Question

The minimum value of k for which $f\left(x\right)=2e^x-k{e}^{-x}+(2k+1)x-3$ is monotonically increasing for all real x is?

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

The correct option is C $0$

Given:

$f\left(x\right)=2e^x-k{e}^{-x}+(2k+1)x-3ismonotonicallyincreasing\forall x\in R$

We know,

f(x) is monotonically increasing if

$f^{\prime }\left(x\right)\ge 0$

and

$e^x>0$

Solution:

$f\left(x\right)=2e^x-k{e}^{-x}+(2k+1)x-3$

$\Rightarrow f^{\prime }\left(x\right)=2e^x+k{e}^{-x}+(2k+1)$

We know,

$f^{\prime }\left(x\right)\ge 0$

$\Rightarrow 2e^x+k{e}^{-x}+(2k+1)\ge 0$

$\Rightarrow e^{-x}(2e^{2x}+k+(2k+1\left)e^x\right)\ge 0$

$\Rightarrow 2e^{-x}(e^{2x}+\frac{k}{2}+(k+\frac{1}{2})e^x)\ge 0$

as exponential function is always positive,

$\therefore (e^{2x}+\frac{k}{2}+(k+\frac{1}{2})e^x)\ge 0$

$\Rightarrow e^{2x}+\frac{k}{2}+k{e}^x+\frac{1}{2}{e}^x\ge 0$

$\Rightarrow e^x(e^x+\frac{1}{2})+k(e^x+\frac{1}{2})\ge 0$

$\Rightarrow (e^x+k)(e^x+\frac{1}{2})\ge 0$

$\Rightarrow (e^x+k)\ge 0$

$\Rightarrow k\ge 0$

Minimum value of k is 0.