Problem 62 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \(f(0)\) so that the function \(f(x)=\frac{\cos ^{-1}\left(1-\\{x\\}^{2}\right) \sin ^{-1}(1-\\{x\\})}{\\{x\\}-\\{x\\}^{3}}, x \neq 0(\\{x\\}\) denotes fractional part of \(x\) ) becomes continuous at \(x=0\) is (A) \(\frac{\pi}{\sqrt{2}}\) (B) \(\frac{\pi}{4}\) (C) \(\frac{\pi}{2}\) (D) None of these

Step-by-Step Solution

Answer

The answer is (D) None of these.

Step 1: Identify the Continuity Condition

A function is continuous at a point if the left-hand limit, right-hand limit, and the value of the function at that point are all equal. Therefore, we need to ensure that \( \lim_{{x \to 0}} f(x) = f(0) \).

Step 2: Understand the Function and Fractional Part

The function given is \( f(x)=\frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3} \). Here, \( \{x\} \) denotes the fractional part of \( x \), which means \( \{x\} = x - \lfloor x \rfloor \). For any small positive or negative values approaching zero, \(\{x\} = x\).

Step 3: Simplify the Expression Near Zero

For small \( x \) around zero, let \( \{x\} = x \). Substitute this into the expression to get:\[f(x)=\frac{\cos^{-1}(1-x^2) \sin^{-1}(1-x)}{x(1-x^2)}\]

Step 4: Evaluate the Limit as x Approaches Zero

We need to compute \( \lim_{{x \to 0}} \frac{\cos^{-1}(1-x^2) \sin^{-1}(1-x)}{x(1-x^2)} \). We can use series expansions:- \( \cos^{-1}(1-x^2) \approx \sqrt{2x^2} = \sqrt{2}x \)- \(\sin^{-1}(1-x) \approx \sqrt{2x} \)So, \( f(x) \approx \frac{\sqrt{2}x \times \sqrt{2x}}{x} \) which simplifies to:\[ f(x) \approx \frac{2x^{3/2}}{x} = 2\sqrt{x} \] as \(x > 0 \).

Step 5: Calculate Final Limit and Compare

The expression \( 2\sqrt{x} \) approaches zero as \( x \to 0\). Therefore, the limit as \( x \to 0 \) of \( f(x) \) is zero. Hence, for the function to be continuous at \( x=0 \), we require \( f(0) = 0 \).

Step 6: Compare the Options

None of the given options match zero. Therefore, the correct answer considering these steps and computation is: (D) None of these.

Key Concepts

Fractional Part of a Number

The fractional part of a number is a concept that many students find interesting. It refers to the portion of a number that is not an integer. For a real number \( x \), the fractional part, denoted as \( \{x\} \), is computed by subtracting the largest integer less than or equal to \( x \) from \( x \) itself. This can be mathematically expressed as \( \{x\} = x - \lfloor x \rfloor \).

Essentially, \( \{x\} \) picks out everything past the decimal point. So, for example, if \( x = 7.85 \), then \( \{x\} = 0.85 \). This fractional part is always between 0 and 1, inclusive on zero and exclusive on one.

When dealing with functions involving the fractional part, it's important to remember how it behaves, especially as \( x \) approaches an integer. The fractional part \( \{x\} \) helps in studying behavior of functions at discontinuities or cusps, due to its unique properties.

Inverse Trigonometric Functions

Inverse trigonometric functions are the reverse operations of the basic trigonometric functions. These are used when you want to obtain an angle from the trigonometric ratio. There are six main inverse trigonometric functions: \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), \( \tan^{-1}(x) \), \( \cot^{-1}(x) \), \( \sec^{-1}(x) \), and \( \csc^{-1}(x) \).

In the context of the exercise, we specifically deal with \( \sin^{-1} \) and \( \cos^{-1} \). The \( \sin^{-1}(x) \) function, for instance, gives us an angle \( \theta \) such that \( \sin(\theta) = x \). Its principal values are between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\). Similarly, the \( \cos^{-1}(x) \) function yields an angle \( \theta \) such that \( \cos(\theta) = x \) with its principal range within \(0\) to \(\pi\).

These functions are notably useful in various scenarios, especially when finding angles where particular conditions are given using trigonometric identities. They play a critical role in mathematics, particularly in calculus and geometry, by helping to bridge the gap between algebraic expressions and geometric angles.

Limits and Continuity

Understanding limits and continuity is essential for analyzing the behavior of functions. In calculus, a function is deemed continuous at a point if its limit at that point coincides with its function value. Specifically, for a function \( f(x) \) to be continuous at \( x = a \), the following should hold:

• The limit of \( f(x) \) as \( x \to a \) exists.
• The function \( f(a) \) is defined.
• The limit \( \lim_{{x \to a}} f(x) = f(a) \).

In simpler terms, the graph of the function should not break at \( a \).

Calculating limits often involves finding the value that a function approaches as the input approaches some value. This might include techniques like substitution, factoring, or employing L'Hôpital's Rule both analytically and using series expansions.

In the case of small \( x \), analyzing continuity includes ensuring the smooth transition around points like zero, employing the behavior of the fractional part of \( x \) and dealing with inverse trigonometric functions. Understanding these limits helps in determining the necessary condition for a function's continuity at specific points, which is crucial in fields like physics, engineering, and beyond.