Question

Let \(f(x)>0\) for all \(x\in\mathbb{R}\) and \(f(x)\) is bounded. If \[ \lim_{n\rightarrow\infty}\sum_{r=1}^{n}a^{r-1}\int_{(r-1)a}^{ra}\frac{f(x)\,dx}{f(x)+f(2ra-a-x)}=\frac{3}{5}, \] where \(0<a<1\), then the value(s) of \(a\) is/are:

• A. $\frac{5}{11}$
• B. $\frac{7}{11}$
• C. $\frac{6}{11}$
• D. $\frac{1}{11}$

Hint:

Notice how the denominator argument $2ra - a - x$ is precisely engineered to match the boundary sum of King's property for that specific interval block. Recognizing this layout allows you to immediately evaluate the integral as half the interval width ($\Delta/2 = a/2$), bypassing the integration steps entirely.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem combines properties of definite integrals (specifically the "King's Rule") and the sum of an infinite geometric series. The structure of the integral \( \int_A^B \frac{f(x)}{f(x)+f(A+B-x)} dx \) is a standard identity.
Step 2: Key Formula or Approach:
1. King's Property: \( \int_A^B \frac{f(x)}{f(x) + f(A+B-x)} dx = \frac{B - A}{2} \).
2. Infinite GP sum: \( \sum_{r=1}^{\infty} a^{r-1} = \frac{1}{1 - a} \) for \( |a|<1 \).
Step 3: Detailed Explanation:
Consider the integral in the summation for a fixed \( r \):
The lower bound is \( A = (r-1)a \) and upper bound is \( B = ra \).
Sum of bounds: \( A + B = (r-1)a + ra = 2ra - a \).
The integrand is of the form \( \frac{f(x)}{f(x) + f(A+B-x)} \).
According to the property, the value of the integral is \( \frac{B - A}{2} \):
\[ \int_{(r-1)a}^{ra} \frac{f(x) dx}{f(x) + f(2ra - a - x)} = \frac{ra - (r-1)a}{2} = \frac{a}{2} \]
Now substitute this constant back into the summation:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} a^{r-1} \left( \frac{a}{2} \right) = \frac{a}{2} \sum_{r=1}^{\infty} a^{r-1} \]
Using the GP sum formula for \( 0<a<1 \):
\[ \frac{a}{2} \cdot \frac{1}{1 - a} = \frac{3}{5} \]
\[ 5a = 6(1 - a) \implies 5a = 6 - 6a \]
\[ 11a = 6 \implies a = \frac{6}{11} \]
Step 4: Final Answer:
The value of a is \( 6/11 \), which is option (C).