The value of α for which 4α ∫₋₁² e⁻ᵃ|ˣ| dx = 5, is : (A) logₑ 2 (B) logₑ √2 (C) logₑ (4/3) (D) logₑ

Q: Why can't we just integrate $e^{-\alpha|x|}$ directly without splitting?

That is a common instinct, but it is a trap! The modulus function $|x|$ changes its algebraic definition at $x=0$. If you try to integrate across $x=0$ without splitting, you are essentially treating the function as if it were the same on both sides, which it is not. The slope flips sign at the origin. Always respect the boundaries of piecewise functions; splitting the integral is the only way to ensure mathematical rigor.

Q: Why do we reject the root $t = -\frac{3}{2}$?

Great question! We substituted $t = e^{-\alpha}$. The range of the exponential function $e^u$ is always $(0, \infty)$ for any real $u$. Since $e^{-\alpha}$ must be positive, a negative value like $-\frac{3}{2}$ is physically and mathematically impossible in this context. Always check the domain of your substitution variables before accepting a root.

Q: Is there a symmetry shortcut here since the function is even?

You have a sharp eye! The function $f(x) = e^{-\alpha|x|}$ is indeed an even function. However, symmetry properties like $\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx$ only apply when the limits are symmetric about the origin. Here, our limits are $-1$ to $2$, which are asymmetric. Because the interval is not symmetric, the 'even function' property does not simplify the limits, so splitting the integral is the only robust path forward.

Question

The value of $\alpha$ for which $4\alpha \int_{-1}^{2} e^{-\alpha|x|} dx = 5$, is :

Options:

(A) $\log_e 2$

(B) $\log_e \sqrt{2}$

(C) $\log_e (\frac{4}{3})$

(D) $\log_e (\frac{3}{2})$

Sigma Percentile · Accepted Answer

Welcome, future engineers. Today, we are going to dissect a problem that perfectly bridges the gap between pure calculus and algebraic intuition. We are tasked with finding the value of $\alpha$ for the equation:

$$4\alpha \int_{-1}^{2} e^{-\alpha|x|} dx = 5$$

At first glance, this looks like a standard integration problem, but it is actually a test of your ability to handle piecewise functions and transform transcendental equations into solvable algebra.

### The Anatomy of the Modulus

The first thing that should catch your eye is the term $|x|$ inside the exponential. The modulus function is the gatekeeper of this problem. It is defined as $|x| = -x$ when $x < 0$ and $|x| = x$ when $x \geq 0$.

This means our function $y = e^{-\alpha|x|}$ is essentially a 'tent' function. It rises from the left, peaks at the y-axis, and decays to the right.

Because the definition of the function changes at $x = 0$, we cannot simply integrate from $-1$ to $2$ in one go. We must split the integral at the point of discontinuity, which is $x = 0$. This gives us two distinct regions: the left region from $-1$ to $0$ and the right region from $0$ to $2$.

### The Calculus of Two Worlds

Let us tackle the left region first. In the interval $[-1, 0]$, $x$ is negative, so $|x| = -x$. Our integral becomes:

$$\int_{-1}^{0} e^{-\alpha(-x)} dx = \int_{-1}^{0} e^{\alpha x} dx$$

Integrating this, we get:

$$\left[ \frac{e^{\alpha x}}{\alpha} \right]_{-1}^{0} = \frac{e^0 - e^{-\alpha}}{\alpha} = \frac{1 - e^{-\alpha}}{\alpha}$$

Now, let us move to the right region, the interval $[0, 2]$. Here, $x$ is positive, so $|x| = x$. Our integral is:

$$\int_{0}^{2} e^{-\alpha x} dx = \left[ \frac{e^{-\alpha x}}{-\alpha} \right]_{0}^{2}$$

Substituting the limits, we get:

$$\frac{e^{-2\alpha} - e^0}{-\alpha} = \frac{1 - e^{-2\alpha}}{\alpha}$$

### The Algebraic Transformation

Now, we bring these two pieces together. Our original equation was:

$$4\alpha \left( \int_{-1}^{0} e^{\alpha x} dx + \int_{0}^{2} e^{-\alpha x} dx \right) = 5$$

Substituting our results, we get:

$$4\alpha \left( \frac{1 - e^{-\alpha}}{\alpha} + \frac{1 - e^{-2\alpha}}{\alpha} \right) = 5$$

Notice the elegance here: the $\alpha$ in the denominator cancels perfectly with the $\alpha$ outside the bracket. We are left with:

$$4(1 - e^{-\alpha} + 1 - e^{-2\alpha}) = 5$$

Simplifying this, we get $4(2 - e^{-\alpha} - e^{-2\alpha}) = 5$, which expands to $8 - 4e^{-\alpha} - 4e^{-2\alpha} = 5$. Rearranging everything to one side, we arrive at:

$$4e^{-2\alpha} + 4e^{-\alpha} - 3 = 0$$

This is the moment where the problem shifts from calculus to algebra. By substituting $t = e^{-\alpha}$, we transform this into a quadratic equation:

$$4t^2 + 4t - 3 = 0$$

### The Final Victory

Factoring this quadratic is straightforward. We split the middle term to get $4t^2 + 6t - 2t - 3 = 0$, which factors into:

$$(2t - 1)(2t + 3) = 0$$

This gives us two potential roots: $t = 1/2$ and $t = -3/2$. However, we must pause and think. Since $t = e^{-\alpha}$, and the exponential function is always positive, we must reject the negative root.

Thus, $e^{-\alpha} = 1/2$. Taking the natural logarithm on both sides, we get $-\alpha = \ln(1/2)$, which is $-\ln 2$.

Therefore, **$\alpha = \ln 2$**. You have successfully navigated the modulus, the integration, and the algebraic transformation. This is the essence of JEE Advanced mathematics: breaking down a complex problem into manageable, elegant steps.

Animated Solution for Mathematics - Definite Integration: The value of $α$ for which $4 α \int_{- 1}^{2} e^{- α ∣ x ∣} d x = 5$ , is :

Select Answer:

Visualized Solution

The Definite Integral Setup

Breaking the Modulus

Splitting the Integral at $x = 0$

Evaluating the Left Region

Evaluating the Right Region

Combining the Evaluated Parts

Simplifying the Equation

Transforming to a Quadratic

Solving the Quadratic Equation

Selecting the Valid Root

Finding $α$

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Anatomy of the Modulus

The Calculus of Two Worlds

The Algebraic Transformation

The Final Victory

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of α for which 4α∫−12​e−α∣x∣dx=5, is :