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

Concept: The localized integral segments can be simplified by applying King's Property ($\int_p^q g(x)dx = \int_p^q g(p+q-x)dx$) to each sub-interval block. This collapses the fractional integrand into a constant, transforming the infinite summation loop into a standard geometric progression series. Step 1: Evaluate the general internal definite integral segment.
Let us consider the integral segment for a specific index step $r$: \[ I_r = \int_{(r-1)a}^{ra} \frac{f(x)}{f(x) + f(2ra - a - x)}\,dx \] Calculate the sum of the upper and lower integration limits for this interval block: \[ \text{Sum} = (r - 1)a + ra = ra - a + ra = 2ra - a \] Apply King's Property to the integral by replacing the variable $x$ with $(2ra - a - x)$: \[ I_r = \int_{(r-1)a}^{ra} \frac{f(2ra - a - x)}{f(2ra - a - x) + f(x)}\,dx \] Summing these two matching representations of $I_r$ cancels out the function terms in the integrand: \[ 2I_r = \int_{(r-1)a}^{ra} \frac{f(x) + f(2ra - a - x)}{f(x) + f(2ra - a - x)}\,dx = \int_{(r-1)a}^{ra} 1 \cdot dx \] \[ 2I_r = \Big[ x \Big]_{(r-1)a}^{ra} = ra - (ra - a) = a \quad \Rightarrow \quad I_r = \frac{a}{2} \quad \cdots (1) \]

Step 2: Substitute the evaluated integral back into the summation loop.
Substitute our constant value $I_r = \frac{a}{2}$ from equation (1) back into the primary infinite limit sum equation: \[ \lim_{n \rightarrow \infty} \sum_{r=1}^{n} a^{r-1} \cdot \left( \frac{a}{2} \right) = \frac{3}{5} \] Pull out the constant multiplier terms from the summation loop: \[ \frac{a}{2} \cdot \left( \lim_{n \rightarrow \infty} \sum_{r=1}^{n} a^{r-1} \right) = \frac{3}{5} \quad \cdots (2) \]

Step 3: Solve the resulting infinite geometric series expansion.
The terms inside the summation expand into a standard infinite geometric progression series: \[ \sum_{r=1}^{\infty} a^{r-1} = 1 + a + a^2 + a^3 + \dots \] We are given the constraint that the parameter lies strictly within the fractional range $0 < a < 1$. Applying the infinite $G.P.$ sum formula ($S_{\infty} = \frac{1}{1-a}$): \[ \frac{a}{2} \cdot \left( \frac{1}{1 - a} \right) = \frac{3}{5} \quad \Rightarrow \quad \frac{a}{2(1 - a)} = \frac{3}{5} \]

Step 4: Cross-multiply the fractions to solve for parameter $a$.
Cross-multiply the denominators to form a single linear algebraic equation: \[ 5a = 6(1 - a) \quad \Rightarrow \quad 5a = 6 - 6a \] Collect all terms containing $a$ on the left side: \[ 11a = 6 \quad \Rightarrow \quad a = \frac{6}{11} \] This matches option (C) perfectly.