Question

The solution set of the inequality ${4^{ - x + \dfrac{1}{2}}} - 7.\left( {{2^{ - x}}} \right) - 4 < 0$, for all $x$ belongs to $\mathbb{R}$, is

1. $\left( { - \infty ,2} \right)$
2. $\left( { - 2,\infty } \right)$
3. $\left( { - \infty ,\infty } \right)$
4. $\left( {2,\infty } \right)$

Answer 1

First we are to convert the given inequality into the form so that we can determine the intervals in which the required term lies. We can do this by factorising the left side of the inequality by the middle term. On determining the interval of the term of the left, we have to further operate the term as in this question, we have $x$ not directly in the term but as an exponent term. On correctly operating, we will get the required answer.

Complete step-by-step solution:
Given, ${4^{ - x + \dfrac{1}{2}}} - 7.\left( {{2^{ - x}}} \right) - 4 < 0$,
We can write, $4 = {2^2}$.
So, replacing it, we get,
${2^{2( - x + \dfrac{1}{2})}} - 7.\left( {{2^{ - x}}} \right) - 4 < 0$
$\Rightarrow {2^{ - 2x + 1}} - 7.\left( {{2^{ - x}}} \right) - 4 < 0$
Now, we know, we can write, ${2^{x + y}} = {2^x}{.2^y}$.
Using this property we get,
$\Rightarrow {2^{ - 2x}}.2 - 7.\left( {{2^{ - x}}} \right) - 4 < 0$
Now, let us assume, ${2^{ - x}} = t$.
Therefore,
$\Rightarrow {\left( t \right)^2}.2 - 7.\left( t \right) - 4 < 0$
$\Rightarrow 2{t^2} - 7t - 4 < 0$
Now, factorising the term by middle term, we get,
$\Rightarrow 2{t^2} - 8t + t - 4 < 0$
Now, taking $2t$ common from first two terms, we get,
$\Rightarrow 2t\left( {t - 4} \right) + t - 4 < 0$
Now, taking $\left( {t - 4} \right)$ common from left hand side, we get,
$\Rightarrow \left( {t - 4} \right)\left( {2t + 1} \right) < 0$
Here, the two factors we got are, $\left( {t - 4} \right)$ and $\left( {2t + 1} \right)$.
So, $t - 4 = 0$
$\Rightarrow t = 4$
And, $2t + 1 = 0$
$\Rightarrow t = - \dfrac{1}{2}$
Therefore, by using the method of intervals, we can say that,

So, since the interval is less than zero, so we are to take the negative portion from the real line, so we get,
$- \dfrac{1}{2} < t < 4$
$\Rightarrow - \dfrac{1}{2} < {2^{ - x}} < 4$
It is clearly visible that ${2^{ - x}} > 0$, since any exponent of a positive number is always positive.
Therefore, we can write,
$\Rightarrow 0 < {2^{ - x}} < 4$
Now, we know, we can write $0$ as ${2^{ - \infty }}$ and $4$ as ${2^2}$.
Therefore,
$\Rightarrow {2^-\infty } < {2^{ - x}} < {2^2}$
We know, if${a^x} < {a^y}$ or ${a^x} > {a^y}$, then, $x < y$ or $x > y$ respectively.
So,
$\Rightarrow - \infty < \- x < 2$
Now, taking negative over the inequality, we know, taking negative over the inequality changes the sign of the inequality.
Therefore, $\Rightarrow - 2 < x < \infty$
Therefore, we can say, $x \in \left( { - 2,\infty } \right)$.
The correct option is 2.

Note: Taking the inequality and taking the intervals are the most vulnerable part where we make mistakes. So, we must keep in mind, if the inequality is greater than zero then we must take the positive parts as intervals and vice-versa. And, when we solve inequalities, we must keep in mind, negation changes the sign of inequality and takin reciprocal over the inequality also changes the sign of inequality.