<s>[INST] Inequality $9^x-k\cdot 3^x-k+3\leq 0$ is satisfied for at least one real $x$ 
If the Inequality  $9^x-k\cdot 3^x-k+3\leq 0$ is satisfied for  at least one real $x$ in $(0,1)\;,$ Then $k\in $

$\bf{My\; Try::}$ Let $3^x=y>0\;,$ Then inequality is 
$$y^2-ky-k+3\leq 0\Rightarrow y^2-ky-k+3\leq 0\forall y\in (1,3)$$
So $$y^2-ky+\frac{k^2}{4}-\frac{k^2}{4}-k+3\leq 0$$
So $$\bigg(y-\frac{k}{2}\bigg)^2-\bigg(\frac{k^2+4k-12}{4}\bigg)\leq 0$$
Now above $\displaystyle 1<y<3\Rightarrow 1-\frac{k}{2}<y-\frac{k}{2}<3-\frac{k}{2}\Rightarrow \left(3-\frac{k}{2}\right)^2<\left(y-\frac{k}{2}\right)^2<\left(3-\frac{k}{2}\right)^2$
Now how can i solve it,Help required, Thanks
 [/INST] Set $t=\mathrm 3^x$. The given equation has at least one root in $(0,1)$ if and only if the equation $\;p(t)=t^2-kt+3-t=0\;$ has at least one root in $(1,3)$.
The first condition is this equation has real roots , i.e.
\begin{align}&\Delta=k^2-4(3-t)=k^2+4t-12=(k+2)^2-16\ge 0\\
&\iff k\ge 2\enspace\text{or}\enspace k\le-6.
\end{align}
Now observe that if $k\le -6$, the equation has real roots with the same sign (their product $3-k$ is positive) and this sign is negative (their sum $k$ is negative. Hence, if there is one root in $(1,3)$, we have $k\ge 2$.
Furthermore, in this case, $1$ separates the roots, since $p(1)=4-2k\le 0$. On the other hand, $p(3)=9-3k$, so:


*

*if $k>3$,  $3$ also separates the roots, hence one root is $<1$, the other is $>3$, and there's no root in $(1,3)$.

*if $2<k<3$, $3$ does not separate the roots, but $1$ does, hence there's one root $<1$ and the other one in $(1,3)$.

*if $k=2$, $1$ is a double root. Ik $k=3$, the roots are $0$ and $3$. Hence in both cases, there's no root in $(0,1)$.


As a conclusion, if the given equation has a root in $(0,1)$, necessarily $k\in (2,3)$.
</s>