The value of k for which the equation e 2 x = k √ x has exactly one solution, is

Question

The value of k for which the equation e 2 x = k √ x has exactly one solution, is (A) (1/√ e ) (B) √ e  (C) 2 √ e  (D) (2/√ e )

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/867580778471ffaef8f7159dab062f90-.png" /> k text can not be negative ( dy / dx )| P =2 e 2 x 1 ∴ ( dy / dx )| P =( k /2 √ x l ) 2 e 2 x 1=( k /2 √ x 1)( text but e 2 x 1= k √ x 1) 2 k √ x 1=( k /2 √ x 1) ; x 1=(1/4) ; ∴ e (1/2)=( k /2) ; k =2 √ e

Q. The value of $k$ for which the equation $e^{2 x} = k x$ has exactly one solution, is

$\frac{1}{e}$

$e$

$2 e$

$\frac{2}{e}$

$k can not be negative$
$\frac{d y}{d x} ∣_{P} = 2 e^{2 x_{1}}$
$∴ \frac{d y}{d x} ∣_{P} = \frac{k}{2 x _{l}}$
$2 e^{2 x_{1}} = \frac{k}{2 x _{1}} (but e^{2 x_{1}} = k x_{1})$
$2 k x_{1} = \frac{k}{2 x _{1}}; x_{1} = \frac{1}{4};$
$∴ e^{\frac{1}{2}} = \frac{k}{2}; k = 2 e$

Q. The value of k for which the equation e2x=kx​ has exactly one solution, is

e​1​

e​

2e​

e​2​

k can not be negative dxdy​∣P​=2e2x1​ ∴dxdy​∣P​=2xl​​k​ 2e2x1​=2x1​​k​( but e2x1​=kx1​​) 2kx1​​=2x1​​k​;x1​=41​; ∴e21​=2k​;k=2e​

Q. The value of $k$ for which the equation $e^{2 x} = k x$ has exactly one solution, is

$\frac{1}{e}$

$e$

$2 e$

$\frac{2}{e}$

$k can not be negative$
$\frac{d y}{d x} ∣_{P} = 2 e^{2 x_{1}}$
$∴ \frac{d y}{d x} ∣_{P} = \frac{k}{2 x _{l}}$
$2 e^{2 x_{1}} = \frac{k}{2 x _{1}} (but e^{2 x_{1}} = k x_{1})$
$2 k x_{1} = \frac{k}{2 x _{1}}; x_{1} = \frac{1}{4};$
$∴ e^{\frac{1}{2}} = \frac{k}{2}; k = 2 e$