Problem 16 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(f(x)=\left(1+b^{2}\right) x^{2}+2 b x+1\) and \(m(b)\) the minimum value of \(f(x)\) for a given \(b\). As \(b\) varies, the range of \(m(b)\) is \(\left.\begin{array}{ll}\text { (A) }[0,1] & \text { (B) }\left(0, \frac{1}{2}\right]\end{array}\right]\) (C) \(\left[\frac{1}{2}, 1\right]\) (D) \((0,1]\)

Step-by-Step Solution

Answer

(D) (0,1].

Step 1: Identify the Form of f(x)

We observe that \(f(x) = (1 + b^2)x^2 + 2bx + 1\) is a quadratic function of the form \(ax^2 + bx + c\), where \(a = 1 + b^2\), \(b = 2b\), and \(c = 1\).

Step 2: Find the Vertex of the Quadratic Function

The vertex form of a quadratic \(ax^2 + bx + c\) has a vertex at \(x = -\frac{b}{2a}\). Substituting \(b = 2b\) and \(a = 1 + b^2\), we find the vertex occurs at \(x = -\frac{2b}{2(1 + b^2)} = -\frac{b}{1 + b^2}\).

Step 3: Calculate f(x) at the Vertex

The minimum value \(m(b)\) of \(f(x)\) is at this vertex. Plug \(x = -\frac{b}{1 + b^2}\) into \(f(x)\):\[f\left(-\frac{b}{1 + b^2}\right) = (1 + b^2)\left(-\frac{b}{1 + b^2}\right)^2 + 2b\left(-\frac{b}{1 + b^2}\right) + 1\].

Step 4: Simplify Expression for m(b)

Simplify the expression:\[\begin{align*}f\left(-\frac{b}{1 + b^2}\right) &= (1 + b^2)\left(\frac{b^2}{(1 + b^2)^2}\right) - \frac{2b^2}{1+b^2} + 1\&= \frac{b^2}{1 + b^2} - \frac{2b^2}{1 + b^2} + 1\&= \frac{b^2 - 2b^2 + (1 + b^2)}{1 + b^2}\&= \frac{-b^2 + 1 + b^2}{1 + b^2}\&= \frac{1}{1 + b^2}.\end{align*}\]

Step 5: Determine the Range of m(b)

The value \(m(b) = \frac{1}{1 + b^2}\) is always positive and reaches its maximum when \(b = 0\), which makes \(m(0) = 1\). As \(|b|\) increases, \(m(b)\) approaches 0. Therefore, the range of \(m(b)\) is \((0, 1]\).

Step 6: Match Solution with the Given Options

Comparing with the options provided, \((0, 1]\) corresponds to option (D). Therefore, the correct choice is (D) \((0,1]\).

Key Concepts

Vertex of a Quadratic Function

Understanding the vertex of a quadratic function is crucial for analyzing its graph and behavior. For a quadratic function in standard form, \( f(x) = ax^2 + bx + c \), the vertex is the high or low turning point on its parabolic graph.
The vertex can be found using the formula: \( x = -\frac{b}{2a} \). This formula helps locate the x-coordinate of the vertex. Substituting this value back into the equation gives you the y-coordinate. Thus, the vertex is \( \left( -\frac{b}{2a}, f\left( -\frac{b}{2a} \right) \right) \).
In our specific function, \( f(x) = (1 + b^2)x^2 + 2bx + 1 \), with \( a = 1 + b^2 \) and \( b = 2b \), applying the vertex formula gives the vertex at \( x = -\frac{b}{1 + b^2} \).
This gives insight into where the function's minimum or maximum value will occur. With quadratic functions, this point usually represents the minimum value when \( a > 0 \) and a maximum when \( a < 0 \).

Minimum Value of a Quadratic

The minimum value of a quadratic function occurs at its vertex when the parabola opens upwards, which is determined by the leading coefficient \( a \) being positive. In the case of our function \( f(x) = (1 + b^2)x^2 + 2bx + 1 \), \( a = 1 + b^2 \), this condition is met since the expression is always positive.
To find the minimum value of \( f(x) \), we evaluate the function at the vertex found earlier. Substituting \( x = -\frac{b}{1 + b^2} \) into \( f(x) \) gives us:

• First, \( (1 + b^2)\left( -\frac{b}{1 + b^2} \right)^2 = \frac{b^2}{1+b^2} \)
• Second, the linear term simplifies: \( 2b\left( -\frac{b}{1 + b^2} \right) = -\frac{2b^2}{1 + b^2} \)
• Add the constant term 1.

The simplified expression results in the minimum value: \( m(b) = \frac{1}{1 + b^2} \).
This value depends solely on \( b \) and showcases how the quadratic behaves at its lowest point depending on the value of \( b \).

Completing the Square

Completing the square is a technique used to rewrite a quadratic expression, making it easier to identify its properties, such as the vertex. The process involves rearranging the terms of the quadratic to fit the form \( (x - d)^2 + e \).
For the quadratic \( f(x) = (1 + b^2)x^2 + 2bx + 1 \), we want to express it in vertex form: \( (1 + b^2)(x - \text{something})^2 + \text{something else} \).
Let's go through this process:

• Start with \( (1 + b^2)x^2 + 2bx \). Divide everything by \( a = 1+b^2 \) to help complete the square.
• Rewrite \( x^2 + \frac{2b}{1+b^2}x \) by adding and subtracting \( \left(\frac{b}{1+b^2}\right)^2 \) inside the expression.
• Now, the expression becomes \( \left(x + \frac{b}{1+b^2}\right)^2 \). Subtract what was added and adjust the constant term.

This rewrites the quadratic in a form where the vertex can be identified at \( \left(-\frac{b}{1+b^2}, \frac{1}{1+b^2}\right) \).
The process of completing the square clarifies the vertex, helping students understand the geometry of quadratics better.