Problem 86 - A Complete Resource Book in Mathematics for JEE Main

Question

Let the function \(f\) be defined as \(f(x)=\left\\{\begin{aligned} \frac{P(x)}{x-2}, & x \neq 2 \\ 7, & x=2 \end{aligned}\right.\) where \(P(x)\) is a polynomial such that \(P^{\prime \prime \prime}(x)\) is identically equal to 0 and \(P(3)=9 .\) If \(f(x)\) is continuous at \(x=2\), then (A) \(P(x)=2 x^{2}-x-6\) (B) \(P(x)=2 x^{2}+x-6\) (C) \(P(x)=2 x^{2}-x+6\) (D) None of these

Step-by-Step Solution

Answer

The polynomial is (A) \( P(x)=2x^2-x-6 \).

Step 1: Understand the Properties of P(x)

The function \( P(x) \) is a polynomial whose third derivative is identically zero, i.e., \( P'''(x) = 0 \). This implies \( P(x) \) must be a polynomial of degree at most 2. Therefore, we can express \( P(x) \) as \( P(x) = ax^2 + bx + c \).

Step 2: Apply Continuity Condition

The function \( f(x) \) must be continuous at \( x = 2 \). This means the limit \( \lim_{x \to 2} f(x) = f(2) \). Therefore, we find \( \lim_{x \to 2} \frac{P(x)}{x-2} = 7 \).

Step 3: Use L'Hospital's Rule for the Limit

To evaluate \( \lim_{x \to 2} \frac{P(x)}{x-2} \), substitute \( P(x) = ax^2 + bx + c \) and apply L'Hospital's rule if direct substitution results in an indeterminate form:\[P(2) = a(2)^2 + b(2) + c\]. The limit becomes 0/0, so apply L'Hospital's:\[ \lim_{x \to 2} \frac{P(x)}{x-2} = \lim_{x \to 2} \frac{P'(x)}{1} = 7 \].

Step 4: Differentiate P(x) and Apply Limit

Differentiate \( P(x) \) to get \( P'(x) = 2ax + b \). Setting \( P'(2) = 7 \), we solve for constants:\[ 2a(2) + b = 7 \Rightarrow 4a + b = 7 \].

Step 5: Use Given Condition P(3) = 9

Substitute \( x = 3 \) into \( P(x) = ax^2 + bx + c \), with known values. Thus,\[ 9 = a(3)^2 + b(3) + c \Rightarrow 9a + 3b + c = 9 \].

Step 6: Solve the System of Equations

Now solve the system of equations:1. \( 4a + b = 7 \)2. \( 9a + 3b + c = 9 \).Solve for \( a \), \( b \), and \( c \) based on these constraints.

Step 7: Verify the Polynomial

From equations, express \( b \) in terms of \( a \), and substitute these into \( 9a + 3b + c = 9 \). Solve the equations:1. \( b = 7 - 4a \)2. \( 9 + 3(7 - 4a) + c = 9 \)After solving, verify solution by substitution in the equation for \( f(x) \).

Key Concepts

Polynomials

Polynomials are specific types of mathematical expressions that consist of variables, coefficients, and exponents combined using addition, subtraction, and multiplication. For example, a polynomial in variable \(x\) is typically written as \(P(x) = ax^2 + bx + c\). Here, \(a\), \(b\), and \(c\) are coefficients, and \(x\) is the variable. Polynomials are essential in calculus and other fields of mathematics due to their well-defined behaviors and characteristics:

• **Degree:** The highest power of the variable in a polynomial, which in turn determines its general shape and the number of roots it can have.
• **Derivative:** The polynomial's rate of change, which is also a polynomial of lower degree.

In this exercise, we know that the third derivative of \(P(x)\) is zero, signifying that it's a quadratic polynomial. A quadratic polynomial simplifies many calculus problems, especially those involving continuity and limits.

L'Hôpital's Rule

L'Hôpital's rule is a fundamental tool in calculus used to find limits that yield indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When a direct substitution results in an indeterminate form, L'Hôpital's rule provides a structured approach to resolving it by differentiating the numerator and denominator separately.To apply L'Hôpital's rule:

• Ensure the original limit gives a \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form.
• Differentiate the numerator and denominator.
• Evaluate the limit of the resulting fraction.

In this problem, the function \(f(x)\) is described using \(P(x)\) in the denominator, yielding a \(\frac{0}{0}\) form as \(x\) approaches 2. By applying L'Hôpital's rule, directly differentiating provides a straightforward path to determining the function’s behavior at points of discontinuity and finding the limit at \(x = 2\). This is essential to guaranteeing the function’s continuity.

Limits in Calculus

Limits are a foundational concept in calculus, representing the value a function approaches as the input approaches a particular point. Limits are critical for defining derivatives, integrals, and ensuring continuity of functions.There are several key points to understand about limits:

• **Existence:** A limit exists if the function approaches a specific value from both the left and the right side as the input nears the point of interest.
• **Continuity:** A function is continuous at a point if the limit exists and equals the function's value at that point. For \(f(x)\) to be continuous at \(x=2\), \(\lim_{x \to 2} f(x) = f(2)\).
• **Indeterminate Forms:** Limits that remove indeterminacy, such as through L'Hôpital's rule, help confirm continuous function behavior.

In this exercise, examining the limit \(\lim_{x \to 2}\frac{P(x)}{x-2} = 7\) ensures the function \(f(x)\) behaves predictably around \(x = 2\). This continuity requirement leads us to find suitable coefficients for \(P(x)\). Mastering limits helps in understanding both the immediate behavior and broader stability of functions at crucial points.