Question:
The maximum value of $ \frac{log x}{x} $ is
The maximum value of $ \frac{log x}{x} $ is
Updated On: Jun 14, 2022
- $ 1 $
- $ \frac {2}{e} $
- $ e $
- $ \frac{1}{e} $
Hide Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Let $y = \frac{log\,x}{x}$
On differentiating both sides w.r.t, $x$, we get
$\frac{dy}{dx} = \frac{x\frac{d}{dx} log\,x - log\,x \frac{d}{dx} \,x}{x^2}$
$ = \frac{x\cdot \frac{1}{x} - log\,x\cdot 1}{x^2}$
$\therefore \frac{dy}{dx} = \frac{1 - log\,x }{x^2} \,\,...(i)$
For maximum or minimum values,
$\frac{dy}{dx} = 0$
$\Rightarrow \frac{1- log \,x}{x^2} = 0$
$\Rightarrow 1 - log \,x = 0$
$\Rightarrow log\,x = 0$
$ x = e$
Now, differentiate E $(i)$ w.r.t. $'x'$ we get
$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - log\,x) \cdot 2x}{x^4}$
$ = \frac{-x - 2x + 2x \,log\,x}{x^4}$
$ = \frac{-3x + 2x\,log\,x}{x^4}$
$\left(\frac{d^{2}y}{dx^{2}}\right)_{x =e} = \frac{-3e + 2e\, log \,x}{e^{4}} $
$ = \frac{-3e+2e}{e^{4}}$
$= -\frac{e}{e^{4}} = -\frac{1}{e^{3}} < 0 $
$ \therefore y = \frac{log x}{x}$ is maximum at $x = e $
$ \Rightarrow y_{max} = \frac{log\, e}{e} = \frac{1}{e} $
On differentiating both sides w.r.t, $x$, we get
$\frac{dy}{dx} = \frac{x\frac{d}{dx} log\,x - log\,x \frac{d}{dx} \,x}{x^2}$
$ = \frac{x\cdot \frac{1}{x} - log\,x\cdot 1}{x^2}$
$\therefore \frac{dy}{dx} = \frac{1 - log\,x }{x^2} \,\,...(i)$
For maximum or minimum values,
$\frac{dy}{dx} = 0$
$\Rightarrow \frac{1- log \,x}{x^2} = 0$
$\Rightarrow 1 - log \,x = 0$
$\Rightarrow log\,x = 0$
$ x = e$
Now, differentiate E $(i)$ w.r.t. $'x'$ we get
$\frac{d^2y}{dx^2} = \frac{x^2(-\frac{1}{x}) - (1 - log\,x) \cdot 2x}{x^4}$
$ = \frac{-x - 2x + 2x \,log\,x}{x^4}$
$ = \frac{-3x + 2x\,log\,x}{x^4}$
$\left(\frac{d^{2}y}{dx^{2}}\right)_{x =e} = \frac{-3e + 2e\, log \,x}{e^{4}} $
$ = \frac{-3e+2e}{e^{4}}$
$= -\frac{e}{e^{4}} = -\frac{1}{e^{3}} < 0 $
$ \therefore y = \frac{log x}{x}$ is maximum at $x = e $
$ \Rightarrow y_{max} = \frac{log\, e}{e} = \frac{1}{e} $
Was this answer helpful?
0
0
Top Questions on Application of derivatives
- Evaluate the integral: \[ \int_{-\pi}^{\pi} x^2 \sin(x) \, dx \]
- VITEEE - 2024
- Mathematics
- Application of derivatives
- For the function $f(x) = x^3 - 6x^2 + 12x - 3$, $x = 2$ is:}
- KCET - 2024
- Mathematics
- Application of derivatives
- If \( \frac{dy}{dx} - y \log_e 2 = 2^{\sin x} (\cos x - 1) \log_e 2 \), then \( y \) is:
- BITSAT - 2024
- Mathematics
- Application of derivatives
- The solution of the differential equation \( (x + 1)\frac{dy}{dx} - y = e^{3x}(x + 1)^2 \) is:
- BITSAT - 2024
- Mathematics
- Application of derivatives
- If the area bounded by the curves \( y = ax^2 \) and \( x = ay^2 \) (where \( a>0 \)) is 3 sq. units, then the value of \( a \) is:
- BITSAT - 2024
- Mathematics
- Application of derivatives
View More Questions
Questions Asked in AMUEEE exam
- The ratio of the half-life time $ (t_{1/2}) $ , to the three quarter life-time, $ (t_{3/4}) $ , for a reaction that is second order
- AMUEEE - 2018
- Order of Reaction
- $ 0.002\, M $ solution of a weak acid has an equivalent conductance $ (\wedge) 60\, ohm^{-1}\, cm^{2}\, eq^{-1} $ . What will be the $ pH $ ? (Given : $ \wedge = 400 \, ohm^{-1} \, cm^{2} \, eq^{-1} $ ]
- AMUEEE - 2018
- Electrochemistry
- The electrode potential, $ E^{\circ} $ , for the reduction of $ MnO^{-}_{4} $ to $ Mn^{2+} $ in acidic medium is $ +1.51\, V $ . Which of the following metal(s) will be oxidised? The reduction reactions and standard electrode potentials for $ Zn^{2+}, Ag^{+} $ , and $ Au^{+} $ are given as $ Zn^{2+} (aq) +2e \to Zn(s), E^{\circ} = -0.762\, V $ $ Ag+ (aq) +e {\rightleftharpoons} Ag (s), E^{\circ}=+0.80 \,V $ $ Au^{+} (aq) +e {\rightleftharpoons} Au(s), E^{\circ}+1.69\,V $
- AMUEEE - 2018
- Electrochemistry
- Three particles, two with masses $ m $ and one with mass $ M $ , might be arranged in any of the four configurations shown below. Rank the configurations according to the magnitude of the gravitational force on $ M $ , least to greatest
- AMUEEE - 2018
- Gravitation
- The $ EAN $ value $ y \left[Ti\left(\sigma-C_{6}H_{5}\right)_{2}\left(\pi-C_{5}H_{5}\right)_{2}\right]^{0} $ is
- AMUEEE - 2018
- coordination compounds
View More Questions
Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
Read More: Application of Derivatives