Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if $\lim_{x \to c} f(x) = f(c)$. This implies that the left-hand limit (LHL), the right-hand limit (RHL), and the function value at $c$ must all exist and be equal.
• Definition of Fractional Part Function: The fractional part of $x$, denoted by $\{x\}$, is defined as $\{x\} = x - [x]$, where $[x]$ is the greatest integer less than or equal to $x$.
• Standard Limits:
  • $\lim_{y \to 0} \frac{\sin^{-1} y}{y} = 1$
  • $\lim_{y \to 0} \frac{\tan^{-1} y}{y} = 1$
  • $\lim_{y \to 0} \frac{\cos^{-1} y}{y}$ is undefined.
  • L'Hôpital's Rule: If $\lim_{x \to c} \frac{g(x)}{h(x)}$ is of the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to c} \frac{g(x)}{h(x)} = \lim_{x \to c} \frac{g'(x)}{h'(x)}$, provided the latter limit exists.
• Properties of Inverse Trigonometric Functions:
  • $\cos^{-1}(1) = 0$
  • $\sin^{-1}(0) = 0$
  • $\cos^{-1}(y) + \sin^{-1}(y) = \frac{\pi}{2}$ for $y \in [-1, 1]$.

Step-by-Step Solution

The function is given by f(x) = \left\{ {\matrix{ {{{{{\cos }^{ - 1}}(1 - {{\{ x\} }^2}){{\sin }^{ - 1}}(1 - \{ x\} )} \over {\{ x\} - {{\{ x\} }^3}}},} & {x \ne 0} \cr {\alpha ,} & {x = 0} \cr } } \right. For $f(x)$ to be continuous at $x=0$, we must have $\lim_{x \to 0} f(x) = f(0) = \alpha$. This requires the left-hand limit (LHL) and the right-hand limit (RHL) at $x=0$ to be equal to $\alpha$.

Step 1: Analyze the behavior of {x} near x = 0.

• For $x \to 0^+$, $x$ is a small positive number. Thus, $[x] = 0$, and $\{x\} = x - 0 = x$.
• For $x \to 0^-$, $x$ is a small negative number. Let $x = -\epsilon$ where $\epsilon > 0$ is small. Then $[x] = [- \epsilon] = -1$. So, $\{x\} = x - [-1] = x - (-1) = x + 1$.

Step 2: Calculate the Right-Hand Limit (RHL).

We need to find $\lim_{x \to 0^+} f(x)$. As $x \to 0^+$, $\{x\} = x$. Substituting this into the function definition for $x \ne 0$: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\cos^{-1}(1 - x^2) \sin^{-1}(1 - x)}{x - x^3}$ The denominator can be factored as $x(1 - x^2)$. $\lim_{x \to 0^+} \frac{\cos^{-1}(1 - x^2) \sin^{-1}(1 - x)}{x(1 - x^2)}$ As $x \to 0^+$, $1-x^2 \to 1$ and $1-x \to 1$. Therefore, $\cos^{-1}(1-x^2) \to \cos^{-1}(1) = 0$ and $\sin^{-1}(1-x) \to \sin^{-1}(1) = \frac{\pi}{2}$. This leads to an indeterminate form of $\frac{0 \cdot \frac{\pi}{2}}{0}$, which is $\frac{0}{0}$.

We can rewrite the limit expression: $\lim_{x \to 0^+} \left( \frac{\cos^{-1}(1 - x^2)}{1} \cdot \frac{\sin^{-1}(1 - x)}{1 - x^2} \cdot \frac{1 - x^2}{x(1 - x^2)} \right)$ $= \lim_{x \to 0^+} \cos^{-1}(1 - x^2) \cdot \lim_{x \to 0^+} \frac{\sin^{-1}(1 - x)}{1 - x^2} \cdot \lim_{x \to 0^+} \frac{1}{x}$ This approach seems complicated. Let's try to use standard limits and L'Hôpital's Rule.

Consider the term $\frac{\sin^{-1}(1-x)}{x(1-x^2)}$. As $x \to 0^+$, $1-x \to 1$. Let $y = 1-x$. As $x \to 0^+$, $y \to 1^-$. The expression becomes: $\lim_{x \to 0^+} \frac{\cos^{-1}(1 - x^2) \sin^{-1}(1 - x)}{x(1 - x^2)} = \lim_{x \to 0^+} \cos^{-1}(1 - x^2) \cdot \lim_{x \to 0^+} \frac{\sin^{-1}(1 - x)}{1 - x} \cdot \lim_{x \to 0^+} \frac{1 - x}{x(1 - x^2)}$ This is also not straightforward.

Let's go back to the original form and apply L'Hôpital's Rule directly to the indeterminate part. $\lim_{x \to 0^+} \frac{\cos^{-1}(1 - x^2) \sin^{-1}(1 - x)}{x(1 - x^2)}$ As $x \to 0^+$, $\cos^{-1}(1-x^2) \to 0$ and $\sin^{-1}(1-x) \to \frac{\pi}{2}$. The expression is $\frac{0 \cdot \pi/2}{0}$. We can rewrite the expression as: $\lim_{x \to 0^+} \frac{\cos^{-1}(1 - x^2)}{1} \cdot \frac{\sin^{-1}(1 - x)}{x(1 - x^2)}$ Let's consider the behavior of $\sin^{-1}(1-x)$ as $x \to 0^+$. Let $u = 1-x$. As $x \to 0^+$, $u \to 1^-$. $\sin^{-1}(u) \approx \sin^{-1}(1) - \frac{1}{\sqrt{1-1^2}}(u-1)$ near $u=1$. This is not helpful as the derivative is undefined.

Let's try a different grouping. $\lim_{x \to 0^+} \frac{\cos^{-1}(1 - x^2)}{x} \cdot \frac{\sin^{-1}(1 - x)}{1 - x^2}$ We know $\lim_{x \to 0^+} \frac{\sin^{-1}(1 - x)}{1 - x^2} = \frac{\sin^{-1}(1)}{1} = \frac{\pi}{2}$. So, the RHL becomes: $RHL = \left( \lim_{x \to 0^+} \frac{\cos^{-1}(1 - x^2)}{x} \right) \cdot \frac{\pi}{2}$ Now, let's evaluate $\lim_{x \to 0^+} \frac{\cos^{-1}(1 - x^2)}{x}$. This is of the form $\frac{0}{0}$. We can apply L'Hôpital's Rule. $\lim_{x \to 0^+} \frac{\frac{d}{dx}(\cos^{-1}(1 - x^2))}{\frac{d}{dx}(x)} = \lim_{x \to 0^+} \frac{\frac{-1}{\sqrt{1 - (1 - x^2)^2}} \cdot (-2x)}{1}$ $= \lim_{x \to 0^+} \frac{2x}{\sqrt{1 - (1 - 2x^2 + x^4)}} = \lim_{x \to 0^+} \frac{2x}{\sqrt{1 - 1 + 2x^2 - x^4}}$ $= \lim_{x \to 0^+} \frac{2x}{\sqrt{2x^2 - x^4}} = \lim_{x \to 0^+} \frac{2x}{\sqrt{x^2(2 - x^2)}}$ Since $x \to 0^+$, $x > 0$, so $\sqrt{x^2} = x$. $= \lim_{x \to 0^+} \frac{2x}{x\sqrt{2 - x^2}} = \lim_{x \to 0^+} \frac{2}{\sqrt{2 - x^2}} = \frac{2}{\sqrt{2 - 0}} = \frac{2}{\sqrt{2}} = \sqrt{2}$ So, the RHL is: $RHL = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi \sqrt{2}}{2} = \frac{\pi}{\sqrt{2}}$

Step 3: Calculate the Left-Hand Limit (LHL).

We need to find $\lim_{x \to 0^-} f(x)$. As $x \to 0^-$, $\{x\} = x + 1$. Substituting this into the function definition for $x \ne 0$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\cos^{-1}(1 - (x+1)^2) \sin^{-1}(1 - (x+1))}{(x+1) - (x+1)^3}$ Let $y = x+1$. As $x \to 0^-$, $y \to 1^-$. The limit becomes: $\lim_{y \to 1^-} \frac{\cos^{-1}(1 - y^2) \sin^{-1}(1 - y)}{y - y^3}$ The denominator is $y(1 - y^2)$. $\lim_{y \to 1^-} \frac{\cos^{-1}(1 - y^2) \sin^{-1}(1 - y)}{y(1 - y^2)}$ As $y \to 1^-$, $1 - y^2 \to 0^+$. Let $u = 1 - y^2$. As $y \to 1^-$, $u \to 0^+$. $\cos^{-1}(u) \approx \frac{\pi}{2} - u$ as $u \to 0$. So, $\cos^{-1}(1 - y^2) \approx \frac{\pi}{2} - (1 - y^2)$. This approximation is not quite right. The behavior of $\cos^{-1}(z)$ near $z=1$ is that it approaches 0. Let $z = 1 - y^2$. As $y \to 1^-$, $z \to 0^+$. $\cos^{-1}(z) \approx \sqrt{2(1-z)}$ for $z$ near 1. This is also not correct. Near $z=1$, let $z = \cos \theta$, so $\theta = \cos^{-1} z$. As $z \to 1^-$, $\theta \to 0^+$. $\cos^{-1}(1 - y^2)$. Let $1 - y^2 = \cos \theta$. As $y \to 1^-$, $1 - y^2 \to 0^+$, so $\cos \theta \to 0^+$, which means $\theta \to \frac{\pi}{2}^-$. So, $\cos^{-1}(1 - y^2) \to \frac{\pi}{2}^-$.

Let's look at the terms:

• As $y \to 1^-$, $\cos^{-1}(1 - y^2) \to \cos^{-1}(0) = \frac{\pi}{2}$.
• As $y \to 1^-$, $\sin^{-1}(1 - y) \to \sin^{-1}(0) = 0$.
• As $y \to 1^-$, $y(1 - y^2) \to 1(0) = 0$.

So, the limit is of the form $\frac{\frac{\pi}{2} \cdot 0}{0}$, which is $\frac{0}{0}$. Let's rewrite the expression and use L'Hôpital's Rule. $\lim_{y \to 1^-} \frac{\cos^{-1}(1 - y^2) \sin^{-1}(1 - y)}{y(1 - y^2)}$ We can split this into: $\lim_{y \to 1^-} \frac{\sin^{-1}(1 - y)}{1 - y} \cdot \lim_{y \to 1^-} \frac{\cos^{-1}(1 - y^2)}{y(1 + y)}$ Let $v = 1 - y$. As $y \to 1^-$, $v \to 0^+$. $\lim_{v \to 0^+} \frac{\sin^{-1}(v)}{v} = 1$ So, the LHL becomes: $LHL = 1 \cdot \lim_{y \to 1^-} \frac{\cos^{-1}(1 - y^2)}{y(1 + y)}$ Now, let's evaluate $\lim_{y \to 1^-} \frac{\cos^{-1}(1 - y^2)}{y(1 + y)}$. As $y \to 1^-$, the denominator $y(1+y) \to 1(1+1) = 2$. The numerator $\cos^{-1}(1 - y^2) \to \cos^{-1}(1 - 1^2) = \cos^{-1}(0) = \frac{\pi}{2}$. So, this limit is $\frac{\pi/2}{2} = \frac{\pi}{4}$.

Therefore, $LHL = 1 \cdot \frac{\pi}{4} = \frac{\pi}{4}$.

Step 4: Compare LHL and RHL and determine the value of $\alpha$.

We found $RHL = \frac{\pi}{\sqrt{2}}$ and $LHL = \frac{\pi}{4}$. For the function to be continuous at $x=0$, we need $LHL = RHL = \alpha$. However, $\frac{\pi}{\sqrt{2}} \ne \frac{\pi}{4}$. Since the left-hand limit and the right-hand limit are not equal, the limit $\lim_{x \to 0} f(x)$ does not exist. For a function to be continuous at a point, the limit must exist and be equal to the function's value at that point. Since the limit does not exist, there is no value of $\alpha$ that can make the function continuous at $x=0$.

Common Mistakes & Tips

• Incorrect handling of {x} near 0: Pay close attention to whether $x \to 0^+$ or $x \to 0^-$, as this determines whether $\{x\} = x$ or $\{x\} = x+1$.
• Algebraic manipulation of inverse trigonometric functions: Be careful when simplifying expressions involving inverse trigonometric functions, especially near their boundaries (e.g., $\cos^{-1}(y)$ near $y=1$).
• Applying L'Hôpital's Rule: Ensure the limit is in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying L'Hôpital's Rule. Also, verify that the derivative of the denominator is not zero at the limit point.
• Recognizing standard limits: Familiarity with standard limits like $\lim_{y \to 0} \frac{\sin^{-1} y}{y} = 1$ can simplify calculations significantly.

Summary

To ensure continuity of the function $f(x)$ at $x=0$, the left-hand limit (LHL), the right-hand limit (RHL), and the function value $f(0) = \alpha$ must all be equal. We analyzed the behavior of the fractional part function $\{x\}$ as $x$ approaches $0$ from the left and right. For $x \to 0^+$, $\{x\} = x$, leading to an RHL of $\frac{\pi}{\sqrt{2}}$. For $x \to 0^-$, $\{x\} = x+1$, leading to an LHL of $\frac{\pi}{4}$. Since the RHL and LHL are not equal ($\frac{\pi}{\sqrt{2}} \ne \frac{\pi}{4}$), the limit of $f(x)$ as $x \to 0$ does not exist. Consequently, there is no value of $\alpha$ that can make the function continuous at $x=0$.

The final answer is \boxed{A}.