Question

If \( f(x) = \int \frac{dx}{x^2 + 2} \) and \( f(\sqrt{2}) = 0 \), then \( f(0) = \)

• A. \( \frac{\pi}{2\sqrt{2}} \)
• B. \( -\frac{\pi}{2\sqrt{2}} \)
• C. \( -\frac{\pi}{4\sqrt{2}} \)
• D. \( \frac{\pi}{4\sqrt{2}} \)

Hint:

When solving integrals involving rational functions, always check if the denominator is of the form \( x^2 + a^2 \), which allows you to use the standard inverse tangent formula. This will help simplify the integration process significantly.

Answer 1

The Correct Option is C

We are given the integral \( f(x) = \int \frac{dx}{x^2 + 2} \). The standard form for the integral of this type is: \[ \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \] For our case, \( a^2 = 2 \), hence \( a = \sqrt{2} \). Thus, the integral becomes: \[ f(x) = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) \] Now, we are given that \( f(\sqrt{2}) = 0 \), so substitute \( x = \sqrt{2} \) into the equation: \[ f(\sqrt{2}) = \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{\sqrt{2}}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}} \tan^{-1}(1) \] Since \( \tan^{-1}(1) = \frac{\pi}{4} \), we have: \[ f(\sqrt{2}) = \frac{1}{\sqrt{2}} \times \frac{\pi}{4} = \frac{\pi}{4\sqrt{2}} \] To satisfy \( f(\sqrt{2}) = 0 \), we must subtract \( \frac{\pi}{4\sqrt{2}} \) from the original function, which implies: \[ f(x) = \frac{1}{\sqrt{2}} \left( \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) - \frac{\pi}{4} \right) \] Finally, evaluate \( f(0) \): \[ f(0) = \frac{1}{\sqrt{2}} \left( \tan^{-1}(0) - \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \left( 0 - \frac{\pi}{4} \right) = -\frac{\pi}{4\sqrt{2}} \] Thus, the correct value of \( f(0) \) is \( -\frac{\pi}{4\sqrt{2}} \).