ has a vertical asymptote at $x=0$. Since the function itself is undefined at $x=0$, it cannot have an extreme value there. We only care about the critical points where the derivative exists and equals zero.

Select Answer:

Visualized Solution

Understanding the Function $f (x)$

Given function: $f (x) = ln ∣ x ∣ + b x^{2} + a x, x \neq = 0$
Extreme values occur at $x = - 1$ and $x = 2$

Condition for Extreme Values

At extreme values, the tangent is horizontal.
Therefore, the first derivative $f^{'} (x)$ must be zero.

Finding the First Derivative $f^{'} (x)$

Differentiating $f (x)$ with respect to $x$ :
$\frac{d}{d x} (ln ∣ x ∣) = \frac{1}{x}$
$\frac{d}{d x} (b x^{2}) = 2 b x$
$\frac{d}{d x} (a x) = a$
Result: $f^{'} (x) = \frac{1}{x} + 2 b x + a$

Applying Condition at $x = - 1$

Condition 1: $f^{'} (- 1) = 0$
Substitute $x = - 1$ : $\frac{1}{- 1} + 2 b (- 1) + a = 0$
Equation 1: $a - 2 b = 1$

Applying Condition at $x = 2$

Condition 2: $f^{'} (2) = 0$
Substitute $x = 2$ : $\frac{1}{2} + 2 b (2) + a = 0$
Equation 2: $a + 4 b = - \frac{1}{2}$

Solving for Constants $a$ and $b$

Subtracting Eq 1 from Eq 2: $(a + 4 b) - (a - 2 b) = - \frac{1}{2} - 1$
$6 b = - \frac{3}{2} \Rightarrow b = - \frac{1}{4}$
Substituting $b$ in Eq 1: $a - 2 (- \frac{1}{4}) = 1$
$a + \frac{1}{2} = 1 \Rightarrow a = \frac{1}{2}$
Statement-2 is true.

The Second Derivative Test

To check if extrema are maxima or minima, we use the second derivative $f^{''} (x)$ .
If $f^{''} (x) < 0$ , the point is a local maximum.
If $f^{''} (x) > 0$ , the point is a local minimum.

Calculating the Second Derivative $f^{''} (x)$

Differentiate $f^{'} (x) = \frac{1}{x} + 2 b x + a$
$f^{''} (x) = \frac{d}{d x} (\frac{1}{x}) + \frac{d}{d x} (2 b x) + \frac{d}{d x} (a)$
$f^{''} (x) = - \frac{1}{x ^{2}} + 2 b$

Analyzing the Sign of $f^{''} (x)$

Substitute $b = - \frac{1}{4}$ : $f^{''} (x) = - \frac{1}{x ^{2}} + 2 (- \frac{1}{4})$
$f^{''} (x) = - \frac{1}{x ^{2}} - \frac{1}{2}$
Since $x^{2} > 0$ for all $x \neq = 0$ , $\frac{1}{x ^{2}}$ is always positive.
Therefore, $- \frac{1}{x ^{2}} - \frac{1}{2}$ is strictly negative for all $x \neq = 0$ .

Conclusion on Local Maxima

Since $f^{''} (x) < 0$ for all valid $x$ , it is negative at $x = - 1$ and $x = 2$ .
Thus, $f (x)$ has local maxima at both $x = - 1$ and $x = 2$ .
Statement-1 is true.
Statement-2 provides the exact values of $a$ and $b$ that make Statement-1 true.
Final Answer: Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

00:00 / 00:00

The Sigma Insight: Maxima and Minima

Solution Diagram

Analyzing the Setup

We are given the function

f (x) = ln ∣ x ∣ + b x^{2} + a x

, where $x eq 0$. This function combines logarithmic and quadratic behaviors, creating a unique landscape of extrema.

Our goal is to determine the constants

a

and

b

and verify the nature of the extrema at

x = - 1

and

x = 2

The Hunt for Critical Points

In calculus, an extreme value (a peak or a valley) occurs where the tangent line is horizontal. This implies that the first derivative,

f^{'} (x)

, must be zero at these points.

We differentiate the function term by term:

f^{'} (x) = \frac{1}{x} + 2 b x + a

This expression represents the slope of the curve at any valid point

x

. Since we know extrema occur at

x = - 1

and

x = 2

, we set

f^{'} (x) = 0

at these specific locations.

The System of Equations

x = - 1

, the condition

f^{'} (- 1) = 0

yields:

\frac{1}{- 1} + 2 b (- 1) + a = 0 ⟹ a - 2 b = 1

x = 2

, the condition

f^{'} (2) = 0

yields:

\frac{1}{2} + 2 b (2) + a = 0 ⟹ a + 4 b = - \frac{1}{2}

We now solve this system of two linear equations. Subtracting the first equation from the second eliminates

a

(a + 4 b) - (a - 2 b) = - \frac{1}{2} - 1

6 b = - \frac{3}{2} ⟹ b = - \frac{1}{4}

Substituting

b = - \frac{1}{4}

back into the first equation:

a - 2 (- \frac{1}{4}) = 1 ⟹ a + \frac{1}{2} = 1 ⟹ a = \frac{1}{2}

The Second Derivative Test

To confirm the nature of these extrema, we use the Second Derivative Test. We differentiate

f^{'} (x)

to find

f^{''} (x)

f^{''} (x) = - \frac{1}{x ^{2}} + 2 b

Substituting our value

b = - \frac{1}{4}

into the expression:

f^{''} (x) = - \frac{1}{x ^{2}} - \frac{1}{2}

Since

x^{2} > 0

for all $x eq 0$, the term

- \frac{1}{x ^{2}}

is always negative. Consequently,

f^{''} (x) < 0

for all

x

in the domain.

Because the second derivative is strictly negative, the function is concave down everywhere. This confirms that the points at

x = - 1

and

x = 2

are indeed local maxima.

Common Traps & FAQs

Wait, why is the derivative of

ln ∣ x ∣

just

1/ x

? Shouldn't the absolute value change something?

That is a fantastic question! It often trips students up. Remember that for

x > 0

ln ∣ x ∣ = ln (x)

, and its derivative is

1/ x

. For

x < 0

ln ∣ x ∣ = ln (- x)

. By the chain rule, the derivative is

\frac{1}{- x} \cdot (- 1)

, which simplifies back to

1/ x

. So, in both domains, the derivative is elegantly

1/ x

! It is a beautiful consistency in calculus.

Do I really need the second derivative test? Can't I just look at the function and guess?

Guessing is the enemy of precision in JEE Advanced! The second derivative test is your mathematical proof. It tells us the concavity of the curve. Without it, you are just hoping; with it, you are certain. In this problem, the second derivative test provides the rigorous proof required to confirm that both points are indeed local maxima.

Why do we reject the case where the derivative is undefined at

x = 0

The problem statement explicitly defines the domain as

x \neq = 0

. Furthermore, the function

ln ∣ x ∣

has a vertical asymptote at

x = 0

. Since the function itself is undefined at

x = 0

, it cannot have an extreme value there. We only care about the critical points where the derivative exists and equals zero.

Select Answer:

Visualized Solution

Understanding the Function $f (x)$

Condition for Extreme Values

Finding the First Derivative $f^{'} (x)$

Applying Condition at $x = - 1$

Applying Condition at $x = 2$

Solving for Constants $a$ and $b$

The Second Derivative Test

Calculating the Second Derivative $f^{''} (x)$

Analyzing the Sign of $f^{''} (x)$

Conclusion on Local Maxima

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Hunt for Critical Points

The System of Equations

The Second Derivative Test

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Select Answer:

Visualized Solution

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Condition for Extreme Values

Finding the First Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Applying Condition at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1

Applying Condition at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2

Solving for Constants .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The Second Derivative Test

Calculating the Second Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′(x)

Analyzing the Sign of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′′(x)

Conclusion on Local Maxima

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Hunt for Critical Points

The System of Equations

The Second Derivative Test

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