Question

If $f(x)=(2k + 1)x - 3 - ke^{-x } + 2e^x$ is monotonically increasing for all $x \in R$, then the least value of $k$ is

• A. 1
• B. 0
• C. $- \frac{1}{2}$
• D. -1

Answer 1

Answer: (B)

0

Answer 2

Given,
$f(x)=(2 k+1) x-3-k e^{-x}+2 e^{x}$
Since, $f(x)$ is monotonically increasing for all $x \in R .$
So, $f'(x) \geq 0$
$(2 k+1)+k e^{-x}+2 e^{x} \geq 0$
$\Rightarrow e^{-x}\left((2 k+1) e^{x}+k+2 e^{2 x}\right) \geq 0$
or $(2 k+1) e^{x}+k+2 e^{2 x} \geq 0$
$\Rightarrow 2 e^{x}\left(e^{x}+k\right)+ l \left(e^{x}+k\right) \geq 0$
$\Rightarrow \left(2 e^{x}+1\right)\left(e^{x}+k\right) \geq 0$
$\Rightarrow e^{x}+k \geq 0$ or $k \geq 0$
Hence, least value of $k$ is zero.