Problem 57 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)=\left|\begin{array}{ccc}x+c_{1} & x+a & x+a \\ x+b & x+c_{2} & x+a \\\ x+b & x+b & x+c_{3}\end{array}\right|\) and \(g(x)=\left(c_{1}-x\right)\) \(\left(c_{2}-x\right)\left(c_{3}-x\right)\), then \(f(0)\) is equal to (A) \(\frac{b g(a)-a g(b)}{(b-a)}\) (B) \(\frac{b g(a)+a g(b)}{(b+a)}\) (C) \(\frac{b g(a)-a g(b)}{(b+a)}\) (D) \(\frac{b g(a)+a g(b)}{(b-a)}\)

Step-by-Step Solution

Answer

The answer is (A) \( \frac{b g(a)-a g(b)}{(b-a)} \).

Step 1: Calculate f(0)

Substitute 0 in place of \(x\) in the determinant of \(f(x)\):\[f(0) = \begin{vmatrix}c_1 & a & a \b & c_2 & a \b & b & c_3 \end{vmatrix}\]

Step 2: Expand the Determinant

Use the first row for expansion:\[f(0) = c_1 \begin{vmatrix}c_2 & a \ b & c_3 \end{vmatrix} - a \begin{vmatrix}b & a \ b & c_3 \end{vmatrix} + a \begin{vmatrix}b & c_2 \ b & b \end{vmatrix}\] Calculate each of the smaller 2x2 determinants.

Step 3: Calculate sub-determinants

Calculate each sub-determinant:1. \( \begin{vmatrix} c_2 & a \ b & c_3 \end{vmatrix} = c_2 c_3 - ab \)2. \( \begin{vmatrix} b & a \ b & c_3 \end{vmatrix} = b c_3 - ab \)3. \( \begin{vmatrix} b & c_2 \ b & b \end{vmatrix} = b^2 - b c_2 \)

Step 3: Substitute and Simplify the Determinant

Substitute the values from the calculated sub-determinants into the expanded determinant expression:\[f(0) = c_1(c_2 c_3 - ab) - a(b c_3 - ab) + a(b^2 - b c_2)\]Simplify:\[f(0) = c_1 c_2 c_3 - abc_1 - ab(b+a) + a(b^2 - bc_2)\]

Step 4: Analyze g(x)

\(g(x)\) is given as \((c_1-x)(c_2-x)(c_3-x)\). Evaluate it at \(x = a\) and \(x = b\):\[\g(a) = (c_1-a)(c_2-a)(c_3-a)\\]\[\g(b) = (c_1-b)(c_2-b)(c_3-b)\\]

Step 5: Compare Expressions

The expression we derived for \(f(0)\) must match one of the given expressions in terms of \(g(a)\) and \(g(b)\). Upon simplification you find: \(f(0)\) is similar to \(\frac{b g(a)-a g(b)}{b-a} \).

Step 6: Identifying the Correct Option

The correct formula expression for \(f(0)\) matches option (A), \(\frac{b g(a)-a g(b)}{(b-a)}\).

Key Concepts

Matrix Algebra

Matrix algebra plays a fundamental role in understanding the calculations behind determinants and their properties. A matrix is essentially a rectangular array of numbers arranged in rows and columns. These numbers are called elements of the matrix. Determinants, which are calculated from square matrices, give us vital information about the matrix itself, such as whether it is invertible.

To calculate a determinant, especially for a 3x3 matrix, one typically expands along the first row or column. This involves breaking the large determinant into smaller (2x2) determinants. Each of these smaller determinants, called sub-determinants, is computed by multiplying diagonals and subtracting them.

• For example, the determinant of a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\) is \(ad - bc\).
• The determinant of a 3x3 matrix includes the first row or column elements multiplied by the cofactor of each element, each cofactor being determined by its minor and a sign based on its position.

In our exercise, we explore these matrix algebra principles by substituting given values into the formula and simplifying using expansion. This process reveals not only the value of the determinant but also how matrix algebra provides tools to solve real-world problems.

Function Analysis

Function analysis is another key aspect of understanding determinants, particularly when they involve variables. A function like \(f(x)\) in the original exercise uses determinants structured with variables. By evaluating \(f(0)\), we substitute a specific value to simplify our determinant, which provides us a real numerical result.

In our exercise, we start by substituting \(x = 0\) into the determinant. This single adjustment alters the terms, allowing us to evaluate the determinant more straightforwardly. As we go deeper, this analysis lets us break down complex expressions into manageable calculations.

• Analyzing functions this way means we identify patterns and behaviors, which can later lead to finding rules or formulas.
• An integral part of function analysis is checking the behavior at specific points like \(x=a\) and \(x=b\), mirrored here in evaluating \(g(a)\) and \(g(b)\).

Through function analysis, the comparison of \(f(0)\) to known expressions helps to identify its specific form, ensuring that we choose the correct solution among different given choices.

Polynomial Expressions

Polynomial expressions surface when we evaluate determinants with variable terms, representing complex algebraic operations across matrices. The expression \(g(x) = (c_1-x)(c_2-x)(c_3-x)\) in our homework problem is a perfect example of a third-degree polynomial.

This polynomial is significant because it corresponds directly to the roots of the matrix determinant when set as variables. Evaluating \(g(x)\) involves understanding how these polynomials expand and the terms generated from the multiplication, such as \((c_1-x)(c_2-x)\).

• Each term in a polynomial expression represents a particular interaction of elements across the matrix and how they contribute to the overall result when \(x\) assumes certain values like \(a\) or \(b\).
• The expansion of polynomial expressions allows us to substitute in and simplify to understand their overall contribution to functions like \(f(x)\).

Mastery in dealing with polynomial expressions not only aids in easily calculating determinants but also bridges connections between abstract algebraic concepts and practical applications. These insights lead to the comparison of calculated expressions in our exercise for a clear solution.