Problem 49 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(a>0\) and discriminant of \(a x^{2}+2 b x+c\) is \(-\) ve, then \(\left|\begin{array}{ccc}a & b & a x+b \\ b & c & b x+c \\ a x+b & b x+c & 0\end{array}\right|\) is equal to (a) \(+\mathrm{ve}\) (b) \(\left(a c-b^{2}\right)\left(a x^{2}+2 b x+c\right)\) (c) \(-\mathrm{ve}\) (d) 0

Step-by-Step Solution

Answer

The determinant evaluates to (b): \((ac-b^2)(ax^2+2bx+c)\).

Step 1: Calculate the Determinant

Let's express the determinant of the given matrix directly. The matrix is\[\begin{bmatrix} a & b & ax + b \ b & c & bx + c \ ax + b & bx + c & 0 \end{bmatrix}\]. Using the formula for the determinant of a 3x3 matrix, \(\Delta = a(ei-fh)-b(di-fg)+c(dh-eg)\), calculate:- Term 1: \(a \times \) determinant of \(\begin{bmatrix} c & bx+c \ bx+c & 0 \end{bmatrix} \) results in \( a \times (0 - (bx+c)^{2}) = -a(bx+c)^{2} \).- Term 2: \(b \times \) determinant of \(\begin{bmatrix} b & ax+b \ bx+c & 0 \end{bmatrix} \) results in \(b \times (0 - (ax+b)(bx+c)) = b(ax+b)(bx+c) \).- Term 3: \( (ax+b) \times \) determinant of \(\begin{bmatrix} b & c \ b & c \end{bmatrix} \) results in \((ax+b)(bc - bc) = 0 \).Add all three terms: \[ -a(bx+c)^{2} + b(ax+b)(bx+c) + 0 \].

Step 2: Simplify the Expression

Factor the terms derived from calculating the determinant:- Combine the terms: \( -a(bx+c)^{2} + b(ax+b)(bx+c) \).- Notice this simplifies to a product of two factors, \(-a(bx+c)(bx+c) + b(ax+b)(bx+c) = (a(ax^2 + 2bx + c))(ac - b^2) \).

Step 3: Use Discriminant Condition

Since the discriminant \(\Delta = (2b)^2 - 4ac = 4b^2 - 4ac < 0\), it implies \(ac - b^2 > 0\). Thus, \((ac - b^2)(ax^2 + 2bx + c)\) is certainly positive since both parts are positive (given that \(a > 0\)).

Step 4: Match with Options

Based on the simplified expression and condition, the determinant evaluates to\( (ac-b^2)(ax^2+2bx+c) \). This matches option (b) from the given choices.

Key Concepts

Discriminant

The discriminant is a key concept in understanding the nature of roots for quadratic equations. In any quadratic equation of the form \( ax^2 + bx + c = 0 \), the discriminant is represented by \( \Delta = b^2 - 4ac \). It helps determine whether the roots are real or complex, and if real, whether they are distinct or repeated.

• If \( \Delta > 0 \), the equation has two distinct real roots.
• If \( \Delta = 0 \), there are two real and identical roots.
• If \( \Delta < 0 \), the equation has two complex roots.

In the exercise, it's given that the discriminant \( \Delta = 4b^2 - 4ac \) is negative, indicating complex roots for the quadratic \( ax^2 + 2bx + c = 0 \). Given this, and since \( a > 0 \), we can deduce that the factor \( ac - b^2 \) is positive.
Hence, the condition of the discriminant greatly influences the determination of resultant properties in problems involving quadratic equations.

Determinant of a matrix

A determinant offers a scalar value that can be computed from a square matrix, providing meaningful properties about the matrix. In a 3x3 matrix, the determinant can define whether the matrix is invertible, and it plays a crucial role in applications like solving linear equations, finding the volume under transformations, and more.
For a matrix \[ \begin{bmatrix} a & b & d \ e & f & g \ h & i & j \end{bmatrix} \], its determinant is given by:\[ \Delta = a(ej - fi) - b(dj - fh) + c(di - eh) \]In our exercise, we apply this formula to compute the determinant of a specific 3x3 matrix. The calculation requires looking closely at the sub-matrices formed by removing the row and column of each element in turn and considering their determinants. This systematic approach simplifies the problem significantly, eventually providing insights aligned with conditions like the discriminant being negative.
The simplification of these calculated terms, when aligned with known sign conditions from the discriminant, leads us to a conclusive option in the solution.

Quadratic equations

Quadratic equations serve as a fundamental concept in algebra that takes the form \( ax^2 + bx + c = 0 \). These are second-degree polynomial equations meaning they have the highest exponent of two. Quadratic equations are renowned for their application across various fields including physics, economics, and engineering for modeling diverse phenomena.
To solve a quadratic equation, several methods can be used:

• Factoring, when applicable.
• Completing the square method.
• Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Understanding the nature of the roots is vital, which is directly linked to the discriminant, as explored earlier. For our exercise, given that the quadratic component \( ax^2 + 2bx + c \) arises with a negative discriminant, it implies roots are complex, leading to specific resultant expressions used in simplifying matrices or related calculations.
Grasping how quadratics integrate with other mathematical concepts like determinants enriches an understanding of interconnected topics within mathematics.