If $x = -1$ and $x = 2$ are extreme points of $f(x) = \alpha \log|x| + \beta x^2 + x$ then Options: (A) $\alpha = 2, \beta = -\frac{1}{2}$ (B) $\alpha = 2, \beta = \frac{1}{2}$ (C) $\alpha = -6, \beta = \frac{1}{2}$ (D) $\alpha = -6, \beta = -\frac{1}{2}$

### Analyzing the Setup We are tasked with finding the parameters $\alpha$ and $\beta$ for the function: $$f(x) = \alpha \ln|x| + \beta x^2 + x$$ The problem states that the function has extreme points at $x = -1$ and $x = 2$. At these points, the slope of the tangent line must be zero. ### The Engine of Calculus: Differentiation To find the extreme points, we calculate the derivative $f'(x)$ and set it to zero. Differentiating $f(x)$ term by term, we obtain: $$f'(x) = \frac{\alpha}{x} + 2\beta x + 1$$ Since the function reaches extreme points at $x = -1$ and $x = 2$, we must satisfy the condition $f'(x) = 0$ at these specific values. ### The Constraints: Applying the Conditions First, we substitute $x = -1$ into the derivative: $$\frac{\alpha}{-1} + 2\beta(-1) + 1 = 0$$ Simplifying this expression leads to $-\alpha - 2\beta + 1 = 0$, which gives us our first linear equation: $$\alpha + 2\beta = 1 \quad \text{--- (1)}$$ Next, we substitute $x = 2$ into the derivative: $$\frac{\alpha}{2} + 2\beta(2) + 1 = 0$$ Multiplying the entire equation by $2$ to clear the fraction, we get $\alpha + 8\beta + 2 = 0$, or: $$\alpha + 8\beta = -2 \quad \text{--- (2)}$$ ### The Resolution: Solving the System We now solve the system of equations: 1) $\alpha + 2\beta = 1$ 2) $\alpha + 8\beta = -2$ Subtracting equation (1) from equation (2) eliminates $\alpha$: $$(\alpha + 8\beta) - (\alpha + 2\beta) = -2 - 1$$ $$6\beta = -3 \quad \Rightarrow \quad \beta = -\frac{1}{2}$$ Finally, substitute $\beta = -\frac{1}{2}$ back into equation (1) to solve for $\alpha$: $$\alpha + 2\left(-\frac{1}{2}\right) = 1$$ $$\alpha - 1 = 1 \quad \Rightarrow \quad \alpha = 2$$ The values that define the topography of the function are **$\alpha = 2$** and **$\beta = -\frac{1}{2}$**.

JEE Main 2014

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: If $x = - 1$ and $x = 2$ are extreme points of $f (x) = α lo g ∣ x ∣ + β x^{2} + x$ then

Q: Is it possible for the derivative to be undefined at an extreme point?

Yes, absolutely! In some functions, like $f(x) = |x|$, the derivative is undefined at the sharp corner (the extreme point). However, in this problem, our function is differentiable at $x = -1$ and $x = 2$. Therefore, the condition $f'(x) = 0$ is the correct and necessary path to follow.

Select Answer:

Visualized Solution

Understanding Extreme Points

Given function: $f (x) = α lo g ∣ x ∣ + β x^{2} + x$
Extreme points occur at $x = - 1$ and $x = 2$
At extreme points, the first derivative $f^{'} (x) = 0$

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

Differentiating $f (x)$ with respect to $x$ :
$\frac{d}{d x} (α lo g ∣ x ∣) = \frac{α}{x}$
$\frac{d}{d x} (β x^{2}) = 2 β x$
$\frac{d}{d x} (x) = 1$
Result: $f^{'} (x) = \frac{α}{x} + 2 β x + 1$

Applying Condition at $x = - 1$

Since $x = - 1$ is an extreme point, $f^{'} (- 1) = 0$
Substitute $x = - 1$ into $f^{'} (x)$ :
$\frac{α}{- 1} + 2 β (- 1) + 1 = 0$

Simplifying First Equation

Simplify: $- α - 2 β + 1 = 0$
Rearrange: $α + 2 β = 1$ --- (Eq. 1)

Applying Condition at $x = 2$

Since $x = 2$ is an extreme point, $f^{'} (2) = 0$
Substitute $x = 2$ into $f^{'} (x)$ :
$\frac{α}{2} + 2 β (2) + 1 = 0$

Simplifying Second Equation

Simplify: $\frac{α}{2} + 4 β + 1 = 0$
Multiply by $2$ : $α + 8 β + 2 = 0$
Rearrange: $α + 8 β = - 2$ --- (Eq. 2)

Solving for $β$

Subtract (Eq. 1) from (Eq. 2):
$(α + 8 β) - (α + 2 β) = - 2 - 1$
$6 β = - 3$
$β = - \frac{3}{6} = - \frac{1}{2}$

Solving for $α$

Substitute $β = - \frac{1}{2}$ into (Eq. 1):
$α + 2 (- \frac{1}{2}) = 1$
$α - 1 = 1$
$α = 2$

Conclusion and Final Takeaway

Final values: $α = 2, β = - \frac{1}{2}$
Key Takeaway: Extreme points are found by setting $f^{'} (x) = 0$ .
Next Challenge: Try to determine if these points are local maxima or minima using the second derivative $f^{''} (x)$ .

00:00 / 00:00

The Sigma Insight: Maxima and Minima

Analyzing the Setup

We are tasked with finding the parameters

α

and

β

for the function:

f (x) = α ln ∣ x ∣ + β x^{2} + x

The problem states that the function has extreme points at

x = - 1

and

x = 2

. At these points, the slope of the tangent line must be zero.

The Engine of Calculus

Differentiation

To find the extreme points, we calculate the derivative

f^{'} (x)

and set it to zero. Differentiating

f (x)

term by term, we obtain:

f^{'} (x) = \frac{α}{x} + 2 β x + 1

Since the function reaches extreme points at

x = - 1

and

x = 2

, we must satisfy the condition

f^{'} (x) = 0

at these specific values.

The Constraints

Applying the Conditions

First, we substitute

x = - 1

into the derivative:

\frac{α}{- 1} + 2 β (- 1) + 1 = 0

Simplifying this expression leads to

- α - 2 β + 1 = 0

, which gives us our first linear equation:

α + 2 β = 1 — (1)

Next, we substitute

x = 2

into the derivative:

\frac{α}{2} + 2 β (2) + 1 = 0

Multiplying the entire equation by

2

to clear the fraction, we get

α + 8 β + 2 = 0

, or:

α + 8 β = - 2 — (2)

The Resolution

Solving the System

We now solve the system of equations:

α + 2 β = 1

α + 8 β = - 2

Subtracting equation (1) from equation (2) eliminates

α

(α + 8 β) - (α + 2 β) = - 2 - 1

6 β = - 3 \Rightarrow β = - \frac{1}{2}

Finally, substitute

β = - \frac{1}{2}

back into equation (1) to solve for

α

α + 2 (- \frac{1}{2}) = 1

α - 1 = 1 \Rightarrow α = 2

The values that define the topography of the function are $α = 2$ and $β = - \frac{1}{2}$ .

Common Traps & FAQs

Wait, why is the derivative of

lo g ∣ x ∣

just

\frac{1}{x}

? Doesn't the absolute value change things?

That is a brilliant observation! It feels like the absolute value should complicate things, right? But think about the chain rule. If

x > 0

\frac{d}{d x} (ln x) = \frac{1}{x}

. If

x < 0

, we are looking at

ln (- x)

. The derivative is

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

, which simplifies back to

\frac{1}{x}

. So, regardless of the sign of

x

, the derivative is consistently

\frac{1}{x}

Why do we set the derivative to zero? Can't we just plug in the points to the original function?

Plugging the points into the original function would only tell us the height of the curve at those points. But we don't know

α

β

yet, so that leads to a dead end. We need the defining characteristic of an extreme point: the slope of the tangent line must be zero. That is why we differentiate!

Is it possible for the derivative to be undefined at an extreme point?

Yes, absolutely! In some functions, like

f (x) = ∣ x ∣

, the derivative is undefined at the sharp corner (the extreme point). However, in this problem, our function is differentiable at

x = - 1

and

x = 2

. Therefore, the condition

f^{'} (x) = 0

is the correct and necessary path to follow.

Animated Solution for Mathematics - Differentiation: If $x = - 1$ and $x = 2$ are extreme points of $f (x) = α lo g ∣ x ∣ + β x^{2} + x$ then

Select Answer:

Visualized Solution

Understanding Extreme Points

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

Applying Condition at $x = - 1$

Simplifying First Equation

Applying Condition at $x = 2$

Simplifying Second Equation

Solving for $β$

Solving for $α$

Conclusion and Final Takeaway

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Engine of Calculus

The Constraints

The Resolution

Similar Questions

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 $b = - 1/4$ .

If $y = a ln x + b x^{2} + x$ has its extremum values at $x = - 1$ and $x = 2$ , then

Determine the points of maxima and minima of the function $f (x) = \frac{1}{8} ln x - b x + x^{2}, x > 0$ , where $b \geq 0$ is a constant.

Let $(2 α, α)$ be the largest interval in which the function $f (t) = \frac{∣ t + 1∣}{t ^{2}}, t < 0$ , is strictly decreasing. Then the local maximum value of the function $g (x) = 2 lo g_{e} (x - 2) + α x^{2} + 4 x - α, x > 2$ , is .........

Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If $lim_{x \to 0} (\frac{f ( x )}{x ^{2}} + 1) = 3$ then f(-1) is equal to :-

Let $f : R \to R$ be defined as $f (x) = ∣ x ∣ + ∣ x^{2} - 1∣$ . The total number of points at which $f$ attains either a local maximum or a local minimum is

If the functions $f (x) = \frac{x ^{3}}{3} + 2 b x + \frac{a x ^{2}}{2}$ and $g (x) = \frac{x ^{3}}{3} + a x + b x^{2}, a \neq = 2 b$ have a common extreme point, then $a + 2 b + 7$ is equal to

If the absolute maximum value of the function $f (x) = (x^{2} - 2 x + 7) e^{(4 x^{3} - 12 x^{2} - 180 x + 31)}$ in the interval $[- 3, 0]$ is $f (α)$ , then :

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=−1 and x=2 are extreme points of f(x)=αlog∣x∣+βx2+x then

Select Answer:

Visualized Solution

Understanding Extreme Points

Finding the 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

Simplifying First Equation

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

Simplifying Second Equation

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

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

Conclusion and Final Takeaway

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Engine of Calculus

The Constraints

The Resolution

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=alnx+bx2+x has its extremum values at x=−1 and x=2, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Determine the points of maxima and minima of the function f(x)=81​lnx−bx+x2,x>0, where b≥0 is a constant.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If limx→0​(x2f(x)​+1)=3 then f(-1) is equal to :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined as f(x)=∣x∣+∣x2−1∣. The total number of points at which f attains either a local maximum or a local minimum is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the functions f(x)=3x3​+2bx+2ax2​ and g(x)=3x3​+ax+bx2,a=2b have a common extreme point, then a+2b+7 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the absolute maximum value of the function f(x)=(x2−2x+7)e(4x3−12x2−180x+31) in the interval [−3,0] is f(α), then :

Animated Solution for Mathematics - Differentiation: If $x = - 1$ and $x = 2$ are extreme points of $f (x) = α lo g ∣ x ∣ + β x^{2} + x$ then

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

Applying Condition at $x = - 1$

Applying Condition at $x = 2$

Solving for $β$

Solving for $α$

If $y = a ln x + b x^{2} + x$ has its extremum values at $x = - 1$ and $x = 2$ , then

Determine the points of maxima and minima of the function $f (x) = \frac{1}{8} ln x - b x + x^{2}, x > 0$ , where $b \geq 0$ is a constant.

Let f(x) be a polynomial of degree 4 having extreme values at x = 1 and x = 2. If $lim_{x \to 0} (\frac{f ( x )}{x ^{2}} + 1) = 3$ then f(-1) is equal to :-

Let $f : R \to R$ be defined as $f (x) = ∣ x ∣ + ∣ x^{2} - 1∣$ . The total number of points at which $f$ attains either a local maximum or a local minimum is

If the functions $f (x) = \frac{x ^{3}}{3} + 2 b x + \frac{a x ^{2}}{2}$ and $g (x) = \frac{x ^{3}}{3} + a x + b x^{2}, a \neq = 2 b$ have a common extreme point, then $a + 2 b + 7$ is equal to

If the absolute maximum value of the function $f (x) = (x^{2} - 2 x + 7) e^{(4 x^{3} - 12 x^{2} - 180 x + 31)}$ in the interval $[- 3, 0]$ is $f (α)$ , then :