$Consider f : [1, \infty) \to R is given by f (x) = {(\log_{e} x)}^{2} - \int_{1}^{e} \frac{f (t)}{t} d t, then which of the$ following is/are correct?

Difficult

A

$f (e) = \frac{5}{6}$
B

$f (x) \geq - 1 \forall x \in [1, \infty)$
C

$Area of the figure bounded by the tangent line of y = f (x) at point (e, f (e))$ $the curve y = f (x) and line x = 1 is e - \frac{1}{e} + 3 .$
D

$Area of the figure bounded by the tangent line of y = f (x) at point$ $(e, f (e)), the curve y = f (x) and line x = 1 is e + \frac{1}{e} - 3 .$

Solution

$\begin{array}{l} Sol. f (x) = \ln^{2} x - \frac{1}{6} \Rightarrow f (e) = \frac{5}{6} \\ f^{'} (x) = \frac{2 \ln x}{x} \geq 0 \forall x \in [1, \infty) \\ f (x) is strictly increasing function \\ f (1) = - \frac{1}{6} < 0 \\ f^{''} (x) = 2 [\frac{1 - ℓ n x}{x^{2}}] \end{array}$
$\begin{array}{l} Equation of tangent: \\ y - \frac{5}{6} = \frac{2}{e} (x - e) \Rightarrow y = \frac{2 x}{e} - \frac{7}{6} \\ Area = \int_{1}^{e} [(\ln^{2} x - \frac{1}{6}) - (\frac{2 x}{e} - \frac{7}{6})] d x = e + \frac{1}{e} - 3 \end{array}$

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