Problem 49 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(S_{1}, S_{2}, \ldots\) be squares such that for each \(n \geq 1\), the length of a side of \(S_{n}\) equals the length of a diagonal of \(S_{n+1} .\) If the length of a side of \(S_{1}\) is \(10 \mathrm{~cm}\), then for which of the following values of \(n\) is the area of \(S_{n}\) less than 1 square \(\mathrm{cm} ?\) (A) 7 (B) 8 (C) 9 (D) 10

Step-by-Step Solution

Answer

The area of \( S_n \) is less than 1 square cm for \( n = 8 \).

Step 1: Understanding the Diagonal Relationship

For a square with side of length \( a \), the length of its diagonal is given by \( a\sqrt{2} \). This implies that the side length of the square \( S_{n+1} \) will be the diagonal length of square \( S_n \).

Step 2: Recursive Side Length Calculation

If the side of \( S_1 \) is 10 cm, then the side of \( S_2 \) is \( 10\sqrt{2} \) cm. Continuing this pattern, for any square \( S_n \), the side length \( a_n \) is given by \( a_n = 10(\sqrt{2})^{n-1} \).

Step 3: Area of Square Calculation

The area of any square \( S_n \) with side length \( a_n \) is \( a_n^2 \). Thus, the area \( A_n \) is given by \( A_n = (10(\sqrt{2})^{n-1})^2 \), or equivalently, \( A_n = 100(2)^{n-1} \).

Step 4: Setting the Area Less Than 1

We need to find the smallest \( n \) such that \( A_n < 1 \). Therefore, we solve the inequality \( 100 \times (2)^{n-1} < 1 \).

Step 5: Solve the Inequality

Divide both sides by 100 to obtain \( (2)^{n-1} < 0.01 \). Taking logarithms on both sides results in \( (n-1) \log 2 < \log 0.01 \). \( \log 2 \approx 0.3010 \) and \( \log 0.01 = -2 \). Solving the inequality gives \( n-1 > \frac{2}{0.3010} \approx 6.644 \), so \( n > 7.644 \). Since \( n \) is an integer, \( n = 8 \).

Key Concepts

Area of Squares

The concept of the area of a square is straightforward. A square is a quadrilateral with four equal sides. To calculate the area of a square, we use the formula \[\text{Area} = \text{side length}^2\]This formula simply involves squaring the length of one of the sides of the square. If the side of a square measures 10 cm, then its area is \[10 \times 10 = 100 \text{ square centimeters}\]In the context of sequences, such as the problem with squares where each subsequent square is constructed using the diagonal of the previous, the area of each square will decrease geometrically. To determine when the area becomes less than 1 square cm, we repeatedly apply the relationship of side lengths and calculate the succeeding squares.

• The area diminishes as we progress through the sequence.
• This decrease follows a geometric sequence.

Understanding how the area shrinks as the sequence progresses is crucial for tackling these kinds of problems effectively.

Diagonal of a Square

The diagonal of a square is a line segment that connects two opposite corners of the square. To calculate the length of a diagonal in a square, you need to know the length of its side. For a square with side length \(a\), the diagonal \(d\) can be calculated using the formula:\[d = a \sqrt{2}\]This relationship comes from the Pythagorean theorem. Consider half of the square to be a right-angled triangle, where the diagonal forms the hypotenuse while each side forms the other two legs. Using the Pythagorean theorem:

• \(a^2 + a^2 = d^2\)
• Rearrange and simplify to obtain \(d = a \sqrt{2}\)

In the sequence problem, this diagonal becomes the side length of the next square. By continually applying this diagonal formula, the progression allows each new square to be built upon the previous, leading to an ever-decreasing sequence of squares. Recognizing how the diagonal calculates allows one to deduce the recursion in the problem.

Logarithms in Inequalities

Logarithms are mathematical operations that help solve equations where variables appear in exponents. When dealing with inequalities involving exponential terms, logarithms become essential in isolating and solving for unknowns.In the problem context, determining when the area of squares becomes less than a certain value requires solving the inequality \(100 \times (2)^{n-1} < 1\). To tackle this, logarithms are utilized:1. Start by dividing the inequality to simplify to \((2)^{n-1} < 0.01\).2. Take the logarithm of both sides.3. Using properties of logarithms, solve for \((n-1) \log 2 < \log 0.01\). - Approximate values: \(\log 2 \approx 0.3010\) and \(\log 0.01 = -2\). By solving this inequality, we find the smallest integer \(n\) such that the specified condition is satisfied.

• Logarithms help convert exponential relationships into linear ones.
• This simplification is pivotal for solving exponential inequalities easily.

Understanding these steps helps demystify the process, making it clear and straightforward to find when the conditions of the problem are met.