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Consider the Inequality $9^x-a\cdot 3^x-a+3\leq 0\;,$ Where $a\in \mathbb{R}$

Then The real values of $a$ so that

$(1)$ The given inequality has at least one real solution.

$(2)$ The given Inequality has at least one positive solution

$(3)$ The given Inequality has at least one negative solution.

$\bf{My\; Try::}$ For $(1)$

We can write Inequality as $$\displaystyle 9^x+3\leq a(3^x+1)\Rightarrow a\geq \frac{9^x+3}{3^x+1} = \frac{9^x-1+4}{3^x+1} = 3^x-1+\frac{4}{3^x+1}$$

So $$a\geq \underbrace{3^x+1+\frac{4}{3^x+1}}_{\bf{Using\; A.M\geq G.M}}-2\geq 4-2=2$$

So In first case we get $a\geq 2$

Now How can I solve $(2)$ and $(3)$ cases, Help required, Thanks

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First denote t=3^x. Notice 9^x=t^2. Then solve inequality for t using vieta's formulas With different requirements: 1. x is real iff t is positive. 2. x is positive iff t is greater than 1 3. x is negative iff t is less than one. For each step add conditions of inequality for parameter a

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Let $3^x=X$, the given inequation becomes $q(X)\leq 0$ with $q(X)=X^2-aX+(-a+3)$

The issue is now the sign of quadratic function $q(X)$ for positive values of $X$, with discussion on discriminant $\Delta=a^2-4(-a+3)=(a-2)(a+6)$.

Let $I$ denote the open interval $(-6,2)$.

If $a \in I$, $\Delta<0$, thus $q(X)$ having no real roots, will always be $>0$. Thus no solution in this case.

If $a \not\in I$, this case splits into two subcases :

  • either $a \leq -6$: by Vieta's formulas, the product of the roots being equal to $-a+3$ is $>0$ with a product which is $a<0$, thus one of them exactly is $>0$. Let us call this solution $X_1$. Then we are back to equation $3^x=X_1$, which itself has a solution $$x=ln(X_1)/ln(3) \ \ \ (*)$$Then condition (1) is fulfilled.

But, if condition (2) has to be fulfilled, i.e. $x \geq 0$, due to (*), $X_1$ has to be $\geq 1$. We have to resort to the explicit expression of the largest root :

$$\dfrac{a+\sqrt{a^2+4a-12}}{2} \geq 1 \ \Leftrightarrow \ \sqrt{a^2+4a-12} \geq 2-a \ \Leftrightarrow \ a \geq -1$$

(by squaring positive quantities). This condition $a \geq -1$ is not compatible with $a\leq-6$. Thus, in this case, (2) has no solution.

  • or $a \geq 2$. I let you treat more or less in the same way this case.

and the solution to condition (3).

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$(3^x)^2-a(3^x)+9\le 0$. Thus $$(a-\sqrt{(a+6)(a+2)})/2\le 3^x\le (a+\sqrt{(a+6)(a+2)}).$$

Thus $(a+6)(a+2)\ge 0$.

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  • $\begingroup$ I dont understand from where is issued number 9 in the first inequality. $\endgroup$ – Jean Marie Aug 3 '16 at 8:01
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    $\begingroup$ Your have given 174 answers in this site. Time to learn MathJax now! $\endgroup$ – user99914 Aug 3 '16 at 8:21

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