If y = a ln x + bx² + x has its extremum values at x = -1 and x = 2, then (A) a = 2, b = -1 (B) a =

Q: Wait, $\ln(-1)$ is undefined! Why are we using $x = -1$ in the derivative?

You have a sharp eye! In pure mathematics, the domain of $\ln x$ is $x > 0$. However, in the context of competitive exams like JEE, we often treat the derivative as a formal algebraic relation. We are looking for the values of $a$ and $b$ that satisfy the *algebraic condition* of the derivative being zero, even if the function itself behaves strangely at that point.

Q: Why can't we just use energy conservation or some other physics principle?

This is a pure calculus problem, not a physics problem! While energy conservation is a powerful tool in mechanics, here we are dealing with the geometry of a curve. The 'energy' of this problem lies in the slope of the tangent, which is governed by the derivative.

Q: How do I know if these are local maxima or minima?

Great question! To determine the nature of the extremum, you would typically use the second derivative test. However, the problem only asks for the constants $a$ and $b$. Once you have the equations, you don't need to classify the extrema to find the constants.

Question

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

Options:

(A) $a = 2, b = -1$

(B) $a = 2, b = -\frac{1}{2}$

(C) $a = -2, b = \frac{1}{2}$

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a landscape defined by the function $y = a \ln x + bx^2 + x$. You are looking for the peaks and valleys—the points where the ground levels off, and the slope of the tangent line becomes perfectly horizontal.

In the language of calculus, these are our local extrema. The problem asks us to find the constants $a$ and $b$ that create these specific features at $x = -1$ and $x = 2$.

### The Derivative: Our Mathematical Compass

To find where the landscape levels off, we need the derivative. The derivative, $\frac{dy}{dx}$, tells us the slope at any point $x$. When we are at an extremum, the slope is zero.

Our first step is to differentiate the function:
$$\frac{dy}{dx} = \frac{d}{dx}(a \ln x + bx^2 + x)$$

Applying the power rule and the derivative of the natural log, we get:
$$\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$$

This expression is our compass. It tells us the slope at any point $x$.

### The Domain Paradox

Now, here is where the plot thickens. We are told an extremum exists at $x = -1$. But wait—the natural log function $\ln x$ is only defined for $x > 0$.

Does this mean the problem is impossible? Not at all! In the world of competitive exams, we treat the derivative as a formal algebraic relation.

We are looking for the values of $a$ and $b$ that satisfy the condition $\frac{dy}{dx} = 0$ at these points, regardless of the domain constraints of the original function.

### Solving the System

With our derivative $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$, we can now set up our system of equations.

At $x = 2$, we have:
$$\frac{a}{2} + 2b(2) + 1 = 0 \Rightarrow \frac{a}{2} + 4b = -1 \Rightarrow a + 8b = -2$$

At $x = -1$, we have:
$$\frac{a}{-1} + 2b(-1) + 1 = 0 \Rightarrow -a - 2b = -1 \Rightarrow a + 2b = 1$$

We now have a system of two linear equations:
1) $a + 8b = -2$
2) $a + 2b = 1$

### Final Calculation

Subtracting Equation (2) from Equation (1) gives us:
$$6b = -3 \Rightarrow b = -\frac{1}{2}$$

Substituting this back into Equation (2), we find:
$$a + 2\left(-\frac{1}{2}\right) = 1 \Rightarrow a - 1 = 1 \Rightarrow a = 2$$

We have arrived at our destination: **$a = 2$** and **$b = -\frac{1}{2}$**. You have successfully navigated the domain paradox and used the power of calculus to define the curve.

Animated Solution for Mathematics - Differentiation: If $y = a ln x + b x^{2} + x$ has its extremum values at $x = - 1$ and $x = 2$ , then

Select Answer:

Visualized Solution

Visual Anchor and Problem Setup

The Derivative Tool

Differentiating the Function

Applying Condition at $x = 2$

Simplifying Equation 1

The Domain Paradox & Condition at $x = - 1$

Simplifying Equation 2

Solving the System: Elimination

Finding the Value of $b$

Finding the Value of $a$

Final Conclusion

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Derivative

The Domain Paradox

Solving the System

Final Calculation

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 $x = - 1$ and $x = 2$ are extreme points of $f (x) = α lo g ∣ x ∣ + β x^{2} + x$ 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 $f (x) = {x^{3} - x^{2} + 10 x - 7, - 2 x + lo g_{2} (b^{2} - 4), x \leq 1 x > 1$ . Then the set of all values of $b$ , for which $f (x)$ has maximum value at $x = 1$ , 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

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 $f (x) = ⎩ ⎨ ⎧ e^{x}, 2 - e^{x - 1}, x - e, 0 \leq x \leq 1 1 < x \leq 2 2 < x \leq 3$ and $g (x) = \int_{0}^{x} f (t) d t, x \in [1, 3]$ then $g (x)$ has

The function $f (x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at

Animated Solution for Mathematics - Differentiation: .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