Problem 155 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The equation \(e^{\sin x}-e^{-\sin x}-4=0\), for \(x\) real, has \([2012]\) (A) infinite number of roots (B) no roots (C) exactly one root (D) exactly four roots

Step-by-Step Solution

Answer

The equation has exactly one root (C).

Step 1: Simplify the Equation

Given the equation: \[ e^{\sin x} - e^{-\sin x} - 4 = 0 \]We can rewrite this by letting \( y = e^{\sin x} \). Thus, \( e^{-\sin x} = \frac{1}{y} \). Substituting these into the original equation gives:\[ y - \frac{1}{y} - 4 = 0 \]

Step 2: Multiply through by y

To eliminate the fraction, multiply through by \( y \):\[ y^2 - 1 - 4y = 0 \]which rearranges to:\[ y^2 - 4y - 1 = 0 \]

Step 3: Solve the Quadratic Equation

This is a simple quadratic equation. Use the quadratic formula:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = -4 \), and \( c = -1 \). Substitute in these values:\[ y = \frac{4 \pm \sqrt{16 + 4}}{2} \]\[ y = \frac{4 \pm \sqrt{20}}{2} \]\[ y = \frac{4 \pm 2\sqrt{5}}{2} \]\[ y = 2 \pm \sqrt{5} \]

Step 4: Analyze Valid Solutions for y

Remember that \( y = e^{\sin x} \) must be positive. Both solutions, \( y = 2 + \sqrt{5} \) and \( y = 2 - \sqrt{5} \) need to be checked:- \( y = 2 + \sqrt{5} \) is positive, valid.- \( y = 2 - \sqrt{5} \) is negative because \( \sqrt{5} \) is approximately 2.236, making \( 2 - \sqrt{5} \) negative, invalid as \( y \) must be positive.

Step 5: Find the Real Roots for x

Since only \( y = 2 + \sqrt{5} \) is valid, solve for \( \sin x \):\[ e^{\sin x} = 2 + \sqrt{5} \]Take the natural logarithm:\[ \sin x = \ln(2 + \sqrt{5}) \]

Step 6: Conclusion on the Number of Roots

The equation \( \sin x = \ln(2 + \sqrt{5}) \) has solutions \( x \) since the sine function is periodic. Since no other valid \( y \) provided a root, there is exactly one root based on the valid \( y \) and its corresponding \( \sin x \).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a variable appears in the exponent of a constant base. A common form is \(a^x\), where \(a\) is a constant and \(x\) is a variable. In the exercise, we initially encounter \(e^{\sin x}\) and its inverse form \(e^{-\sin x}\). Here, \(e\) is Euler's number, approximately 2.718, a mathematical constant that forms the base of natural logarithms.
Understanding exponential functions involves recognizing their characteristics:

• They grow rapidly. A small change in the exponent yields a significant change in value.
• They always return positive values, since raising a positive base to any power does not yield a negative number.
• For \(e^x\), the function is continuously increasing, whereas \(e^{-x}\) decreases.

The exercise used a substitution of \(y = e^{\sin x}\) to facilitate solving the equation. This transformation simplifies the exponential equation into a quadratic, highlighting a common technique in managing intricate equations involving exponentials.

Quadratic Equations

Quadratic equations take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. They are fundamental to solving many algebraic problems and can often be tackled using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The problem provided an equation \(y^2 - 4y - 1 = 0\), an example of a quadratic equation. Using the formula, we found solutions \(y = 2 \pm \sqrt{5}\). This demonstrates solving quadratics:

• Identify coefficients: \(a = 1\), \(b = -4\), \(c = -1\).
• Apply the quadratic formula to find the roots.
• Calculate the discriminant \(b^2 - 4ac\) to determine the number and type of roots. A positive discriminant implies two real roots.

Ultimately, only solutions that make sense in the context are valid, such as ensuring the positivity of \(y = e^{\sin x}\).

Periodic Functions

Periodic functions repeat their values in regular intervals. The sine function, \(\sin x\), is a classic example, repeating every \(2\pi\) radians. In our exercise, understanding the periodic nature of the sine function is crucial where \(\sin x = \ln(2 + \sqrt{5})\).
The periodic behavior means:

• Possible multiple solutions for any given value of \(\sin x\).
• Sine function values range from -1 to 1, repeating its path continually over each interval of \(2\pi\).
• A single complete cycle encompasses rise and fall symmetrically around zero.

By determining that exactly one value of \(\sin x\) is possible within the valid range, periodic functions help establish that there is only one valid root across the infinite periodic cycles of \(\sin x\). This underscores the importance of capturing the periodicity in trigonometric equations to ensure accurate solution finding.