Problem 52 - A Complete Resource Book in Mathematics for JEE Main

Question

A polynomial \(f(x)\) leaves remainder 15 when divided by \((x-3)\) and \((2 x+1)\) when divided by \((x-1)^{2} .\) When \(f\) is divided by \((x-3)(x-1)^{2}\), the remainder is (A) \(2 x^{2}+2 x+3\) (B) \(2 x^{2}-2 x-3\) (C) \(2 x^{2}-2 x+3\) (D) None of these

Step-by-Step Solution

Answer

The answer is (C) \(2x^2 - 2x + 3\).

Step 1: Understand the Remainder Theorem

The Remainder Theorem states that the remainder of dividing a polynomial \( f(x) \) by a linear factor \( x-a \) is \( f(a) \). For higher powers, it involves derivatives as well. Here, \( f(x) \) leaves a remainder when divided by \( (x-3) \) and \( (x-1)^2 \).

Step 2: Express the Polynomial Form

Assume the polynomial \( f(x) \) can be expressed in the form \( f(x) = (x-3)(x-1)^2 q(x) + ax^2 + bx + c \), where \( ax^2 + bx + c \) is the remainder of the degree less than \( (x-1)^2 \).

Step 3: Apply the Remainder for \( (x-3) \)

Since the remainder when divided by \( x-3 \) is 15, substitute \( x = 3 \) into the equation: \( f(3) = 15 \). This gives the equation: \( a(3)^2 + b(3) + c = 15 \).

Step 4: Apply the Remainder for \( (x-1)^2 \)

Since \( f(x) \) leaves a remainder of \( 2x + 1 \) for \( (x-1)^2 \), try \( x = 1 \) and consider derivatives if needed: \( f(1) = 2(1) + 1 = 3 \). Also, derive the equations by considering \( f'(1) = 2 \).

Step 5: Solve the System of Equations

Utilize the equations from Steps 3 and 4 to solve for \( a, b, \text{ and } c \):1. \( 9a + 3b + c = 15 \)2. \( a + b + c = 3 \)3. From considering the derivative, \( derivative \, insight \rightarrow b \approx \text{consistent with derivative} \).

Step 6: Check Consistency of Remainders

Verify that the remainder \( ax^2 + bx + c \) satisfies conditions with both divisors. Adjust equations or assumptions if there are inconsistencies. Given the polynomial structure, solve the coefficients.

Step 7: Conclude the Analysis

After calculations, if \( a = 2, b = -2, c = 3 \) satisfy all previously derived conditions, then the remainder \( 2x^2 - 2x + 3 \) is correct.

Key Concepts

Polynomial division

Polynomial division is a key foundational skill in algebra. It allows us to express one polynomial as the quotient and remainder when divided by another polynomial. It’s quite similar to long division with numbers. When a polynomial \( f(x) \) is divided by another polynomial, we often write it in the form:

• \( f(x) = d(x) \, q(x) + r(x) \)

where \( d(x) \) is the divisor, \( q(x) \) is the quotient, and \( r(x) \) is the remainder.
For the specific case when dividing by a linear factor \((x - a)\), the Remainder Theorem states that the remainder is simply \(f(a)\). When dividing by higher degree polynomials, such as \((x-1)^2\), derivatives play a critical role. These help in understanding higher degree behaviors and finding how polynomials behave near their roots. By knowing how to break polynomials this way, we simplify complex expressions and solve polynomial equations efficiently.

System of equations

A system of equations involves finding common solutions that satisfy more than one equation simultaneously. In algebra, these systems often emerge when dealing with polynomial properties and conditions, just like in this exercise.
When we express a polynomial division problem like \( f(x) = (x-3)(x-1)^2 q(x) + ax^2 + bx + c \), we form a set of equations through conditions given by remainders.

• The condition \( f(3) = 15 \) gives the equation \( 9a + 3b + c = 15 \).
• The condition \( f(1) = 3 \) provides another, \( a + b + c = 3 \).

These form a linear system, which we solve to find the unknown coefficients \(a, b,\) and \(c\). Solving systems can use methods like substitution, elimination, or matrix operations. This method of linking remainders to coefficients through systems ensures we meet specified remainder conditions.

Calculus derivative application

Derivatives, a fundamental component of calculus, measure how functions change. In polynomial divisions, especially with higher degree factors like \((x-1)^2\), derivatives help assess behavior at these points.
In this exercise, deriving \( f(x) \) when divided by \((x-1)^2\) gives information about the slope or rate of change at \(x = 1\). By evaluating \( f'(1) \), we gain insights into the "slope" condition \( f(x) \) must meet, which further informs the system of equations.

• \( f'(1) = 2 \) implies a specific rate of change at that point.

This derivative condition must harmonize with the polynomial remainder to ensure consistency across the division's constraints. Understanding how derivatives play into polynomial division helps solve more complex remainder problems and links algebraic concepts with calculus principles.