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

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Aicharya · Accepted Answer

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, $ \begin{array}{rlrl} & & (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}ight) & \geq 0 \ 	ext { or } & (2 k+1) e^x+k+2 e^{2 x} & \geq 0 \ \Rightarrow & 2 e^x\left(e^x+kight)+1\left(e^x+kight) & \geq 0 \ \Rightarrow & & \left(2 e^x+1ight)\left(e^x+kight) & \geq 0 \ \Rightarrow & & e^x+k \geq 0 	ext { or } k & \geq 0 \end{array} $ Hence, least value of $k$ is zero

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

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