Question

If, \( I_n = \int_{-\pi}^{\pi} \frac{\cos(nx)(1+2^x)}{dx} \), where \( n = 0, 1, 2, \dots \), then which of the following are correct?

A. \( I_n = I_{n+2} \), for all \( n = 0, 1, 2, \dots \) 
B. \( I_n = 0 \), for all \( n = 0, 1, 2, \dots \) 
C. \( \sum_{n=1}^{10} I_n = 2^{10} \) 
D. \( \sum_{n=1}^{10} I_n = 0 \) 
 

• A. A, B and D only
• B. A and C only
• C. B and D only
• D. A, C and D only

Hint:

For integrals over a symmetric interval like \( [-a, a] \), always test the property \( \int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx \). It can dramatically simplify integrands that are neither purely even nor purely odd, as seen with the term \( \frac{1}{1+c^x} \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem involves analyzing a definite integral \(I_n\) over a symmetric interval \( [-\pi, \pi] \). The integrand is a product of an even function (\(\cos(nx)\)) and a function that is neither even nor odd (\(\frac{1}{1+2^x}\)). This structure suggests using a specific property of definite integrals over symmetric intervals to simplify the expression for \(I_n\).

Step 2: Key Formula or Approach:
We use the property of definite integrals over a symmetric interval \( [-a, a] \): \[ \int_{-a}^{a} f(x) dx = \int_{0}^{a} [f(x) + f(-x)] dx \] Let the integrand be \( f(x) = \frac{\cos(nx)}{1+2^x} \).

Step 3: Detailed Explanation:
First, we apply the property to simplify \(I_n\). We need to calculate \(f(x) + f(-x)\). \[ f(-x) = \frac{\cos(n(-x))}{1+2^{-x}} = \frac{\cos(nx)}{1+\frac{1}{2^x}} = \frac{\cos(nx)}{\frac{2^x+1}{2^x}} = \frac{2^x \cos(nx)}{1+2^x} \] Now, we sum the two parts: \[ f(x) + f(-x) = \frac{\cos(nx)}{1+2^x} + \frac{2^x \cos(nx)}{1+2^x} = \frac{(1+2^x)\cos(nx)}{1+2^x} = \cos(nx) \] So the integral becomes: \[ I_n = \int_{0}^{\pi} \cos(nx) dx \] Now we evaluate this simplified integral.
For \(n=0\): \[ I_0 = \int_{0}^{\pi} \cos(0) dx = \int_{0}^{\pi} 1 dx = [x]_0^\pi = \pi \]
For \(n \ge 1\): \[ I_n = \left[ \frac{\sin(nx)}{n} \right]_0^\pi = \frac{\sin(n\pi) - \sin(0)}{n} = \frac{0 - 0}{n} = 0 \] So, we have \( I_0 = \pi \) and \( I_n = 0 \) for all integers \( n \ge 1 \). Now, let's check the given statements:
A. \( I_n = I_{n+2} \), for all \( n=0, 1, 2, ... \): This statement is technically false as written because for \(n=0\), \(I_0 = \pi\) and \(I_{0+2}=I_2=0\), and \( \pi \neq 0 \). However, for all \(n \ge 1\), \(I_n = 0\) and \(I_{n+2}=0\), so the equality holds. It's likely the question intended this for \(n \ge 1\).
B. \( I_n = 0 \), for all \( n=0, 1, 2, ... \): This statement is false because \( I_0 = \pi \neq 0 \). However, the statement is true for all \(n \ge 1\).
C. \( \sum_{n=1}^{10} I_n = 2^{10} \): Since \(I_n = 0\) for \(n \ge 1\), the sum is \( \sum_{n=1}^{10} 0 = 0 \). The statement claims the sum is \(2^{10}\), which is false.
D. \( \sum_{n=1}^{10} I_n = 0 \): As calculated above, \( \sum_{n=1}^{10} I_n = 0 \). This statement is true. Based on a strict interpretation, only statement D is true. However, this is not an option. This indicates a likely error in the problem statement, where the conditions for A and B were intended for \(n \ge 1\). If we make this reasonable assumption (common in competitive exams), then statements A, B, and D are all considered correct in their intended context.

Step 4: Final Answer:
Assuming the likely intent that statements A and B apply for \( n \ge 1 \), the correct statements are A, B, and D.