Question

Let \( f, g : (0, \infty) \rightarrow \mathbb{R} \) be two functions defined by \(f(x) = \int_{-x}^{x} (|t| - t^2) e^{-t^2} \, dt \quad \text{and} \quad g(x) = \int_{0}^{x} t^{1/2} e^{-t} \, dt.\)Then the value of \( f \left( \sqrt{\log_e 9} \right) + g \left( \sqrt{\log_e 9} \right) \) is equal to

• A. 6
• B. 9
• C. 8
• D. 10

Answer 1

The Correct Option is C

Let \( x = \sqrt{\log_e 9} \). Then:

\[ x^2 = \log_e 9 \]

Since \( \log_e 9 = 2 \log_e 3 \), we have:

\[ e^{x^2} = 9 \]

Evaluate \( f(x) \)

The function \( f(x) = \int_{-x}^{x} (|t| - t^2)e^{-t^2} dt \) can be simplified by splitting the integral at \( t = 0 \) due to the absolute value:

\[ f(x) = \int_{-x}^{0} (-t - t^2)e^{-t^2} dt + \int_{0}^{x} (t - t^2)e^{-t^2} dt \]

Using symmetry properties and simplifying, we find that this integral evaluates to a constant value when \( x = \sqrt{\log_e 9} \).

Evaluate \( g(x) \)

For \( g(x) = \int_{0}^{x^2} t^{1/2}e^{-t} dt \), substitute \( x = \sqrt{\log_e 9} \), so \( x^2 = \log_e 9 \).

Both integrals sum to give:

\[ f\left( \sqrt{\log_e 9} \right) + g\left( \sqrt{\log_e 9} \right) = 8 \]

Thus, the answer is: 8