Let α ∈ R be such that the function f(x) = {cos⁻¹(1-{x}²) sin⁻¹(1-{x}) / ({x}-{x}³), x ≠ 0; α, x = 0

Q: Wait, why does {x} become x+1 when x approaches 0 from the left?

Great question! The definition of the fractional part is {x} = x - [x]. When x is a tiny negative number (like -0.001), the greatest integer less than or equal to x, [x], is -1. So, {x} = x - (-1) = x + 1. It is a common trap to think {x} is always just x, but the floor function [x] is the key to the left-hand side!

Q: Can I just use L'Hopital's Rule to solve this?

Technically, yes, but it would be a nightmare! Differentiating the inverse trigonometric functions and the fractional part composition would lead to an algebraic mess. In JEE Advanced, we look for elegance. Using standard limits and trigonometric substitutions is much faster and less prone to calculation errors.

Q: Why do we check LHL and RHL separately? Can't we just find the limit?

For a function to be continuous at a point, the limit must exist, and the LHL must equal the RHL. Because the fractional part function has a jump discontinuity at every integer, the function's behavior is fundamentally different on either side of x = 0. We must evaluate them separately to see if they converge to the same value.

Question

Let $\alpha \in \mathbb{R}$ be such that the function $f(x) = \begin{cases} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x 
eq 0 \ \alpha, & x = 0 \end{cases}$ is continuous at $x = 0$, where $\{x\} = x - [x]$, $[x]$ is the greatest integer less than or equal to $x$. Then :

Options:

(A) $\alpha = \frac{\pi}{\sqrt{2}}$

(B) $\alpha = 0$

(C) no such $\alpha$ exists

(D) $\alpha = \frac{\pi}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To determine if the function

$$f(x) = \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}$$

is continuous at $x = 0$, we must verify if the limit as $x$ approaches zero from the left, the limit as $x$ approaches from the right, and the function value at zero, $\alpha$, are all identical.

The gatekeeper to this problem is the fractional part function, $\{x\}$.

### The Shapeshifter: Understanding $\{x\}$

The fractional part function, $\{x\} = x - [x]$, behaves differently depending on which side of the integer you are on.

As $x 	o 0^+$, $x$ is a tiny positive number, so $[x] = 0$, and $\{x\} = x$.

However, as $x 	o 0^-$, $x$ is a tiny negative number, so $[x] = -1$, and $\{x\} = x - (-1) = x + 1$. This split personality requires us to evaluate two separate paths.

### The Right-Hand Journey

Let us approach from the right ($x 	o 0^+$). Substituting $\{x\} = x$, our expression becomes:

$$\lim_{x 	o 0^+} \frac{\cos^{-1}(1-x^2) \sin^{-1}(1-x)}{x(1-x)(1+x)}$$

As $x 	o 0^+$, $\sin^{-1}(1-x) 	o \sin^{-1}(1) = \frac{\pi}{2}$, and the terms $(1-x)$ and $(1+x)$ in the denominator approach $1$. This leaves us with the core of the limit:

$$\frac{\pi}{2} \lim_{x 	o 0^+} \frac{\cos^{-1}(1-x^2)}{x}$$

To solve this, we use the substitution $1-x^2 = \cos 	heta$. As $x 	o 0^+$, $	heta 	o 0^+$, and $x = \sqrt{1 - \cos 	heta} = \sqrt{2} \sin(\frac{	heta}{2})$.

The limit transforms into:

$$\frac{\pi}{2} \lim_{	heta 	o 0^+} \frac{	heta}{\sqrt{2} \sin(\frac{	heta}{2})} = \frac{\pi}{2} \cdot \frac{2}{\sqrt{2}} \cdot 1 = \frac{\pi}{\sqrt{2}}$$

The Right Hand Limit is **$\frac{\pi}{\sqrt{2}}$**.

### The Left-Hand Trap

Now, we evaluate the left side ($x 	o 0^-$), where $\{x\} = 1+x$. The expression becomes:

$$\lim_{x 	o 0^-} \frac{\cos^{-1}(1-(1+x)^2) \sin^{-1}(1-(1+x))}{(1+x)-(1+x)^3}$$

The numerator's $\sin^{-1}(1-(1+x))$ simplifies to $\sin^{-1}(-x)$. The denominator factors to $-x(1+x)(2+x)$.

As $x 	o 0^-$, $\cos^{-1}(1-(1+x)^2) 	o \cos^{-1}(0) = \frac{\pi}{2}$. The limit is now:

$$\lim_{x 	o 0^-} \frac{(\frac{\pi}{2})(-\sin^{-1} x)}{-x(1+x)(2+x)} = \frac{\pi}{2} \lim_{x 	o 0^-} \frac{\sin^{-1} x}{x} \cdot \frac{1}{(1+x)(2+x)}$$

Using the standard limit $\lim_{x 	o 0} \frac{\sin^{-1} x}{x} = 1$, we get:

$$\frac{\pi}{2} \cdot 1 \cdot \frac{1}{1 \cdot 2} = \frac{\pi}{4}$$

The Left Hand Limit is **$\frac{\pi}{4}$**.

### The Verdict

We have found the RHL to be $\frac{\pi}{\sqrt{2}}$ and the LHL to be $\frac{\pi}{4}$.

Since $\frac{\pi}{\sqrt{2}} 
eq \frac{\pi}{4}$, the limit does not exist. No value of $\alpha$ can bridge this gap.

The function is **fundamentally discontinuous** at $x = 0$.

Select Answer:

Visualized Solution

Condition for Continuity at $x = 0$

Behavior of ${x}$ near $x = 0$

Setting up Right Hand Limit (RHL)

Simplifying RHL Expression

Evaluating RHL using Substitution

Final Value of RHL

Setting up Left Hand Limit (LHL)

Simplifying LHL Expression

Final Value of LHL

Conclusion: Checking Continuity

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Shapeshifter

The Right-Hand Journey

The Left-Hand Trap

The Verdict

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