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

Question

Solve the following equations and tick the correct one. The value of ' \(a\) ' for which the equation \(4 \operatorname{cosec}^{2}[\pi(a+x)]+a^{2}-4 a=0\), has a real solution, if (a) \(a=1\) (b) \(a=2\) (c) \(a=3\) (d) None

Step-by-Step Solution

Answer

The correct answer is (b) \(a=2\) is the value of 'a' for which the given equation has a real solution.

Step 1: Analyze the Given Options

For each option, check if \(a^2 - 4a >= 0\).

Step 2: Check for \(a=1\)

Substitute \(a=1\) into the inequality. This gives \(1^2 - 4*1 = -3\), which is not greater than or equal to 0. So, \(a=1\) is not a valid solution.

Step 3: Check for \(a=2\)

Substitute \(a=2\) into the inequality. This gives \(2^2 - 4*2 = 0\), which is equal to 0. So, \(a=2\) can be a valid solution.

Step 4: Check for \(a=3\)

Substitute \(a=3\) into the inequality. This gives \(3^2 - 4*3 = -3\), which is not greater than or equal to 0. So, \(a=3\) is not a valid solution.

Key Concepts

Cosecant Function

The cosecant function, denoted as \( \operatorname{cosec}\), is one of the six fundamental trigonometric functions. It is closely related to the sine function and is defined as the reciprocal of sine. Thus, for any angle \(x\), the cosecant is given by:

• \( \operatorname{cosec}(x) = \frac{1}{\sin(x)} \)

This definition immediately indicates that the cosecant function is undefined at any angle \(x\) where \(\sin(x) = 0\). Therefore, it is essential to remember that the cosecant function has vertical asymptotes at these points.
In the context of the given equation \(4 \operatorname{cosec}^{2}[\pi(a+x)]+a^{2}-4 a=0\), the term \( \operatorname{cosec}^{2}[\pi(a+x)]\) suggests that the original function involves squares of the cosecant function. Squaring the cosecant ensures that the expression is always positive or zero.
Thus, when solving such an equation, it is crucial to consider where the trigonometric function is defined and its behavior across different values to ensure that only valid solutions are entertained.

Quadratic Equations

A quadratic equation is a second-degree polynomial equation of the form \(ax^2 + bx + c = 0\). Such an equation may have zero, one, or two real solutions; these solutions correspond to the values of \(x\) that satisfy the equation.
In this particular problem, part of the equation takes the form \(a^2 - 4a = 0\), which simplifies to a quadratic equation in terms of \(a\). To find solutions, we remember that one approach is to factorize the quadratic into:

• \((a)(a-4) = 0\)

The solutions to this equation are the values of \(a\), which make the expression equal to zero. Therefore, the values of \(a\) that satisfy this equation are the roots \(a = 0\) and \(a = 4\).
However, since the problem checks for certain values of \(a\), we are interested in testing if each provided option satisfies \(a^2 - 4a \geq 0\) to determine if those specific values provide a real solution.

Real Solutions

In mathematics, a real solution to an equation refers to a solution that is a real number, which can literally be plotted on the number line. In the context of quadratic equations, for example, a real solution exists if the discriminant \((b^2 - 4ac)\) is greater than or equal to zero.
In this exercise, each value of \(a\) is tested in the inequality \(a^2 - 4a \geq 0\) as a part of identifying the real solutions. If the expression evaluates to zero or a positive number, it implies the potential for real solutions to the original problem.

• For \(a=1\), \(1^2 - 4 \times 1 = -3\). This does not meet the condition \(a^2 - 4a \geq 0\), giving no real solutions.
• For \(a=2\), \(2^2 - 4 \times 2 = 0\). This equals zero, indicating real solutions are possible.
• For \(a=3\), \(3^2 - 4 \times 3 = -3\). This is less than zero, indicating no real solutions.

Understanding real solutions helps in solving equations effectively and determining the validity of given options.