Let f(x) = x + _ e x - x _ e x, x ∈ (0, ∞) • Column 1 contains...

f\left(x\right) = x + ℓnx - xℓnx, x > 0$</p><p>${f}^{'}\left(x\right) = 1 + \dfrac{1}{x} - ℓnx - 1$</p><p>${f}^{''}\left(x\right) = - \dfrac{1}{{x}^{2}} - \dfrac{1}{x} = \dfrac{- \left(x + 1\right)}{{x}^{2}}$</p><p>In Column 1,</p><p>(I)$f\left(1\right)f\left({e}^{2}\right) < 0$so true</p><p>(II)${f}^{'}\left(1\right){f}^{'}\left(e\right) < 0$so true</p><p>(III)${f}^{'}\left(x\right)$is always positive for$( 0, 1 )$so it cannot be zero. So (III) is false.</p><p>(IV) Is false</p><p>In Column 2</p><p>As$\lim_{x \to \infty} f\left(x\right) = \lim_{x \to \infty} x\left[1 + \dfrac{ℓnx}{x} - ℓnx\right] = - \infty$</p><p>$∴$(i) is false (ii) is true</p><p>$\lim_{x \to \infty} {f}^{'}\left(x\right) = - \infty$,so (iii) is true.</p><p>$\lim_{x \to \infty} {f}^{''}\left(x\right) = 0$, so (iv) is true.</p><p>(P)${f}^{'}\left(x\right)$is positive in$( 0, 1 )$,so true.</p><p>(Q)${f}^{'}\left(x\right) < 0$,for in \left(e, {e}^{2}\right)$ ,so true.

As  ${f}^{''}\left(x\right) 0$ therefore, R is false, S is true.

Alternate Solution:

$f\left(x\right) = x + ℓnx - xℓnx$

${f}^{'}\left(x\right) = \dfrac{1}{x} - ℓnx = 0$ at $x = {x}_{0}$ where ${x}_{0} \in \left(1, e\right)$

${f}^{''}\left(x\right) = - \dfrac{1}{{x}^{2}} - \dfrac{1}{x} 0 ⇒ f\left(x\right)$ concave down