Problem 117 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(\phi(x)=\int_{0}^{x} g(t) d t\), where the function \(g\) is such that \(-\frac{1}{2} \leq g(t) \leq 0, \forall t \in[0,1] \frac{1}{2} \leq g(t) \leq 1, \forall t \in\) \([1,3] g(t) \leq 1, \forall t \in[3,4]\) 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

The correct option is (A) \(\frac{1}{2} \leq \phi(4) \leq 3\).

Step 1: Understand the integral function

The function \(\phi(x)\) is defined as an integral of \(g(t)\) from 0 to \(x\). This means \(\phi(x)\) accumulates the area under the curve of \(g(t)\) from 0 to \(x\). We need to integrate \(g(t)\) across the specified intervals, considering the bounds provided for \(g(t)\).

Step 2: Evaluate \(\int_0^1 g(t) \, dt\)

In the interval \([0,1]\), \(g(t)\) has values between \(-\frac{1}{2}\) and 0. The most negative the integral can be over this interval is \(-\frac{1}{2}\cdot(1-0) = -\frac{1}{2}\). Thus, the integral \(\int_0^1 g(t) \, dt \) ranges from \(-\frac{1}{2}\) to 0.

Step 3: Evaluate \(\int_1^3 g(t) \, dt\)

In the interval \([1,3]\), \(g(t)\) has values between \(\frac{1}{2}\) and 1. Thus, \(\int_1^3 g(t) \, dt \) could be little as \(\frac{1}{2}\cdot(3-1) = 1\) or as much as \(1\cdot(3-1) = 2\).

Step 4: Evaluate \(\int_3^4 g(t) \, dt\)

In the interval \([3,4]\), \(g(t)\) is less than or equal to 1. Therefore, \(\int_3^4 g(t) \, dt \) could be as little as 0 or as much as \(1\cdot(4-3) = 1\).

Step 5: Sum the integrals for \(\phi(4)\)

Combine the results of the integrals: \(-\frac{1}{2} \leq \int_0^1 g(t) \, dt \leq 0\) , \(1 \leq \int_1^3 g(t) \, dt \leq 2\) , \(0 \leq \int_3^4 g(t) \, dt \leq 1\). Thus, \(-\frac{1}{2} + 1 + 0 \leq \phi(4) \leq 0 + 2 + 1\), giving \(\frac{1}{2} \leq \phi(4) \leq 3\).

Step 6: Choose the correct option

Among the options provided \(\frac{1}{2} \leq \phi(4) \leq 3\) matches the option (A).

Key Concepts

Lower and Upper Bounds

In integral calculus, understanding the concept of lower and upper bounds is crucial.
It helps determine the possible values of an integral based on the function being integrated. In our exercise, the function under consideration is \( g(t) \) with different bounded intervals. For instance, on the interval

• \([0, 1]\), \( g(t) \) ranges between \(-\frac{1}{2}\) and 0.
• For \([1, 3]\), values are between \(\frac{1}{2}\) and 1.
• And for \([3, 4]\), \( g(t) \) is less than or equal to 1.

These bounds define the spectrum of possible areas, visually representing them as the tallest and shortest rectangles under the curve across each segment. By calculating these possible extremes, we define lower and upper limits for the definite integrals over each interval.
This bounded information helps us choose the correct answer option by ensuring the computed total area under the curve is between the determined limits.

Piecewise Function

A piecewise function is a function composed of multiple sub-functions, each defined on a particular interval.
In this problem, the function \( g(t) \) is piecewise, as it takes on different ranges of values depending on which interval \( t \) falls within:

• For \( t \) in \([0, 1]\), \( g(t) \) is constrained by \(-\frac{1}{2} \leq g(t) \leq 0\).
• From \( t \) in \([1, 3]\), \( \frac{1}{2} \leq g(t) \leq 1\).
• In the \([3, 4]\) interval, \( g(t) \leq 1\).

The challenge with piecewise functions is understanding and managing these transitions between intervals. Each sub-function affects the total result of the integral differently, as seen in the summation of their separate integrals. This requires careful attention to ensure all conditions of each piece are correctly applied when computing the area under the curve.

Area Under Curve

The area under a curve in integral calculus is a fundamental concept for finding accumulative values over a specified interval.
When dealing with integrals like \( \phi(x) = \int_{0}^{x} g(t) \, dt \), the solution involves calculating the aggregate area between the curve \( g(t) \) and the x-axis, from the lower limit to the upper limit. Each segment of \( g(t) \) contributes a defined area computed via its lower and upper bounds.

• For \([0, 1]\), the area can be negative reflecting the negative part of the function.
• An area from \([1, 3]\) varies significantly due to the positive values of \( g(t) \).
• And for \([3, 4]\), while it can reach a maximum of 1, it also includes zero when the function remains non-positive entire time.

The final area under the total curve is the sum of these pieces, which leads to the conclusion that \( \phi(4) \) ranges from \( \frac{1}{2} \) to 3.
Understanding these areas in segments and then adding them together is key to solving the problem accurately.