Problem 178 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The equation of the tangent to the curve \(y=x+\frac{4}{x^{2}}\), which is parallel to the \(x\)-axis, is (A) \(y=1\) (B) \(y=2\) (C) \(y=3\) (D) \(y=0\)

Step-by-Step Solution

Answer

The equation of the tangent parallel to the x-axis is \( y = 3 \). Option (C).

Step 1: Differentiate the function

The given function is \( y = x + \frac{4}{x^2} \). To find the slope of the tangent, we first differentiate the function with respect to \( x \). Using the power rule and the derivative of a negative power, the derivative is: \( \frac{dy}{dx} = 1 - \frac{8}{x^3} \).

Step 2: Set the slope to zero for parallelism

For the tangent to be parallel to the \( x \)-axis, the slope \( \frac{dy}{dx} \) must be zero. Therefore, set \( 1 - \frac{8}{x^3} = 0 \).

Step 3: Solve for x

Solving \( 1 = \frac{8}{x^3} \) gives \( x^3 = 8 \), hence \( x = 2 \) after taking the cube root of both sides.

Step 4: Calculate the y-coordinate

Substitute \( x = 2 \) back into the original equation to find the corresponding \( y \)-value. \( y = 2 + \frac{4}{2^2} = 2 + 1 = 3 \).

Step 5: Write down the equation of the tangent line

Since the tangent line is parallel to the \( x \)-axis, its equation is simply \( y = 3 \).

Key Concepts

Differentiation

When we talk about differentiation in calculus, we are often referring to finding the derivative of a function. The derivative represents the rate of change, or the slope of the tangent, of the function at any point on its curve. In this exercise, we are given the function \( y = x + \frac{4}{x^2} \). To find the equation of the tangent line that is parallel to the \(x\)-axis, we must first determine the derivative of this function.

The rules for differentiation, such as the power rule, help us to accomplish this. The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). By applying these rules, we find the derivative: \( \frac{dy}{dx} = 1 - \frac{8}{x^3} \). In simpler terms, the derivative calculates how steep or flat the function is at every point along its curve.

Finding the right point where the slope is zero (flat, aligning with the \(x\)-axis) is crucial for solving this problem.

Parallel Lines

Parallel lines are lines in a plane that do not intersect, no matter how far they are extended. In mathematics, when two lines are parallel, they have the same slope. If a tangent line is parallel to the \(x\)-axis, it means its slope is zero since the \(x\)-axis is horizontal.

In this exercise, we are tasked with finding the tangent line to the curve that is parallel to the \(x\)-axis. Using differentiation, we calculate the slope of the tangent line at any point on the curve \( y = x + \frac{4}{x^2} \). Setting this slope \( \frac{dy}{dx} = 1 - \frac{8}{x^3} \) equal to zero allows us to identify specific points where the curve's tangent is parallel to the \(x\)-axis.

Solving \( 1 - \frac{8}{x^3} = 0 \) gives us \( x = 2 \), which is the \(x\)-coordinate where our tangent line is parallel to the \(x\)-axis.

Equations of Tangent

Once we have determined the point on the curve where the slope is zero, we can write the equation of the tangent line. The equation of a horizontal line (parallel to the \(x\)-axis) takes the form \( y = c \), where \( c \) is a constant. This constant is the \(y\)-coordinate of the point where the slope was found to be zero.

In our exercise, after finding \( x = 2 \) where the slope of the tangent is zero, we substitute back into the original function to find the \(y\)-coordinate. Substituting \( x = 2 \) into \( y = 2 + \frac{4}{2^2} = 2 + 1 = 3 \). Therefore, the \(y\)-coordinate is 3.

Now, we have the equation of the tangent line: \( y = 3 \). At this point, the tangent line doesn't rise or fall but remains constant, demonstrating that it is perfectly parallel to the \(x\)-axis.