Question

`\boxed{\begin{gathered} \rm Let \: \psi_1 : [0, \infty ) \to \mathbb{R} , \psi_2 : [0, \infty ) \to \mathbb{R},f :[0, \infty )\to \mathbb{R} \: and \: g :[0, \infty) \to \mathbb{R} \: be \\ \rm functions \: such \: that \: f(0) = g(0) = 0, \\\\ \rm \psi_(1)(x) = {e}^( - x) + x, \: \: x \geq0, \\ \rm \psi_(2)(x) = {x}^(2) - 2x - 2 {e}^( - x) + 2, \: \: x > 0, \\ \rm f(x) = \int_( - x)^(x) ( |t| - {t}^(2) ) {e}^{ - {t}^(2) } \: dt, \: \: x > 0 \\\\ \rm g(x) = \int_0^{ {x}^(2) } √(t) \: {e}^( - t) \: dt, \: \: x > 0 \end{gathered}}`

Which of the following is True?

`\rm (A) \: \rm \: f ( √( \ln 3 ) )+ g( √( \ln3) ) = (1)/(3)`
(B) For every x>1, there exists an α ∈ (1,x) such that ψ₁(x)=1+ax

(C) For every x>0, there exists a β ∈ (0,x) such that ψ₂(x)=2x(ψ₁(β)-1)

(D) f is an increasing function on the interval
`\bigg [0 , (3)/(2) \bigg]` ​

Answer 1

(A) is false. By symmetry,

`\displaystyle f(x) = \int_(-x)^x (|t|_t^2) e^(-t^2) \, dt = 2 \int_0^x (t-t^2) e^(-t^2) \, dt`

where
`|t|=t` since
`x>0`. Substitute
`s=t^2` to get the equivalent integral,

`\displaystyle f(x) = \int_0^(x^2) (1 - \sqrt s) e^(-s) \, ds`

Then

`\displaystyle f(x) + g(x) = \int_0^(x^2) e^(-s) \, ds`

`\displaystyle f(√(\ln(3))) + g(√(\ln(3))) = \int_0^(\ln(3)) e^(-s) \, ds = \frac23 \\eq \frac13`

(B) is false. Note that
`1+\alpha x` is linear so its derivative is the constant
`\alpha` at every point. We then have

`{\psi_1}'(\alpha) = -e^(-\alpha)+1 = \alpha \implies 1-\alpha = e^(-\alpha)`

But this has no solutions, since the left side is negative for
`\alpha>1` and the right side is positive for all
`\alpha`.

(C) is true. By the same reasoning as in (B), the line
`2x(\psi_1(\beta)-1)` has constant derivative,
`2\psi_1(\beta)-2 = 2e^{-\beta+2\beta-2`. Then

`{\psi_2}'(\beta) = 2\beta-2+2e^(-\beta) = 2e^(-\beta)+2\beta-2`

holds for all values of
`\beta`.

(D) is false. We use the first derivative test. By the fundamental theorem of calculus,

`\displaystyle f(x) = 2 \int_0^x (t-t^2)e^(-t^2)\,dt \implies f'(x) = 2(x-x^2)e^(-x^2)`

Solve for the critical points.

`f'(x) = 0 \implies x-x^2 = 0 \implies x = 0 \text{ or } x = 1`

`e^(-x^2)>0` for all
`x`, so the sign of
`f'` depends on the sign of
`x-x^2`. It's easy to see
`f'>0` for
`x\in(0,1)` and
`f'<0` for
`x\in\left(0,\frac32\right)`