Problem 31 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

The value of a for which \(a x^{2}+\sin ^{-1}\left(x^{2}-2 x+2\right)+\cos ^{-1}\left(x^{2}-2 x+2\right)=0\) has a real solution, is (a) \(\pi / 2\) (b) \(-\pi / 2\) (c) \(2 / \pi\) (d) \(-2 / \pi\)

Step-by-Step Solution

Answer

The correct value of 'a' for the equation to have real solutions is \(-\pi / 2\).

Step 1: Analyze the inverse trigonometric functions

Let us consider \(x^{2}-2x+2\), and we will denote it as \(y\). For real values, \(y\) must lie in the domain of both \(\sin^{-1}\) and \(\cos^{-1}\) functions. Therefore, \(-1\leq y\leq 1\).

Step 2: Apply the properties of inverse trigonometric functions

Remember that \(\sin ^{-1}(y)+\cos ^{-1}(y)=\pi / 2\). We substitute \(x^{2}-2x+2\) for \(y\) in the main equation.

Step 3: Simplify the equation

Substituting \(\sin ^{-1}(y)+\cos ^{-1}(y)=\pi / 2\) into the equation, we then obtain \(ax^{2} + \pi / 2 = 0\).

Step 4: Find the value of 'a'

Rearranging the equation for 'a', we get \(a=-\pi / 2x^{2}\). This value for 'a' would result in real solutions for the equation. However, it is not found among our options.

Step 5: Analyze the proposed options

Looking the options we have, if we set \(a=\pi / 2\) and substitute it into \(ax^{2} + \pi / 2 = 0\), we obtain the equation \(\pi / 2 * x^{2} + \pi / 2 = 0\). Simplifying, this results in \(x^{2}=-1\), which doesn't have real roots. Thus, \(\pi / 2\) cannot be the correct value of 'a'. Using the same process, if \(a=-\pi / 2\), we obtain the equation \(-\pi / 2 * x^{2} + \pi / 2 = 0\). Simplifying, we are left with \(x^{2}=1\) which indeed has real solutions. Therefore, \(a=-\pi / 2\) is indeed the value of 'a' for which the equation will have real solutions.

Key Concepts

Domain of inverse functions

Understanding the domain of inverse trigonometric functions is crucial, especially when solving equations involving arcsine and arccosine. For these functions, the input values (let's call it \( y \)) must lie within a specific range to be valid:

• The domain for \( \sin^{-1}(y) \) and \( \cos^{-1}(y) \) is \( -1 \leq y \leq 1 \).

This means that when you're working with expressions like \( x^{2}-2x+2 \), ensuring that it stays between \(-1\) and \(1\) is essential. Failure to do so would make the inverse trigonometric expressions undefined and, consequently, there would be no real solutions for the equation.

Trigonometric equation solving

When solving equations involving trigonometric functions, it’s important to apply the known properties of these functions to simplify the process. In this problem, a key property is that:

• \( \sin^{-1}(y) + \cos^{-1}(y) = \pi / 2 \)

This trigonometric identity simplifies the equation significantly. By substituting \( y = x^2 - 2x + 2 \) into this identity, the equation transforms into a form that is easier to handle:
\[ a x^2 + \pi/2 = 0 \]
This equation is simpler to solve for \( a \) and allows one to assess possible real solutions effectively.

Real solutions in trigonometry

Finding real solutions in trigonometric equations often involves checking each proposed solution carefully. In this problem, setting different values for \( a \) alters the possibility of real solutions. Testing the options:

• If \( a = \pi / 2 \), solving the resulting equation leads to \( x^2 = -1 \), which has no real solutions because square roots of negative numbers aren’t real.
• Conversely, if \( a = -\pi / 2 \), the equation becomes \( x^2 = 1 \), offering real solutions \( x = 1 \) and \( x = -1 \).

Choosing \( a = -\pi / 2 \) ensures that the trigonometric equation has valid real solutions, confirming it as the correct choice.