Problem 165 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched? Interval Function (A) \((-\infty, \infty)\) \(x^{3}-3 x^{2}+3 x+3\) (B) \([2, \infty)\) \(2 x^{3} 3 x^{2}-12 x+6\) (C) \(\left(-\infty, \frac{1}{3}\right]\) \(3 x^{2}-2 x+1\) (D) \((-\infty,-4]\) \(x^{3}+6 x^{2}+6\)

Step-by-Step Solution

Answer

Pair (C) is incorrectly matched.

Step 1: Determine Increasing Condition

A function is increasing on an interval if its derivative is positive throughout that interval. Therefore, for each function, we need to find its derivative and determine where it is positive.

Step 2: Find the Derivative of Function A

The given function is \( x^3 - 3x^2 + 3x + 3 \). The derivative is \( f'(x) = 3x^2 - 6x + 3 \). To find where this function is increasing, solve \( 3x^2 - 6x + 3 > 0 \). Factor to get \( 3(x^2 - 2x + 1) = 3(x-1)^2 \). Since \( (x-1)^2 \) is always non-negative and is 0 only at \( x = 1 \), the function is increasing everywhere except at \( x = 1 \). So, the interval \((-\infty, \infty)\) is correct for function A.

Step 3: Find the Derivative of Function B

The function given is \( 2x^3 + 3x^2 - 12x + 6 \). Its derivative is \( f'(x) = 6x^2 + 6x - 12 \). Determine where \( 6x^2 + 6x - 12 > 0 \). Factor as \( 6(x^2 + x - 2) = 6(x+2)(x-1) \). Thus, the function is increasing when \( x > 1 \) or \( x < -2 \). The interval \([2, \infty)\) is a subset of \((1, \infty)\), so it is correct.

Step 4: Find the Derivative of Function C

Here, the function is \( 3x^2 - 2x + 1 \). Its derivative is \( f'(x) = 6x - 2 \). The function is increasing when \( 6x - 2 > 0 \), or \( x > \frac{1}{3} \). Given interval \((-\infty, \frac{1}{3}]\) implies the function should be increasing on that interval, which is incorrect as the derivative is negative for \( x < \frac{1}{3} \). This pair is incorrectly matched.

Step 5: Find the Derivative of Function D

The function is \( x^3 + 6x^2 + 6 \). Its derivative is \( f'(x) = 3x^2 + 12x \), or \( 3x(x+4) \). The function increases when \( 3x(x+4) > 0 \). This is true for \( x < -4 \) or \( x > 0 \). The interval \((-\infty, -4]\) is correct as it fits part of the interval where \( f'(x) > 0 \).

Key Concepts

Increasing Functions

In calculus, understanding whether a function is increasing or not is pivotal. A function is described as increasing on an interval if, as you move from left to right on the x-axis, the function's values (y-values) keep rising. This means if you pick any two values for the input, say, \(x_1\) and \(x_2\), where \(x_1 < x_2\), then the output \(f(x_1) < f(x_2)\).

This concept is especially crucial when analyzing graphs and understanding the behavior of functions. Here’s why this idea is central to the problem at hand:

• The condition for a function to be increasing over an interval is that its first derivative must be positive throughout that interval.
• If the derivative is zero or negative, then the function isn't increasing, and any intervals described as increasing would be incorrect.
• Consistently finding and interpreting where the derivative is positive helps in determining the increasing nature of functions across given intervals.

Understanding increasing functions helps not only in determining intervals correctly but also in visualizing how functions behave dynamically over a range.

Derivatives

A derivative represents the rate at which a function is changing at any given point. In simpler terms, it tells us how steep the graph of the function is at a particular x-value. This is especially useful for determining when a function is increasing or decreasing.

When we compute the derivative of a function, such as in the problems provided:

• The derivative of \(x^3 - 3x^2 + 3x + 3\) is \(3x^2 - 6x + 3\). The structure and sign of this quadratic expression dictate where the function increases or decreases. Solving \(3x^2 - 6x + 3 > 0\) reveals positive regions.
• Derivatives like \(6x^2 + 6x - 12\) for other given functions, are solved to find critical intervals where the function behavior shifts.

Steps such as these put the power of calculus in student hands, letting them examine function behaviors piece by piece.

Calculating derivatives and examining their sign across intervals is a central part of interval analysis and guides decisions on where a function is increasing or not.

Interval Analysis

Interval analysis in calculus refers to examining where certain conditions, like positivity or negativity of a derivative, hold over parts of a function's domain. Intervals are key to interpreting and analyzing mathematical functions.

Let's break down the components:

• Once the derivative of a function is calculated, the next task is to solve inequalities to determine where that derivative is positive.
• Interval solutions reveal where a function rises (increasing intervals) or falls (decreasing intervals).
• Using these intervals, we can accurately describe the full behavior of a function over its domain.

In the provided solution steps, careful consideration of intervals helped identify incorrect matches between functions and intervals. For instance, determining that \(6x - 2 > 0\) leads to \(x > \frac{1}{3}\), implies understanding the mismatch in the interval \((-fty, \frac{1}{3}]\), which was crucial in finding the incorrectly matched pair.

Mastering interval analysis equips students with the tools needed for a deeper understanding and accurate predictions about function behaviors across various domains.