# Find least value of a

1. Nov 1, 2012

### utkarshakash

1. The problem statement, all variables and given/known data
Find all the values of the parameter 'a' for which the inequality $4^x-a2^x-a+3 \leq 0$ is satisfied by at least one real x.

2. Relevant equations

3. The attempt at a solution
Let $2^x=t$
Then the inequality changes to
$t^2-at-(a-3)\leq0$
For the required condition D must be greater than or equal to 0.

$a^2+4a-12\geq0$
$(a-2)(a+6)\geq0$
$a\in (-∞,6]U[2,∞)$

2. Nov 1, 2012

### lurflurf

I do not understand what you have done.
I would have written
2<=(4^x+3)/(2^x+1)<=a

3. Nov 1, 2012

### utkarshakash

How can you write this?

4. Nov 1, 2012

### SammyS

Staff Emeritus
I take it that 'D' refers to the discriminant .

What you have so far is correct.

Can 2x ever be negative?

5. Nov 1, 2012

### lurflurf

using your
t=2^x
(t^2+3)-a(t^2+1)<=0
(t^2+3)<=a(t^2+1)<=0
(t^2+3)/(t^2+1)<=a
in fact s=(2^x+1)/2 is a nicer substitution

(t^2+3)/(t^2+1)=2((2^x+1)/2)+1/((2^x+1)/2)-1)=2(s+1/s-1)
since ((2^x+1)/2)(1/((2^x+1)/2))=s/s=1 the minimum occurs when
((2^x+1)/2)=(1/((2^x+1)/2)) or s=1/s

or use calculus if you like

6. Nov 1, 2012

### utkarshakash

You are right. It can never be negative. But substituting 6 for a does not give a -ve t. So that must be included.

7. Nov 1, 2012

### SammyS

Staff Emeritus
That should be $\displaystyle a\in (-\infty,-6]\cup[2,\infty)$ in your Original Post.

8. Nov 2, 2012

### utkarshakash

Yes it should be -6 instead of 6 (that was just a typing mistake) but the answer is still incorrect.

9. Nov 2, 2012

### SammyS

Staff Emeritus
Check a = -6. See what that makes t and thus 2x.

10. Nov 2, 2012

### ehild

Remember t=2x must be positive. What values of a would yield positive solutions for t?

ehild