Problem 86 - A Complete Resource Book in Mathematics for JEE Main 2017

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

Option (A): \( P(x) = 2x^2 - x - 6 \).

Step 1: Analyze the Conditions for Continuity

The function \( f(x) \) is continuous at \( x = 2 \), which means that \( \lim_{x \to 2} f(x) = f(2) = 7 \). We have \( f(x) = \frac{P(x)}{x-2} \) for \( x eq 2 \). To satisfy this condition, \( P(x) \) is divisible by \( x-2 \). This implies \( P(x) = (x-2)Q(x) \) for some polynomial \( Q(x) \).

Step 2: Determine the Degree of Polynomial P(x)

Since \( P'''(x) = 0 \), \( P(x) \) must be a polynomial of degree at most 2. Therefore, we can assume \( P(x) = ax^2 + bx + c \).

Step 3: Express P(x) in Factored Form

Given that \( P(x) = (x-2)Q(x) \), for \( P(x) \) to be a quadratic, \( Q(x) \) must be a linear polynomial, say \( Q(x) = ax + b \). Hence, \( P(x) = a(x-2)x + b(x-2) = ax^2 + bx - 2ax - 2b = ax^2 + (b-2a)x - 2b \).

Step 4: Use the Condition P(3)=9

Substitute \( x = 3 \) into \( P(x) = ax^2 + (b-2a)x - 2b \) and set it equal to 9: \( 9a + (b-2a)3 - 2b = 9 \). Simplifying this equation gives \( 3a + b = 9 \).

Step 5: Use Continuity at x=2 for Further Equations

Since \( \lim_{x \to 2} \frac{P(x)}{x-2} = Q(2) \), we must have \( Q(2) = 7 \). Thus, \( 2a + b = 7 \).

Step 6: Solve the System of Equations

We have two equations: \( 3a + b = 9 \) and \( 2a + b = 7 \). Subtract the second equation from the first: \( (3a + b) - (2a + b) = 9 - 7 \), which gives \( a = 2 \). Substitute \( a = 2 \) into \( 2a + b = 7 \): \( 4 + b = 7 \), so \( b = 3 \).

Step 7: Form the Polynomial P(x)

With \( a = 2 \) and \( b = 3 \), substitute into the expression \( ax^2 + (b-2a)x - 2b \) to find \( P(x) = 2x^2 - 1x - 6 \).

Step 8: Verify Which Option Matches

The polynomial obtained is \( P(x) = 2x^2 - x - 6 \). Check which option this corresponds to, which is option (A).

Key Concepts

Polynomials

Polynomials are mathematical expressions that consist of variables, coefficients, and exponents. They are the building blocks in algebra and calculus, making them crucial in mathematical studies. A polynomial in one variable, say \( x \), can be expressed as:

• \( P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \)

Here, \( a_n, a_{n-1}, \ldots, a_1, a_0 \) are constants, and \( n \), which must be a non-negative integer, represents the degree of the polynomial. Polynomials are powerful due to their properties and operations, such as addition, subtraction, multiplication, and division. They are continuous and smooth, which means they can be graphed without lifting the pencil from the paper. Understanding the behavior of these expressions helps in solving more complex mathematical problems.

Limits

Limits are fundamental to calculus and analysis, dealing with the behavior of a function as it comes arbitrarily close to a specific point, but not necessarily at that point. The limit of a function \( f(x) \) as \( x \) approaches a point \( c \) is represented by \( \lim_{x \to c} f(x) \). This concept helps to understand how functions behave near specific values, even when they are not defined at those values. For example, to determine the continuity of the function \( f(x) \), we need it to satisfy three conditions at \( x = c \):

• The function \( f(x) \) is defined at \( x = c \).
• \( \lim_{x \to c} f(x) \) exists.
• \( \lim_{x \to c} f(x) = f(c) \).

In scenarios like this exercise, the continuity of functions at certain points is verified through finding limits, as we solved for \( \lim_{x \to 2} f(x) = 7 \) to confirm the polynomial's behavior.

Calculus

Calculus is a broad field of mathematics, which, at its core, deals with change and motion. It primarily involves two major branches: Differential Calculus and Integral Calculus. Each branch has its focus:

• Differential Calculus: It is concerned with the concept of the derivative, which measures how a function changes as its input changes. Differentiation allows us to find rates of change, such as velocities and slopes of curves. In the exercise, it helps determine degrees through derivatives.
• Integral Calculus: This branch focuses on accumulation and calculates total sizes, like areas and volumes, using integration. Although not directly applied here, it's crucial in understanding functions over intervals.

Calculus provides the language and tools necessary to model and solve real-world situations accurately. Every time we compute a derivative or evaluate an integral, we engage with calculus, helping us predict and understand the behavior of complex systems.

Degree of Polynomial

The degree of a polynomial is the highest power of the variable in the polynomial expression. It gives an immediate sense of the polynomial's general form and behavior. For instance, a polynomial \( P(x) = ax^n + \ldots + a_0 \) would have a degree of \( n \) if \( a_n eq 0 \). The degree provides critical insights into the following:

• The number of possible roots (real and complex).
• The general growth rate or steepness of the polynomial graph.
• The highest number of times the graph of the polynomial can change direction.

In this exercise, knowing that \( P'''(x) = 0 \) tells us \( P(x) \) must be of degree 2, as higher derivatives would not exist. Recognizing this helps us narrow down the possibilities and understand the roots and behavior of the polynomial more completely.