Problem 40 - Calculus for JEE ADVANCED

Question

\(\frac{d y}{d x}=5 x^{2}(x-1)(x-3)=0\).

Step-by-Step Solution

Answer

The solutions are \(x = 0\), \(x = 1\), and \(x = 3\).

Step 1: Identify the equation

The given equation is \(\frac{d y}{d x} = 5x^2(x-1)(x-3) = 0\). This is a differential equation that involves finding the values of \(x\) that satisfy this equation.

Step 2: Set the equation to zero

We know that for the product of several terms to be zero, at least one of the factors must be zero. Therefore, set the equation \(5x^2(x-1)(x-3) = 0\) to zero and solve for \(x\).

Step 3: Solve each factor for zero

The factors are \(5x^2\), \((x-1)\), and \((x-3)\). Set each factor equal to zero: 1. \(5x^2 = 0\)2. \(x-1 = 0\)3. \(x-3 = 0\).Solve each equation for \(x\).

Step 4: Solve for \(x\) from the first factor

The equation \(5x^2 = 0\) simplifies to \(x^2 = 0\). Taking the square root of both sides gives \(x = 0\).

Step 5: Solve for \(x\) from the second factor

The equation \(x-1 = 0\) simplifies directly to \(x = 1\).

Step 6: Solve for \(x\) from the third factor

The equation \(x-3 = 0\) simplifies directly to \(x = 3\).

Step 7: List the solutions

The solutions where the derivative \(\frac{d y}{d x} = 0\) are \(x = 0\), \(x = 1\), and \(x = 3\).

Key Concepts

Derivative

In the world of calculus, the concept of a derivative is fundamental. It represents the rate at which a function is changing at any given point. If you think of a graph, the derivative gives you the slope of the line tangent to the curve of the function at a specific point. For a function like our example, where the derivative is expressed as \(\frac{d y}{d x}=5 x^{2}(x-1)(x-3)\), it shows how fast the values of \(y\) are changing as \(x\) changes.

• The derivative is used to find critical points that tell us where the rate of change is zero. This is vital for finding maxima and minima in functions.
• Derivatives can also tell us about the concavity of a function, helping in understanding the function’s overall shape.

In our example, we are solving \(\frac{d y}{d x} = 0\), which is the condition to find zero slope points - the stationary points where changes are momentarily halted.

Factorization

Factorization is a mathematical method used to express a polynomial as a product of its factors. This approach simplifies solving equations, especially when dealing with polynomials like those in differential equations. In our example, \(5x^2(x-1)(x-3)\), factorization plays a key role.

• Each term in the factored equation represents a potential point where the whole expression can equal zero. By setting each factor to zero, we identify potential solutions.
• Factorization is essential for breaking down complex polynomial expressions into more manageable pieces.

This process leads directly to simpler equations that are easier to solve, as we see with the individual factors \(5x^2\), \(x-1\), and \(x-3\). Breaking these down allows us to find the roots of the equation.

Roots of Equations

Finding the roots of an equation means identifying the values of \(x\) where the expression equals zero. These roots are crucial in solving differential equations because they signify points where the derivative equals zero, linking closely to the concept of stationary points.

• In our exercise, each solved factor gives us a root: from \(5x^2 = 0\) we get \(x = 0\), from \(x-1 = 0\) we get \(x = 1\), and from \(x-3 = 0\) we get \(x = 3\).
• Roots are not just solutions; they help understand the behavior of the function itself - particularly in terms of where the function levels out, changes direction, or crosses the x-axis.

Each root gives insight into the behavior of the graph of the function, thus telling us a lot about its characteristics without even needing to graph it.