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}}$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/52_to_54_solu_math_pap1.jpg"/></p><p>(I)$f\left(1\right)f\left({e}^{2}\right) < 0$, so true.</p><p>(II)${f}^{'}\left(1\right){f}^{'}\left(e\right) < 0$, so true.</p><p>(III) Graph of${f}^{'}\left(x\right)$, so (III) is false.</p><p>(IV) Is false.</p><p>As$\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:

$f\left(x\right) = x + ℓnx - xℓnx$

${f}^{'}\left(x\right) = \dfrac{1}{x} - ℓnx = 0$ 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