Problem 73 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\cos ^{-1} x+\left(\sin ^{-1} y\right)^{2}=\frac{p \pi^{2}}{4}\) and \(\left(\cos ^{-1} x\right)\left(\sin ^{-1} y\right)^{2}=\frac{\pi^{2}}{16}\), (A) \(0 \leq p \leq \frac{4}{\pi}+1\) (B) \(p=2\) is the only integral value of \(p\) (C) \(p=0,1,2\) (integral values). (D) \(p=1\) is the only integral value of \(p\)

Step-by-Step Solution

Answer

C: Integral values of \( p \) are 0, 1, 2.

Step 1: Analyze Given Equations

We have two equations to work with. The first equation is \( \cos^{-1}(x) + \left(\sin^{-1}(y)\right)^2 = \frac{p \pi^2}{4} \) and the second is \( \left(\cos^{-1}(x)\right)\left(\sin^{-1}(y)\right)^2 = \frac{\pi^2}{16} \). We need to determine the possible values of \( p \).

Step 2: Simplify the Equations

For simplicity, let \( a = \cos^{-1}(x) \) and \( b = \sin^{-1}(y) \). Plug these into our equations to get: \( a + b^2 = \frac{p \pi^2}{4} \) and \( a b^2 = \frac{\pi^2}{16} \).

Step 3: Solve for \( a \) in Terms of \( b \)

From the second equation \( a b^2 = \frac{\pi^2}{16} \), we can solve for \( a \): \( a = \frac{\pi^2}{16b^2} \).

Step 4: Substitute \( a \) Back into the First Equation

Substitute \( a = \frac{\pi^2}{16b^2} \) back into the first equation: \( \frac{\pi^2}{16b^2} + b^2 = \frac{p\pi^2}{4} \).

Step 5: Simplify the Substituted Equation

Rearrange the equation: \( \frac{\pi^2}{16b^2} + b^2 = \frac{p\pi^2}{4} \). Multiply through by \(16b^2\) to clear the fraction: \( \pi^2 + 16b^4 = 4pb^2 \pi^2 \).

Step 6: Rearrange and Simplify further

This simplifies to: \( 16b^4 - 4pb^2 \pi^2 + \pi^2 = 0 \). Let \( z = b^2 \), then \( 16z^2 - 4pz\pi^2 + \pi^2 = 0 \).

Step 7: Solve Quadratic in \( z \)

This is a quadratic equation in \( z \): it can be solved using the quadratic formula \( z = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), where \( A = 16\), \( B = -4p\pi^2 \), and \( C = \pi^2 \).

Step 8: Determine Values of \( p \)

For \( z \) to be a real number, the discriminant must be non-negative. So \( (-4p \pi^2)^2 - 4 \times 16 \times \pi^2 \geq 0 \). Solve this inequality to find bounds for \( p \).

Step 9: Simplify Discriminant Condition

Simplifying gives \((16p^2 \pi^4 - 64\pi^2) \geq 0 \), implying \( p^2 \geq 4/\pi^2 \). This suggests possible integer values of \( p \).

Step 10: Identify Integer Values of \( p ",

Considering the constraints \(0 \leq \sin^{-1}(y) \leq \pi/2\) and \(0 \leq \cos^{-1}(x) \leq \pi\), \(p\) could be 0, 1 or 2 as potential integral values that satisfy the given \(p^2 \geq 4/\pi^2\).

Step 11: Conclusion: Evaluate Options

Based on step 10 and considering all possibilities, the correct answer that includes these values is option C. For integral values of \(p\), the options are 0, 1, and 2.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are fundamental when solving trigonometric equations, as they allow us to determine angles from known trigonometric values. These functions include:

• \( \sin^{-1}(x) \) - also known as arcsine, used to find the angle whose sine is \( x \).
• \( \cos^{-1}(x) \) - or arccosine, used to determine the angle whose cosine is \( x \).

In the given exercise, the inverse functions are utilized by expressing specific variables as angles using these, which helps transform the equations into manageable algebraic expressions. It's important to remember that:

• The range for \( \sin^{-1}(x) \) is \([-\pi/2, \pi/2]\).
• For \( \cos^{-1}(x) \), the range is \([0, \pi]\).

These ranges imply limitations on possible values of \( x \) and \( y \), which are between -1 and 1. Understanding these constraints can be vital when determining feasible solutions for the equations.

Quadratic Equation Solver

Solving quadratic equations is a common task in algebra and trigonometry since these equations often appear when working with trigonometric identities and formulas. A typical quadratic equation has the form \( ax^2 + bx + c = 0 \). In our equation: \[ 16z^2 - 4pz\pi^2 + \pi^2 = 0 \]we used the quadratic formula:\[ z = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \] Here, \(A = 16\), \(B = -4p\pi^2\), and \(C = \pi^2\).

The quadratic formula is a reliable way to find the roots of any quadratic equation, assuming the discriminant \( B^2 - 4AC \) is non-negative. This non-negativity is crucial since it ensures real solutions, which is necessary for the exercise requirements.The positive or zero value of the discriminant indicates that the quadratic has real roots. For our equation to have real solutions, it's essential that \[ 16p^2\pi^4 - 64\pi^2 \geq 0 \]Thus, understanding how to derive solutions from a quadratic equation is crucial for tackling these types of trigonometric problems.

Inequalities in Trigonometry

Inequalities often play a significant role in trigonometric solutions, helping us to determine the range of possible values for our variables. For this exercise, we analyzed the inequality:\[16p^2\pi^4 - 64\pi^2 \geq 0\]This inequality arises from ensuring the discriminant of our quadratic equation remains non-negative. Solving this inequality allows us to pinpoint potential integer values for \( p \) that satisfy all given conditions.

When solving inequalities, here’s a helpful guide:

• Identify the correct range: The inequality should be simplified to find restrictions on \( p \).
• Derive potential solutions: Factor larger expressions when possible, or use techniques like the quadratic formula for complex inequalities.
• Verify constraints: Check that these mathematical solutions adhere to any given domain constraints from trigonometric functions.

In this case, understanding how the inequality provides permissible ranges for \( p \) ensures all potential solutions are explored, as evidenced by identifying 0, 1, and 2 as integral solutions.