Problem 138 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)\) be a forth degree polynomial having extreme values at \(x=1\) and \(x=2\). If \(\lim _{x \rightarrow 0}\left[1+\frac{f(x)}{x^{2}}\right]=3\), then \(f(2)\) is equal to (A) \(-4\) (B) 0 (C) 4 (D) \(-8\)

Step-by-Step Solution

Answer

The value of \(f(2)\) is 0.

Step 1: Understanding the Polynomial's Characteristics

Given the polynomial function \(f(x)\) is of degree 4 and has extreme values at \(x = 1\) and \(x = 2\). This implies that \(f'(x)\), the derivative of the polynomial, has roots at \(x = 1\) and \(x = 2\). Thus, the derivative can be expressed as \(f'(x) = (x-1)(x-2)g(x)\) where \(g(x)\) is a polynomial of degree 2, because the degree of \(f(x)\) is 4.

Step 2: Using Given Limit Equation

We are given \(\lim_{x \to 0}\left[1 + \frac{f(x)}{x^2}\right] = 3\). Simplifying, this leads to \(\lim_{x \to 0} \frac{f(x)}{x^2} = 2\), hence \(f(x) = 2x^2 + ax + b\) near zero, because the lowest degree of \(x^2\) in the polynomial is consistent with the given limit condition.

Step 3: Expand the Fourth Degree Polynomial

Consider the zeroes and derivative conditions, and express \(f(x) = a(x-1)^2(x-2)^2 + bx^2\). This satisfies the polynomial being fourth degree, and accommodates the critical points \(x=1\) and \(x=2\). Expanding the given \((x-1)^2(x-2)^2\) would yield a fourth degree polynomial.

Step 4: Match Coefficients Using Limit

We set \(x=0\) in the function obtained in the expanded expression to find the constant term using the limit condition \(f(0) = 2(0)^2 + a(0) + b = 0\), therefore \(b=0\). Evaluate matching coefficients for terms to match \(f(x)\) with \(2x^2 + ax\) and solve for relevant coefficients.

Step 5: Evaluate \(f(2)\)

Finally, substitute \(x = 2\) into \(f(x) = a(x-1)^2(x-2)^2\) to simplify since it becomes \(f(2) = 0\), because \(f(x)\) achieves an extreme value (potential local min or max) at \(x=2\).

Key Concepts

Extreme Values

Extreme values in polynomial functions are important points where the function either reaches a maximum or a minimum. In the context of a fourth-degree polynomial:

• An extreme value means the slope of the function at those points is zero.
• This occurs when the first derivative, denoted as \( f'(x) \), equals zero at that point.

For the polynomial discussed, it has extreme values at \( x=1 \) and \( x=2 \). This tells us that these values make the derivative of the function zero. Therefore, the points \( x = 1 \) and \( x = 2 \) are critical points where the polynomial potentially reaches local maximum or minimum values.
These characteristics are helpful for solving mathematical problems and understanding the behavior of graphs.Checking for extreme values involves calculating the derivative and finding points where it equals zero. In our problem, these points are known and coincide with the roots we derive from the derivative equation. E.g., the derivative can be expressed as \( f'(x) = (x-1)(x-2)g(x) \), considering \( g(x) \) is a polynomial of degree 2. This is crucial in analyzing and plotting polynomial functions.

Derivative of Polynomial

The derivative of a polynomial gives us information about the slope of the tangent at any point on the curve defined by the function. For a fourth-degree polynomial function:

• The first derivative \( f'(x) \) provides insights into where the slope of the curve is zero, i.e., critical points or extreme values.
• It simplifies the process of finding maximums and minimums of the function.

In our exercise, the polynomial \( f(x) \) has roots in its derivative at \( x=1 \) and \( x=2 \) because these are where extreme values occur. The expression for the derivative was derived as \( f'(x) = (x-1)(x-2)g(x) \). Here, \( g(x) \) is another polynomial helping complete the function to a specific degree, confirming its original degree of four.
The approach taken in solving such exercises helps one understand various characteristics of polynomial graphs, the nature of their turning points, and how these points are integral to graph analysis.

Limit of a Function

In calculus, the concept of a limit examines the behavior of a function as it approaches a certain point. Limits are foundational in understanding how functions behave when inputs tend towards specific values.
For polynomial functions:

• Limits can be used to predict the behavior of functions at points that are not directly evaluable.
• They offer insights into the end behavior or asymptotic tendencies of polynomials.

In this problem, the limit provided, \( \lim_{x \rightarrow 0} \left[ 1 + \frac{f(x)}{x^2} \right] = 3 \), suggests the behavior as \( x \) approaches zero. Breaking it down, this limit leads to \( \lim_{x \rightarrow 0} \frac{f(x)}{x^2} = 2 \).
This helps establish an expression for \( f(x) \), namely that near zero, \( f(x) = 2x^2 + ax + b \). Calculating such limits allow for assumptions regarding coefficients and functional forms in problem-solving, supporting the full definition of the function \( f(x) \).
Thus, understanding limits is crucial for exploring continuity and behavior, particularly in polynomial contexts where graphs may not otherwise convey clear information.