Problem 40 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

The number of solutions of the equation \(1+x^{2}+2 x \sin \left(\cos ^{-1} y\right)=0\) is (a) 1 (b) 2 (c) 3 (d) 4 .

Step-by-Step Solution

Answer

The quadratic equation has two solutions, hence the number of solutions of the given equation is 2. Therefore, option (b) is correct.

Step 1: Convert the sine of inverse cosine into an expression

Using the identity \( \sin(\cos^{-1}(y)) = \sqrt{1 - y^2} \), the expression simplifies to \( 1 + x^2 + 2x\sqrt{1 - y^2} = 0 \)

Step 2: Put the equation into quadratic form

Rearranging the above equation, we can write it in the form of a quadratic equation: \(x^2 + 2x\sqrt{1 - y^2} + 1 = 0\)

Step 3: Finding the roots x

The roots of the quadratic equation \(ax^2 + bx + c = 0\) are given by \(x = [-b ± \sqrt{b^2 - 4ac}]/(2a)\). Substituting the coefficients into this formula gives \(x = [-2\sqrt{1 - y^2} ± \sqrt{(2\sqrt{1 - y^2})^2 - 4*1*1}]/2\).

Step 4: Simplifying the roots

Upon simplification, we have two solutions for x: \(x_1 = \sqrt{1 - y^2} - 1\) and \(x_2 = -\sqrt{1 - y^2} - 1\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find angles given their trigonometric values. When dealing with these functions, we often encounter expressions like \( \cos^{-1}(y) \), which represents the angle whose cosine is \( y \). In the exercise, we used this concept by replacing \( \sin(\cos^{-1}(y)) \) with \( \sqrt{1 - y^2} \). This substitution relies on the Pythagorean identity, which states that for any angle \( \theta \), the sine can be expressed as the square root of one minus the square of the cosine of \( \theta \), i.e., \( \sin^2(\theta) + \cos^2(\theta) = 1 \). Understanding this connection is essential as it enables us to convert between trigonometric and inverse trigonometric functions, simplifying complex expressions for easier computation.

Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. These equations can have up to two solutions or roots because they are second-degree polynomials. In the given exercise, the equation \( 1 + x^2 + 2x\sqrt{1 - y^2} = 0 \) was rearranged into a quadratic form. This step is crucial in solving because the method to find roots of quadratic equations applies directly once the equation is in standard quadratic form. The coefficients \( a \), \( b \), and \( c \) must be identified for further calculation using the quadratic formula. Knowing how to manipulate and rearrange equations into quadratic form is an invaluable skill as it simplifies many algebraic equations that might initially appear complex.

Roots of Equations

The roots, or solutions, of an equation \( ax^2 + bx + c = 0 \) are determined using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula calculates the values of \( x \) that satisfy the equation. For quadratic equations, these solutions can be real or complex depending on the discriminant \( b^2 - 4ac \). In our exercise, after substituting the coefficients into this formula, the discriminant influenced the number of real solutions. Simplifying the discriminant provided two distinct solutions: \( x_1 = \sqrt{1 - y^2} - 1 \) and \( x_2 = -\sqrt{1 - y^2} - 1 \).Understanding how to find roots is a fundamental concept as it enables us to determine all possible solutions to quadratic equations, a common task in algebra and trigonometry problems.