Problem 7 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\sum_{i=1}^{2 n} \sin ^{-1} x_{i}=n \pi\), then \(\sum_{i=1}^{2 n} x_{i}\) is equal to (A) \(n\) (B) \(2 n\) (C) \(\frac{n(n+1)}{2}\) (D) none of these

Step-by-Step Solution

Answer

(D) none of these

Step 1: Understand the Given Equation

The problem states \( \sum_{i=1}^{2n} \sin^{-1} x_i = n \pi \). This implies that the sum of the inverse sines of \(x_i\) equals \(n\pi\), where the sum is taken over \(2n\) terms. Each \(\sin^{-1} x_i\) is evaluated to be within the principal range \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

Step 2: Sum of Arcsin Properties

Since \( \sin^{-1} x_i \) for each \(x_i\) is between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), the maximum possible contribution from each \( \sin^{-1} x_i \) to the sum \(n\pi\) is \(\frac{\pi}{2}\). This means exactly half of the terms must be \(\sin^{-1}(-1) = -\frac{\pi}{2}\) and the other half \(\sin^{-1}(1) = \frac{\pi}{2}\) to reach \(n\pi\).

Step 3: Consider Values of x_i

For \( \sin^{-1} x_i = \frac{\pi}{2}\), it implies \( x_i = 1 \). Likewise, for \( \sin^{-1} x_i = -\frac{\pi}{2}\), it implies \( x_i = -1 \). Since there are \(2n\) terms, half of them, i.e., \(n\) of \( x_i\) must be 1, and the other \(n\) values must be -1.

Step 4: Calculate \(\sum_{i=1}^{2n} x_i\)

The sum can be calculated as \(n \times 1 + n \times (-1) = n - n = 0\). Thus, the sum of \(x_i\) terms is 0.

Step 5: Evaluate Given Options

Comparing the calculated sum (0) with the options, neither \(n\), \(2n\), nor \(\frac{n(n+1)}{2}\) equals 0. Therefore, the correct choice is \(\text{none of these}\).

Key Concepts

Sum of Arcsine

Arcsine, denoted as \( \sin^{-1} \), is an inverse trigonometric function that reverses the sine function. When we talk about the sum of arcsine values, it's important to remember that each value \( \sin^{-1} x_i \) can range only within the principal range of \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). Therefore, when calculating a sum like \( \sum_{i=1}^{2n} \sin^{-1} x_i = n\pi \), it's necessary to consider how each arcsine contributes to reaching this total.

Since \( \sum_{i=1}^{2n} \sin^{-1} x_i = n\pi \) means the arcsines must somehow add up to multiples of \(\pi\), the options of \( \pm \frac{\pi}{2} \) per arcsine function imply very specific values of \( x_i \). Each term can contribute either \(\frac{\pi}{2}\) for \( \sin^{-1}(1) \) or \(-\frac{\pi}{2}\) for \( \sin^{-1}(-1) \). This property allows us to split them equally to sum up to exactly \( n\pi \).

Thus, combining arcsine values properly is key to utilizing the sum properties effectively. Think of these contributions like adding distinct blocks until the right total is reached.

Inverse Functions Properties

Inverse functions essentially reverse the roles of the inputs and outputs of the original function. For trigonometric functions like sine and its inverse, the conditions become more specific.

The property of the inverse functions dictates that if \( y = \sin^{-1} x \), then its inverse should satisfy \( x = \sin y \). This "undoes" the original function's effect.

• For \( \sin \theta \), the range is \([-1, 1]\), thus for \( \sin^{-1} x \), the output (\(y\)) is restricted to \([-\frac{\pi}{2}, \frac{\pi}{2}]\).


• The principal nature of arcsine confines its output to these values to maintain a well-defined, one-to-one function.

This property helps in calculations where specific outputs are desired, as in our given exercise where each \(x_i\) needed to be precisely \(1\) or \(-1\) for the sum to meet specific criteria (\(n\pi\)). Understanding this property allows us to predict values and sums more reliably.

Principal Range

When dealing with inverse trigonometric functions like the arcsine, understanding the concept of a principal range is crucial. This range determines the possible outputs that ensure the function remains well-defined and does not repeat over different intervals.

The principal range of \( \sin^{-1} x \) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), which means the function will only provide outputs between these limits. This is because the sine function, being periodic, can have the same sine value for different angles, but the inverse sine is restricted to this range.

• The principal range ensures that each \(x\) can map back to exactly one \(y\) within the range.


• This allows us to avoid ambiguity and helps to easily work with angles, particularly when summing multiple arcsine values as in the original exercise.

Grasping this concept is especially important for understanding why only specific values (\(1\) and \(-1\)) were needed to achieve our set sum \( n\pi \), by exploiting the limits of what arcsine can produce within its principal range.