Problem 117 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(\phi(x)=\int_{0}^{x} g(t) d t\), where the function \(g\) is such that \(-\frac{1}{2} \leq g(t) \leq 0, \forall t \in[0,1] \frac{1}{2} \leq g(t) \leq 1, \forall t \in\) 1 \([1,3] g(t) \leq 1, \forall t \in[3,4]\) Then, \(\phi(4)\) satisfies the inequality (A) \(\frac{1}{2} \leq \phi(4) \leq 3\) (B) \(0 \leq \phi(4) \leq 2\) (C) \(\phi(4) \leq 3\) (D) None of these

Step-by-Step Solution

Answer

Option (C), \(\phi(4) \leq 3\), is correct.

Step 1: Break down integration intervals

First, we analyze the behavior of the function \(g(t)\) over the different segments of \([0,4]\) as described. The integral \( \phi(x) = \int_{0}^{x} g(t) \, dt \) will be computed piecewise. We will separately consider three segments: \([0,1]\), \([1,3]\), and \([3,4]\).

Step 2: Integrate over [0, 1]

In this interval, we know \(-\frac{1}{2} \leq g(t) \leq 0\). The maximum possible value for the integral from 0 to 1 is \(0\) since \(g(t)\) could be 0 throughout, which results in an integral of \(0\). The minimum is when \(g(t) = -\frac{1}{2}\), so \[ \int_{0}^{1} g(t)\, dt \geq \int_{0}^{1} -\frac{1}{2} \, dt = -\frac{1}{2}. \]

Step 3: Integrate over [1, 3]

Here, \(\frac{1}{2} \leq g(t) \leq 1\). The maximum possible value for the integral over this range is \[ \int_{1}^{3} 1\, dt = 2. \]The minimum is when \(g(t) = \frac{1}{2}\), giving \[ \int_{1}^{3} \frac{1}{2}\, dt = 1. \]

Step 4: Integrate over [3, 4]

In this segment, \(g(t) \leq 1\). The maximum value for the integral is \[ \int_{3}^{4} 1\, dt = 1. \]Since \(g(t)\) can be 0, the minimum value is \[ \int_{3}^{4} 0\, dt = 0. \]

Step 5: Add up all integral segments

Let's sum up the contributions from each interval. The maximum sum, when \(g(t)\) is maximum, is: \[ (-\frac{1}{2}) + 2 + 1 = 2.5 \]And the minimum sum considering the minima is: \[ 0 + 1 + 0 = 1. \] Hence, \[ 1 \leq \phi(4) \leq 2.5. \]

Step 6: Compare with given options

From the calculated bounds, \(\phi(4)\) satisfies \(1 \leq \phi(4) \leq 2.5\). Comparing these with the provided choices, only option (C), \(\phi(4) \leq 3\), encompasses the possible range of \(\phi(4)\).

Key Concepts

Piecewise Functions

A piecewise function is essentially a combination of multiple sub-functions, each of which applies to a specific interval of the domain. Think of it as having different rules for different sections or segments of your number line. In our given problem, the function \( g(t) \) is piecewise defined over the interval \([0, 4]\). Each sub-function operates within specific bounds and rules on set intervals, which is why we tackle each interval separately when solving the integral.

Key characteristics of piecewise functions include:

• Different rules or expressions apply to different parts of the function's domain.
• Each segment is carefully defined within a specific interval.
• They allow complex functions to be broken down into simpler, more manageable pieces.

Understanding piecewise functions helps in analyzing how a function behaves over different segments, such as identifying which parts contribute most to the integral. This trait is crucial when considering how to approach specific problems that involve varied conditions like the one in the exercise.

Integration Over Intervals

Integration over intervals involves summing up the function values across a certain range. For piecewise functions, like in our exercise, this needs to be carried out separately for each defined interval. This is why we integrate over \([0, 1]\), \([1, 3]\), and \([3, 4]\) individually.

By evaluating the integral over each interval separately, you ensure that each segment respects the prescribed behavior of \( g(t) \). Here, the aim is to find the accumulated value of the function from the beginning to the end of the specified interval.

Important points to remember:

• Each piecewise part needs separate attention, ensuring specific characteristics of the interval are acknowledged.
• Always consider both minimum and maximum bounds when integrating, as these dictate the range of potential integral values.
• The calculated integrals in each interval must then be summed to form the overall definite integral of the function over the entire domain.

When dealing with integration over separate intervals, it's like solving a puzzle by focusing on each section individually before assembling the complete picture.

Bounds of Integration

When calculating definite integrals, it's vital to understand and apply the correct bounds of integration. These bounds define the start and end points for your calculations, effectively setting the limits for which parts of the function's domain you're integrating over.

In our exercise, each interval – \([0, 1]\), \([1, 3]\), and \([3, 4]\) – has specific bounds within which the function \( g(t) \) is considered. These bounds determine the contribution of each section to the entire definite integral.

Key aspects of bounds of integration include:

• They demarcate the endpoints of the intervals over which the function is integrated.
• The bounds help identify where one piecewise definition ends and another begins.
• Assessing both the lower and upper bounds in terms of maximum or minimum possible values reflects variations in the integral's outcome.

Understanding the bounds thoroughly ensures that when calculating integrals, no section is overlooked, and the integral accurately represents the total accumulation of the function's values over its entire domain.