Question

Let $(2^{1-a} + 2^{1+a}), f(a), (3^a + 3^{-a})$ be in A.P. and $\alpha$ be the minimum value of $f(a)$. Then the value of the integral $\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}$ is :

• A. $\frac{1}{2} \log_e \left(\frac{4}{3}\right)$
• B. $\frac{1}{4} \log_e \left(\frac{4}{3}\right)$
• C. $\frac{1}{2} \log_e \left(\frac{8}{5}\right)$
• D. $\frac{1}{4} \log_e \left(\frac{8}{5}\right)$

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
First find $f(a)$ using A.P. properties, find its minimum $\alpha$, then evaluate the definite integral. 
: Key Formula or Approach: 
Arithmetic Mean: $f(a) = \frac{(2 \cdot 2^{-a} + 2 \cdot 2^a) + (3^a + 3^{-a})}{2} = (2^a + 2^{-a}) + \frac{1}{2}(3^a + 3^{-a})$. 
AM-GM: $x + 1/x \ge 2$ for $x>0$. 
Step 2: Detailed Explanation: 
$\alpha = \min f(a) = \min (2^a + 2^{-a}) + \frac{1}{2} \min (3^a + 3^{-a}) = 2 + \frac{1}{2}(2) = 3$. 
Limits: $\log_e(3-1) = \log_e 2$ and $\log_e 3$. 
Integral $I = \int_{\log 2}^{\log 3} \frac{dx}{e^{2x} - e^{-2x}} = \int_{\log 2}^{\log 3} \frac{e^{2x} dx}{e^{4x} - 1}$. 
Let $e^{2x} = t \implies 2 e^{2x} dx = dt$. 
When $x = \log 2, t = 4$. When $x = \log 3, t = 9$. 
$I = \frac{1}{2} \int_{4}^{9} \frac{dt}{t^2 - 1} = \frac{1}{2} \cdot \frac{1}{2} [ \log | \frac{t-1}{t+1} | ]_{4}^{9}$. 
$I = \frac{1}{4} [ \log(8/10) - \log(3/5) ] = \frac{1}{4} [ \log(4/5) - \log(3/5) ] = \frac{1}{4} \log(4/3)$. 
Step 3: Final Answer: 
The integral value is $\frac{1}{4} \log_e (4/3)$.