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

\(f(x)\) has minimum at \(x=0\)
And \(f(x)_{min}=f(0)\)
\(f(x)_{min}=0\)

So, the answer is \(0\).

Answer 2

To find the minimum value of the function f(x), let's start by analysing the integrand \(\frac{e^{(t-\cos(t))}}{1+t^{2023}}\).
1. \(e^{(t-\cos(t))}\) is always positive for all t in R.
2. \(1 + t^{2023}\) is also always positive.

So, \(\frac{e^{(t-\cos(t))}}{1+t^{2023}}\) is non-negative for all t in R.

Now, we want to find the minimum value of f(x), which occurs when the integrand is at its minimum value. Since the integrand is non-negative, the minimum value of the integral occurs when the integrand is at its minimum value, which is 0.
The minimum value of f(x) is achieved when x is the smallest possible value, which is when x = 0.
Thus, the minimum value of f(x) is 0, attained at x = 0.

So, the answer is 0.