Problem 91 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the ordinate \(x=a\) divides the area bounded by \(x\)-axis, part of the curve \(y=1+\frac{8}{x^{2}}\) and the ordinates \(x=2\), \(x=4\), into two equal parts, then \(a\) is equal to (A) \(\sqrt{2}\) (B) \(2 \sqrt{2}\) (C) 3 (D) None of these

Step-by-Step Solution

Answer

(B) \(2\sqrt{2}\)

Step 1: Compute the total area

The total area under \(y = 1 + \frac{8}{x^2}\) from \(x=2\) to \(x=4\) is:
\(A = \int_2^4 \left(1 + \frac{8}{x^2}\right) dx = \left[x - \frac{8}{x}\right]_2^4 = \left(4 - 2\right) - \left(2 - 4\right) = 2 + 2 = 4\)

Step 2: Set up the equation for equal division

The ordinate \(x = a\) divides the area into two equal parts, so:
\(\int_2^a \left(1 + \frac{8}{x^2}\right) dx = \frac{A}{2} = 2\)

Step 3: Evaluate the integral from 2 to a

\(\int_2^a \left(1 + \frac{8}{x^2}\right) dx = \left[x - \frac{8}{x}\right]_2^a = \left(a - \frac{8}{a}\right) - \left(2 - 4\right) = a - \frac{8}{a} + 2\)

Step 4: Solve for a

Setting this equal to 2:
\(a - \frac{8}{a} + 2 = 2\)
\(a - \frac{8}{a} = 0\)
\(a^2 = 8\)
\(a = 2\sqrt{2}\)
Since \(2 < a < 4\), we have \(a = 2\sqrt{2} \approx 2.83\), which lies in \([2, 4]\).
The answer is (B) \(2\sqrt{2}\).

Key Concepts

Definite Integrals

A definite integral is a powerful tool for determining the area under a curve between two specified points on the x-axis. In this problem, to find the area between the curve \(y=1+\frac{8}{x^2}\) and the x-axis from \(x=2\) to \(x=4\), we compute the definite integral of the function \(y\) between these bounds. The process involves a few essential steps:

• Firstly, understand the functional form: \( y = 1 + \frac{8}{x^2} \) which combines a constant function with an inverse square function.
• Next, to find the total area, perform the integration of \(y\) from \(x=2\) to \(x=4\). This gives the total area under the curve for the specified bounds.
• The definite integral is solved by evaluating the antiderivative at the upper and lower limits of integration and subtracting these evaluations.

Definite integrals are crucial when trying to calculate exact areas, especially when the curve is not a simple geometric shape. They allow us to "sum up" an infinite number of infinitesimally small areas to find the total area enclosed by a curve, an axis, and any other boundary.

Areas under Curves

The concept of determining the area under curves is fundamental in integral calculus. The area under a curve like \(y=1+\frac{8}{x^2}\) represents the accumulation of values of \(y\) over an interval on the x-axis. Here's how you'll generally approach finding this area:

• The curve provides the upper boundary of the region whose area you want to measure.
• Below this curve and above the x-axis between the ordinates \(x=2\) and \(x=4\), lies the area we are interested in.
• Visualize splitting this area as integral calculations allow for specifying exact sections. The goal is sometimes to find when this area can be evenly divided, like the problem asks.

In many real-world applications, areas under curves relate to finding quantities such as distance (in physics) where the rate (speed/velocity) changes over time. By mastering the integral calculus approach to areas like these, one can solve a broad range of mathematical and practical problems.

Ordinate Division

Ordinate division is the concept of splitting areas under curves at specific vertical lines (ordinates), creating sections of defined proportions. This is precisely what the exercise seeks by asking where the ordinate \(x=a\) divides the area under the curve between \(x=2\) and \(x=4\) into two equal parts.

• The idea is to determine where to "cut" the region so that the areas on each side of the ordinate \(x=a\) are equal.
• This requires two definite integrals. One integral calculates the area from \(x=2\) to \(x=a\), and another from \(x=a\) to \(x=4\).
• The ordinate \(x=a\) is found by setting these two integral calculations equal and solving for \(a\), ensuring both areas are identical.

The dividing ordinate is often not intuitively obvious and is precisely calculated through integral calculus methods. This technique has broad applications in data analytics, engineering, and resource allocation where balanced distribution is desired.