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What is the least value of $\mathrm{k}$ such that the function $\mathrm{x}^{2}+\mathrm{kx}+1$ is strictly increasing on $(1,2)$
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Verified Answer
The correct answer is:
$-2$
Let $f(x)=x^{2}+k x+1$
$f^{\prime}(x)=2 x+k$
$\mathrm{f}(\mathrm{x})$ is strictly increasing on $(1,2)$
if $f^{\prime}(x)>0$ for $x \in(1,2)$
$\Rightarrow 2 \mathrm{x}+\mathrm{k}>0 \quad$ for $\mathrm{x} \in(1,2)$
$\Rightarrow \mathrm{k}>-2 \mathrm{x} \quad$ for $\mathrm{x} \in(1,2)$
Now, $1 < x < 2 \Rightarrow 2 < 2 x < 4$
$\begin{array}{l}
\Rightarrow-2>-2 x>-4 \\
\Rightarrow-4 < -2 x < -2 \\
\Rightarrow k \geq-2
\end{array}$
Hence least value of $\mathrm{k}=-2$.
$f^{\prime}(x)=2 x+k$
$\mathrm{f}(\mathrm{x})$ is strictly increasing on $(1,2)$
if $f^{\prime}(x)>0$ for $x \in(1,2)$
$\Rightarrow 2 \mathrm{x}+\mathrm{k}>0 \quad$ for $\mathrm{x} \in(1,2)$
$\Rightarrow \mathrm{k}>-2 \mathrm{x} \quad$ for $\mathrm{x} \in(1,2)$
Now, $1 < x < 2 \Rightarrow 2 < 2 x < 4$
$\begin{array}{l}
\Rightarrow-2>-2 x>-4 \\
\Rightarrow-4 < -2 x < -2 \\
\Rightarrow k \geq-2
\end{array}$
Hence least value of $\mathrm{k}=-2$.
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