Problem 123 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ y=2 \sin ^{-1} \frac{x-2}{\sqrt{6}}-\sqrt{2+4 x-x^{2}} $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x <= 2\) and \(x >= 2\), with the function being defined as \(y = \Theta + 2-x\) for \(x <= 2\) and \(y = \Theta - x+2\) for \(x >= 2\), where \(\Theta = 2\sin^{-1}(\frac{x-2}{\sqrt{6}})\).

Step 1: Simplify the square root

The expression \(\sqrt{2+4x-x^2}\) can be simplified if we rearrange the terms under the square root to obtain a perfect square. In doing so, we can rewrite it as \(\sqrt{(2-x)^2}\), which simplifies to \(|2-x|\). For \(x<2\), it becomes \(-(2-x)\), for \(x>2\) it becomes \(x-2\). Since both scenarios are possible, we will keep it as an absolute value. So our equation becomes: \(y=2 \sin^{-1}\frac{x-2}{\sqrt{6}} - |2-x|\).

Step 2: Transform the inverse sine function

Let's individually consider the sine inverse component. We know that \(\sin^{-1}(z) = y \), implies that \(\sin(y) = z \). So if we let \(z = \frac{x-2}{\sqrt{6}}\), then \(y = 2 \sin^{-1}z = \Theta \). Now we can solve both parts of the equation individually.

Step 3: Solve for x

Since for the inverse sine, the value of z (our \(\frac{x-2}{\sqrt{6}}\)) must lie between -1 to 1. Solving for x, we get \( -\sqrt{6}+2 <= x <= \sqrt{6}+2 \). And combining this with the transformed equation, we get \(y = \Theta -|2-x|\). Combining both, we can now write, for \(x <= 2\), \(y = \Theta + 2-x\) and for \(x >= 2\), \(y = \Theta - x+2\). Hence, finally, we get 2 ranges of x that satisfy the given equation.

Key Concepts

Simplifying Square Root Expressions

Understanding how to simplify square root expressions is crucial for solving various algebraic problems. To simplify a square root expression like \(\sqrt{2+4x-x^2}\), we should try to manipulate the terms under the square root sign into a perfect square.

A perfect square is an expression that can be written as \(a^2\), where \(a\) is a real number. By rearranging \(2+4x-x^2\), we can rewrite it as \(\sqrt{(2-x)^2}\), which simplifies further to \(\left| 2-x \right|\). This simplification is possible because the square root of a squared quantity is the absolute value of that quantity.

When dealing with absolute value during this simplification, we always consider two scenarios: one where \(x\) is less than 2, leading to \( -(2-x)\), and one where \(x\) is greater than or equal to 2, resulting in \(x-2\). Thus, the simplification of square root expressions requires understanding the properties of both perfect squares and absolute values.

Properties of Absolute Value

The absolute value measures the magnitude of a number or expression, regardless of its sign. It's denoted by vertical bars, like this: \( |x| \). Understanding its properties is fundamental in algebra.

Firstly, the absolute value of a real number is always non-negative. This means \( |x| \geq 0\) for any real number \(x\). Secondly, the absolute value of a product is the product of the absolute values \( |ab| = |a| \cdot |b| \). Lastly, for the equation \( |x| = a \), where \(a \geq 0\), there are two possible solutions for \(x\): \(x = a\) or \(x = -a\).

In the context of simplifying square root expressions, we employ the property that \( \sqrt{x^2} = |x| \). This property is crucial when we want to rewrite expressions under the square root in forms like \( (2-x)^2 \), leading to \( |2-x| \). Understanding the scenarios where the absolute value becomes positive or negative helps us accurately simplify these expressions.

Domain of Inverse Sine Function

Understanding the domain of the inverse sine function, or \( \sin^{-1} \), is important when solving trigonometric equations. The function \( \sin^{-1}(x) \) (also denoted as arcsin) is defined for all real numbers \(x\) such that \( -1 \leq x \leq 1 \). This is because the sine function, of which \( \sin^{-1} \) is the inverse, has a range of \( [-1, 1] \).

When solving equations that involve \( \sin^{-1} \), it's important to ensure that the argument of the inverse sine is within this interval. If not, the expression does not represent a valid input for the function and thus is outside its domain.

In our exercise, \( \sin^{-1}\frac{x-2}{\sqrt{6}} \) implies that \( \frac{x-2}{\sqrt{6}} \) must lie between -1 and 1 for the expression to be valid. By solving the inequality \( -\sqrt{6}+2 \leq x \leq \sqrt{6}+2 \) we are effectively finding the values of \(x\) that keep the argument of the inverse sine within its domain. It's essential to check the domain constraints when dealing with inverse trigonometric functions to avoid invalid solutions.