Problem 103 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the function \(f(x)=\left(1-\frac{\sqrt{21-4 b-b^{2}}}{b+1}\right) x^{3}\) \(+5 x+\sqrt{16}\) increases for all \(x\), then (A) \(b \in(-1,2)\) (B) \(b \in(-7,3)-\\{-1\\}\) (C) \(b \in(-7,-1) \cup(2,3)\) (D) None of these

Step-by-Step Solution

Answer

Option C matches: \(b \in (-7,-1) \cup (2,3)\).

Step 1: Understanding the Problem

We are given a function \(f(x) = \left(1-\frac{\sqrt{21-4b-b^2}}{b+1}\right)x^3 + 5x + \sqrt{16}\), which needs to be increasing for all \(x\). To determine the value of \(b\), we need to ensure that the derivative \(f'(x)\) is positive for all \(x\).

Step 2: Find the Derivative

Calculate the derivative of \(f(x)\). The derivative is given by \(f'(x) = 3 \left(1-\frac{\sqrt{21-4b-b^2}}{b+1}\right)x^2 + 5\). For \(f(x)\) to be increasing for all \(x\), \(f'(x) > 0\) for all \(x\).

Step 3: Ensure Positive Coefficient for \(x^3\) Term

Since \(x^3\) has the highest power, the condition for \(f'(x)\) to be positive for all \(x\) boils down to consulting the leading term: \(1-\frac{\sqrt{21-4b-b^2}}{b+1} > 0\). Solve this inequality.

Step 4: Solve the Inequality

First, solve \(\sqrt{21-4b-b^2} < b+1\). Squaring both sides, we get \(21-4b-b^2 < (b+1)^2\), leading to the equation \(21-4b-b^2 < b^2 + 2b + 1\). This simplifies to \(-2b^2 - 6b + 20 > 0\). Factor the quadratic inequality to find the valid intervals for \(b\).

Step 5: Factor and Determine Valid Solutions

Solve \(-2(b^2 + 3b - 10) > 0\) by finding the roots of \(b^2 + 3b - 10 = 0\) through the quadratic formula: \(b = \frac{-3 \pm \sqrt{3^2 + 40}}{2}\). Calculate the roots to find \(b = 2\) and \(b = -5\), leading to the intervals \((-\infty, -5) \cup (2,+\infty)\). Check which of the given options align with these results.

Key Concepts

The Derivative and Its Role

The concept of the derivative is central in determining where a function is increasing or decreasing. For a function to be increasing on an interval, its derivative must be positive throughout that interval. In the given exercise, we are tasked with ensuring that the function \(f(x) = \left(1-\frac{\sqrt{21-4b-b^2}}{b+1}\right)x^3 + 5x + \sqrt{16}\) is increasing for all values of \(x\).
To do this, we first compute the derivative, resulting in \(f'(x) = 3 \left(1-\frac{\sqrt{21-4b-b^2}}{b+1}\right)x^2 + 5\). This expression tells us how swiftly or slowly \(f(x)\) ascends as \(x\) increases.
We require \(f'(x) > 0\) for all \(x\). Hence, ensuring the term \(1-\frac{\sqrt{21-4b-b^2}}{b+1}\) is greater than zero guarantees that the coefficient of \(x^2\) is positive. This condition turns our attention to solving for \(b\).

Quadratic Inequalities Explained

Quadratic inequalities can seem daunting at first, yet they boil down to finding where a quadratic expression is greater or less than zero. In our exercise, after simplifying the initial condition, we face the quadratic inequality \(-2b^2 - 6b + 20 > 0\).
To solve, the first step is factoring or using the quadratic formula to identify the roots of \(b^2 + 3b - 10 = 0\). Roots are the points where the expression equals zero, which helps identify the intervals where the expression is positive or negative.
Here, using the quadratic formula \(b = \frac{-3 \pm \sqrt{3^2 + 40}}{2}\), we find that the roots are \(b = 2\) and \(b = -5\). This gives the intervals \((-fty, -5)\) and \((2, +fty)\) as solutions where our quadratic term is positive.

• Understand the problem turns into a simple check of intervals between roots.
• Roots function as boundaries of our solution space.

Intervals and Their Purpose

Intervals describe continuous ranges of values and are critical in conveying where functions satisfy certain conditions, like being positive or negative. With the roots \(b = 2\) and \(b = -5\), we identified intervals from our problem.
Intervals like \((-fty, -5)\) and \((2, +fty)\) reflect where the quadratic aspect of the function in the inequality problem is greater than zero.
This approach employs open intervals (denoted with parentheses) to show that the inequality is strict \((> 0)\), ensuring the function increases across all values of \(x\) as required.

• Recognize interval notation: parentheses \(()\) for strict inequalities and brackets \([]\) for inclusive boundaries.
• Link the derived intervals to given answer choices to pinpoint which satisfy the condition.

Understanding these intervals precisely is key in selecting the correct answer for problems, ensuring efficiency and clarity during problem-solving.