Question

$\left( a \right)$ Find all the integral values of $x$ for which
$\left( 5x-1 \right)<{{\left( x+1 \right)}^{2}}<\left( 7x-3 \right)$
$\left( b \right)$ Solve $\left| \dfrac{12x}{4{{x}^{2}}+9} \right|\le 1$.

Answer 1

Hint: For part $\left( a \right)$, let us consider $\left( 5x-1 \right)$ as I, ${{\left( x+1 \right)}^{2}}$ as II and $\left( 7x-3 \right)$ as III. First, we will solve $I$ and $II$ and then we will solve $II$ and $III$. Finally take the common solution from both the solved inequation. For part $\left( b \right)$, we must know that if for any function $f\left( x \right)$, it Is given \[\left| f\left( x \right) \right|\le 1\], then we can convert this inequation to \[-1\le f\left( x \right)\le 1\].

Complete step-by-step solution -
$\left( a \right)$ In this part of the question, it is given,
$\left( 5x-1 \right)<{{\left( x+1 \right)}^{2}}<\left( 7x-3 \right)$
Let us consider $\left( 5x-1 \right)$ as I, ${{\left( x+1 \right)}^{2}}$ as II and $\left( 7x-3 \right)$ as III.
Let us first solve equation $I$ and equation $II$.
 \[\begin{align}
  & \left( 5x-1 \right)<{{\left( x+1 \right)}^{2}} \\
 & \Rightarrow 5x-1<{{x}^{2}}+1+2x \\
 & \Rightarrow {{x}^{2}}-3x+2>0 \\
 & \Rightarrow {{x}^{2}}-x-2x+2>0 \\
 & \Rightarrow x\left( x-1 \right)-2\left( x-1 \right)>0 \\
 & \Rightarrow \left( x-2 \right)\left( x-1 \right)>0 \\
 & \Rightarrow x\in \left( -\infty ,1 \right)\cup \left( 2,\infty \right)...............\left( 1 \right) \\
\end{align}\]
Now, let us solve the equation $II$ and equation$III$.
\[\begin{align}
  & {{\left( x+1 \right)}^{2}}<\left( 7x-3 \right) \\
 & \Rightarrow {{x}^{2}}+1+2x<7x-3 \\
 & \Rightarrow {{x}^{2}}-5x+4<0 \\
 & \Rightarrow {{x}^{2}}-x-4x+4<0 \\
 & \Rightarrow x\left( x-1 \right)-4\left( x-1 \right)<0 \\
 & \Rightarrow \left( x-4 \right)\left( x-1 \right)<0 \\
 & \Rightarrow x\in \left( 1,4 \right).............\left( 2 \right) \\
\end{align}\]
From solution $\left( 1 \right)$ and solution $\left( 2 \right)$, we have two possible solution sets of $x$. Since both the solutions are the possible values of $x$, we will take out those values of $x$ which are common in both the solutions. Hence, we will take the intersection of the two solution sets.
Common values of $x$ in both the solutions are,
$x\in \left( 2,4 \right)$
In the question, we are asked to find the integral values of $x$. Since we have found that $x\in \left( 2,4 \right)$, the only possible integral value of $x$ is $3$.
$\left( b \right)$ In this part of the question, we are given $\left| \dfrac{12x}{4{{x}^{2}}+9} \right|\le 1$.
Before proceeding with the question, we must know that if for any function $f\left( x \right)$, it is given \[\left| f\left( x \right) \right|\le 1\], then we can convert this inequation to \[-1\le f\left( x \right)\le 1\].
In this question, we have $f\left( x \right)=\dfrac{12x}{4{{x}^{2}}+9}$ . So, we obtain an inequation,
$-1\le \dfrac{12x}{4{{x}^{2}}+9}\le 1$
Let us solve the left part of the inequation first.
$\begin{align}
  & -1\le \dfrac{12x}{4{{x}^{2}}+9} \\
 & \Rightarrow -4{{x}^{2}}-9\le 12x \\
 & \Rightarrow 4{{x}^{2}}+12x+9\ge 0 \\
 & \Rightarrow 4{{x}^{2}}+6x+6x+9\ge 0 \\
 & \Rightarrow 2x\left( 2x+3 \right)+3\left( 2x+3 \right)\ge 0 \\
 & \Rightarrow \left( 2x+3 \right)\left( 2x+3 \right)\ge 0 \\
 & \Rightarrow {{\left( 2x+3 \right)}^{2}}\ge 0 \\
 & \Rightarrow x\in R..........\left( 3 \right) \\
\end{align}$
Now, we will solve the right part of the inequation.
$\begin{align}
  & \dfrac{12x}{4{{x}^{2}}+9}\le 1 \\
 & \Rightarrow 12x\le 4{{x}^{2}}+9 \\
 & \Rightarrow 4{{x}^{2}}-12x+9\ge 0 \\
 & \Rightarrow 4{{x}^{2}}-12x+9\ge 0 \\
 & \Rightarrow 4{{x}^{2}}-12x+9\ge 0 \\
 & \Rightarrow 4{{x}^{2}}-6x-6x+9\ge 0 \\
 & \Rightarrow 2x\left( 2x-3 \right)-3\left( 2x-3 \right)\ge 0 \\
 & \Rightarrow \left( 2x-3 \right)\left( 2x-3 \right)\ge 0 \\
 & \Rightarrow {{\left( 2x-3 \right)}^{2}}\ge 0 \\
 & \Rightarrow x\in R............\left( 4 \right) \\
\end{align}$
From solution $\left( 3 \right)$ and solution $\left( 4 \right)$, we have two possible solution sets of $x$. Since both the solutions are the possible values of $x$, we will take out those values of $x$ which are common in both the solutions. Hence, we will take the intersection of the two solution sets.
From the intersection of solution $\left( 3 \right)$ and solution $\left( 4 \right)$, we obtain $x\in R$.
So, we have $3$ as the only possible integral value in $\left( a \right)$ and the solution of the inequation in $\left( b \right)$ is $x\in R$.

Note: There is a possibility that one may commit mistakes while finding out the common values of $x$ from the two obtained solution sets. To avoid this mistake, one must analyse the two solution sets on a number line and then he/she can easily find the common values of $x$ in the two obtained solution sets.