The set of all values of k for which (tan⁻¹ x)³ + (cot⁻¹ x)³ = kπ³, x ∈ R, is the interval : (A) [1/

Q: Wait, why can't we just treat $\tan^{-1} x$ and $\cot^{-1} x$ as independent variables and find the range of $a^3 + b^3$?

That is a classic trap! If they were independent, you could vary them freely. But they are linked by the constraint $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$. Because $x$ is the same for both, they are locked in a dance—when one increases, the other must decrease. You must use this constraint to reduce the expression to a single variable.

Q: Why is the range of $t = \tan^{-1} x$ restricted to $(-\frac{\pi}{2}, \frac{\pi}{2})$?

Great question! This comes from the definition of the inverse trigonometric functions. The function $y = \tan x$ is periodic and not one-to-one over its entire domain. To define an inverse, we restrict the domain of $\tan x$ to the principal branch, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$. Consequently, the range of $\tan^{-1} x$ is strictly limited to this interval.

Q: My calculation shows the range of $k$ is $[\frac{1}{32}, \frac{7}{8})$, but the options are all closed intervals. Did I make a mistake?

You are mathematically correct! The upper bound is indeed an open interval because $\tan^{-1} x$ never actually reaches $\frac{\pi}{2}$ or $-\frac{\pi}{2}$. However, in many JEE questions, the options provided are simplified or represent the closure of the interval. Always choose the option that encompasses your derived range.

Question

The set of all values of $k$ for which $(	an^{-1} x)^3 + (\cot^{-1} x)^3 = k\pi^3, x \in \mathbb{R}$, is the interval :

Options:

(A) $[\frac{1}{32}, \frac{7}{8}]$

(B) $[\frac{1}{24}, \frac{13}{16}]$

(C) $[\frac{1}{48}, \frac{13}{16}]$

(D) $[\frac{1}{32}, \frac{9}{8}]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex, intimidating equation: $(	an^{-1} x)^3 + (\cot^{-1} x)^3 = k\pi^3$. At first glance, it looks like a chaotic mess of cubes and inverse functions.

In the world of JEE mathematics, complexity is often just a mask for a beautiful, underlying simplicity. Our goal is to find the set of all values of $k$ for which this equation holds true for any real $x$.

### The Golden Key

The first step is to recognize that we are not dealing with two independent variables. We have a powerful, elegant identity in our toolkit:

$$	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$$

This is our golden key. It tells us that $\cot^{-1} x$ is not a stranger; it is simply $\frac{\pi}{2} - 	an^{-1} x$.

By using this, we can collapse the entire problem into a single variable. Let us define $t = 	an^{-1} x$. Now, our expression $S$ becomes:

$$S = t^3 + \left( \frac{\pi}{2} - t ight)^3$$

### The Algebraic Transformation

We have an expression in the form of $a^3 + b^3$. While we could expand it blindly, let us use the identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$.

Here, $a = t$ and $b = \frac{\pi}{2} - t$. Notice the magic: $a+b = t + (\frac{\pi}{2} - t) = \frac{\pi}{2}$.

Substituting this into our identity, we get:

$$S = \left( \frac{\pi}{2} ight)^3 - 3t \left( \frac{\pi}{2} - t ight) \left( \frac{\pi}{2} ight)$$

This simplifies beautifully to:

$$S = \frac{\pi^3}{8} - \frac{3\pi}{2}t \left( \frac{\pi}{2} - t ight)$$

Expanding this, we arrive at a quadratic expression:

$$S = \frac{3\pi}{2}t^2 - \frac{3\pi^2}{4}t + \frac{\pi^3}{8}$$

### Completing the Square

To understand the behavior of this quadratic, we must complete the square. Let us factor out $\frac{3\pi}{2}$:

$$S = \frac{3\pi}{2} \left[ t^2 - \frac{\pi}{2}t ight] + \frac{\pi^3}{8}$$

Adding and subtracting $\left( \frac{\pi}{4} ight)^2$, we get:

$$S = \frac{3\pi}{2} \left[ \left( t - \frac{\pi}{4} ight)^2 - \frac{\pi^2}{16} ight] + \frac{\pi^3}{8}$$

Simplifying this, we find:

$$S = \frac{3\pi}{2} \left( t - \frac{\pi}{4} ight)^2 - \frac{3\pi^3}{32} + \frac{4\pi^3}{32}$$

This results in:

$$S = \frac{3\pi}{2} \left( t - \frac{\pi}{4} ight)^2 + \frac{\pi^3}{32}$$

### The Domain Trap

Now, we must be careful. We are not working in the realm of all real numbers for $t$. Since $t = 	an^{-1} x$, we are strictly bound by the domain $t \in \left( -\frac{\pi}{2}, \frac{\pi}{2} ight)$.

The minimum value of the squared term $\left( t - \frac{\pi}{4} ight)^2$ is $0$, which occurs at $t = \frac{\pi}{4}$. Since $\frac{\pi}{4}$ is within our domain, the minimum value of $S$ is $\frac{\pi^3}{32}$.

For the maximum, we look at the boundaries. As $t$ approaches $-\frac{\pi}{2}$, the term $\left( t - \frac{\pi}{4} ight)^2$ approaches:

$$\left( -\frac{\pi}{2} - \frac{\pi}{4} ight)^2 = \left( -\frac{3\pi}{4} ight)^2 = \frac{9\pi^2}{16}$$

Plugging this back into our expression for $S$:

$$S 	o \frac{3\pi}{2} \left( \frac{9\pi^2}{16} ight) + \frac{\pi^3}{32} = \frac{27\pi^3}{32} + \frac{\pi^3}{32} = \frac{28\pi^3}{32} = \frac{7\pi^3}{8}$$

### The Final Victory

We have determined that the range of $S$ is $\left[ \frac{\pi^3}{32}, \frac{7\pi^3}{8} ight)$. Since $S = k\pi^3$, we divide the entire inequality by $\pi^3$ to find the range of $k$:

$$k \in \left

Animated Solution for Mathematics - Inverse Trigonometric Functions: The set of all values of $k$ for which $(tan^{- 1} x)^{3} + (cot^{- 1} x)^{3} = k π^{3}, x \in R$ , is the interval :

Select Answer:

Visualized Solution

Identify the Core Expression

Recall the Fundamental Identity

Apply Algebraic Identity $a^{3} + b^{3}$

Substitute the Sum Identity

Reduce to a Single Variable

Formulate the Quadratic Expression

Complete the Square

Analyze the Domain of $t$

Find the Minimum Value

Find the Maximum Value

Determine the Interval for $k$

Final Conclusion

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Golden Key

The Algebraic Transformation

Completing the Square

The Domain Trap

The Final Victory

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Animated Solution for Mathematics - Inverse Trigonometric Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The set of all values of k for which (tan−1x)3+(cot−1x)3=kπ3,x∈R, is the interval :