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. $a \in {\text{(0, 4)}}$
C. $a \in {\text{ }}(4,\infty )$
D. None of These

Answer 1

Hint: As we know that simple second-degree polynomials, which are what quadratic equations are, typically have the form $y = a{x^2} + bx + c$.
To determine the value of $a$, the given equation must first be transformed into a quadratic equation. After that, compute the quadratic formula and evaluate the discriminant.

Formula Used:
The quadratic formula that can be used with the quadratic equation is as follows: $t = \dfrac{{ - b \pm \sqrt D }}{{2a}}$. And, the discriminant is represented by: $D = {b^2} - 4ac$.

Complete step by step Solution:
In the question, the equation ${4^x} - (a - 3){2^x} + a - 4 = 0\;$is given,
The given equation can be written as:
${({2^x})^2}\;-{\text{ }}\left( {a{\text{ }}-{\text{ }}3} \right){\text{ }}{2^x}\;-{\text{ }}{2^x}\; + {\text{ }}a{\text{ }}-{\text{ }}4{\text{ }} = {\text{ }}0$
Consider the value of ${2^x}$for $t$, we have:
${(t)^2}\;-{\text{ }}\left( {a{\text{ }}-{\text{ }}3} \right){\text{ t}}\;-{\text{ t}}\; + {\text{ }}a{\text{ }}-{\text{ }}4{\text{ }} = {\text{ }}0 \\$
${t^2} - (a - 3)t + (a - 4) = 0 \\$
To determine the discriminant of the above equation, use the formula of discriminate$D = {b^2} - 4ac$, then:
$D = {\left\{ { - (a - 3)} \right\}^2} - 4(1)(a - 4) \\$
$\Rightarrow D = {a^2} - 6a + 9 - 4a + 16 \\$
$\Rightarrow D = {a^2} - 10a + 25 \\$
$\Rightarrow {(a - 5)^2} \geqslant 0 \\$
Now, apply the quadratic formula $t = \dfrac{{ - b \pm \sqrt D }}{{2a}}$on the obtained equation,
 $t = \dfrac{{ - \left\{ { - (a - 3)} \right\} \pm \sqrt {{{(a - 5)}^2}} }}{{2(1)}} \\$
 $t = \dfrac{{(a - 3) \pm (a - 5)}}{2} \\$
 $t = \dfrac{{2a - 8}}{2} \\$
 $t = a - 4\,or\,1 \\$
Apply the following conditions to determine the interval,
$\because {2^x} = a - 4 \\$
 $\Rightarrow {2^x} = 1\,\,\,\,...(i) \\$
$\because x \leqslant 0,\,{2^x} > 0\,\,\,\,...(ii) \\$
Now, using the above conditions, we obtain:
$0 < a - 4 \leqslant 1 \\$
 $4 < a \leqslant 5 \\$
 $a \in (4,5] \\$

Hence, the correct option is (A).

Note: While solving the question, it should be noted that the Bhaskar formula, commonly referred to as the quadratic formula, is one of several approaches to solving second degree polynomial equations: $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ And the delta stands in for the discriminant, $D = {b^2} - 4ac$. The discriminant may be zero, positive or negative. Examples include: When ${b^2} - 4ac$ is greater than $0$, the equation has two separate real solutions; when ${b^2} - 4ac$ is equal to $0$, there is only one real solution and when ${b^2} - 4ac$ is less than $0$, there are no real solutions to the equation.