Problem 70 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\tan ^{-1} y=4 \tan ^{-1} x\), then \(y\) is finite if (A) \(x^{2} \neq 3+2 \sqrt{2}\) (B) \(x^{2} \neq 3-2 \sqrt{2}\) (C) \(x^{4} \neq 6 x^{2}-1\) (D) \(x^{4} \neq 6 x^{2}+1\)

Step-by-Step Solution

Answer

Option (A): \(x^2 \neq 3 + 2\sqrt{2}\).

Step 1: Understand the Problem

We are given the equation \( \tan^{-1} y = 4 \tan^{-1} x \) and need to determine the condition under which \( y \) remains finite. The options are expressed as conditions on \( x^2 \).

Step 2: Use the Tangent Addition Formula

To solve \( \tan^{-1} y = 4 \tan^{-1} x \), we can use the tangent of multiple angles formula: \( \tan 4\theta = \frac{4\tan\theta - 4\tan^3\theta}{1 - 6\tan^2\theta + \tan^4\theta} \). Here, \( \theta = \tan^{-1} x \), hence, \( \tan \theta = x \).

Step 3: Express \( y \, \text{In Terms of} \, x\)

Apply the formula to get \( y = \frac{4x - 4x^3}{1 - 6x^2 + x^4} \). To avoid making \( y \) undefined or infinite, the denominator \( 1 - 6x^2 + x^4 \) must not be zero.

Step 4: Set the Denominator Equal to Zero

Equating the denominator to zero gives \( x^4 - 6x^2 + 1 = 0 \). Solve this to find values of \( x \) that make the denominator zero.

Step 5: Solve for Critical Values of \( x^2 \)

Replace \( x^2 \) with \( t \) so it becomes \( t^2 - 6t + 1 = 0 \). Use the quadratic formula: \( t = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm 2\sqrt{2} \). So, the critical values are \( t = 3 + 2\sqrt{2} \) and \( t = 3 - 2\sqrt{2} \).

Step 6: Relate to the Given Options

Given the critical conditions \( x^2 = 3 + 2\sqrt{2} \) and \( x^2 = 3 - 2\sqrt{2} \), we determine that \( y \) is finite when \( x^2 eq 3 + 2\sqrt{2} \).

Step 7: Determine the Correct Answer

The correct condition for \( y \) to remain finite is that \( x^2 eq 3 + 2\sqrt{2} \). This corresponds to option (A).

Key Concepts

Tangent Addition Formula

The Tangent Addition Formula is an essential tool that simplifies the process of finding the tangent of the sum of multiple angles. This formula can be extended to cover cases such as the tangent of twice, thrice, or four times an angle. Here is the generalized formula used for tangent multiples:

• For two angles: \( \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \)
• For four times an angle: \( \tan(4\theta) = \frac{4\tan \theta - 4\tan^3 \theta}{1 - 6\tan^2 \theta + \tan^4 \theta} \)

When dealing with inverse trigonometric functions, such as \( \tan^{-1} \), this formula allows us to break down complex equations into simpler, more manageable parts. In the exercise at hand, we apply it to resolve the expression \( \tan^{-1} y = 4 \tan^{-1} x \) by substituting \( \tan \theta = x \), allowing us to express \( y \) in terms of \( x \). The formula effectively helps in understanding relationships between angles and ensuring the solution maintains mathematical validity by avoiding undefined or infinite results.

Trigonometric Equations

Trigonometric equations incorporate trigonometric functions and are solved by finding angles that satisfy the equation. These equations often involve inverses, like \( \tan^{-1} \), which play a significant role in converting present values into angles. Solving such equations frequently requires creativity and a deep understanding of various trigonometric identities and formulas.
One approach to solving these equations is to rearrange them into a more familiar form, using identities like:

• \( \cos^2\theta + \sin^2\theta = 1 \)
• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)

In the example of \( \tan^{-1} y = 4 \tan^{-1} x \), we transform the equation into a solvable form using the tangent addition formula. By expressing \( y \) in terms of \( x \), we can effectively solve the equation, ensuring that the denominator does not equal zero, thereby maintaining a finite solution for \( y \).
Trigonometric equations require careful manipulation and a keen eye for when to apply certain identities, so practice makes perfect in this area.

Quadratic Formula

The quadratic formula is a fundamental tool used to solve quadratic equations, which take the form of \( ax^2 + bx + c = 0 \). It is expressed as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is reliable for determining the roots of any quadratic equation, which are the values that make the equation equal to zero.
In our given problem, we encounter a need to find critical values for \( x^2 \), which leads to setting \( x^4 - 6x^2 + 1 = 0 \). By substituting \( x^2 \) with \( t \), the expression is transformed into \( t^2 - 6t + 1 = 0 \). We then apply the quadratic formula to find:

• \( t = 3 + 2\sqrt{2} \)
• \( t = 3 - 2\sqrt{2} \)

The formula not only provides us with correct solutions but also ensures that no critical values make \( y \) infinite, reinforcing the importance of the quadratic formula in solving such mathematical problems. Understanding how to apply it correctly is crucial for solving many types of equations encountered in mathematics.