The maximum value of (logex/x), if x 0 is

Question

The maximum value of (logex/x), if x > 0 is (A) e (B) 1 (C) (1/e) (D) - (1/e)

Tardigrade Admin · Accepted Answer

(y = (loge x/x)) â ((dy/dx)=(1 - loge x/x2)) For maxima, dy/dx = 0 â 1 - loge x = 0 â x = e dy/dx changes sign from positive to negative at x = e â´ y textmax = 1/e

Q. The maximum value of $\frac{l o g _{e} x}{x}$ , if x > 0 is

$e$

$1$

$\frac{1}{e}$

$- \frac{1}{e}$

$(y = \frac{l o g _{e} x}{x})$

$\Rightarrow (\frac{d y}{d x} = \frac{1 - l o g _{e} x}{x ^{2}})$

For maxima, $d y / d x = 0$

$\Rightarrow 1 - l o g_{e} x = 0$

$\Rightarrow x = e$

$d y / d x$ changes sign from positive to negative at $x = e$

$∴ y_{max} = 1/ e$

Q. The maximum value of xloge​x​, if x > 0 is

e

1

e1​

−e1​

(y=xloge​ x​) ⇒(dxdy​=x21−loge​x​) For maxima, dy/dx=0 ⇒1−loge​x=0 ⇒x=e dy/dx changes sign from positive to negative at x=e ∴ymax​=1/e

Q. The maximum value of $\frac{l o g _{e} x}{x}$ , if x > 0 is

$e$

$1$

$\frac{1}{e}$

$- \frac{1}{e}$

$(y = \frac{l o g _{e} x}{x})$

$\Rightarrow (\frac{d y}{d x} = \frac{1 - l o g _{e} x}{x ^{2}})$

For maxima, $d y / d x = 0$

$\Rightarrow 1 - l o g_{e} x = 0$

$\Rightarrow x = e$

$d y / d x$ changes sign from positive to negative at $x = e$

$∴ y_{max} = 1/ e$