I came across a question in my textbook which is something like this
The least positive value of a for which $4^x-a\times2^x-a+3 \le 0$ is satisfied by at least one real value of $x$ is ______
So to solve this question I used this approach:
- Converted it into Quadratic by assuming "$2^x$" as '$t$'
- Since the "leading coefficient in the inequality != zero". Hence the equation is quadratic.
- Now the leading coefficient is greater than $0$. And the inequality condition is $\le 0$.Hence, discriminant must be greater than or equal to zero.
- I got range of a which comes out to be $a \in (-\infty, -6] \cup [2,\infty)$
- I think the answer should be $2$ because it is the least positive value. But, I have some kind of confusion in my mind. As I have assumed "$2^x$' as '$t$'.. I'm not sure about the answer.. Please help me after this
The interval where the discriminant is nonnegative is $a \in (-\infty, -6] \cup [2, \infty)$.
For $a =2$, we have $t^2-2t+1 = (t-1)^2$ as the quadratic equation, and it is equal to zero when $t = 2^x = 1$, which is certainly satisfied by some real value of $x$.
For $0 < a < 2$, by your discriminant and leading coefficient argument, $t^2 - at - a+3 > 0$ for all $t$.
Hence the least positive value of $a$ must be $2$.