Question

Let $y = \log_e \left( \frac{1 - x^2}{1 + x^2} \right), -1 < x < 1$. Then at $x = \frac{1}{2}$, the value of $225(y' - y'')$ is equal to

• A. 732
• B. 746
• C. 742
• D. 736

Answer 1

Answer: (D)

736

Answer 2

**Given: **$y = \log_e \left( \frac{1 - x^2}{1 + x^2} \right)$

Step 1. First derivative $\frac{dy}{dx}$:$\frac{dy}{dx} = y' = \frac{-4x}{1 - x^4}$

Step 2. Second derivative $\frac{d^2y}{dx^2}$:$y'' = \frac{-4(1 + 3x^4)}{(1 - x^4)^2}$

Step 3. Calculate $y' - y''$ at $x = \frac{1}{2}$: $y' - y'' = \frac{-4x}{1 - x^4} + \frac{4(1 + 3x^4)}{(1 - x^4)^2}$

Step 4. Substitute $x = \frac{1}{2}$ and simplify to find $225(y' - y'')$: After calculations: $225(y' - y'') = 736$

The Correct Answer is:736