Which of the following must be the true value of the statements below in that order. Statement I: For x R , let tan - 1 ( x ) ( - 2 , 2 ) . Then the minimum value of the function f : R R , defined by f ( x ) = 0 x tan - 1 x e ( t - cos t ) 1 +

Question

$Which of the following must be the true value of the statements below in that order.$

$Statement I: For x \in R, let \tan^{- 1} (x) \in (- \frac{π}{2}, \frac{π}{2}) . Then the minimum value of the function$
$f : R \to R, defined by f (x) = \int_{0}^{x \tan^{- 1} x} \frac{e^{(t - \cos t)}}{1 + t^{2023}} d t is zero.$

$Statement II: If f (x) is a periodic function with period T, then$ $\int_{a + n T}^{b + n T} f (x) d x = \int_{a}^{b} f (x) d x, n \in Z, a, b \in R .$

$Statement III: If ψ (x) \leq ϕ (x) for a \leq x \leq b, then$ $\int_{a}^{b}$ $| ψ (x) |$ $d x \leq \int_{a}^{b}$ $| ϕ (x) |$ $d x .$

Smart Achievers · Accepted Answer

(1)

$Given, f (x) = \int_{0}^{x \tan^{- 1} x} \frac{e^{t - \cos t}}{1 + t^{2023}} d t$

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

$For x < 0, \tan^{- 1} x \in (- \frac{π}{2}, 0)$

$For x \geq 0, \tan^{- 1} x \in [0, \frac{π}{2}] \Rightarrow x \tan^{- 1} \forall x \in R$

$And \frac{x}{1 + x^{2}} + \tan^{- 1} x =$ $d x \leq \int_{a}^{b}$

$So, f (x) is minimum at x = 0$

$Here minimum value is f (0) = \int_{0}^{0} = 0 .$

$ψ (x) and ϕ (x) can be negative functions$

Solution

1 TTF

2 FTF

3 FTT

4 TTT

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