Problem 87 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(g(x)=\int_{0}^{x} f(t) d t\), where \(f\) is such that \(\frac{1}{2} \leq f(t) \leq 1\) for \(t \in[0,1]\) and \(0 \leq f(t) \leq \frac{1}{2}\) for \(t \in[1,2]\). Then, (A) \(-\frac{3}{2} \leq g(2) \leq \frac{1}{2}\) (B) \(\frac{3}{2} \leq g(2) \leq \frac{5}{2}\) (C) \(\frac{1}{2} \leq g(2) \leq \frac{3}{2}\) (D) None of these

Step-by-Step Solution

Answer

(C) \(\frac{1}{2} \leq g(2) \leq \frac{3}{2}\).

Step 1: Understanding the Problem

We have a function \( g(x) = \int_{0}^{x} f(t) \, dt \) and need to evaluate \( g(2) \). The function \( f(t) \) is bounded such that \( \frac{1}{2} \leq f(t) \leq 1 \) for \( t \in [0, 1] \) and \( 0 \leq f(t) \leq \frac{1}{2} \) for \( t \in [1, 2] \). We must estimate the range in which \( g(2) \) falls.

Step 2: Breaking Down the Integral

The integral \( g(2) \) can be split into two parts: \( g(2) = \int_{0}^{1} f(t) \, dt + \int_{1}^{2} f(t) \, dt \). We need to find bounds for each of these integrals given the bounds for \( f(t) \).

Step 3: Calculating \( \int_{0}^{1} f(t) \, dt \)

For \( t \in [0, 1] \), \( \frac{1}{2} \leq f(t) \leq 1 \). Thus, \( \int_{0}^{1} \frac{1}{2} \, dt \leq \int_{0}^{1} f(t) \, dt \leq \int_{0}^{1} 1 \, dt \). Evaluating these gives: \( \frac{1}{2}(1-0) = \frac{1}{2} \leq \int_{0}^{1} f(t) \, dt \leq 1(1-0) = 1 \).

Step 4: Calculating \( \int_{1}^{2} f(t) \, dt \)

For \( t \in [1, 2] \), \( 0 \leq f(t) \leq \frac{1}{2} \). Thus, \( \int_{1}^{2} 0 \, dt \leq \int_{1}^{2} f(t) \, dt \leq \int_{1}^{2} \frac{1}{2} \, dt \). Evaluating these gives: \( 0(2-1) = 0 \leq \int_{1}^{2} f(t) \, dt \leq \frac{1}{2}(2-1) = \frac{1}{2} \).

Step 5: Finding Bounds for \( g(2) \)

Combining the bounds from both intervals, we have \( \frac{1}{2} = \frac{1}{2} + 0\leq g(2) = \int_{0}^{1} f(t) \, dt + \int_{1}^{2} f(t) \, dt \leq 1 + \frac{1}{2} = \frac{3}{2} \).

Step 6: Comparison with Given Options

The calculated bounds for \( g(2) \) are \( \frac{1}{2} \leq g(2) \leq \frac{3}{2} \). Comparing this to the options, only option (C) matches the calculated bounds.

Key Concepts

Definite Integration

Definite integration is a fundamental concept in calculus used to determine the area under a curve within a specific interval. It is represented by:

• \( \int_{a}^{b} f(t) \, dt \)

In the context of the given exercise, we need to evaluate the definite integral of a function \( f(t) \) from 0 to 2 in order to compute \( g(2) \). This process consists of summing the infinite number of infinitesimal slices beneath the curve of \( f(t) \) between the lower bound \( a = 0 \) and the upper bound \( b = 2 \).
This helps us find the exact accumulation of the area under the curve over a specified interval, providing a numerical value instead of an expression. In this problem, definite integration helps us evaluate the net accumulated value of \( f(t) \) from 0 to 2, using the sub-intervals [0,1] and [1,2].

Function Range

Understanding the range of a function is crucial in evaluating integrals. The range of a function refers to the set of possible output values produced by the function. For the given exercise, the function \( f(t) \) has two distinct ranges based on the domain.
For the interval \([0,1]\):

• The range is \( \frac{1}{2} \leq f(t) \leq 1 \).

For the interval \([1,2]\):

• The range is \( 0 \leq f(t) \leq \frac{1}{2} \).

Identifying these ranges of \( f(t) \) allows us to determine upper and lower bounds for the integrals \( \int_{0}^{1} f(t) \, dt \) and \( \int_{1}^{2} f(t) \, dt \). This step is crucial in computing the bounds for the entire integral \( g(2) \). By knowing where \( f(t) \) lies, we estimate how large or small the integral can be, ultimately helping us to understand the possible values for \( g(2) \).

Integral Bounds

Integral bounds define the limits over which we perform the integration. In this exercise, the bounds are from \( 0 \) to \( 2 \), represented by \( \int_{0}^{2} f(t) \, dt \). These bounds help confine the problem to a specific section of \( f(t) \), effectively controlling the portion of the function we are interested in.
To solve the problem, we split the whole integral into two parts based on the change in the function's range:

• \( g(2) = \int_{0}^{1} f(t) \, dt + \int_{1}^{2} f(t) \, dt \)

Each part of the split integral has specific bounds, which allow us to apply the given ranges of \( f(t) \) to get estimates for each part's contribution to \( g(2) \). By achieving the bounds for \( \int_{0}^{1} f(t) \, dt \) and \( \int_{1}^{2} f(t) \, dt \), we eventually determine the overall bounds for \( g(2) \), which is \( \frac{1}{2} \leq g(2) \leq \frac{3}{2} \). This process illustrates how integral bounds structure and guide the evaluation process in calculus.