Problem 103 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=x^{2}+a x+b\) be a quadratic polynomial, where \(a\) and \(b\) are integers. If for a given integer \(n\), \(f(n) f(n+1)=f(m)\) for some integer \(m\), then the value of \(m\) is (A) \(n(n+1)+a n+b\) (B) \(n(n+1)+a+b n\) (C) \(n(n+1)+a+b\) (D) None of these

Step-by-Step Solution

Answer

Option (A) \(m = n(n+1) + an + b\) satisfies the condition.

Step 1: Express the Functions

Let's start by writing down the expressions for \(f(n)\) and \(f(n+1)\). We have:\[f(n) = n^2 + an + b\]and\[f(n+1) = (n+1)^2 + a(n+1) + b = n^2 + 2n + 1 + an + a + b\] Thus, \[f(n+1) = n^2 + (a+2)n + (a+b+1)\].

Step 2: Set Equation for f(m)

According to the problem, we have the condition:\[f(n)f(n+1) = f(m)\]First, find the product \(f(n)f(n+1)\).

Step 3: Simplify the Product

Calculate\[f(n)f(n+1) = (n^2 + an + b)(n^2 + (a+2)n + (a+b+1))\]Instead of expanding completely, the problem suggests that \(f(n)f(n+1)\) equals some \(f(m)\). We will find suitable assumptions about \(m\) leading to equality.

Step 4: Trial and Simplification

Assume \(f(m) = m^2 + am + b\) needs to match with the polynomial format after the simplification of \(f(n)f(n+1)\).Let's assume \(m = n(n+1) + an + b\), check if it suits the formation of polynomials.Expand and compare with possible expressions for \(m\).

Step 5: Matching Choices

Given options:- (A) \(n(n+1)+an+b\)- (B) \(n(n+1)+a+bn\)- (C) \(n(n+1)+a+b\)- (D) None of theseWe suspect option A matches

Key Concepts

Integer Solutions

When solving quadratic polynomials, one often searches for integer solutions. This means we look for integer values that satisfy the equation of the polynomial, such as having zero as an outcome when plugged into the polynomial function.
Integer solutions are crucial as they offer concrete results that are often easier to visualize and verify. In problems involving quadratic polynomials, where you're given conditions or constraints with integers like in our exercise, these solutions help us determine specific values of variables. They play a key role in solving polynomial equations by providing precise points that meet the polynomial's conditions.
One common method to find integer solutions involves factoring the polynomial or using strategies like trying possible values and checking for outcomes that result in integers.

Polynomial Equations

Polynomial equations, especially quadratic ones, are mathematical expressions involving variables raised to powers and constants. A quadratic polynomial typically looks like this:

• In standard form: \( ax^2 + bx + c \)

In a polynomial equation, our goal is to find values of the variables that make the equation true. This often involves balancing the equation or setting it equal to another expression, which might also be a polynomial in multi-step problems.
In the particular problem we're analyzing, you encounter conditions like \( f(n)f(n+1) = f(m) \). Here, each-term follows a polynomial structure. Understanding how these equations form and interact helps us recognize patterns or simplifications that can point to solutions.
Typical steps include setting the polynomial equal to zero or another specific target expression and then simplifying to identify possible roots or solutions.

Factoring

Factoring is a valuable tool in solving polynomial equations. It turns a complex polynomial into easier-to-handle expressions that can reveal solutions or simplifications more clearly.
The process involves finding expressions that, when multiplied, recreate the original polynomial. For quadratic polynomials, this often takes the form of determining factors like \((x + m)(x + n) = 0\), where \(m\) and \(n\) are numbers matched to the values needed.

• Factoring usually requires experimentation or known techniques like the quadratic formula or patterns like perfect squares.
• Understanding the structure of polynomials helps identify potential factorizations quickly.

In our task, factoring helps verify expressions for \(f(m)\) against \(f(n)f(n+1)\). By logically breaking down \(f(n)\) and \(f(n+1)\) into factors, comparing with potential solutions becomes manageably straightforward. This verification guides us to the correct choice among options.