Problem 143 - A Complete Resource Book in Mathematics for JEE Main

Question

$$ \begin{array}{ll} \text { Column-I } & \text { Column-II } \\ \hline \text { I. Let } f(x)=\left(1+b^{2}\right) x^{2}+2 b x+1 & \text { (A) }(0,1] \\ \text { and } m(b) \text { the minimum value of } \\ f(x) \text { for a given } b . \text { As } b \text { varies, the } \\ \text { range of } m(b) \text { is } \\ \text { II. The set of values of } x \text { for which } & \text { (B) }(0,1) \\\ \log (1+x)

Step-by-Step Solution

Answer

I matches with A, II matches with B, III matches with C, IV matches with D.

Step 1: Analyzing function f(x) in Column I

The function given is \( f(x) = (1+b^2)x^2 + 2bx + 1 \). To find the minimum value for given \( b \), we complete the square or use calculus. The derivative \( f'(x) = 2(1+b^2)x + 2b \) is set to zero to find the critical point: \( x = -\frac{b}{1+b^2} \). Evaluating \( f(x) \) at this point gives the minimum value \( m(b) = 1 - \frac{b^2}{1+b^2} = \frac{1}{1+b^2} \). As \( b \) varies from \( -\infty \) to \( \infty \), \( m(b) \) approaches values from 0 to 1, thus ranges over \((0,1]\).

Step 2: Analyzing inequality in Column II

The inequality is \( \log(1+x) < x \). Consider the Taylor expansion of \( \log(1+x) \), which is \( x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots \). For \( x > 0 \), \( \log(1+x) \) is approximated by \( x - \frac{x^2}{2} \), which is less than \( x \). For small positive \( x \), this approximation holds true. Thus, by inspection or graphically, \( x \) must be in \((0,1)\) for the inequality to hold for all \( x > 0 \).

Step 3: Analyzing polynomial condition in Column III

Given the condition \( \frac{a_0}{n+1} + \frac{a_1}{n} + \frac{a_2}{n-1} + \ldots + a_n = 0 \), it suggests the existence of a root between \(0\) and \(\infty\) due to mean value considerations. This means the polynomial \( a_0 x^n + a_1 x^{n-1} + \ldots + a_n = 0 \) must have at least one root in the interval \((0, \infty)\).

Step 4: Solving the equation in Column IV

Given \( 27a + 9b + 3c + d = 0 \), it factors into \(((3x+1)^3) = 0\), suggesting that \( x = -\frac{1}{3} \) satisfies \(4ax^3 + 3bx^2 + 2cx + d = 0\). As this implies an interval, and given signs and linear conditions of polynomials, the root \( -\frac{1}{3} \) situates the root in the larger interval \((0, 3)\), where conditions allow a real root by Intermediate Value Theorem or Rolle's theorem.

Key Concepts

Quadratic Functions

A quadratic function is a polynomial function of degree 2, typically in the form \( ax^2 + bx + c \). This type of function graphs as a parabola, which can either open upwards or downwards based on the coefficient of \( x^2 \). For quadratic functions, identifying key features such as the vertex and axis of symmetry is crucial in optimization tasks. The vertex, found at \( x = -\frac{b}{2a} \), represents either the maximum or minimum point of the parabola. In optimization, particularly mathematical optimization, this point is critical, as it provides us with the function's extreme values. In our problem, the function \( f(x) = (1+b^2)x^2 + 2bx + 1 \) is a quadratic function where we aim to find its minimum. By completing the square or using derivatives, we calculate this minimum and determine how it varies as the parameter \( b \) changes.

Derivatives

The derivative of a function gives us the rate at which the function’s value changes, providing insight into the function’s slopes and critical points. With quadratic functions, the first derivative is a linear expression. For the function \( f(x) = (1+b^2)x^2 + 2bx + 1 \), the first derivative \( f'(x) = 2(1+b^2)x + 2b \) tells us how the slope of \( f(x) \) changes over \( x \). Setting \( f'(x) = 0 \) allows us to find the critical point, a point where the function may reach a minimum or maximum. This is essential for determining the minimum value of our quadratic function over any real number \( b \). Derivatives are a powerful tool in calculus that allow us to perform such analysis quickly and efficiently, offering precise interpretations of a function's behavior over its domain.

Inequality Analysis

Analyzing inequalities often involves finding ranges of values that satisfy a particular condition. In the given exercise with the inequality \( \log(1+x) < x \), we must find the values for \( x \) that make the inequality true. Given the properties of logarithmic functions and their Taylor approximations, it is important to understand the behavior of the function near zero. We approximate \( \log(1+x) \) using a power series and determine that for small values of \( x \), the inequality holds true. The effective range for \( x \) is thus determined to be \((0,1)\), as this is where \( x \) is positive and satisfies the inequality. Understanding graphical representations of the functions can also offer helpful insights in analyzing such inequalities.

Polynomial Roots

Finding the roots of a polynomial means identifying the values of \( x \) for which the polynomial equals zero. In problems involving polynomial equations, conditions given, such as the sum \( \frac{a_0}{n+1} + \frac{a_1}{n} + \cdots + a_n = 0 \), hint at the presence of roots over certain intervals. The roots of a polynomial can often be found using methods such as factoring, graphing, or applying calculus-based theorems like the Intermediate or Mean Value Theorems. In analyzing the equation \( a_0 x^n + a_1 x^{n-1} + \cdots + a_n = 0 \), it's inferred that there is at least one root in the interval \((0, \infty)\), supported by theoretical means value proposition suggesting info on distribution of values across its range.