Question

Let \(\{x\}\) denote the fractional part of \(x\), and \(f(x) = \frac{\cos^{-1}(1 - \{x\}^2) \sin^{-1}(1 - \{x\})}{\{x\} - \{x\}^3}, \quad x \neq 0\).If \(L\) and \(R\) respectively denote the left-hand limit and the right-hand limit of \(f(x)\) at \(x = 0\), then  \(\frac{32}{\pi^2} \left(L^2 + R^2\right)\) is equal to \(\_\_\_\_\_\_\_\_\).

Answer 1

Correct Answer: 18

The objective is to compute the value of the expression \( \frac{32}{\pi^2}(L^2 + R^2) \). Here, \( L \) represents the left-hand limit and \( R \) represents the right-hand limit of the function \( f(x) = \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3} \) as \( x \) approaches 0. The notation \( \{x\} \) denotes the fractional part of \( x \).

Concepts Utilized:

1. Fractional Part Function (\(\{x\}\)): Defined as \( \{x\} = x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). For limits at an integer, its behavior is critical:

• As \( x \to 0^+ \), \( \lfloor x \rfloor = 0 \), hence \( \{x\} = x \).
• As \( x \to 0^- \), \( \lfloor x \rfloor = -1 \), hence \( \{x\} = x - (-1) = 1+x \).

2. Fundamental Calculus Limits: The following standard limits are employed:

\[ \lim_{\theta \to 0} \frac{\sin^{-1}(\theta)}{\theta} = 1 \] \[ \lim_{y \to 0^+} \frac{\cos^{-1}(1-y)}{\sqrt{y}} = \sqrt{2} \]

The latter limit can be verified using L'Hôpital's rule or a trigonometric substitution.

Detailed Solution:

Step 1: Evaluation of the Right-Hand Limit (R) at \( x = 0 \).

For \( x \to 0^+ \), \( \{x\} = x \). Thus, \( R \) is:

\[ R = \lim_{x \to 0^+} \frac{\cos^{-1}(1-x^2) \sin^{-1}(1-x)}{x - x^3} \]

Factoring the denominator and separating terms yields:

\[ R = \lim_{x \to 0^+} \frac{\cos^{-1}(1-x^2)}{x} \cdot \lim_{x \to 0^+} \frac{\sin^{-1}(1-x)}{1-x^2} \]

In the second limit, as \( x \to 0^+ \), \( \sin^{-1}(1-x) \to \sin^{-1}(1) = \frac{\pi}{2} \) and \( 1-x^2 \to 1 \), making the second limit \( \frac{\pi}{2} \).

For the first limit, let \( y = x^2 \). As \( x \to 0^+ \), \( y \to 0^+ \) and \( x = \sqrt{y} \). The limit transforms to:

\[ \lim_{y \to 0^+} \frac{\cos^{-1}(1-y)}{\sqrt{y}} \]

This is a known standard limit equal to \( \sqrt{2} \).

Combining these results, \( R = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}} \).

Step 2: Evaluation of the Left-Hand Limit (L) at \( x = 0 \).

For \( x \to 0^- \), \( \{x\} = 1+x \). Let \( x = -h \), where \( h \to 0^+ \). Then \( \{x\} = \{-h\} = 1-h \).

\[ L = \lim_{h \to 0^+} \frac{\cos^{-1}(1-(1-h)^2) \sin^{-1}(1-(1-h))}{(1-h) - (1-h)^3} \]

Simplifying the terms:

\( \sin^{-1}(1-(1-h)) = \sin^{-1}(h) \)

\( \cos^{-1}(1-(1-h)^2) = \cos^{-1}(1 - (1-2h+h^2)) = \cos^{-1}(2h-h^2) \)

\( (1-h) - (1-h)^3 = (1-h)(1-(1-h)^2) = (1-h)(2h-h^2) = h(1-h)(2-h) \)

The expression for L becomes:

\[ L = \lim_{h \to 0^+} \frac{\cos^{-1}(2h-h^2) \sin^{-1}(h)}{h(1-h)(2-h)} \]

Separating the limits:

\[ L = \left( \lim_{h \to 0^+} \frac{\sin^{-1}(h)}{h} \right) \cdot \left( \lim_{h \to 0^+} \frac{\cos^{-1}(2h-h^2)}{(1-h)(2-h)} \right) \]

Evaluating each limit:

• \( \lim_{h \to 0^+} \frac{\sin^{-1}(h)}{h} = 1 \) (standard limit).
• For the second limit, substituting \( h = 0 \) gives \( \frac{\cos^{-1}(0)}{(1-0)(2-0)} = \frac{\pi/2}{2} = \frac{\pi}{4} \).

Thus, \( L = 1 \cdot \frac{\pi}{4} = \frac{\pi}{4} \).

Final Calculation and Outcome:

Step 3: Substitution of L and R into the target expression.

We have \( L = \frac{\pi}{4} \) and \( R = \frac{\pi}{\sqrt{2}} \).

Calculating the squares:

\[ L^2 = \left(\frac{\pi}{4}\right)^2 = \frac{\pi^2}{16} \] \[ R^2 = \left(\frac{\pi}{\sqrt{2}}\right)^2 = \frac{\pi^2}{2} \]

Summing the squares:

\[ L^2 + R^2 = \frac{\pi^2}{16} + \frac{\pi^2}{2} = \frac{\pi^2 + 8\pi^2}{16} = \frac{9\pi^2}{16} \]

Computing the final expression value:

\[ \frac{32}{\pi^2} (L^2 + R^2) = \frac{32}{\pi^2} \left( \frac{9\pi^2}{16} \right) \] \[ = \frac{32}{16} \times \frac{\pi^2}{\pi^2} \times 9 = 2 \times 1 \times 9 = 18 \]

The value of the expression is determined to be 18.