Problem 16 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \(\begin{array}{ll}\text { (A) }[0,1] & \text { (B) }\left(0, \frac{1}{2}\right]\end{array}\) (C) \(\left[\frac{1}{2}, 1\right]\) (D) \((0,1]\)

Step-by-Step Solution

Answer

The range of \(m(b)\) is (0,1]. Option (D) is correct.

Step 1: Identify the Structure of a Quadratic Function

The given function is a quadratic function, which can be written as \( f(x) = ax^2 + bx + c \). In this exercise, you need to determine the minimum value of this function for varying \( b \).

Step 2: Determine the Axis of Symmetry

For a quadratic function \( ax^2 + bx + c \), the vertex (extreme point) can be found at \( x = -\frac{b}{2a} \). Here, \( a = 1 + b^2 \) and \( b = 2b \), so the x-coordinate of the vertex is \( x = -\frac{2b}{2(1 + b^2)} = -\frac{b}{1 + b^2} \).

Step 3: Evaluate the Function at the Vertex

Substitute \( x = -\frac{b}{1 + b^2} \) into \( f(x) \) to find \( m(b) \), the minimum value of \( f(x) \). Calculate \( f\left(-\frac{b}{1 + b^2}\right) = (1 + b^2)\left(-\frac{b}{1 + b^2}\right)^2 + 2b\left(-\frac{b}{1 + b^2}\right) + 1 \).

Step 4: Simplify the Expression for the Minimum Value

Solve the expression: \( f\left(-\frac{b}{1 + b^2}\right) = \left(1 + b^2\right) \cdot \frac{b^2}{(1 + b^2)^2} - \frac{2b^2}{1 + b^2} + 1 \). This simplifies to \( \frac{b^2 + b^4 - 2b^2 (1 + b^2) + (1 + b^2)^2}{(1 + b^2)^2} \).

Step 5: Further Simplification

The expression becomes \( \frac{1 - b^2}{1 + b^2} \), which is the simplified form of the minimum point value.

Step 6: Determine the Range of \( m(b) \)

Since \( \frac{1 - b^2}{1 + b^2} \) yields values between \( 0 \) and \( 1 \) due to the subtraction and addition of squared terms, the range of \( m(b) \) as \( b \) varies is \( (0, 1] \). The minimum value of this expression, when \( b = 0 \), is \( 1 \).

Step 7: Conclusion: Identify the Correct Option

Based on the calculations, as \( b \) varies, the range of \( m(b) \) covered is \((0, 1]\), hence the answer is (D).

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a crucial point where the parabola changes direction.
For a quadratic function of the form \( ax^2 + bx + c \), the vertex can be found using the vertex formula:

• The x-coordinate of the vertex is \( x = -\frac{b}{2a} \).

This formula helps locate the axis of symmetry, which is a vertical line that passes through the vertex and divides the parabola into two mirror images.
Knowing the x-coordinate allows us to find the corresponding y-coordinate by substituting this value back into the function.
In our exercise with the function \( f(x) = (1 + b^2)x^2 + 2bx + 1 \):

• Here, \( a = 1 + b^2 \) and \( b = 2b \).
• The formula gives us \( x = -\frac{b}{1 + b^2} \).

This x-coordinate helps in evaluating the function at its minimum point.

Minimum Value of a Function

Finding the minimum value of a quadratic function is a vital study topic.
Once the vertex x-coordinate is found, substitute it back into the original function to find the y-coordinate — which represents the minimum value if the parabola opens upwards.
For the quadratic function \( f(x) = (1 + b^2)x^2 + 2bx + 1 \), substitution gives:

• \( f\left(-\frac{b}{1 + b^2}\right) = (1 + b^2)\left(-\frac{b}{1 + b^2}\right)^2 + 2b\left(-\frac{b}{1 + b^2}\right) + 1 \).

Simplifying, we find \( m(b) = \frac{1 - b^2}{1 + b^2} \).
This simplification shows how to get the minimum value based on the varied parameter \( b \).

Range of a Function

The range of a function refers to all possible output values (y-values) that the function can produce.
It's essential to consider this when analyzing functions like quadratic ones, which can have different ranges based on their shape and orientation.
For the given exercise, we determined the range of the function's minimum value, \( m(b) \), as \((0, 1]\).
This means \( m(b) \) can take values greater than \( 0 \) and up to \( 1 \).

• This conclusion arises from analyzing the expression \( \frac{1 - b^2}{1 + b^2} \).
• Since \( b^2 \) is always non-negative, \( \frac{1 - b^2}{1 + b^2} \) remains less than or equal to 1 as \( b \) changes.

The range gives insight into how \( m(b) \) behaves as \( b \) varies, capturing the broad scope of possible minimum values for various \( b \).
Understanding the range is integral for a complete grasp of the behavior of quadratic functions across different conditions.