Problem 44 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If a curve passes through the point \(\left(2, \frac{7}{2}\right)\) and has slope \(\left(1-\frac{1}{x^{2}}\right)\) at any point \((x, y)\) on it, then the ordinate of the point on the curve whose abscissa is \(-2\) is (a) \(-\frac{3}{2}\) (b) \(\frac{3}{2}\) (c) \(\frac{5}{2}\) (d) \(-\frac{5}{2}\)

Step-by-Step Solution

Answer

The ordinate of the point is \(-\frac{3}{2}\), option (a).

Step 1: Identify the Differential Equation

The problem provides a slope, which is the derivative of the curve, as \( \frac{dy}{dx} = 1 - \frac{1}{x^2} \). Our task is to solve this differential equation to find the general expression of the curve.

Step 2: Integrate to Find the Curve Equation

Integrate \( \frac{dy}{dx} = 1 - \frac{1}{x^2} \) with respect to \( x \):\[ y = \int \left(1 - \frac{1}{x^2}\right)\, dx = \int 1\, dx - \int \frac{1}{x^2}\, dx = x + \frac{1}{x} + C\]Here, \( C \) is the constant of integration.

Step 3: Apply Initial Condition to Determine C

The curve passes through the point \( \left(2, \frac{7}{2}\right) \). Plug this into the equation \( y = x + \frac{1}{x} + C \):\[ \frac{7}{2} = 2 + \frac{1}{2} + C \]Solve for \( C \):\[ \frac{7}{2} = \frac{4}{2} + \frac{1}{2} + C \]\[ \frac{7}{2} = \frac{5}{2} + C \]\[ C = \frac{7}{2} - \frac{5}{2} = 1 \]},

Step 4: Write the Particular Solution

With \( C = 1 \), the equation of the curve is:\[ y = x + \frac{1}{x} + 1 \]

Step 5: Find the Ordinate for x = -2

Substitute \( x = -2 \) into \( y = x + \frac{1}{x} + 1 \):\[ y = -2 + \left(-\frac{1}{2}\right) + 1 \]\[ y = -2 - \frac{1}{2} + 1 \]\[ y = -\frac{4}{2} - \frac{1}{2} + \frac{2}{2} \]\[ y = -\frac{3}{2} \]

Step 6: Choose the Correct Answer

The ordinate of the point on the curve where the abscissa is \(-2\) is \(-\frac{3}{2}\), which corresponds to option (a).

Key Concepts

Calculus

Calculus is a branch of mathematics that helps us understand changes between values that are related by a function. Central to calculus are the concepts of derivatives and integrals. They allow us to analyze continuous change. For example:

• Derivatives help us understand how a function's output changes as its input changes. For curves, this is often represented as the slope.
• Integrals help us find areas under curves or reconstruct functions from their slopes.

Calculus is fundamental for solving differential equations, such as finding a curve's equation given its slope. It's the tool for understanding how different rates of change come together to shape curves and surfaces. Understanding calculus gives us insights into how systems evolve over time.

Initial Conditions

Initial conditions are like starting points that determine the specific path a solution takes. When we solve a differential equation, we're often left with a family of solutions. Each potential curve in this family aligns with the general solution of the differential equation.
However, to pinpoint one particular curve from this family, we need additional information, which is provided by initial conditions. In this exercise, the curve passes through a given point:

• The point is \((2, \frac{7}{2})\) in this case. Inserting these coordinates into the general solution helps us find the exact value of integration constant \((C)\).
• This results in a specific solution, tailored to the initial condition.

Using initial conditions ensures that the curve correctly models the real-world scenario or the specific problem being analyzed.

Integration

Integration is the process of finding an integral, which represents the accumulation of quantities. In the context of differential equations, integration allows us to go from the derivative (slope) of a function, back to the function itself. It's the reverse process of differentiation.
In the given problem, the slope of the curve is expressed as a differential equation:

• \(\frac{dy}{dx} = 1 - \frac{1}{x^{2}}\) is integrated to obtain the general solution \(y = x + \frac{1}{x} + C\).
• The integral \(\int(1 - \frac{1}{x^2})\,dx\) leads to terms \(x\) and \(\frac{1}{x}\).

Here, \(C\) is an arbitrary constant known as the constant of integration. This constant is crucial, as it adjusts the solution to match specific conditions.

Slope of a Curve

The slope of a curve at any given point represents how steep the curve is at that specific point. In mathematical terms, it's the derivative of the curve's equation.

• For instance, in our problem, the slope is given by \(1 - \frac{1}{x^2}\).
• This tells us how the curve rises or falls as \(x\) changes.

Having the slope means we can construct the curve through integration, going backward from the rate of change to the original function. Understanding the slope helps predict how a curve behaves at different points, making it easier to model or interpret dynamic systems.