Question

If all real values of $x$ obtained from the equation $4^x-(a-3)2^x+a-4=0$ are non-positive, then

• A. $a\in (4,5]$
• B. $a\in (0,4)$
• C. $a\in (4,\infty )$
• D. $Noneofthese$

Answer 1

The correct option is A

$a\in (4,5]$

Step 1:Reduce the given equation into a quadratic equation.

We have given an equation $4^x-(a-3)2^x+a-4=0$, $x\le 0$ and$2^x>0$.

We can rewrite the given equation as${\left(2^x\right)}^2-(a-3)2^x+(a-4)=0$

let us take $2^x=t$

$t^2-(a-3)t+(a-4)=0$

Step 2: Find the discriminant of the above quadratic equation.

$$\begin{gathered} D={\{-(a-3\left)\right\}}^2-4\left(1\right)(a-4);\mathit{D}\mathbf{=}{\mathit{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathit{a}\mathit{c} \\ \Rightarrow a^2-6a+9-4a+16 \\ \Rightarrow a^2-10a+25 \\ \Rightarrow {(a-5)}^2\ge 0 \end{gathered}$$

Step 3: Applying the quadratic formula in the above quadratic equation.

$$\begin{gathered} t=\frac{-\{-(a-3\left)\right\}\pm \sqrt{{(a-5)}^2}}{2\left(1\right)};\mathit{t}\mathbf{=}\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{\mathbf{D}}}{\mathbf{2}\mathbf{a}} \\ t=\frac{(a-3)\pm (a-5)}{2} \\ t=\frac{2a-8}{2}=a-4or1 \end{gathered}$$

Step 4: Apply the conditions to find the interval, in which ‘a’ lies.

$$\begin{gathered} \therefore 2^x=a-4or{2}^x=1.......\left(1\right) \\ \because x\le 0and{2}^x>0.......\left(2\right) \\ from\left(1\right)and\left(2\right) \\ 0<a-4\le 1adding4allsides, \\ 4<a\le 5 \\ i.e.a\in (4,5] \end{gathered}$$

Hence, the correct option is$\left(\mathit{A}\right)\mathit{a}\mathbf{\in }\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{5}\mathbf{]}$.