Let the functions f : R → R and g : R → R be defined by f(x) = e^(x-1) – e^(-|x-1|) and g(x) = 1/2(e

Q: Wait, why do we split the integral at $x=1$? Can't we just integrate from $0$ to the intersection point?

That is a fantastic question! If you try to integrate from $0$ to the intersection point using a single formula, you are assuming the lower boundary function $f(x)$ remains the same throughout. But look closely at the definition of $f(x)$. For $x \le 1$, $f(x) = 0$. For $x > 1$, $f(x) = e^{x-1} - e^{1-x}$. Because the 'floor' of your area changes its mathematical identity at $x=1$, you must treat the two regions separately. It is like walking on a path that suddenly changes from concrete to grass; you have to adjust your stride!

Q: How did $f(x)$ become zero for $x \le 1$? I thought exponential functions were never zero.

You are absolutely right that $e^u$ is never zero. However, $f(x)$ is not just an exponential; it is a difference of two exponentials. When $x \le 1$, the term $|x-1|$ becomes $1-x$. So, $f(x) = e^{x-1} - e^{-(1-x)}$. Notice that $-(1-x)$ is just $x-1$. Thus, $f(x) = e^{x-1} - e^{x-1}$, which is identically zero. It is a beautiful algebraic cancellation that simplifies the problem immensely!

Q: Why did we reject the negative root when solving $t^2 = 3$?

We defined our substitution as $t = e^{x-1}$. Since the exponential function $e^u$ is always positive for any real $u$, $t$ must be strictly positive. Furthermore, since we are looking for the intersection point where $x > 1$, $x-1 > 0$, which implies $t = e^{x-1} > e^0 = 1$. Therefore, $t = -\sqrt{3}$ is physically impossible in the context of this problem. Always check the domain of your substitutions!

Question

Let the functions $f: \mathbb{R} 	o \mathbb{R}$ and $g: \mathbb{R} 	o \mathbb{R}$ be defined by $f(x) = e^{x-1} - e^{-|x-1|}$ and $g(x) = \frac{1}{2}(e^{x-1} + e^{1-x})$. Then the area of the region in the first quadrant bounded by the curves $y = f(x)$, $y = g(x)$ and $x = 0$ is

Options:

(A) $(2 - \sqrt{3}) + \frac{1}{2}(e - e^{-1})$

(B) $(2 + \sqrt{3}) + \frac{1}{2}(e - e^{-1})$

(C) $(2 - \sqrt{3}) + \frac{1}{2}(e + e^{-1})$

(D) $(2 + \sqrt{3}) + \frac{1}{2}(e + e^{-1})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a calculus problem; we are peeling back the layers of a function to reveal the elegant geometry hidden beneath.

We are tasked with finding the area bounded by $f(x) = e^{x-1} - e^{-|x-1|}$, $g(x) = \frac{1}{2}(e^{x-1} + e^{1-x})$, and the $y$-axis ($x=0$). At first glance, this looks like a standard area-under-the-curve problem.

However, in the world of JEE Advanced, the devil is always in the details—specifically, in that absolute value sign.

### Decoding the Piecewise Nature

The function $f(x) = e^{x-1} - e^{-|x-1|}$ is a wolf in sheep's clothing. The modulus $|x-1|$ is a signal, a warning light that tells us to stop and find the critical point where the expression inside the modulus vanishes. That happens at $x=1$.

Let us analyze the two worlds created by this hinge point:

1. **The Region $x \le 1$:** Here, $x-1$ is negative, so $|x-1| = -(x-1) = 1-x$. Substituting this into our function, we get $f(x) = e^{x-1} - e^{-(1-x)} = e^{x-1} - e^{x-1} = 0$.

The complex exponential expression collapses into a flat, horizontal line lying perfectly on the $x$-axis. Our lower boundary for the first part of the area is simply $y=0$.

2. **The Region $x > 1$:** Here, $x-1$ is positive, so $|x-1| = x-1$. Our function becomes $f(x) = e^{x-1} - e^{-(x-1)} = e^{x-1} - e^{1-x}$. This is a growing, curved function that will serve as our lower boundary for the second part of the area.

### The Hyperbolic Dance

Now, let us turn our attention to $g(x) = \frac{1}{2}(e^{x-1} + e^{1-x})$. If you have spent time with hyperbolic functions, you might recognize this as a shifted version of $\cosh(x-1)$.

It is a beautiful, symmetric curve that reaches its minimum value of $1$ at $x=1$. Because $g(x)$ is always positive and sits above $f(x)$ in our region of interest, it will act as our upper boundary.

We are essentially calculating the area trapped between this hyperbolic curve and the piecewise floor we just defined.

### The Intersection

To find the right-hand limit of our area, we must find where $f(x)$ and $g(x)$ meet. We set them equal:

$$e^{x-1} - e^{1-x} = \frac{1}{2}(e^{x-1} + e^{1-x})$$

Let $t = e^{x-1}$. Since $x > 1$, we know $t > 1$. Our equation transforms into:

$$t - \frac{1}{t} = \frac{1}{2}\left(t + \frac{1}{t}\right)$$

Multiplying by $2$ and rearranging, we get $2t - \frac{2}{t} = t + \frac{1}{t}$, which simplifies to $t = \frac{3}{t}$, or $t^2 = 3$. Since $t > 1$, we take the positive root: $t = \sqrt{3}$.

Thus, $e^{x-1} = \sqrt{3}$, which gives us our upper limit: $x = 1 + \ln(\sqrt{3})$.

### The Integration

We are ready to calculate the total area $A$. Because the lower boundary changes at $x=1$, we split the integral into two acts:

**Act 1 ($0$ to $1$):** The area is $\int_0^1 (g(x) - 0) dx$. Integrating $\frac{1}{2}(e^{x-1} + e^{1-x})$, we get:

$$A_1 = \frac{1}{2}\left[e^{x-1} - e^{1-x}\right]_0^1 = \frac{1}{2}\left((e^0 - e^0) - (e^{-1} - e^1)\right) = \frac{1}{2}\left(e - \frac{1}{e}\right)$$

**Act 2 ($1$ to $1 + \ln(\sqrt{3})$):** The area is $\int_1^{1+\ln(\sqrt{3})} (g(x) - f(x)) dx$. Simplifying the integrand $g(x) - f(x)$ gives us $-\frac{1}{2}e^{x-1} + \frac{3}{2}e^{1-x}$.

Integrating this, we get:

$$\left[-\frac{1}{2}e^{x-1} - \frac{3}{2}e^{1-x}\right]_1^{1+\ln(\sqrt{3})} = 2 - \sqrt{3}$$

### The Grand Finale

Adding our two acts together, the total area is:

$$A = (2 - \sqrt{3}) + \frac{1}{2}\left(e - \frac{1}{e}\right)$$

Look at what we have achieved. We navigated the piecewise trap, solved a transcendental equation using clever substitution, and carefully partitioned our integral.

This is the essence of JEE Advanced mathematics: not just calculating, but understanding the narrative of the functions. You have done well. Keep this clarity of thought, and no problem will ever be too daunting.

Select Answer:

Visualized Solution

The Bounded Region

Analyzing $f (x)$

$f (x)$ for $x \leq 1$

$f (x)$ for $x > 1$

Graphing $g (x)$

Finding the Intersection

Equating $f (x)$ and $g (x)$

Solving for $x$

The Integration Limits

Setting Up the Area Integral

Evaluating $A_{1}$

Simplifying Integrand for $A_{2}$

Evaluating $A_{2}$

Total Area

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Decoding the Piecewise Nature

The Hyperbolic Dance

The Intersection

The Integration

The Grand Finale

Similar Questions

The area (in sq. units) of the region bounded by the curves $y = 2^{x}$ and $y = ∣ x + 1∣$ , in the first quadrant is :

The area of the region enclosed by the curves $y = e^{x}$ , $y = ∣ e^{x} - 1∣$ and y-axis is:

The area bounded by the curves $y = ∣ x^{2} - 1∣$ and $y = 1$ is

Area of the region bounded by the curve $y = e^{x}$ and lines $x = 0$ and $y = e$ is

The area bounded by the curves $y = ∣ x ∣ - 1$ and $y = - ∣ x ∣ + 1$ is

The area enclosed by the curves $y = lo g_{e} (x + e^{2})$ , $x = lo g_{e} (\frac{2}{y})$ and $x = lo g_{e} 2$ , above the line $y = 1$ is

Let $S$ be the area of the region enclosed by $y = e^{- x^{2}}, y = 0, x = 0$ and $x = 1$ ; then

Compute the area of the region bounded by the curves $y = e x ln x$ and $y = \frac{l n x}{e x}$ where $ln e = 1$ .

Sketch the region bounded by the curves $y = x^{2}$ and $y = \frac{2}{1 + x ^{2}}$ . Find the area.

Select Answer:

Visualized Solution

The Bounded Region

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x≤1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>1

Graphing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Finding the Intersection

Equating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x