Considering only the principal values of inverse functions, the set A = {x ≥ 0 : tan⁻¹(2x) + tan⁻¹(3

Q: Wait, why do we need to check the condition $AB < 1$? Can't we just solve the algebra?

That is a brilliant question! The identity $\tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$ is only valid when $AB 1$, the sum actually equals $\pi + \tan^{-1}\left(\frac{A+B}{1-AB}\right)$. Ignoring this condition is a classic trap in JEE Advanced that can lead to extraneous solutions or missing valid ones. Always verify your result against the domain of the identity!

Q: Why do we reject $x = -1$? It is a valid root of the quadratic equation, right?

Mathematically, yes, $x = -1$ is a root of the quadratic $6x^2 + 5x - 1 = 0$. However, the problem statement explicitly defines set $A$ as $\{x \ge 0 : \dots\}$. This is a constraint imposed by the problem setter. In competitive exams, you must respect the domain constraints as much as the algebraic steps. Since $-1$ is not greater than or equal to zero, it is physically and logically excluded from our set.

Q: Is there a graphical way to see why there is only one solution?

Absolutely! If you plot $f(x) = \tan^{-1}(2x) + \tan^{-1}(3x)$, you will see that it is a strictly increasing function for $x \ge 0$. Since it is strictly increasing, it can cross the horizontal line $y = \pi/4$ at most once. This confirms that our algebraic finding of a single solution is consistent with the geometric behavior of the function.

Question

Considering only the principal values of inverse functions, the set $A = \{x \ge 0 : 	an^{-1}(2x) + 	an^{-1}(3x) = \pi/4\}$

Options:

(A) is an empty set

(B) Contains more than two elements

(C) Contains two elements

(D) is a singleton

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are not just solving an equation; we are embarking on a journey through the landscape of inverse trigonometric functions. The problem before us, $	an^{-1}(2x) + 	an^{-1}(3x) = \frac{\pi}{4}$, is a classic JEE Advanced challenge.

It tests your algebraic dexterity, but more importantly, it tests your discipline. Let us break this down step by step.

### The Boundary of Reality

Before we touch a single variable, we must look at the constraint: $x \ge 0$. In the world of competitive exams, constraints are not mere suggestions; they are the boundaries of our universe.

We are searching for a value of $x$ that satisfies our equation while remaining in the non-negative domain. Imagine standing on the $x$-axis, looking only to the right. This is our search space.

If our algebra leads us to a negative number, we must have the courage to reject it.

### The Golden Key

How do we merge two inverse tangent functions? We reach into our mathematical toolkit and pull out the addition identity:

$$	an^{-1}(A) + 	an^{-1}(B) = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$$

This is our golden key. However, remember the fine print: this identity holds true only when the product $AB < 1$. If $AB$ were greater than $1$, the identity would shift by $\pi$.

We will keep this in mind as we proceed. Let us set $A = 2x$ and $B = 3x$. Substituting these into our identity, we get:

$$	an^{-1}\left(\frac{2x + 3x}{1 - (2x)(3x)}ight) = \frac{\pi}{4}$$

### The Algebraic Metamorphosis

Now, the magic happens. We simplify the expression inside the inverse tangent. The numerator, $2x + 3x$, becomes $5x$. The denominator, $1 - (2x)(3x)$, becomes $1 - 6x^2$.

Our equation now stands as:

$$	an^{-1}\left(\frac{5x}{1 - 6x^2}ight) = \frac{\pi}{4}$$

To isolate $x$, we apply the tangent function to both sides. Since $	an(	an^{-1}(	heta)) = 	heta$, the left side sheds its inverse operator, and the right side becomes $	an(\frac{\pi}{4})$. We know that $	an(\frac{\pi}{4}) = 1$.

Thus, we arrive at the beautiful, clean algebraic equation:

$$\frac{5x}{1 - 6x^2} = 1$$

### The Final Verdict

Cross-multiplying gives us $5x = 1 - 6x^2$. Rearranging this into the standard quadratic form, we get:

$$6x^2 + 5x - 1 = 0$$

Factoring this quadratic is a breeze: we need two numbers that multiply to $-6$ and add to $5$. Those numbers are $6$ and $-1$. We rewrite the equation as $6x^2 + 6x - x - 1 = 0$, which factors into $(6x - 1)(x + 1) = 0$.

This yields two potential roots: $x = \frac{1}{6}$ and $x = -1$. Now, we return to our boundary. We recall our constraint $x \ge 0$. The root $x = -1$ is immediately rejected. The root $x = \frac{1}{6}$ is accepted.

Finally, we verify our identity condition:

$$AB = (2x)(3x) = 6x^2 = 6\left(\frac{1}{6}ight)^2 = 6\left(\frac{1}{36}ight) = \frac{1}{6}$$

Since $\frac{1}{6} < 1$, our identity was valid. We have found exactly one solution: **$x = \frac{1}{6}$**. Therefore, set $A$ is a singleton. You have navigated the traps, respected the constraints, and arrived at the truth. Well done!

Animated Solution for Mathematics - Inverse Trigonometric Functions: Considering only the principal values of inverse functions, the set $A = {x \geq 0 : tan^{- 1} (2 x) + tan^{- 1} (3 x) = π /4}$

Select Answer:

Visualized Solution

Analyze the Equation and Constraint $x \geq 0$

The Inverse Tangent Addition Formula

Substitute into the Formula

Simplify the Argument

Remove the Inverse Tangent

Formulate the Quadratic Equation

Factorize the Quadratic Equation

Find the Roots

Verify Constraints and Conclude

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

The Boundary of Reality

The Golden Key

The Algebraic Metamorphosis

The Final Verdict

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