Problem 19 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\), then \(f(x)\) is differentiable on (A) \((-\infty, \infty)\) (B) \((-\infty, \infty) \backslash\\{0\\}\) (C) \((-\infty, \infty) \backslash\\{-1,1\\}\) (D) None of these

Step-by-Step Solution

Answer

The function is differentiable on \((- inf, inf) \ \{-1, 1\}\).

Step 1: Understanding the Function

The given function is \(f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}\right)\). We need to find where this function is differentiable. Recognize that the inverse cosine function \(\cos^{-1}(u)\) is defined and differentiable when \(-1 \le u \le 1\).

Step 2: Setting the Range of the Argument

Calculate the domain where \(-1 \le \frac{2x}{1+x^2} \le 1\). This inequality will help us determine the intervals where \(f(x)\) is differentiable.

Step 3: Solving the Inequality

First, solve \(\frac{2x}{1+x^2} \le 1\):\( 2x \le 1 + x^2 \rightarrow x^2 - 2x + 1 \ge 0 \rightarrow (x-1)^2 \ge 0\). Since \((x-1)^2\) is always \(\ge 0\), this inequality holds for all \(x\).

Step 4: Solving the Second Inequality

Now solve \(-1 \le \frac{2x}{1+x^2}\): \(- \frac{2x}{1+x^2} \le 1\) gives \( -2x \le 1 + x^2 \rightarrow x^2 + 2x + 1 \ge 0 \rightarrow (x+1)^2 \ge 0\). This inequality is also true for all \(x\), except when \((x+1)eq 0\).

Step 5: Identifying Problematic Points

Combine the solutions; \(-1 \le \frac{2x}{1+x^2} \le 1\) holds for all real numbers except when \(x+1=0\) or \((x-1)=0\), i.e., \(x = 1\) and \(x = -1\). The function might not be differentiable at these points.

Step 6: Conclusion on Differentiability

Since \(f(x)\) is an inverse cosine function, it is differentiable except at \(x=-1\) and \(x=1\), which yields the domain of differentiation as \((-\infty, \infty) \backslash \{-1, 1\}\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions play a crucial role in calculus, particularly due to their ability to "undo" trigonometric operations. They help to determine an angle given a trigonometric ratio. In our exercise, we explore the inverse cosine function, denoted as \(\cos^{-1}(u)\). By definition, \(\cos^{-1}(u)\) gives us an angle whose cosine is \(u\). It's vital to remember that inverse trigonometric functions are only defined within a specific range.

• For the cosine function, the domain is \([-1, 1]\) because cosine values are restricted to this range.
• The range for \(\cos^{-1}(u)\) functions is \([0, \pi]\) since these are the angles over which cosine is non-negative and non-repeating.

Understanding these properties helps in solving exercises involving inverse trigonometric functions since they impact not only the differentiability but also the possible values \(u\) can take.

Domain and Range

The concept of domain and range is crucial when dealing with any function, especially inverse trigonometric functions. The domain refers to all the possible input values for which the function is defined, while the range refers to all possible outputs. In our exercise involving \(f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}\right)\), identifying the valid domain is fundamental.

• The expression \(\frac{2x}{1+x^2}\) needs to be within the interval \([-1, 1]\).
• Solving this provides us with an inequality \(-1 \le \frac{2x}{1+x^2} \le 1\).

Such inequality determines where the function within the inverse context is defined. This is why determining the domain accurately informs us about where the function is smooth and differentiable.

Inequality Solving

Inequalities are mathematical expressions that involve the comparison of two quantities. In the exercise, solving inequalities is key to finding the domain of differentiability for the function \(f(x)\). The inequalities \(\frac{2x}{1+x^2} \le 1\) and \(-1 \le \frac{2x}{1+x^2}\) were addressed.

• For \(\frac{2x}{1+x^2} \le 1\), we eventually derived \((x-1)^2 \ge 0\), holding true for all real \(x\).
• For \(-1 \le \frac{2x}{1+x^2}\), it simplified to \((x+1)^2 \ge 0\), also always true.

However, when solving, points where these inequalities might be equal, specifically \(x = -1\) and \(x = 1\), brought challenges. At these points, the smoothness (differentiability) of the function might be disrupted, suggesting the importance of solving inequalities correctly to identify such breaks.

Mathematical Analysis

Mathematical analysis involves breaking down complex functions or expressions to better understand their properties such as limits, continuity, and differentiability. When determining differentiability as in the exercise, it's essential to analyze the expression \(\frac{2x}{1+x^2}\).

• Identify critical points: Solving \(\frac{2x}{1+x^2} = 1\) or \(-1\) yields the points where potential discontinuities in differentiability may occur.
• Determine differentiability: By analyzing \((x-1)^2\) and \((x+1)^2\), it becomes clear these points must be excluded from differentiable intervals.

Effectively using mathematical analysis helps us conclude that \(f(x)\) is differentiable in the domain \(( - \infty, \infty ) \backslash \{-1, 1\}\). This sort of analytical approach is key to mastering higher-level mathematical concepts and exercises.