Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1461 Upvotes
Verified Answer
The correct answer is:
0
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,
$$
\begin{array}{rlrl}
& & (2 k+1)+k e^{-x}+2 e^x & \geq 0 \\
\Rightarrow & e^{-x}\left((2 k+1) e^x+k+2 e^{2 x}\right) & \geq 0 \\
\text { or } & (2 k+1) e^x+k+2 e^{2 x} & \geq 0 \\
\Rightarrow & 2 e^x\left(e^x+k\right)+1\left(e^x+k\right) & \geq 0 \\
\Rightarrow & & \left(2 e^x+1\right)\left(e^x+k\right) & \geq 0 \\
\Rightarrow & & e^x+k \geq 0 \text { or } k & \geq 0
\end{array}
$$
Hence, least value of $k$ is zero.
$$
f(x)=(2 k+1) x-3-k e^{-x}+2 e^x
$$
Since, $f(x)$ is monotonically increasing for all $x \in R$.
So,
$$
\begin{array}{rlrl}
& & (2 k+1)+k e^{-x}+2 e^x & \geq 0 \\
\Rightarrow & e^{-x}\left((2 k+1) e^x+k+2 e^{2 x}\right) & \geq 0 \\
\text { or } & (2 k+1) e^x+k+2 e^{2 x} & \geq 0 \\
\Rightarrow & 2 e^x\left(e^x+k\right)+1\left(e^x+k\right) & \geq 0 \\
\Rightarrow & & \left(2 e^x+1\right)\left(e^x+k\right) & \geq 0 \\
\Rightarrow & & e^x+k \geq 0 \text { or } k & \geq 0
\end{array}
$$
Hence, least value of $k$ is zero.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.