Problem 5 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\sin \theta, \sin \phi\) and \(\cos \theta\) are in G.P., then the roots of the equation \(x^{2}+2 x \cot \phi+1=0\) are always (A) real (B) imaginary (C) equal (D) greater than 1

Step-by-Step Solution

Answer

The roots are always equal (C).

Step 1: Understanding the Problem

We need to find the nature of the roots of the quadratic equation \(x^2 + 2x \cot \phi + 1 = 0\). Given that \(\sin \theta\), \(\sin \phi\), and \(\cos \theta\) are in geometric progression (G.P.), our first goal is to express the relations between these trigonometric functions and use them to determine \(\cot \phi\).

Step 2: Geometric Progression Condition

For three numbers to be in G.P., the square of the middle term is equal to the product of the other two. Thus, for \(\sin \theta\), \(\sin \phi\), and \(\cos \theta\) to be in G.P.: \[(\sin \phi)^2 = \sin \theta \cdot \cos \theta\] Since \(\sin^2 \theta + \cos^2 \theta = 1\), we have another relation to use later.

Step 3: Using Trigonometric Identities

To find \(\cot \phi\), let's express \(\cot \phi\) using the equation from G.P. \[\cot \phi = \frac{\cos \phi}{\sin \phi}\] We need to find expressions for these terms using the G.P. relation: \[(\sin \phi)^2 = \sin \theta \cdot \cos \theta\], use the identity \(\sin \phi = t\) for simplifying later calculations.

Step 4: Finding \(\cot \phi\)

Since \((\sin \phi)^2 = \sin \theta \cdot \cos \theta\), assume \(\sin \phi = t\). Then, \(t^2 = (\sin \theta \cos \theta)\), meaning \(t^2 = (t/a) \cdot (a^2/t)\) for some common ratio \(a\). Resolve the equation for \(t\) and then find \(\cot \phi\) accordingly.

Step 5: Examining Root Conditions

For the quadratic equation \(x^2 + 2x \cot \phi + 1 = 0\) to have real roots, the discriminant of the quadratic must be zero or positive. Calculate the discriminant: \[\Delta = (2 \cot \phi)^2 - 4 \cdot 1 = 4 \cot^2 \phi - 4\] For real roots, \(\Delta \geq 0\), simplifying gives \(\cot^2 \phi \geq 1\).

Step 6: Conclusion Based on Discriminant

Since \(\cot \phi\) must satisfy \(\cot^2 \phi \geq 1\), the expression indicates it can be either \(\cot \phi = 1\) or \(\cot \phi = -1\). Thus, the quadratic equation has equal roots always because only under these conditions \(\Delta = 0\).

Key Concepts

Geometric Progression

A Geometric Progression (G.P.) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In the context of this problem, we have trigonometric identities where

• \(\sin \theta\), \(\sin \phi\), and \(\cos \theta\) are in G.P.
• This implies that \((\sin \phi)^2 = \sin \theta \cdot \cos \theta\).

To solve for \(\cot \phi\), we examine this relationship further by employing the identity \(\sin^2 \theta + \cos^2 \theta = 1\). G.P. is useful in simplifying the relationship between trigonometric functions as shown by the given equation. Understanding G.P. allows one to establish a connection between complex trigonometric relationships, simplifying further calculations, such as finding \(\cot \phi\).

Quadratic Equation Roots

When dealing with quadratic equations, understanding the nature of roots is crucial. The roots of a quadratic equation \(x^2 + bx + c = 0\) depend on the discriminant \(\Delta\), calculated as \(b^2 - 4ac\). For our problem, we have the equation:

• \(x^2 + 2x \cot \phi + 1 = 0\).

The task is then to assess the discriminant:

• \(\Delta = (2 \cot \phi)^2 - 4\).
• Simplifying yields \(\Delta = 4 \cot^2 \phi - 4\).

For the roots to be real, \(\Delta\) must be greater than or equal to zero. From this, it follows that \(\cot^2 \phi \geq 1\), but for the equation to always have equal roots, \(\Delta\) must be exactly zero, implying that \(\cot \phi = 1\) or \(-1\). These conditions ensure the roots are real and equal, providing insight into the conditions for equal roots in a quadratic equation.

Trigonometry in JEE Main

Trigonometry is an essential part of the curriculum for exams like the Joint Entrance Examination (JEE) Main. It often involves the application of identities and various relations to solve complex problems efficiently. In this problem, we see:

• How trigonometric functions are interconnected.
• The use of identities such as \(\sin^2 \theta + \cos^2 \theta = 1\) to simplify expressions and solve problems.

An understanding of how to manipulate these functions and progressions can greatly aid in identifying solutions quickly. The problem using G.P. relations, combined with understanding discriminants of quadratic equations, showcases a typical style of question you might face in JEE Main. Apt knowledge of Trigonometric Identities and Algebra is crucial for solving such questions correctly and efficiently in competitive exams.