Let (a, b) ⊂ (0, 2π) be the largest interval for which sin⁻¹(sin θ) - cos⁻¹(sin θ) 0, θ ∈ (0, 2π),

Q: Wait, why do we just assume $x=3$ in the second equation? Can't $x$ be a range of values?

Great observation! Look closely at the argument $u = x^2 - 6x + 10$. We can rewrite this as $(x-3)^2 + 1$. Since $(x-3)^2$ is always non-negative, the minimum value of $u$ is $1$. However, the domain of $\sin^{-1}(u)$ and $\cos^{-1}(u)$ is strictly $[-1, 1]$. If $u$ must be $\le 1$ and we know $u ≥ 1$, the only possible value is $u=1$. This forces $(x-3)^2 = 0$, meaning $x=3$ is the only valid input!

Q: Why did we replace $\cos^{-1}(\sin \theta)$ with $\frac{\pi}{2} - \sin^{-1}(\sin \theta)$?

Because it simplifies the expression! By converting everything into terms of $\sin^{-1}(\sin \theta)$, we turn a complex inequality involving two different inverse functions into a linear inequality in one variable. Always look for identities that reduce the number of distinct functions in your equation.

Q: How do I know if I should use the principal value branch for $\sin^{-1}(\sin \theta)$?

In the interval $(0, 2\pi)$, the function $\sin^{-1}(\sin \theta)$ behaves differently depending on the quadrant. However, since we are solving an inequality, we can simply use the property that $\sin^{-1}(\sin \theta) > \frac{\pi}{4}$ is equivalent to $\sin \theta > \sin(\frac{\pi}{4})$ within the relevant range. The unit circle is your best friend here!

Q: Is it always true that $\sin^{-1}(u) + \cos^{-1}(u) = \frac{\pi}{2}$?

Yes, absolutely! This is a standard identity for all $u$ in the domain $[-1, 1]$. It is one of the most powerful tools in your JEE toolkit because it allows you to collapse two complex terms into a single constant.

Q: I got $\alpha = \frac{\pi}{12}$, but what if I made a sign error in the $\alpha - \beta$ substitution?

Sign errors are the most common trap in JEE! Always double-check your substitution. We had $\beta = \alpha - \frac{\pi}{2}$. Plugging this into $9\alpha + 3\beta + \frac{\pi}{2} = 0$ gives $9\alpha + 3(\alpha - \frac{\pi}{2}) + \frac{\pi}{2} = 0$. If you distribute the $3$ correctly, you get $12\alpha - \frac{3\pi}{2} + \frac{\pi}{2} = 0$, which simplifies perfectly to $12\alpha = \pi$. Keep your algebra clean and the signs will follow!

Question

Let $(a, b) \subset (0, 2\pi)$ be the largest interval for which $\sin^{-1}(\sin 	heta) - \cos^{-1}(\sin 	heta) > 0, 	heta \in (0, 2\pi)$, holds. If $\alpha x^2 + \beta x + \sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0$ and $\alpha - \beta = b - a$, then $\alpha$ is equal to :

Options:

(A) $\frac{\pi}{48}$

(B) $\frac{\pi}{16}$

(C) $\frac{\pi}{8}$

(D) $\frac{\pi}{12}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that looks like a tangled knot of trigonometry and algebra, but beneath the surface, it is a beautiful, elegant dance.

Imagine you are standing on the edge of a unit circle, watching the values of $\sin(	heta)$ oscillate as $	heta$ sweeps through the interval $(0, 2\pi)$. Our goal is to find the region where the difference between $\sin^{-1}(\sin 	heta)$ and $\cos^{-1}(\sin 	heta)$ is positive.

### Simplifying the Inequality

At first glance, the inequality $\sin^{-1}(\sin 	heta) - \cos^{-1}(\sin 	heta) > 0$ seems daunting. But remember the identity that binds these two functions together: $\sin^{-1}(z) + \cos^{-1}(z) = \frac{\pi}{2}$.

This is the key that unlocks the door. By substituting $\cos^{-1}(\sin 	heta) = \frac{\pi}{2} - \sin^{-1}(\sin 	heta)$, our inequality transforms into:

$$\sin^{-1}(\sin 	heta) - \left(\frac{\pi}{2} - \sin^{-1}(\sin 	heta)ight) > 0$$

Simplifying this, we get $2\sin^{-1}(\sin 	heta) > \frac{\pi}{2}$, which leads us to the clean, manageable statement: $\sin^{-1}(\sin 	heta) > \frac{\pi}{4}$. Applying the sine function to both sides, we find the condition $\sin 	heta > \sin(\frac{\pi}{4})$, or simply $\sin 	heta > \frac{1}{\sqrt{2}}$.

### Mapping the Interval

Now, visualize the unit circle. Where is the sine value greater than $\frac{1}{\sqrt{2}}$? This occurs in the first and second quadrants, specifically between $\frac{\pi}{4}$ and $\frac{3\pi}{4}$.

Thus, our interval $(a, b)$ is $(\frac{\pi}{4}, \frac{3\pi}{4})$. With $a = \frac{\pi}{4}$ and $b = \frac{3\pi}{4}$, the difference $b - a$ is simply $\frac{\pi}{2}$. We have successfully conquered the first half of our journey!

### The Hidden Constraint

Now, look at the second equation: $\alpha x^2 + \beta x + \sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0$. This looks like a nightmare, but look at the argument of the inverse trigonometric functions: $u = x^2 - 6x + 10$.

If we complete the square, we get $u = (x-3)^2 + 1$. Here is the **"Aha!" moment**: the domain of $\sin^{-1}(u)$ and $\cos^{-1}(u)$ is strictly $[-1, 1]$.

Since $(x-3)^2$ is always $\ge 0$, the minimum value of $u$ is $1$. For the function to be defined, we must have $u \le 1$. The only way for $u \ge 1$ and $u \le 1$ to both be true is if $u = 1$. This forces $(x-3)^2 = 0$, which means $x$ must be $3$.

### The Final Victory

With $x=3$, the equation becomes $\alpha(3)^2 + \beta(3) + \sin^{-1}(1) + \cos^{-1}(1) = 0$. Since $\sin^{-1}(1) + \cos^{-1}(1) = \frac{\pi}{2}$, we have $9\alpha + 3\beta + \frac{\pi}{2} = 0$.

We also know from the problem statement that $\alpha - \beta = b - a = \frac{\pi}{2}$, which means $\beta = \alpha - \frac{\pi}{2}$. Substituting this into our equation:

$$9\alpha + 3\left(\alpha - \frac{\pi}{2}ight) + \frac{\pi}{2} = 0$$

$$12\alpha - \frac{3\pi}{2} + \frac{\pi}{2} = 0$$

$$12\alpha - \pi = 0$$

$$\alpha = \frac{\pi}{12}$$

And there it is. Through careful observation and the application of fundamental identities, we have navigated the complexity to find the elegant solution of **$\alpha = \frac{\pi}{12}$**.

Select Answer:

Visualized Solution

Introduction to the Inequality

Using the Identity $sin^{- 1} z + cos^{- 1} z = \frac{π}{2}$

Simplifying the Inequality

Isolating $sin θ$

Finding the Interval $(a, b)$

Calculating $b - a$

Analyzing the Second Equation

Domain Constraint of Inverse Trig Functions

Finding the Unique Value of $x$

Substituting $x = 3$

Simplifying the Equation

Using the Given Relation

Solving for $α$

Final Calculation

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

Simplifying the Inequality

Mapping the Interval

The Hidden Constraint

The Final Victory

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Select Answer:

Visualized Solution

Introduction to the Inequality

Using the Identity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin−1z+cos−1z=2π​

Simplifying the Inequality

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

Finding the Interval .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a,b)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b−a

Analyzing the Second Equation

Domain Constraint of Inverse Trig Functions

Finding the Unique Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=3

Simplifying the Equation

Using the Given Relation

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

Final Calculation

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

Simplifying the Inequality

Mapping the Interval

The Hidden Constraint

The Final Victory

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