For what value of x, x2 ℓ n(1 / x) is maximum-

Question

For what value of x, x2 ℓ n(1 / x) is maximum- (A) e-1 / 2 (B) e1 / 2 (C) e (D) e-1

Tardigrade Admin · Accepted Answer

f(x)=x2 ℓ n((1/x))=-x2 ℓ ln x f prime(x)=-[2 x ℓ ln x+x]=-x(2 ln x+1) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/02a703df9ac5713aedb5c30b8b22e3af-.png" /> ⇒ maximum at x=(1/√ e )

Q. For what value of $x, x^{2} ℓ n (1/ x)$ is maximum-

$e^{- 1/2}$

$e^{1/2}$

$e$

$e^{- 1}$

$f (x) = x^{2} ℓ n (\frac{1}{x}) = - x^{2} ℓ ln x$
$f^{'} (x) = - [2 x ℓ ln x + x] = - x (2 ln x + 1)$

$\Rightarrow$ maximum at $x = \frac{1}{e}$

Q. For what value of x,x2ℓn(1/x) is maximum-

e−1/2

e1/2

e

e−1

f(x)=x2ℓn(x1​)=−x2ℓlnx f′(x)=−[2xℓlnx+x]=−x(2lnx+1) ⇒ maximum at x=e​1​

Q. For what value of $x, x^{2} ℓ n (1/ x)$ is maximum-

$e^{- 1/2}$

$e^{1/2}$

$e$

$e^{- 1}$

$f (x) = x^{2} ℓ n (\frac{1}{x}) = - x^{2} ℓ ln x$
$f^{'} (x) = - [2 x ℓ ln x + x] = - x (2 ln x + 1)$

$\Rightarrow$ maximum at $x = \frac{1}{e}$