Problem 9 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(a, b, c\) be positive real numbers, such that \(b x^{2}+\) \(\left(\sqrt{(a+c)^{2}+4 b^{2}}\right) x+(a+c) \geq 0, \forall x \in R\), then \(a, b, c\) are in: (A) G.P. (B) A.P. (C) H.P. (D) None of these

Step-by-Step Solution

Answer

Option B: A.P. (Arithmetic Progression)

Step 1: Identify the Expression

The given inequality is \(b x^2 + \sqrt{(a+c)^2 + 4b^2} x + (a+c) \geq 0\) for all \(x \in \mathbb{R}\). We need to consider this expression as a quadratic in \(x\): \(A = b\), \(B = \sqrt{(a+c)^2 + 4b^2}\), and \(C = (a+c)\).

Step 2: Condition for Non-Negativity

For the quadratic expression to be non-negative for all real \(x\), its discriminant must be less than or equal to zero. The discriminant \(\Delta\) of the quadratic \(Ax^2 + Bx + C\) is calculated as \(\Delta = B^2 - 4AC\). Here, \(A = b, B = \sqrt{(a+c)^2 + 4b^2}, C = a+c\).

Step 3: Calculate the Discriminant

Substitute the values of \(A, B, C\) into the discriminant formula: \[\Delta = \left(\sqrt{(a+c)^2 + 4b^2}\right)^2 - 4b(a+c)\]. Simplify it to \[\Delta = (a+c)^2 + 4b^2 - 4b(a+c)\].

Step 4: Simplify the Discriminant

Rewrite the expression: \((a+c)^2 - 4b(a+c) + 4b^2 = (a+c-2b)^2\). Thus, \(\Delta = (a+c-2b)^2\).

Step 5: Analyze the Discriminant for Non-Positivity

For the quadratic to be always non-negative, \((a+c-2b)^2 \leq 0\). Since a square of any real number is non-negative, \((a+c-2b)^2 = 0\), hence \(a+c = 2b\).

Step 6: Conclusion

From \(a+c = 2b\), it follows that \(a, b, c\) must be in Arithmetic Progression (A.P.) where \((a, b, c) = \left(a, \frac{a+c}{2}, c\right)\).

Key Concepts

Arithmetic Progression (A.P.)

An arithmetic progression, often abbreviated as A.P., is a sequence of numbers in which the difference between consecutive terms is constant. This difference is known as the "common difference." For example, in the sequence 2, 4, 6, 8, 10, the common difference is 2 because each term increases by 2.

• To denote A.P., we generally use the form: a, a+d, a+2d,..., where \( d \) is the common difference.
• A property of numbers in an A.P. is that the middle term is the average of the terms on either side.

In the problem statement, the condition given was \( a+c = 2b \). This implies \( b \) is exactly in the middle of \( a \) and \( c \), verifying the nature of an arithmetic progression: \( a, b, c \) are in A.P. because the difference between \( b \) and \( a \) is equal to the difference between \( c \) and \( b \).

Discriminant analysis

The discriminant of a quadratic expression helps determine the nature of the roots of the quadratic equation. Given a quadratic equation \( Ax^2 + Bx + C = 0 \), its discriminant \( \Delta \) is calculated as \( \Delta = B^2 - 4AC \). Here is how the discriminant guides us:

• If \( \Delta > 0 \), the quadratic equation has two distinct real roots.
• If \( \Delta = 0 \), it has exactly one real root or a repeated root.
• If \( \Delta < 0 \), the roots are complex and do not exist on the real number line.

In our problem, for the quadratic to be non-negative for all real numbers, its discriminant must satisfy \( \Delta \leq 0 \). The solution set manipulated the discriminant \((a+c-2b)^2\) to zero, aligning perfectly with \( \Delta = 0\). This condition ensures no real roots cut the x-axis, backing up our inequality.

Quadratic Expressions

Quadratic expressions are central in algebra and take the form \( Ax^2 + Bx + C \). These expressions graph as parabolas, which can open upwards or downwards depending on the sign of \( A \). A positive \( A \) generates a parabola opening upwards, while a negative \( A \) opens downwards. These expressions are fundamental as they describe motion, areas, and more.

• The vertex of the parabola is a key feature and can be found using \(-\frac{B}{2A}\) for its x-coordinate.
• The y-intercept is easily identifiable as \( C \), the constant term.

Quadratic expressions exhibit diverse behaviors through their roots, extrema, and concavity, significantly impacting mathematical modeling and problem-solving. In our original problem, ensuring all real values of \( x \) yield a non-negative result highlights how crucial evaluating a quadratic's discriminant is in guaranteeing its behavior across the real number line.