Question

For \(x ∈ R\), let \(tan^{-1} (x) ∈ \) (\(-\frac{\pi}{2},\frac{\pi}{2}\)). Then the minimum value of function \(f: R \rightarrow R\) defined by \(f(x) =\) \(\int_{0}^{x\,tan^{-1}x}\frac{e^{(t-cos\,t)}}{1+t^{2023}}dt\)  is

Answer 1

Correct Answer: 0

Given: The function \( f(x) \) is defined as:

\(f(x) = \int_0^{x \tan^{-1} x} \frac{e^{t - \cos t}}{1 + t^{2023}} dt\)

The first derivative \( f'(x) \) is:

\(f'(x) = \frac{e^{x \tan^{-1} x - \cos(x \tan^{-1} x)}}{1 + (x \tan^{-1} x)^{2023}} \left( \frac{x}{1 - x^2 + \tan^{-1} x} \right)\)

Step 1: Finding Critical Points

For the maximum or minimum, we set \( f'(x) = 0 \).

\(\Rightarrow x = 0\)

Step 2: Checking Behavior Around \( x = 0 \)

We examine the sign of \( f'(x) \) around \( x = 0 \). We can see that:

\(\text{For} \, x < 0, f'(x) < 0 \quad \text{and} \quad \text{For} \, x > 0, f'(x) > 0\)

Conclusion

Since the derivative changes from negative to positive as \( x \) crosses 0, \( f(x) \) has a minimum at \( x = 0 \).

Final Answer:

The function \( f(x) \) has a minimum at \( x = 0 \), and \( f(x_{\text{min}}) = f(0) = 0 \).