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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:

  1. Converted it into Quadratic by assuming "$2^x$" as '$t$'
  2. Since the "leading coefficient in the inequality != zero". Hence the equation is quadratic.
  3. 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.
  4. I got range of a which comes out to be $a \in (-\infty, -6] \cup [2,\infty)$
  5. 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
player3236
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Kumar Shuvam
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1 Answers1

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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 $2$.

player3236
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  • Sir, can you again clarify what you have written in third step! How, a can belong in this interval 0<a<1 ? – Kumar Shuvam Nov 17 '20 at 06:09
  • I wrote the third step to show that such $a$ is impossible, but it seems that I have made a typo. – player3236 Nov 17 '20 at 06:43
  • What is ''discriminant and leading coefficient argument''..Sorry, to bother you, sir. – Kumar Shuvam Nov 17 '20 at 07:06
  • That is basically your step 3. For $0<a<2$, the discriminant is negative, so there are no real roots. Since the leading coefficient is positive, the whole graph is above the $x$-axis. – player3236 Nov 17 '20 at 11:37