Problem 3 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\phi(x)=\lim _{n \rightarrow \infty} \frac{x^{n}-x^{-n}}{x^{n}+x^{-n}}, 0

Step-by-Step Solution

Answer

The correct answer is (B).

Step 1: Determine \( \phi(x) \)

The function \( \phi(x) \) is defined as \( \phi(x) = \lim _{n \rightarrow \infty} \frac{x^{n}-x^{-n}}{x^{n}+x^{-n}} \). Since \( 0 < x < 1 \), as \( n \rightarrow \infty \), \( x^n \rightarrow 0 \) and \( x^{-n} \rightarrow \infty \). Therefore, \( \phi(x) = \lim_{n \rightarrow \infty} \frac{0 - x^{-n}}{0 + x^{-n}} = -1 \).

Step 2: Set up the Integral

We need to find \( \int \sin^{-1}(x) \phi(x) \, dx \). Substituting \( \phi(x) = -1 \), the integral becomes \( \int -\sin^{-1}(x) \, dx \).

Step 3: Integrate \( -\sin^{-1}(x) \)

The integral \( \int -\sin^{-1}(x) \, dx \) can be solved using integration by parts. Let \( u = \sin^{-1}(x) \) and \( dv = dx \), then \( du = \frac{1}{\sqrt{1-x^2}} \, dx \) and \( v = x \).

Step 4: Apply Integration by Parts

Using integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:\( \int \sin^{-1}(x) \, dx = x \sin^{-1}(x) - \int x \frac{1}{\sqrt{1-x^2}} \, dx \).

Step 5: Solve the Remaining Integral

The integral \( \int x \frac{1}{\sqrt{1-x^2}} \, dx \) can be solved with substitution \( t = 1-x^2 \). Then \( dt = -2x \, dx \), resulting in:\( \int \frac{x}{\sqrt{1-x^2}} = -\frac{1}{2} \int \frac{1}{\sqrt{t}} \, dt = -\sqrt{1-x^2} + C \).

Step 6: Combine the Results

Substituting back, we get:\( \int \sin^{-1}(x) \, dx = x \sin^{-1}(x) - (-\sqrt{1-x^2}) = x \sin^{-1}(x) + \sqrt{1-x^2} + C \).

Step 7: Interpret the Solution

Therefore, \( \int \sin^{-1}(x) \phi(x) \, dx = -(x \sin^{-1}(x) + \sqrt{1-x^2}) + C \), which matches option \( B \).

Key Concepts

Calculus

Calculus is a branch of mathematics focused on change and movement. It's divided into two main branches: differential calculus and integral calculus. Integral calculus deals with the concept of integrals, which allows us to compute areas under curves, among other things.
In the given exercise, the aim was to evaluate a definite integral which involved calculus concepts. This particular problem used integration by parts, a technique often seen in calculus when dealing with specific types of integrals. The process of finding an antiderivative involves reversing differentiation and can provide solutions to wide-ranging problems, such as determining total accumulated change given a rate of change.
Understanding calculus, especially integrals, is vital in various applications in physics, engineering, and beyond. It provides tools to model and solve complex real-world problems.

Limits

The concept of limits is fundamental in calculus and is essential in understanding how functions behave as inputs approach a certain value. In the problem given, the limit is used to find the function \( \phi(x) \), defined by the expression \( \phi(x) = \lim _{n \rightarrow \infty} \frac{x^{n}-x^{-n}}{x^{n}+x^{-n}} \).
Limits help determine the value a function approaches as the input tends toward a particular point. In the context of \( \phi(x) \), as \( n \) goes to infinity, \( x^n \to 0 \) because \( 0 < x < 1 \), while \( x^{-n} \to \infty \). This behavior highlights how limits allow us to rigorously evaluate expressions involving infinity, helping us determine the value of \( \phi(x) \) as \(-1\).
Learning about limits equips students with the ability to analyze function behavior around specific points and infinity, underscoring their importance in calculus.

Integration by Parts

Integration by parts is a technique in calculus derived from the product rule of differentiation. It is particularly useful for integrals resulting from products of functions, such as the integral involving \( \sin^{-1}(x) \) in the exercise.
The integration by parts formula is \( \int u \, dv = uv - \int v \, du \). The trick is to sensibly choose \( u \) and \( dv \) such that the resulting integral becomes easier to solve. In the given solution, \( u = \sin^{-1}(x) \) and \( dv = dx \), leading to \( du = \frac{1}{\sqrt{1-x^2}} \, dx \) and \( v = x \).
After applying integration by parts, a simpler integral remains to be solved, often through basic techniques or substitution. This method is powerful when conventional direct integration is too complicated. Mastering integration by parts enables students to tackle a wide range of complex integrals confidently.