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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

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    $\begingroup$ Just check your initial statement "Then $k \in \ldots$" - I'll delete this comment once clarification is given. $\endgroup$ – Kevin Nov 15 '16 at 15:48
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    $\begingroup$ Why do you have $\forall y\in(1,3)$? You want it only for at least one $y$, not all. $\endgroup$ – Thomas Andrews Nov 15 '16 at 16:09
  • $\begingroup$ Try finding the opposite condition.. i.e. $k$ for which the equation is always $\ge 0$ for $x \in (0,1)$ , then take complement of that set $\endgroup$ – Dhanvi Sreenivasan Nov 15 '16 at 16:28
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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 $k\in (2,3)$.

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Given: the inequality $9^x-k3^x -k + 3 \leq 0$ has at least one real x in $(0,1)$.

Transform the above inequality by replacing $3^x$ by $y$ to get

$y^2 - ky -k + 3 \leq 0$ which has atleast one real $y$ in $(3^0, 3^1) = (1,3)$. Now consider the equality $y^2 - ky -k + 3 = 0$. Note that the coefficient of $y^2$ is greater than $0$ which implies that the curve traced by the inequality will be convex. In general, one should remember the following rules for a curve traced by general quadratic equation $f(x) = ax^2 + bx + c = 0, \, a\neq 0$ which has a discriminant of $D = b^2 - 4ac$,

  • $D > 0$
    • $a>0 \Rightarrow$ convex with two distinct real roots $x_1$ and $x_2$ (say $x_1<x_2$) and $f(x) \leq 0 \, \forall \, x\in(x_1,x_2)$ and $f(x)\geq0$ elsewhere.
    • $a<0 \Rightarrow$ concave with two distinct real roots $x_1$ and $x_2$ (say $x_1<x_2$) and $f(x) \geq 0 \, \forall \, x\in(x_1,x_2)$ and $f(x)\leq0$ elsewhere.
  • $D = 0$
    • $a>0 \Rightarrow$ convex with two equal real roots $x_1$ and $f(x_1) = 0$ and $f(x)>0$ elsewhere.
    • $a<0 \Rightarrow$ concave with two equal real roots $x_1$ and $f(x_1) = 0$ and $f(x)<0$ elsewhere.
  • $D < 0$
    • $a>0 \Rightarrow$ convex with no real root and $f(x) > 0 \, \forall \, x$.
    • $a<0 \Rightarrow$ concave with no real root and $f(x) < 0 \, \forall \, x$.

Note that convex means curve opening up and concave means curve opening down.

Now if the inequality $y^2 - ky -k + 3 \leq 0$ with $a=1>0$ has to have atleast one real y in $(1,3)$, then the equality $y^2 - ky -k + 3 = 0$ should have a discriminant $D \geq 0$ (by eliminating the wrong choices from above).

$D\geq0 \\ \Rightarrow k^2+4k-12\geq0 \\ \Rightarrow (k+6)(k-2) \geq0 \\ \Rightarrow k \in (-\infty,-6)\cup(2,\infty)$.

And the roots of the equation will be $y_1 = \frac{k - \sqrt{(k+6)(k-2)}}{2}$ and $y_2 = \frac{k + \sqrt{(k+6)(k-2)}}{2}$.

Now one of them should lie in $(1,3)$ and other should be greater than $0$ $(y=3^x>0 \, \forall \, x)$. So we need to check two cases:

CASE 1

$y_1 \in (1,3)$ and $y_2 > 0$

$y_1 > 1 \\ \Rightarrow \frac{k - \sqrt{(k+6)(k-2)}}{2} > 1 \\ \Rightarrow (k-2) > \sqrt{(k+6)(k-2)}$

which is never satisfied for $k \in (-\infty,-6)\cup(2,\infty)$. Now, switch to case 2.

CASE 2

$y_1 > 0$ and $y_2 \in (1,3)$

$y_1 > 0 \\ \Rightarrow \frac{k - \sqrt{(k+6)(k-2)}}{2} > 0 \\ \Rightarrow k > \sqrt{(k+6)(k-2)}$.

For $k>2$, both sides will be positive and can be squared without changing inequality which will give,

$k^2 > (k+6)(k-2) \\ \Rightarrow 4k-12 < 0 \\ \Rightarrow k<3$.

For $k<-6$, the inequality is not satisfied.

This further reduces the possibilities of $k$ to $(2,3)$.

Finally, $y_2 \in (1,3)$

$\frac{k + \sqrt{(k+6)(k-2)}}{2} > 1 \\ \Rightarrow (k-2) > -\sqrt{(k+6)(k-2)}$

which is always true for $k>2$.

$\frac{k + \sqrt{(k+6)(k-2)}}{2} < 3 \\ \Rightarrow (k-6) > -\sqrt{(k+6)(k-2)}$.

Both sides will be negative for $k \in (2,3)$ therefore can be squared with change of inequality

$\Rightarrow (k-6) > -\sqrt{(k+6)(k-2)} \\ \Rightarrow (k-6)^2 > (k+6)(k-2) \\ \Rightarrow -12k + 36 > 4k -12 \\ \Rightarrow 16k - 48 < 0 \\ \Rightarrow k<3$.

Therefore, the final solution is $k \in (2,3)$.

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