Let g(x) = ∫₀ˣ f(t) dt, where f is such that 1/2 ≤ f(t) ≤ 1 for t ∈ [0, 1] and 0 ≤ f(t) ≤ 1/2 for t

Q: Wait, why do we split the integral at $t=1$? Can't we just integrate $f(t)$ directly?

Great question! We split the integral because the function $f(t)$ is defined piecewise. It behaves differently in the first half of the interval than in the second. Think of it like calculating the total cost of a trip where the price per kilometer changes halfway through. You cannot use a single formula for the whole distance; you must calculate the cost for the first part and the second part separately and then add them up. The splitting property $\int_{0}^{2} f(t) dt = \int_{0}^{1} f(t) dt + \int_{1}^{2} f(t) dt$ is our mathematical tool to handle this change in behavior.

Q: I got the range $[1/2, 3/2]$, but the options are different. Did I make a mistake?

Not at all! You are likely correct. The range you found, $[1/2, 3/2]$, is the tightest possible bound for $g(2)$. However, in many JEE problems, the options might provide a larger interval that *contains* your calculated range. If your range is $[0.5, 1.5]$, and one of the options is $[0, 2]$, your answer is correct because every value in your range is also in the option's range. Always check if your interval is a subset of the given options.

Q: Why can't we just use the average of the bounds to find the value of $g(2)$?

That is a common trap! The problem asks for the *range* of values, not a single value. Because $f(t)$ is not a fixed function but rather a function that can take any value within the given bounds, $g(2)$ can also take any value within the resulting range. We are finding the 'envelope' of all possible values, not the average. By bounding the integral, we are essentially finding the minimum and maximum possible area that the curve could enclose.

Question

Let $g(x) = \int_{0}^{x} f(t) dt$, where $f$ is such that $\frac{1}{2} \le f(t) \le 1$ for $t \in [0, 1]$ and $0 \le f(t) \le \frac{1}{2}$ for $t \in [1, 2]$. Then $g(2)$ satisfies the inequality

Options:

(A) $-\frac{3}{2} \le g(2) < \frac{1}{2}$

(B) $0 \le g(2) < 2$

(C) $\frac{3}{2} < g(2) \le \frac{5}{2}$

(D) $2 < g(2) < 4$

Sigma Percentile · Accepted Answer

### The Detective's Approach to Calculus

Imagine you are standing before a graph. You have a function $f(t)$, but it is elusive. You do not know its exact shape, but you know its secrets.

You know exactly where it is allowed to roam. In the first part of its journey, from $t=0$ to $t=1$, it is trapped between the horizontal lines $y = \frac{1}{2}$ and $y = 1$.

In the second part, from $t=1$ to $t=2$, it is forced to drop, confined between $y = 0$ and $y = \frac{1}{2}$. Your mission is to determine the possible values of the total area under this curve, defined as:

$$g(2) = \int_{0}^{2} f(t) dt$$

This is a detective story where we use the constraints to corner the truth.

### Phase 1: The Art of Splitting

The first step in our investigation is to recognize that we cannot treat the entire journey as one. The function $f(t)$ changes its rules at $t=1$.

In calculus, when the rules change, we change our strategy. We use the additive property of definite integrals to break the problem into two manageable pieces:

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

By splitting the integral, we isolate the behavior of $f(t)$ in each region. This allows us to apply the constraints effectively, dividing a complex task into two simpler sub-tasks.

### Phase 2: Cornering the Function

Now, let us look at the first interval, $t \in [0, 1]$. We are told that $\frac{1}{2} \le f(t) \le 1$.

Geometrically, this means the area under the curve is trapped within a rectangle of width 1 and height between $\frac{1}{2}$ and 1. When we integrate this inequality, we calculate the area of these bounding rectangles:

$$\int_{0}^{1} \frac{1}{2} dt = \frac{1}{2} \quad 	ext{and} \quad \int_{0}^{1} 1 dt = 1$$

Thus, the area of the first part, $\int_{0}^{1} f(t) dt$, must lie somewhere in the interval $[\frac{1}{2}, 1]$.

### Phase 3: The Second Act

Next, we turn our attention to the second interval, $t \in [1, 2]$. Here, the function is restricted to $0 \le f(t) \le \frac{1}{2}$.

Again, we integrate the inequality over the interval of length 1. The lower bound is:

$$\int_{1}^{2} 0 dt = 0$$

The upper bound is:

$$\int_{1}^{2} \frac{1}{2} dt = \frac{1}{2}$$

So, the area of the second part, $\int_{1}^{2} f(t) dt$, is confined to the interval $[0, \frac{1}{2}]$.

### Phase 4: The Grand Summation

We have successfully cornered both parts of the integral. To find the total area $g(2)$, we simply add the bounds of the two intervals.

The minimum possible value for $g(2)$ is the sum of the minimums:

$$\frac{1}{2} + 0 = \frac{1}{2}$$

The maximum possible value is the sum of the maximums:

$$1 + \frac{1}{2} = \frac{3}{2}$$

Therefore, we have determined that $g(2)$ must reside in the interval $[\frac{1}{2}, \frac{3}{2}]$.

### Phase 5: The Final Verdict

Finally, we look at our options. We found that $g(2) \in [0.5, 1.5]$.

We need to select the option that encompasses this range. Option B, which states $0 \le g(2) < 2$, is the correct choice because our calculated range $[0.5, 1.5]$ is entirely contained within $

Animated Solution for Mathematics - Definite Integration: 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 $g (2)$ satisfies the inequality

Select Answer:

Visualized Solution

Defining the Target Integral $g (2)$

Splitting the Integral at $t = 1$

Bounding the First Interval $[0, 1]$

Integrating the First Interval

Bounding the Second Interval $[1, 2]$

Integrating the Second Interval

Summing the Inequalities

Matching with the Options

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Detective's Approach to Calculus

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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