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

Question

The equation \(e^{\sin x}-e^{-\sin x}-4=0\), for \(x\) real, has \(\mathbf{[ 2 0 1 2}]\) (A) infinite number of roots (B) no roots (C) exactly one root (D) exactly four roots

Step-by-Step Solution

Answer

(C) exactly one root

Step 1: Simplify the Equation

Start by simplifying the given equation \( e^{\sin x} - e^{-\sin x} - 4 = 0 \). This can be rewritten as \( e^{\sin x} - \frac{1}{e^{\sin x}} = 4 \). Let \( y = e^{\sin x} \), then the equation becomes \( y - \frac{1}{y} = 4 \).

Step 2: Relate to a Quadratic Equation

Multiply the entire equation by \( y \) to eliminate the fraction: \( y^2 - 1 = 4y \). Rearrange into a standard quadratic form: \( y^2 - 4y - 1 = 0 \).

Step 3: Solve the Quadratic Equation

Use the quadratic formula to solve \( y^2 - 4y - 1 = 0 \). The quadratic formula is \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -4, c = -1 \). Substituting these gives \( y = \frac{4 \pm \sqrt{16 + 4}}{2} \), yielding solutions \( y = 2 \pm \sqrt{5} \).

Step 4: Evaluate Validity of Solutions

Since \( y = e^{\sin x} \), and \( y > 0 \), both \( y = 2 + \sqrt{5} \) and \( y = 2 - \sqrt{5} \) need to be checked if they can correspond to a real \( \sin x \). Calculating, \( 2 + \sqrt{5} > 1 \), valid for \( y = e^z \), but \( 2 - \sqrt{5} < 0 \), which is not valid since \( e^{\sin x} > 0 \).

Step 5: Determine the Number of Roots

Only \( y = 2 + \sqrt{5} \) is feasible, meaning \( e^{\sin x} = 2 + \sqrt{5} \). Determine \( \sin x \) by taking the natural logarithm, resulting in a single value for \( \sin x \). Since \( \sin x \) can repeat periodically (sine function), but \( e^{\sin x} = 2 + \sqrt{5} \) matches only once per cycle due to its specific value, the overall equation has exactly one solution.

Key Concepts

Quadratic equations

Quadratic equations are fundamental in algebra and have the general form:

• \( ax^2 + bx + c = 0 \)

where \( a, b, \) and \( c \) are constants, and \( x \) represents the variable to solve for. A key feature of quadratics is their parabolic graph shape.
To solve a quadratic equation, you can use the quadratic formula:
\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula provides the roots (solutions) of the equation based on the discriminant \( b^2 - 4ac \). The discriminant can tell us about the roots:

• If \( b^2 - 4ac > 0 \), there are two real roots.
• If \( b^2 - 4ac = 0 \), there is one real root.
• If \( b^2 - 4ac < 0 \), there are no real roots.

In the context of our exercise, the equation \( y^2 - 4y - 1 = 0 \) is a quadratic equation used to find \( y \) values that relate back to \( e^{\sin x} \). With a discriminant of \( 20 \), there are two potential roots for \( y \). However, only the one that yields a valid \( e^{\sin x} > 0 \) remains as a feasible solution.

Exponential equations

Exponential equations involve expressions where the unknown is in the exponent. A classic form is \( a^x = b \). To solve such equations, logarithms are often used, as they allow us to "bring down" the exponent.
In our problem, \( e^{\sin x} \) and \( e^{-\sin x} \) represent exponential terms with the natural base \( e \). We begin by setting \( y = e^{\sin x} \), transforming the initial exponential equation to a more manageable one involving \( y - \frac{1}{y} = 4 \).
Upon solving the quadratic form \( y^2 - 4y - 1 = 0 \), we use the roots to determine feasible values for \( e^{\sin x} \). The solution that fits the real scenario, \( y = 2 + \sqrt{5} \), corresponds to a valid exponential equation.

• Logarithms can be used to solve for \( \sin x \) by transforming \( e^{\sin x} = 2 + \sqrt{5} \) to \( \sin x = \ln(2 + \sqrt{5}) \).

This process highlights how combining logarithmic and quadratic approaches can solve complex exponential equations.

Sine function

The sine function, denoted as \( \sin x \), is a periodic function that oscillates between -1 and 1 with a period of \( 2\pi \). It is one of the primary trigonometric functions and is essential in oscillatory motions, waves, and circular phenomena.
In the context of this equation \( e^{\sin x} = 2 + \sqrt{5} \), understanding the sine function is crucial. The equation is mainly transformed to solve for \( \sin x \) using logarithmic properties. However, as \( \sin x \) repeats its values every cycle of \( 2\pi \), we must ensure that our solution corresponds to valid conditions for a single occurrence.

• Despite being periodic, the specific constraints given by \( e^{\sin x} \) ensure the solution does not repeat within a single cycle.

This means that, although \( \sin x \) could theoretically produce multiple values due to periodicity, the equation permits only one valid \( \sin x \), hence the equation has exactly one root.