JEE quadratics Part1

Algebra Level 5

Consider the inequality $9^{x}-a\cdot 3^{x}-a+3\leq 0$ ,where $a$ is a real parameter. The given inequality has at least one negative solution for $a \in$

(2,3)

(3,\infty)

(2,\infty)

(-2,3)

(-\infty,2)

1 solution

Kunal Verma
Apr 10, 2015

Let's take $3^{x}$ to be $t$ .

Thus the equation is :-

$t^{2}\ - \ a.t \ - \ a \ + \ 3 \le \ 0$

This is untrue when $D \ < \ 0$ as the coefficient of $t^{2}\ = \ 1 \ > \ 0$

i.e $a^{2}\ - \ 4( \ 3 \ - a) \ < \ 0$

Thus $-4 \ < \ a \ < 2$ where the solution doesn't exist.

But since $t \in \ (0, \infty )$ and $a$ has a $minus \ sign$ , this inequality is also not true for all negative real numbers for a.

Thus, our solution is $a \in \ (2, \infty )$

Very smart problem!

Indeed nice!! try and many small flaws like it should be $3^{x}=t$

@Calvin Lin sir plz edit the answer it is $(2,3)$ . to verify this i am posting my solution.

Given that $9^{x}-a.3^{x}-a+3\leq 0$

let $3^{x}=t$

$t^{2}-at-a+3\leq 0$ or $t^{2}+3\leq a(t+1)$ where $t\in~ R^{+} ~\forall ~x\in ~R$ ................................ $(1)$

let $f_{1}(t)=t^{2}+3$ and $f_{2}(t)=a(t+1)$

for $x<0$ , $t\in~(0,1)$ . This means $(1)$ should have at least one solution in $t\in~(0,1)$ . From $(1)$ it is clear that $a\in~R^{+}$ as LHS is always positive..Now $f_{2}(t)=a(t+1)$ represents a straight line. It should meet the curve $f_{1}(t)=t^2+3$ , at least once in $t\in~(0,1)$ ,

$f_{1}(0)=3,~f_{1}(1)=4,~f_{2}(0)=a,~f_{2}(1)=2a$

if $f_{1}(0)=f_{2}(0)$ then $a=3$ if $f_{1}(1)=f_{2}(1)$ then $a=2$ . Hence the required range is $a\in~(2,3)$

Tanishq Varshney - 6 years, 2 months ago

@Calvin Lin sir plz see this

Tanishq Varshney - 6 years, 2 months ago

When $a = 4, x = -1$ , we have $9^x - 4 \times 3^x - a + 3 = \frac{1}{9} - \frac{4}{3} - 4 + 3 = - 2 \frac{2}{9}$ , which satisfies your condition. Hence, $a = 4$ should lie in the solution set, so I disagree with updating the answer to $(2,3)$ .

It seems like you are trying to solve for $f_1 (t) = f_2(t)$ , which is very different from the case of $f_1 (t) \leq f_2 (t)$ .

Thanks. I have updated Kunal's solution so that $t = 3 ^ x$ instead.

Calvin Lin Staff - 6 years, 2 months ago

JEE quadratics Part1

1 solution

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